Integrand size = 22, antiderivative size = 240 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^3} \, dx=-\frac {(5-4 x) \sqrt {1+2 x}}{62 \left (2+3 x+5 x^2\right )^2}+\frac {\sqrt {1+2 x} (67+120 x)}{1922 \left (2+3 x+5 x^2\right )}-\frac {3}{961} \sqrt {\frac {1}{434} \left (15082+2705 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\frac {3}{961} \sqrt {\frac {1}{434} \left (15082+2705 \sqrt {35}\right )} \arctan \left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )}+10 \sqrt {1+2 x}}{\sqrt {10 \left (-2+\sqrt {35}\right )}}\right )+\frac {3}{961} \sqrt {\frac {1}{434} \left (-15082+2705 \sqrt {35}\right )} \text {arctanh}\left (\frac {\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {1+2 x}}{5+\sqrt {35}+10 x}\right ) \] Output:
-1/62*(5-4*x)*(1+2*x)^(1/2)/(5*x^2+3*x+2)^2+(1+2*x)^(1/2)*(67+120*x)/(9610 *x^2+5766*x+3844)-3/417074*(6545588+1173970*35^(1/2))^(1/2)*arctan(((20+10 *35^(1/2))^(1/2)-10*(1+2*x)^(1/2))/(-20+10*35^(1/2))^(1/2))+3/417074*(6545 588+1173970*35^(1/2))^(1/2)*arctan(((20+10*35^(1/2))^(1/2)+10*(1+2*x)^(1/2 ))/(-20+10*35^(1/2))^(1/2))+3/417074*(-6545588+1173970*35^(1/2))^(1/2)*arc tanh((20+10*35^(1/2))^(1/2)*(1+2*x)^(1/2)/(5+35^(1/2)+10*x))
Result contains complex when optimal does not.
Time = 1.70 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.60 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^3} \, dx=\frac {\frac {217 \sqrt {1+2 x} \left (-21+565 x+695 x^2+600 x^3\right )}{2 \left (2+3 x+5 x^2\right )^2}+3 \sqrt {217 \left (15082+961 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} \left (-2-i \sqrt {31}\right )} \sqrt {1+2 x}\right )+3 \sqrt {217 \left (15082-961 i \sqrt {31}\right )} \arctan \left (\sqrt {\frac {1}{7} i \left (2 i+\sqrt {31}\right )} \sqrt {1+2 x}\right )}{208537} \] Input:
Integrate[(1 + 2*x)^(3/2)/(2 + 3*x + 5*x^2)^3,x]
Output:
((217*Sqrt[1 + 2*x]*(-21 + 565*x + 695*x^2 + 600*x^3))/(2*(2 + 3*x + 5*x^2 )^2) + 3*Sqrt[217*(15082 + (961*I)*Sqrt[31])]*ArcTan[Sqrt[(-2 - I*Sqrt[31] )/7]*Sqrt[1 + 2*x]] + 3*Sqrt[217*(15082 - (961*I)*Sqrt[31])]*ArcTan[Sqrt[( I/7)*(2*I + Sqrt[31])]*Sqrt[1 + 2*x]])/208537
Time = 0.65 (sec) , antiderivative size = 350, normalized size of antiderivative = 1.46, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {1164, 1235, 27, 1197, 27, 1483, 27, 1142, 25, 1083, 217, 1103}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(2 x+1)^{3/2}}{\left (5 x^2+3 x+2\right )^3} \, dx\) |
| \(\Big \downarrow \) 1164 |
| \(\displaystyle \frac {1}{62} \int \frac {20 x+17}{\sqrt {2 x+1} \left (5 x^2+3 x+2\right )^2}dx-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 1235 |
| \(\displaystyle \frac {1}{62} \left (\frac {1}{217} \int \frac {21 (40 x+59)}{\sqrt {2 x+1} \left (5 x^2+3 x+2\right )}dx+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{62} \left (\frac {3}{31} \int \frac {40 x+59}{\sqrt {2 x+1} \left (5 x^2+3 x+2\right )}dx+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 1197 |
| \(\displaystyle \frac {1}{62} \left (\frac {6}{31} \int \frac {2 (20 (2 x+1)+39)}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{62} \left (\frac {12}{31} \int \frac {20 (2 x+1)+39}{5 (2 x+1)^2-4 (2 x+1)+7}d\sqrt {2 x+1}+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 1483 |
