Integrand size = 27, antiderivative size = 641 \[ \int \frac {x^4 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\frac {7 \sqrt {c+d x^3}}{3 d^{5/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}+\frac {x^2 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}+\frac {5 \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3 \sqrt {3} d^{5/3}}-\frac {5 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{9 d^{5/3}}+\frac {5 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^{5/3}}-\frac {7 \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{2\ 3^{3/4} d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {7 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{3 \sqrt [4]{3} d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \]
-5/9*c^(1/6)*arctanh(1/3*(c^(1/3)+d^(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/d^ (5/3)+5/9*c^(1/6)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/d^(5/3)+5/9*c^(1/6) *arctan(c^(1/6)*(c^(1/3)+d^(1/3)*x)*3^(1/2)/(d*x^3+c)^(1/2))/d^(5/3)*3^(1/ 2)+1/3*x^2*(d*x^3+c)^(1/2)/d/(-d*x^3+8*c)+7/3*(d*x^3+c)^(1/2)/d^(5/3)/(d^( 1/3)*x+c^(1/3)*(1+3^(1/2)))+7/9*c^(1/3)*(c^(1/3)+d^(1/3)*x)*EllipticF((d^( 1/3)*x+c^(1/3)*(1-3^(1/2)))/(d^(1/3)*x+c^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I) *2^(1/2)*((c^(2/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/(d^(1/3)*x+c^(1/3)*(1+3^ (1/2)))^2)^(1/2)*3^(3/4)/d^(5/3)/(d*x^3+c)^(1/2)/(c^(1/3)*(c^(1/3)+d^(1/3) *x)/(d^(1/3)*x+c^(1/3)*(1+3^(1/2)))^2)^(1/2)-7/6*3^(1/4)*c^(1/3)*(c^(1/3)+ d^(1/3)*x)*EllipticE((d^(1/3)*x+c^(1/3)*(1-3^(1/2)))/(d^(1/3)*x+c^(1/3)*(1 +3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)-1/2*2^(1/2))*((c^(2/3)-c^(1/3)*d^(1 /3)*x+d^(2/3)*x^2)/(d^(1/3)*x+c^(1/3)*(1+3^(1/2)))^2)^(1/2)/d^(5/3)/(d*x^3 +c)^(1/2)/(c^(1/3)*(c^(1/3)+d^(1/3)*x)/(d^(1/3)*x+c^(1/3)*(1+3^(1/2)))^2)^ (1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.15 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.26 \[ \int \frac {x^4 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\frac {80 c x^2 \left (c+d x^3\right )+10 c x^2 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+7 d x^5 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{240 c d \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \]
(80*c*x^2*(c + d*x^3) + 10*c*x^2*(-8*c + d*x^3)*Sqrt[1 + (d*x^3)/c]*Appell F1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), (d*x^3)/(8*c)] + 7*d*x^5*(-8*c + d*x^3) *Sqrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), (d*x^3)/(8*c )])/(240*c*d*(8*c - d*x^3)*Sqrt[c + d*x^3])
Time = 1.02 (sec) , antiderivative size = 641, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {967, 27, 1054, 2009}
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.
