Integrand size = 25, antiderivative size = 197 \[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\frac {24 \sqrt {x} (2+3 x)}{\sqrt {2+5 x+3 x^2}}+\frac {2 x^{5/2} (74+95 x)}{3 \sqrt {2+5 x+3 x^2}}+20 \sqrt {x} \sqrt {2+5 x+3 x^2}-\frac {64}{3} x^{3/2} \sqrt {2+5 x+3 x^2}+\frac {24 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}}-\frac {20 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}} \]
2/3*x^(5/2)*(74+95*x)/(3*x^2+5*x+2)^(1/2)-24*(2+3*x)*x^(1/2)/(3*x^2+5*x+2) ^(1/2)+24*(1+x)^(3/2)*(1/(1+x))^(1/2)*EllipticE(x^(1/2)/(1+x)^(1/2),1/2*I* 2^(1/2))*2^(1/2)*((2+3*x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)-20*(1+x)^(3/2)* (1/(1+x))^(1/2)*EllipticF(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/2))*2^(1/2)*((2+3 *x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)-64/3*x^(3/2)*(3*x^2+5*x+2)^(1/2)+20*x ^(1/2)*(3*x^2+5*x+2)^(1/2)
Result contains complex when optimal does not.
Time = 21.15 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.79 \[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {-2 \left (72+120 x+22 x^2-4 x^3+x^4\right )-72 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+12 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{3 \sqrt {x} \sqrt {2+5 x+3 x^2}} \]
(-2*(72 + 120*x + 22*x^2 - 4*x^3 + x^4) - (72*I)*Sqrt[2]*Sqrt[1 + x^(-1)]* Sqrt[3 + 2/x]*x^(3/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (12*I )*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*EllipticF[I*ArcSinh[Sqrt[ 2/3]/Sqrt[x]], 3/2])/(3*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2])
Time = 0.41 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.10, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {1233, 27, 1236, 27, 1236, 27, 1240, 1503, 1413, 1456}
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.
| \(\displaystyle \int \frac {(2-5 x) x^{7/2}}{\left (3 x^2+5 x+2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1233 |
| \(\displaystyle \frac {2}{3} \int -\frac {5 x^{3/2} (48 x+37)}{\sqrt {3 x^2+5 x+2}}dx+\frac {2 (95 x+74) x^{5/2}}{3 \sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 x^{5/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {10}{3} \int \frac {x^{3/2} (48 x+37)}{\sqrt {3 x^2+5 x+2}}dx\) |
| \(\Big \downarrow \) 1236 |
| \(\displaystyle \frac {2 x^{5/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {10}{3} \left (\frac {2}{15} \int -\frac {9 \sqrt {x} (45 x+32)}{2 \sqrt {3 x^2+5 x+2}}dx+\frac {32}{5} \sqrt {3 x^2+5 x+2} x^{3/2}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 x^{5/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {10}{3} \left (\frac {32}{5} x^{3/2} \sqrt {3 x^2+5 x+2}-\frac {3}{5} \int \frac {\sqrt {x} (45 x+32)}{\sqrt {3 x^2+5 x+2}}dx\right )\) |
| \(\Big \downarrow \) 1236 |
| \(\displaystyle \frac {2 x^{5/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {10}{3} \left (\frac {32}{5} x^{3/2} \sqrt {3 x^2+5 x+2}-\frac {3}{5} \left (\frac {2}{9} \int -\frac {9 (9 x+5)}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx+10 \sqrt {x} \sqrt {3 x^2+5 x+2}\right )\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 x^{5/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {10}{3} \left (\frac {32}{5} x^{3/2} \sqrt {3 x^2+5 x+2}-\frac {3}{5} \left (10 \sqrt {x} \sqrt {3 x^2+5 x+2}-2 \int \frac {9 x+5}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx\right )\right )\) |
| \(\Big \downarrow \) 1240 |
| \(\displaystyle \frac {2 x^{5/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {10}{3} \left (\frac {32}{5} x^{3/2} \sqrt {3 x^2+5 x+2}-\frac {3}{5} \left (10 \sqrt {x} \sqrt {3 x^2+5 x+2}-4 \int \frac {9 x+5}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\right )\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle \frac {2 x^{5/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {10}{3} \left (\frac {32}{5} x^{3/2} \sqrt {3 x^2+5 x+2}-\frac {3}{5} \left (10 \sqrt {x} \sqrt {3 x^2+5 x+2}-4 \left (5 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+9 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )\right )\right )\) |
| \(\Big \downarrow \) 1413 |
| \(\displaystyle \frac {2 x^{5/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {10}{3} \left (\frac {32}{5} x^{3/2} \sqrt {3 x^2+5 x+2}-\frac {3}{5} \left (10 \sqrt {x} \sqrt {3 x^2+5 x+2}-4 \left (9 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {5 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}\right )\right )\right )\) |
| \(\Big \downarrow \) 1456 |
| \(\displaystyle \frac {2 x^{5/2} (95 x+74)}{3 \sqrt {3 x^2+5 x+2}}-\frac {10}{3} \left (\frac {32}{5} x^{3/2} \sqrt {3 x^2+5 x+2}-\frac {3}{5} \left (10 \sqrt {x} \sqrt {3 x^2+5 x+2}-4 \left (\frac {5 (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2} \sqrt {3 x^2+5 x+2}}+9 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )\right )\right )\) |