| \(\displaystyle \frac {1}{62} \left (\frac {12}{31} \left (\frac {\int \frac {5 \left (39 \sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}-\left (39-4 \sqrt {35}\right ) \sqrt {2 x+1}\right )}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {\int \frac {5 \left (\left (39-4 \sqrt {35}\right ) \sqrt {2 x+1}+39 \sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}\right )}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{62} \left (\frac {12}{31} \left (\frac {5 \int \frac {39 \sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}-\left (39-4 \sqrt {35}\right ) \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \int \frac {\left (39-4 \sqrt {35}\right ) \sqrt {2 x+1}+39 \sqrt {\frac {2}{5} \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {1}{62} \left (\frac {12}{31} \left (\frac {5 \left (\frac {1}{10} \sqrt {150820+27050 \sqrt {35}} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\frac {1}{10} \left (39-4 \sqrt {35}\right ) \int -\frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\frac {1}{10} \sqrt {150820+27050 \sqrt {35}} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (39-4 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{62} \left (\frac {12}{31} \left (\frac {5 \left (\frac {1}{10} \sqrt {150820+27050 \sqrt {35}} \int \frac {1}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (39-4 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\frac {1}{10} \sqrt {150820+27050 \sqrt {35}} \int \frac {1}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{10} \left (39-4 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 1083 |
| \(\displaystyle \frac {1}{62} \left (\frac {12}{31} \left (\frac {5 \left (\frac {1}{10} \left (39-4 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\frac {1}{5} \sqrt {150820+27050 \sqrt {35}} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\frac {1}{10} \left (39-4 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}-\frac {1}{5} \sqrt {150820+27050 \sqrt {35}} \int \frac {1}{-2 x+10 \left (2-\sqrt {35}\right )-1}d\left (10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{62} \left (\frac {12}{31} \left (\frac {5 \left (\frac {1}{10} \left (39-4 \sqrt {35}\right ) \int \frac {\sqrt {10 \left (2+\sqrt {35}\right )}-10 \sqrt {2 x+1}}{5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{5} \sqrt {\frac {150820+27050 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\frac {1}{10} \left (39-4 \sqrt {35}\right ) \int \frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}}d\sqrt {2 x+1}+\frac {1}{5} \sqrt {\frac {150820+27050 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{62} \left (\frac {12}{31} \left (\frac {5 \left (\frac {1}{5} \sqrt {\frac {150820+27050 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}-\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )-\frac {1}{10} \left (39-4 \sqrt {35}\right ) \log \left (5 (2 x+1)-\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}+\frac {5 \left (\frac {1}{5} \sqrt {\frac {150820+27050 \sqrt {35}}{10 \left (\sqrt {35}-2\right )}} \arctan \left (\frac {10 \sqrt {2 x+1}+\sqrt {10 \left (2+\sqrt {35}\right )}}{\sqrt {10 \left (\sqrt {35}-2\right )}}\right )+\frac {1}{10} \left (39-4 \sqrt {35}\right ) \log \left (5 (2 x+1)+\sqrt {10 \left (2+\sqrt {35}\right )} \sqrt {2 x+1}+\sqrt {35}\right )\right )}{2 \sqrt {14 \left (2+\sqrt {35}\right )}}\right )+\frac {\sqrt {2 x+1} (120 x+67)}{31 \left (5 x^2+3 x+2\right )}\right )-\frac {(5-4 x) \sqrt {2 x+1}}{62 \left (5 x^2+3 x+2\right )^2}\) |
