Integrand size = 25, antiderivative size = 155 \[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {74 \sqrt {x} (2+3 x)}{3 \sqrt {2+5 x+3 x^2}}-\frac {2 \sqrt {x} (30+37 x)}{\sqrt {2+5 x+3 x^2}}-\frac {74 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {2+5 x+3 x^2}}+\frac {30 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {2+5 x+3 x^2}} \]
74/3*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)-2*(30+37*x)*x^(1/2)/(3*x^2+5*x+2) ^(1/2)-74/3*(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)+30*(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)
Result contains complex when optimal does not.
Time = 21.14 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.90 \[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {148+190 x+74 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} E\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+16 i \sqrt {2} \sqrt {1+\frac {1}{x}} \sqrt {3+\frac {2}{x}} x^{3/2} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right ),\frac {3}{2}\right )}{3 \sqrt {x} \sqrt {2+5 x+3 x^2}} \]
(148 + 190*x + (74*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^(3/2)*Ellip ticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] + (16*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*Sqr t[x]*Sqrt[2 + 5*x + 3*x^2])
Time = 0.32 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {1234, 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.
| \(\displaystyle \int \frac {(2-5 x) \sqrt {x}}{\left (3 x^2+5 x+2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 1234 |
| \(\displaystyle -2 \int -\frac {37 x+30}{2 \sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {2 \sqrt {x} (37 x+30)}{\sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {37 x+30}{\sqrt {x} \sqrt {3 x^2+5 x+2}}dx-\frac {2 \sqrt {x} (37 x+30)}{\sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1240 |
| \(\displaystyle 2 \int \frac {37 x+30}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}-\frac {2 \sqrt {x} (37 x+30)}{\sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1503 |
| \(\displaystyle 2 \left (30 \int \frac {1}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+37 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}\right )-\frac {2 \sqrt {x} (37 x+30)}{\sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1413 |
| \(\displaystyle 2 \left (37 \int \frac {x}{\sqrt {3 x^2+5 x+2}}d\sqrt {x}+\frac {15 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}\right )-\frac {2 \sqrt {x} (37 x+30)}{\sqrt {3 x^2+5 x+2}}\) |
| \(\Big \downarrow \) 1456 |
| \(\displaystyle 2 \left (\frac {15 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} \operatorname {EllipticF}\left (\arctan \left (\sqrt {x}\right ),-\frac {1}{2}\right )}{\sqrt {3 x^2+5 x+2}}+37 \left (\frac {\sqrt {x} (3 x+2)}{3 \sqrt {3 x^2+5 x+2}}-\frac {\sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\arctan \left (\sqrt {x}\right )|-\frac {1}{2}\right )}{3 \sqrt {3 x^2+5 x+2}}\right )\right )-\frac {2 \sqrt {x} (37 x+30)}{\sqrt {3 x^2+5 x+2}}\) |
(-2*Sqrt[x]*(30 + 37*x))/Sqrt[2 + 5*x + 3*x^2] + 2*(37*((Sqrt[x]*(2 + 3*x) )/(3*Sqrt[2 + 5*x + 3*x^2]) - (Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*Ell ipticE[ArcTan[Sqrt[x]], -1/2])/(3*Sqrt[2 + 5*x + 3*x^2])) + (15*Sqrt[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/Sqrt[2 + 5 *x + 3*x^2])
3.11.67.3.1 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*(a + b*x + c*x^2)^(p + 1)*( (f*b - 2*a*g + (2*c*f - b*g)*x)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*Simp[g *(2*a*e*m + b*d*(2*p + 3)) - f*(b*e*m + 2*c*d*(2*p + 3)) - e*(2*c*f - b*g)* (m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1 ] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
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 = 0.21 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.69
| method | result | size |
| default | \(-\frac {21 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )-37 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {6}\, \sqrt {-x}\, E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+666 x^{2}+540 x}{9 \sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(107\) |
| elliptic | \(\frac {\sqrt {x \left (3 x^{2}+5 x +2\right )}\, \left (-\frac {2 x \left (10+\frac {37 x}{3}\right ) \sqrt {3}}{\sqrt {x \left (x^{2}+\frac {5}{3} x +\frac {2}{3}\right )}}+\frac {10 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{\sqrt {3 x^{3}+5 x^{2}+2 x}}+\frac {37 \sqrt {6 x +4}\, \sqrt {3+3 x}\, \sqrt {-6 x}\, \left (\frac {E\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )}{3}-F\left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )\right )}{3 \sqrt {3 x^{3}+5 x^{2}+2 x}}\right )}{\sqrt {x}\, \sqrt {3 x^{2}+5 x +2}}\) | \(182\) |
-1/9*(21*(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))-37*(6*x+4)^(1/2)*(3+3*x)^(1/2)*6^(1/2)*(-x)^(1/2)*Ell ipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))+666*x^2+540*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 = 82, normalized size of antiderivative = 0.53 \[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\frac {2 \, {\left (85 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right ) - 333 \, \sqrt {3} {\left (3 \, x^{2} + 5 \, x + 2\right )} {\rm weierstrassZeta}\left (\frac {28}{27}, \frac {80}{729}, {\rm weierstrassPInverse}\left (\frac {28}{27}, \frac {80}{729}, x + \frac {5}{9}\right )\right ) - 27 \, \sqrt {3 \, x^{2} + 5 \, x + 2} {\left (37 \, x + 30\right )} \sqrt {x}\right )}}{27 \, {\left (3 \, x^{2} + 5 \, x + 2\right )}} \]
2/27*(85*sqrt(3)*(3*x^2 + 5*x + 2)*weierstrassPInverse(28/27, 80/729, x + 5/9) - 333*sqrt(3)*(3*x^2 + 5*x + 2)*weierstrassZeta(28/27, 80/729, weiers trassPInverse(28/27, 80/729, x + 5/9)) - 27*sqrt(3*x^2 + 5*x + 2)*(37*x + 30)*sqrt(x))/(3*x^2 + 5*x + 2)
\[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=- \int \left (- \frac {2 \sqrt {x}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\right )\, dx - \int \frac {5 x^{\frac {3}{2}}}{3 x^{2} \sqrt {3 x^{2} + 5 x + 2} + 5 x \sqrt {3 x^{2} + 5 x + 2} + 2 \sqrt {3 x^{2} + 5 x + 2}}\, dx \]
-Integral(-2*sqrt(x)/(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**(3/2)/(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)
\[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} \sqrt {x}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=\int { -\frac {{\left (5 \, x - 2\right )} \sqrt {x}}{{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(2-5 x) \sqrt {x}}{\left (2+5 x+3 x^2\right )^{3/2}} \, dx=-\int \frac {\sqrt {x}\,\left (5\,x-2\right )}{{\left (3\,x^2+5\,x+2\right )}^{3/2}} \,d x \]