| \(\displaystyle \int \frac {x^4 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 967 |
| \(\displaystyle \frac {x^2 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {\int \frac {x \left (7 d x^3+4 c\right )}{2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{3 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {x^2 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {\int \frac {x \left (7 d x^3+4 c\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{6 d}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {x^2 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {\int \left (\frac {60 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}-\frac {7 x}{\sqrt {d x^3+c}}\right )dx}{6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {x^2 \sqrt {c+d x^3}}{3 d \left (8 c-d x^3\right )}-\frac {-\frac {14 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {7 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {10 \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{\sqrt {3} d^{2/3}}+\frac {10 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{3 d^{2/3}}-\frac {10 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^{2/3}}-\frac {14 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}}{6 d}\) |
(x^2*Sqrt[c + d*x^3])/(3*d*(8*c - d*x^3)) - ((-14*Sqrt[c + d*x^3])/(d^(2/3 )*((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)) - (10*c^(1/6)*ArcTan[(Sqrt[3]*c^(1/ 6)*(c^(1/3) + d^(1/3)*x))/Sqrt[c + d*x^3]])/(Sqrt[3]*d^(2/3)) + (10*c^(1/6 )*ArcTanh[(c^(1/3) + d^(1/3)*x)^2/(3*c^(1/6)*Sqrt[c + d*x^3])])/(3*d^(2/3) ) - (10*c^(1/6)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(3*d^(2/3)) + (7*3^( 1/4)*Sqrt[2 - Sqrt[3]]*c^(1/3)*(c^(1/3) + d^(1/3)*x)*Sqrt[(c^(2/3) - c^(1/ 3)*d^(1/3)*x + d^(2/3)*x^2)/((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)^2]*Ellipti cE[ArcSin[((1 - Sqrt[3])*c^(1/3) + d^(1/3)*x)/((1 + Sqrt[3])*c^(1/3) + d^( 1/3)*x)], -7 - 4*Sqrt[3]])/(d^(2/3)*Sqrt[(c^(1/3)*(c^(1/3) + d^(1/3)*x))/( (1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)^2]*Sqrt[c + d*x^3]) - (14*Sqrt[2]*c^(1/ 3)*(c^(1/3) + d^(1/3)*x)*Sqrt[(c^(2/3) - c^(1/3)*d^(1/3)*x + d^(2/3)*x^2)/ ((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*c^( 1/3) + d^(1/3)*x)/((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)], -7 - 4*Sqrt[3]])/( 3^(1/4)*d^(2/3)*Sqrt[(c^(1/3)*(c^(1/3) + d^(1/3)*x))/((1 + Sqrt[3])*c^(1/3 ) + d^(1/3)*x)^2]*Sqrt[c + d*x^3]))/(6*d)
3.5.6.3.1 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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)* ((c + d*x^n)^q/(b*n*(p + 1))), x] - Simp[e^n/(b*n*(p + 1)) Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*( q - 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] && IntBino mialQ[a, b, c, d, e, m, n, p, q, x]
Int[(((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((e_) + (f_.)*(x_)^(n _)))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*((e + f*x^n)/(c + d*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && IGtQ[n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 4.57 (sec) , antiderivative size = 877, normalized size of antiderivative = 1.37
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(877\) |
| default | \(\text {Expression too large to display}\) | \(1741\) |