Input:
Int[(1 + 2*x)^(3/2)/(2 + 3*x + 5*x^2)^3,x]
Output:
-1/62*((5 - 4*x)*Sqrt[1 + 2*x])/(2 + 3*x + 5*x^2)^2 + ((Sqrt[1 + 2*x]*(67 + 120*x))/(31*(2 + 3*x + 5*x^2)) + (12*((5*((Sqrt[(150820 + 27050*Sqrt[35] )/(10*(-2 + Sqrt[35]))]*ArcTan[(-Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x ])/Sqrt[10*(-2 + Sqrt[35])]])/5 - ((39 - 4*Sqrt[35])*Log[Sqrt[35] - Sqrt[1 0*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10))/(2*Sqrt[14*(2 + Sqrt[ 35])]) + (5*((Sqrt[(150820 + 27050*Sqrt[35])/(10*(-2 + Sqrt[35]))]*ArcTan[ (Sqrt[10*(2 + Sqrt[35])] + 10*Sqrt[1 + 2*x])/Sqrt[10*(-2 + Sqrt[35])]])/5 + ((39 - 4*Sqrt[35])*Log[Sqrt[35] + Sqrt[10*(2 + Sqrt[35])]*Sqrt[1 + 2*x] + 5*(1 + 2*x)])/10))/(2*Sqrt[14*(2 + Sqrt[35])])))/31)/62
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m - 1)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* c)) Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int QuadraticQ[a, b, c, d, e, m, p, x]
Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)), x_Symbol] :> Simp[2 Subst[Int[(e*f - d*g + g*x^2)/(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; Fr eeQ[{a, b, c, d, e, f, g}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(f*(b*c*d - b^2*e + 2 *a*c*e) - a*g*(2*c*d - b*e) + c*(f*(2*c*d - b*e) - g*(b*d - 2*a*e))*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^m *(a + b*x + c*x^2)^(p + 1)*Simp[f*(b*c*d*e*(2*p - m + 2) + b^2*e^2*(p + m + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3)) - g*(a*e*(b*e - 2*c*d* m + b*e*m) - b*d*(3*c*d - b*e + 2*c*d*p - b*e*p)) + c*e*(g*(b*d - 2*a*e) - f*(2*c*d - b*e))*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p] )
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[a/c, 2]}, With[{r = Rt[2*q - b/c, 2]}, Simp[1/(2*c*q*r) In t[(d*r - (d - e*q)*x)/(q - r*x + x^2), x], x] + Simp[1/(2*c*q*r) Int[(d*r + (d - e*q)*x)/(q + r*x + x^2), x], x]]] /; FreeQ[{a, b, c, d, e}, x] && N eQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NegQ[b^2 - 4*a*c]
Time = 2.39 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.18
| method | result | size |
| pseudoelliptic | \(-\frac {3 \left (-86800 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (x^{3}+\frac {139}{120} x^{2}+\frac {113}{120} x -\frac {7}{200}\right ) \sqrt {1+2 x}+\left (5 x^{2}+3 x +2\right )^{2} \left (\sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (329 \sqrt {5}-218 \sqrt {7}\right ) \left (\ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )-\ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )\right ) \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+4836 \left (\sqrt {5}\, \sqrt {7}+\frac {140}{39}\right ) \left (\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )-\arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )\right )\right )\right )}{834148 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}\, \left (5 x^{2}+3 x +2\right )^{2}}\) | \(282\) |