(2*x^(5/2)*(74 + 95*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (10*((32*x^(3/2)*Sqrt[ 2 + 5*x + 3*x^2])/5 - (3*(10*Sqrt[x]*Sqrt[2 + 5*x + 3*x^2] - 4*(9*((Sqrt[x ]*(2 + 3*x))/(3*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/( 1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2])) + (5* (1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(Sqrt[2] *Sqrt[2 + 5*x + 3*x^2]))))/5))/3
3.11.64.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) ^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c *(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f *(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | | !ILtQ[m + 2*p + 3, 0])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[g*(d + e*x)^m*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), x] + Simp[1/(c*(m + 2*p + 2)) Int[(d + e*x)^(m - 1 )*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m *(c*e*f + c*d*g - b*e*g) + e*(p + 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[ {a, b, c, d, e, f, g, p}, x] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (Intege rQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p]) && !(IGtQ[m, 0] && EqQ[f, 0])
Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Simp[2 Subst[Int[(f + g*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, b, c, f, g}, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b ^2 - 4*a*c, 2]}, Simp[(2*a + (b - q)*x^2)*(Sqrt[(2*a + (b + q)*x^2)/(2*a + (b - q)*x^2)]/(2*a*Rt[(b - q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]))*EllipticF [ArcTan[Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[x*((b - q + 2*c*x^2)/(2*c*Sqrt[a + b*x^2 + c*x^4 ])), x] - Simp[Rt[(b - q)/(2*a), 2]*(2*a + (b - q)*x^2)*(Sqrt[(2*a + (b + q )*x^2)/(2*a + (b - q)*x^2)]/(2*c*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[ArcTan [Rt[(b - q)/(2*a), 2]*x], -2*(q/(b - q))], x] /; PosQ[(b - q)/a]] /; FreeQ[ {a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo l] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[d Int[1/Sqrt[a + b*x^2 + c*x^4] , x], x] + Simp[e Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b + q) /a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]
Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.59
| method | result | size |
| default | \(\frac {\frac {16 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-4 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-\frac {2 x^{4}}{3}+\frac {8 x^{3}}{3}+\frac {172 x^{2}}{3}+40 x}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(117\) |
| elliptic | \(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {2 x \left (-\frac {506}{81}-\frac {695 x}{81}\right ) \sqrt {3}}{\sqrt {x \left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )}}-\frac {2 x \sqrt {3 x^{3}+5 x^{2}+2 x}}{9}+\frac {34 \sqrt {3 x^{3}+5 x^{2}+2 x}}{27}-\frac {20 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}-\frac {12 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{\sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(219\) |
2/3*(8*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6*x+4 )^(1/2),I*2^(1/2))-6*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*Ellipt icE(1/2*(6*x+4)^(1/2),I*2^(1/2))-x^4+4*x^3+86*x^2+60*x)/x^(1/2)/(3*x^2+5*x +2)^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.35 \[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (36 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - {\left (x^{3} - 4 \, x^{2} - 86 \, x - 60\right )} \sqrt {3 \, x^{2} + 5 \, x + 2} \sqrt {x}\right )}}{3 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \]
2/3*(36*sqrt(3)*(3*x^2 + 5*x + 2)*weierstrassZeta(28/27, 80/729, weierstra ssPInverse(28/27, 80/729, x + 5/9)) - (x^3 - 4*x^2 - 86*x - 60)*sqrt(3*x^2 + 5*x + 2)*sqrt(x))/(3*x^2 + 5*x + 2)
\[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=- \int \left (- \frac {2 x^{\frac {7}{2}}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {5 x^{\frac {9}{2}}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\, dx \]
-Integral(-2*x**(7/2)/(3*x**2*sqrt(3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5 *x + 2) + 2*sqrt(3*x**2 + 5*x + 2)), x) - Integral(5*x**(9/2)/(3*x**2*sqrt (3*x**2 + 5*x + 2) + 5*x*sqrt(3*x**2 + 5*x + 2) + 2*sqrt(3*x**2 + 5*x + 2) ), x)
\[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} x^{\frac {7}{2}}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} x^{\frac {7}{2}}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(2-5 x) x^{7/2}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\int \frac {x^{7/2}\,\left (5\,x-2\right )}{{\left (3\,x^2+5\,x+2\right )}^{3/2}} \,d x \]