| derivativedivides | \(\frac {\frac {600 \left (1+2 x \right )^{\frac {7}{2}}}{961}-\frac {410 \left (1+2 x \right )^{\frac {5}{2}}}{961}+\frac {1280 \left (1+2 x \right )^{\frac {3}{2}}}{961}-\frac {1638 \sqrt {1+2 x}}{961}}{\left (5 \left (1+2 x \right )^{2}+3-8 x \right )^{2}}-\frac {3 \left (1645 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-1090 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{4170740}-\frac {3 \left (-2418 \sqrt {5}\, \sqrt {7}+\frac {\left (1645 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-1090 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{208537 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {3 \left (1645 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-1090 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{4170740}+\frac {3 \left (2418 \sqrt {5}\, \sqrt {7}-\frac {\left (1645 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-1090 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{208537 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(435\) |
| default | \(\frac {\frac {600 \left (1+2 x \right )^{\frac {7}{2}}}{961}-\frac {410 \left (1+2 x \right )^{\frac {5}{2}}}{961}+\frac {1280 \left (1+2 x \right )^{\frac {3}{2}}}{961}-\frac {1638 \sqrt {1+2 x}}{961}}{\left (5 \left (1+2 x \right )^{2}+3-8 x \right )^{2}}-\frac {3 \left (1645 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-1090 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right )}{4170740}-\frac {3 \left (-2418 \sqrt {5}\, \sqrt {7}+\frac {\left (1645 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-1090 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{208537 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {3 \left (1645 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-1090 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right )}{4170740}+\frac {3 \left (2418 \sqrt {5}\, \sqrt {7}-\frac {\left (1645 \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}-1090 \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}\right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{10}\right ) \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right )}{208537 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(435\) |
| trager | \(\frac {\left (600 x^{3}+695 x^{2}+565 x -21\right ) \sqrt {1+2 x}}{1922 \left (5 x^{2}+3 x +2\right )^{2}}-\frac {6 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right ) \ln \left (-\frac {-7377527808 x \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{5}-45663063488 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{3} x +5415288816 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2} \sqrt {1+2 x}+16335954432 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{3}-67162035120 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right ) x +92971296325 \sqrt {1+2 x}+61800326272 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )}{3472 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2} x +17965 x +3844}\right )}{961}+\frac {3 \operatorname {RootOf}\left (\textit {\_Z}^{2}+188356 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2}+1636397\right ) \ln \left (\frac {-526966272 \operatorname {RootOf}\left (\textit {\_Z}^{2}+188356 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2}+1636397\right ) \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{4} x -5894695136 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+188356 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2}+1636397\right ) x +167873953296 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2} \sqrt {1+2 x}-1166853888 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2} \operatorname {RootOf}\left (\textit {\_Z}^{2}+188356 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2}+1636397\right )-16234968544 \operatorname {RootOf}\left (\textit {\_Z}^{2}+188356 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2}+1636397\right ) x -1423656866023 \sqrt {1+2 x}-5723070208 \operatorname {RootOf}\left (\textit {\_Z}^{2}+188356 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2}+1636397\right )}{3472 \operatorname {RootOf}\left (1722112 \textit {\_Z}^{4}+14961344 \textit {\_Z}^{2}+36585125\right )^{2} x +12199 x -3844}\right )}{208537}\) | \(458\) |
| risch | \(\frac {\left (600 x^{3}+695 x^{2}+565 x -21\right ) \sqrt {1+2 x}}{1922 \left (5 x^{2}+3 x +2\right )^{2}}-\frac {141 \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{119164}+\frac {327 \ln \left (-\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+\sqrt {5}\, \sqrt {7}+5+10 x \right ) \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{417074}-\frac {705 \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \left (2 \sqrt {5}\, \sqrt {7}+4\right )}{59582 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {327 \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \sqrt {5}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}}{208537 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {234 \arctan \left (\frac {-\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \sqrt {5}\, \sqrt {7}}{6727 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {141 \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right ) \sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{119164}-\frac {327 \ln \left (\sqrt {5}\, \sqrt {7}+\sqrt {2 \sqrt {5}\, \sqrt {7}+4}\, \sqrt {5}\, \sqrt {1+2 x}+5+10 x \right ) \sqrt {7}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}}{417074}-\frac {705 \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \left (2 \sqrt {5}\, \sqrt {7}+4\right )}{59582 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {327 \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \sqrt {5}\, \left (2 \sqrt {5}\, \sqrt {7}+4\right ) \sqrt {7}}{208537 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}+\frac {234 \arctan \left (\frac {\sqrt {5}\, \sqrt {2 \sqrt {5}\, \sqrt {7}+4}+10 \sqrt {1+2 x}}{\sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\right ) \sqrt {5}\, \sqrt {7}}{6727 \sqrt {10 \sqrt {5}\, \sqrt {7}-20}}\) | \(643\) |
Input:
int((1+2*x)^(3/2)/(5*x^2+3*x+2)^3,x,method=_RETURNVERBOSE)
Output:
-3/834148/(10*5^(1/2)*7^(1/2)-20)^(1/2)*(-86800*(10*5^(1/2)*7^(1/2)-20)^(1 /2)*(x^3+139/120*x^2+113/120*x-7/200)*(1+2*x)^(1/2)+(5*x^2+3*x+2)^2*((10*5 ^(1/2)*7^(1/2)-20)^(1/2)*(329*5^(1/2)-218*7^(1/2))*(ln(-(2*5^(1/2)*7^(1/2) +4)^(1/2)*5^(1/2)*(1+2*x)^(1/2)+5^(1/2)*7^(1/2)+5+10*x)-ln(5^(1/2)*7^(1/2) +(2*5^(1/2)*7^(1/2)+4)^(1/2)*5^(1/2)*(1+2*x)^(1/2)+5+10*x))*(2*5^(1/2)*7^( 1/2)+4)^(1/2)+4836*(5^(1/2)*7^(1/2)+140/39)*(arctan((5^(1/2)*(2*5^(1/2)*7^ (1/2)+4)^(1/2)-10*(1+2*x)^(1/2))/(10*5^(1/2)*7^(1/2)-20)^(1/2))-arctan((5^ (1/2)*(2*5^(1/2)*7^(1/2)+4)^(1/2)+10*(1+2*x)^(1/2))/(10*5^(1/2)*7^(1/2)-20 )^(1/2)))))/(5*x^2+3*x+2)^2
Leaf count of result is larger than twice the leaf count of optimal. 365 vs. \(2 (177) = 354\).
Time = 0.09 (sec) , antiderivative size = 365, normalized size of antiderivative = 1.52 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^3} \, dx=\frac {6 \, \sqrt {\frac {1}{434}} {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )} \sqrt {18935 \, \sqrt {\frac {5}{7}} + 15082} \arctan \left (\frac {14}{961} \, \sqrt {\frac {1}{434}} {\left (\sqrt {\frac {1}{434}} \sqrt {18935 \, \sqrt {\frac {5}{7}} - 15082} {\left (7 \, \sqrt {\frac {5}{7}} + 2\right )} + \sqrt {2 \, x + 1} {\left (39 \, \sqrt {\frac {5}{7}} - 20\right )}\right )} \sqrt {18935 \, \sqrt {\frac {5}{7}} + 15082}\right ) - 6 \, \sqrt {\frac {1}{434}} {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )} \sqrt {18935 \, \sqrt {\frac {5}{7}} + 15082} \arctan \left (\frac {14}{961} \, \sqrt {\frac {1}{434}} {\left (\sqrt {\frac {1}{434}} \sqrt {18935 \, \sqrt {\frac {5}{7}} - 15082} {\left (7 \, \sqrt {\frac {5}{7}} + 2\right )} - \sqrt {2 \, x + 1} {\left (39 \, \sqrt {\frac {5}{7}} - 20\right )}\right )} \sqrt {18935 \, \sqrt {\frac {5}{7}} + 15082}\right ) + 3 \, \sqrt {\frac {1}{434}} {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )} \sqrt {18935 \, \sqrt {\frac {5}{7}} - 15082} \log \left (42 \, \sqrt {\frac {1}{434}} \sqrt {2 \, x + 1} \sqrt {18935 \, \sqrt {\frac {5}{7}} - 15082} {\left (218 \, \sqrt {\frac {5}{7}} + 235\right )} + 28830 \, x + 20181 \, \sqrt {\frac {5}{7}} + 14415\right ) - 3 \, \sqrt {\frac {1}{434}} {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )} \sqrt {18935 \, \sqrt {\frac {5}{7}} - 15082} \log \left (-42 \, \sqrt {\frac {1}{434}} \sqrt {2 \, x + 1} \sqrt {18935 \, \sqrt {\frac {5}{7}} - 15082} {\left (218 \, \sqrt {\frac {5}{7}} + 235\right )} + 28830 \, x + 20181 \, \sqrt {\frac {5}{7}} + 14415\right ) + {\left (600 \, x^{3} + 695 \, x^{2} + 565 \, x - 21\right )} \sqrt {2 \, x + 1}}{1922 \, {\left (25 \, x^{4} + 30 \, x^{3} + 29 \, x^{2} + 12 \, x + 4\right )}} \] Input:
integrate((1+2*x)^(3/2)/(5*x^2+3*x+2)^3,x, algorithm="fricas")
Output:
1/1922*(6*sqrt(1/434)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*sqrt(18935*sqr t(5/7) + 15082)*arctan(14/961*sqrt(1/434)*(sqrt(1/434)*sqrt(18935*sqrt(5/7 ) - 15082)*(7*sqrt(5/7) + 2) + sqrt(2*x + 1)*(39*sqrt(5/7) - 20))*sqrt(189 35*sqrt(5/7) + 15082)) - 6*sqrt(1/434)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*sqrt(18935*sqrt(5/7) + 15082)*arctan(14/961*sqrt(1/434)*(sqrt(1/434)*sq rt(18935*sqrt(5/7) - 15082)*(7*sqrt(5/7) + 2) - sqrt(2*x + 1)*(39*sqrt(5/7 ) - 20))*sqrt(18935*sqrt(5/7) + 15082)) + 3*sqrt(1/434)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*sqrt(18935*sqrt(5/7) - 15082)*log(42*sqrt(1/434)*sqrt( 2*x + 1)*sqrt(18935*sqrt(5/7) - 15082)*(218*sqrt(5/7) + 235) + 28830*x + 2 0181*sqrt(5/7) + 14415) - 3*sqrt(1/434)*(25*x^4 + 30*x^3 + 29*x^2 + 12*x + 4)*sqrt(18935*sqrt(5/7) - 15082)*log(-42*sqrt(1/434)*sqrt(2*x + 1)*sqrt(1 8935*sqrt(5/7) - 15082)*(218*sqrt(5/7) + 235) + 28830*x + 20181*sqrt(5/7) + 14415) + (600*x^3 + 695*x^2 + 565*x - 21)*sqrt(2*x + 1))/(25*x^4 + 30*x^ 3 + 29*x^2 + 12*x + 4)
\[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^3} \, dx=\int \frac {\left (2 x + 1\right )^{\frac {3}{2}}}{\left (5 x^{2} + 3 x + 2\right )^{3}}\, dx \] Input:
integrate((1+2*x)**(3/2)/(5*x**2+3*x+2)**3,x)
Output:
Integral((2*x + 1)**(3/2)/(5*x**2 + 3*x + 2)**3, x)
\[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^3} \, dx=\int { \frac {{\left (2 \, x + 1\right )}^{\frac {3}{2}}}{{\left (5 \, x^{2} + 3 \, x + 2\right )}^{3}} \,d x } \] Input:
integrate((1+2*x)^(3/2)/(5*x^2+3*x+2)^3,x, algorithm="maxima")
Output:
integrate((2*x + 1)^(3/2)/(5*x^2 + 3*x + 2)^3, x)
Leaf count of result is larger than twice the leaf count of optimal. 640 vs. \(2 (177) = 354\).
Time = 0.81 (sec) , antiderivative size = 640, normalized size of antiderivative = 2.67 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^3} \, dx=\text {Too large to display} \] Input:
integrate((1+2*x)^(3/2)/(5*x^2+3*x+2)^3,x, algorithm="giac")
Output:
3/3576409550*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(-14 0*sqrt(35) + 2450) - sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 2 *(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 420*(7/5)^(3/4)*sqrt(140*sqrt(3 5) + 2450)*(2*sqrt(35) - 35) + 9555*sqrt(31)*(7/5)^(1/4)*sqrt(-140*sqrt(35 ) + 2450) + 19110*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450))*arctan(5/7*(7/5)^ (3/4)*((7/5)^(1/4)*sqrt(1/35*sqrt(35) + 1/2) + sqrt(2*x + 1))/sqrt(-1/35*s qrt(35) + 1/2)) + 3/3576409550*sqrt(31)*(210*sqrt(31)*(7/5)^(3/4)*(2*sqrt( 35) + 35)*sqrt(-140*sqrt(35) + 2450) - sqrt(31)*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 2*(7/5)^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 420*(7/5)^(3/ 4)*sqrt(140*sqrt(35) + 2450)*(2*sqrt(35) - 35) + 9555*sqrt(31)*(7/5)^(1/4) *sqrt(-140*sqrt(35) + 2450) + 19110*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450)) *arctan(-5/7*(7/5)^(3/4)*((7/5)^(1/4)*sqrt(1/35*sqrt(35) + 1/2) - sqrt(2*x + 1))/sqrt(-1/35*sqrt(35) + 1/2)) + 3/7152819100*sqrt(31)*(sqrt(31)*(7/5) ^(3/4)*(140*sqrt(35) + 2450)^(3/2) + 210*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqr t(35) + 2450)*(2*sqrt(35) - 35) - 420*(7/5)^(3/4)*(2*sqrt(35) + 35)*sqrt(- 140*sqrt(35) + 2450) + 2*(7/5)^(3/4)*(-140*sqrt(35) + 2450)^(3/2) + 9555*s qrt(31)*(7/5)^(1/4)*sqrt(140*sqrt(35) + 2450) - 19110*(7/5)^(1/4)*sqrt(-14 0*sqrt(35) + 2450))*log(2*(7/5)^(1/4)*sqrt(2*x + 1)*sqrt(1/35*sqrt(35) + 1 /2) + 2*x + sqrt(7/5) + 1) - 3/7152819100*sqrt(31)*(sqrt(31)*(7/5)^(3/4)*( 140*sqrt(35) + 2450)^(3/2) + 210*sqrt(31)*(7/5)^(3/4)*sqrt(140*sqrt(35)...
Time = 5.82 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.02 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^3} \, dx=\frac {\frac {1638\,\sqrt {2\,x+1}}{24025}-\frac {256\,{\left (2\,x+1\right )}^{3/2}}{4805}+\frac {82\,{\left (2\,x+1\right )}^{5/2}}{4805}-\frac {24\,{\left (2\,x+1\right )}^{7/2}}{961}}{\frac {112\,x}{25}-\frac {86\,{\left (2\,x+1\right )}^2}{25}+\frac {8\,{\left (2\,x+1\right )}^3}{5}-{\left (2\,x+1\right )}^4+\frac {7}{25}}+\frac {\sqrt {217}\,\mathrm {atan}\left (\frac {\sqrt {217}\,\sqrt {-15082-\sqrt {31}\,961{}\mathrm {i}}\,\sqrt {2\,x+1}\,432{}\mathrm {i}}{5656566125\,\left (\frac {94176}{808080875}+\frac {\sqrt {31}\,16848{}\mathrm {i}}{808080875}\right )}+\frac {864\,\sqrt {31}\,\sqrt {217}\,\sqrt {-15082-\sqrt {31}\,961{}\mathrm {i}}\,\sqrt {2\,x+1}}{175353549875\,\left (\frac {94176}{808080875}+\frac {\sqrt {31}\,16848{}\mathrm {i}}{808080875}\right )}\right )\,\sqrt {-15082-\sqrt {31}\,961{}\mathrm {i}}\,3{}\mathrm {i}}{208537}-\frac {\sqrt {217}\,\mathrm {atan}\left (\frac {\sqrt {217}\,\sqrt {-15082+\sqrt {31}\,961{}\mathrm {i}}\,\sqrt {2\,x+1}\,432{}\mathrm {i}}{5656566125\,\left (-\frac {94176}{808080875}+\frac {\sqrt {31}\,16848{}\mathrm {i}}{808080875}\right )}-\frac {864\,\sqrt {31}\,\sqrt {217}\,\sqrt {-15082+\sqrt {31}\,961{}\mathrm {i}}\,\sqrt {2\,x+1}}{175353549875\,\left (-\frac {94176}{808080875}+\frac {\sqrt {31}\,16848{}\mathrm {i}}{808080875}\right )}\right )\,\sqrt {-15082+\sqrt {31}\,961{}\mathrm {i}}\,3{}\mathrm {i}}{208537} \] Input:
int((2*x + 1)^(3/2)/(3*x + 5*x^2 + 2)^3,x)
Output:
((1638*(2*x + 1)^(1/2))/24025 - (256*(2*x + 1)^(3/2))/4805 + (82*(2*x + 1) ^(5/2))/4805 - (24*(2*x + 1)^(7/2))/961)/((112*x)/25 - (86*(2*x + 1)^2)/25 + (8*(2*x + 1)^3)/5 - (2*x + 1)^4 + 7/25) + (217^(1/2)*atan((217^(1/2)*(- 31^(1/2)*961i - 15082)^(1/2)*(2*x + 1)^(1/2)*432i)/(5656566125*((31^(1/2) *16848i)/808080875 + 94176/808080875)) + (864*31^(1/2)*217^(1/2)*(- 31^(1/ 2)*961i - 15082)^(1/2)*(2*x + 1)^(1/2))/(175353549875*((31^(1/2)*16848i)/8 08080875 + 94176/808080875)))*(- 31^(1/2)*961i - 15082)^(1/2)*3i)/208537 - (217^(1/2)*atan((217^(1/2)*(31^(1/2)*961i - 15082)^(1/2)*(2*x + 1)^(1/2)* 432i)/(5656566125*((31^(1/2)*16848i)/808080875 - 94176/808080875)) - (864* 31^(1/2)*217^(1/2)*(31^(1/2)*961i - 15082)^(1/2)*(2*x + 1)^(1/2))/(1753535 49875*((31^(1/2)*16848i)/808080875 - 94176/808080875)))*(31^(1/2)*961i - 1 5082)^(1/2)*3i)/208537
Time = 0.20 (sec) , antiderivative size = 1653, normalized size of antiderivative = 6.89 \[ \int \frac {(1+2 x)^{3/2}}{\left (2+3 x+5 x^2\right )^3} \, dx =\text {Too large to display} \] Input:
int((1+2*x)^(3/2)/(5*x^2+3*x+2)^3,x)
Output:
( - 32700*sqrt(sqrt(35) - 2)*sqrt(14)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2 *sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x**4 - 39240*sqrt(sq rt(35) - 2)*sqrt(14)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sq rt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x**3 - 37932*sqrt(sqrt(35) - 2)*sqrt( 14)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt (35) - 2)*sqrt(2)))*x**2 - 15696*sqrt(sqrt(35) - 2)*sqrt(14)*atan((sqrt(sq rt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2) ))*x - 5232*sqrt(sqrt(35) - 2)*sqrt(14)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) - 49350*sqrt(sqrt( 35) - 2)*sqrt(10)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt( 5))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x**4 - 59220*sqrt(sqrt(35) - 2)*sqrt(10) *atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35 ) - 2)*sqrt(2)))*x**3 - 57246*sqrt(sqrt(35) - 2)*sqrt(10)*atan((sqrt(sqrt( 35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))* x**2 - 23688*sqrt(sqrt(35) - 2)*sqrt(10)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x - 7896*sqrt(sqr t(35) - 2)*sqrt(10)*atan((sqrt(sqrt(35) + 2)*sqrt(2) - 2*sqrt(2*x + 1)*sqr t(5))/(sqrt(sqrt(35) - 2)*sqrt(2))) + 32700*sqrt(sqrt(35) - 2)*sqrt(14)*at an((sqrt(sqrt(35) + 2)*sqrt(2) + 2*sqrt(2*x + 1)*sqrt(5))/(sqrt(sqrt(35) - 2)*sqrt(2)))*x**4 + 39240*sqrt(sqrt(35) - 2)*sqrt(14)*atan((sqrt(sqrt(...