1/3*x^2*(d*x^3+c)^(1/2)/d/(-d*x^3+8*c)-7/9*I/d^2*3^(1/2)*(-c*d^2)^(1/3)*(I *(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2 )^(1/3))^(1/2)*((x-1/d*(-c*d^2)^(1/3))/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2 )/d*(-c*d^2)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c *d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*((-3/2/d*(-c* d^2)^(1/3)+1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*EllipticE(1/3*3^(1/2)*(I*(x+1/2 /d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3) )^(1/2),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/d *(-c*d^2)^(1/3)))^(1/2))+1/d*(-c*d^2)^(1/3)*EllipticF(1/3*3^(1/2)*(I*(x+1/ 2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d*(-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3 ))^(1/2),(I*3^(1/2)/d*(-c*d^2)^(1/3)/(-3/2/d*(-c*d^2)^(1/3)+1/2*I*3^(1/2)/ d*(-c*d^2)^(1/3)))^(1/2)))+10/27*I/d^4*2^(1/2)*sum(1/_alpha*(-c*d^2)^(1/3) *(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3)))/(-c*d^2)^(1 /3))^(1/2)*(d*(x-1/d*(-c*d^2)^(1/3))/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2) ^(1/3)))^(1/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3) ))/(-c*d^2)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-c*d^2)^(1/3)*_alpha*3^(1/2)* d-I*3^(1/2)*(-c*d^2)^(2/3)+2*_alpha^2*d^2-(-c*d^2)^(1/3)*_alpha*d-(-c*d^2) ^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-c*d^2)^(1/3)-1/2*I*3^(1/2)/d* (-c*d^2)^(1/3))*3^(1/2)*d/(-c*d^2)^(1/3))^(1/2),-1/18/d*(2*I*(-c*d^2)^(1/3 )*3^(1/2)*_alpha^2*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.73 (sec) , antiderivative size = 2397, normalized size of antiderivative = 3.74 \[ \int \frac {x^4 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\text {Too large to display} \]
-1/108*(36*sqrt(d*x^3 + c)*d*x^2 + 252*(d*x^3 - 8*c)*sqrt(d)*weierstrassZe ta(0, -4*c/d, weierstrassPInverse(0, -4*c/d, x)) + 5*(d^3*x^3 - 8*c*d^2 - sqrt(-3)*(d^3*x^3 - 8*c*d^2))*(c/d^10)^(1/6)*log(3125/3*((d^11*x^9 + 318*c *d^10*x^6 + 1200*c^2*d^9*x^3 + 640*c^3*d^8 + sqrt(-3)*(d^11*x^9 + 318*c*d^ 10*x^6 + 1200*c^2*d^9*x^3 + 640*c^3*d^8))*(c/d^10)^(5/6) + 6*(2*c*d^2*x^7 + 160*c^2*d*x^4 + 320*c^3*x - 6*(5*c*d^8*x^5 + 32*c^2*d^7*x^2 - sqrt(-3)*( 5*c*d^8*x^5 + 32*c^2*d^7*x^2))*(c/d^10)^(2/3) - (7*c*d^5*x^6 + 152*c^2*d^4 *x^3 + 64*c^3*d^3 + sqrt(-3)*(7*c*d^5*x^6 + 152*c^2*d^4*x^3 + 64*c^3*d^3)) *(c/d^10)^(1/3))*sqrt(d*x^3 + c) - 36*(5*c*d^7*x^7 + 64*c^2*d^6*x^4 + 32*c ^3*d^5*x)*sqrt(c/d^10) + 18*(c*d^4*x^8 + 38*c^2*d^3*x^5 + 64*c^3*d^2*x^2 - sqrt(-3)*(c*d^4*x^8 + 38*c^2*d^3*x^5 + 64*c^3*d^2*x^2))*(c/d^10)^(1/6))/( d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - 5*(d^3*x^3 - 8*c*d^2 - sqrt(-3)*(d^3*x^3 - 8*c*d^2))*(c/d^10)^(1/6)*log(-3125/3*((d^11*x^9 + 31 8*c*d^10*x^6 + 1200*c^2*d^9*x^3 + 640*c^3*d^8 + sqrt(-3)*(d^11*x^9 + 318*c *d^10*x^6 + 1200*c^2*d^9*x^3 + 640*c^3*d^8))*(c/d^10)^(5/6) - 6*(2*c*d^2*x ^7 + 160*c^2*d*x^4 + 320*c^3*x - 6*(5*c*d^8*x^5 + 32*c^2*d^7*x^2 - sqrt(-3 )*(5*c*d^8*x^5 + 32*c^2*d^7*x^2))*(c/d^10)^(2/3) - (7*c*d^5*x^6 + 152*c^2* d^4*x^3 + 64*c^3*d^3 + sqrt(-3)*(7*c*d^5*x^6 + 152*c^2*d^4*x^3 + 64*c^3*d^ 3))*(c/d^10)^(1/3))*sqrt(d*x^3 + c) - 36*(5*c*d^7*x^7 + 64*c^2*d^6*x^4 + 3 2*c^3*d^5*x)*sqrt(c/d^10) + 18*(c*d^4*x^8 + 38*c^2*d^3*x^5 + 64*c^3*d^2...
\[ \int \frac {x^4 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\int \frac {x^{4} \sqrt {c + d x^{3}}}{\left (- 8 c + d x^{3}\right )^{2}}\, dx \]
\[ \int \frac {x^4 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\int { \frac {\sqrt {d x^{3} + c} x^{4}}{{\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \]
\[ \int \frac {x^4 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\int { \frac {\sqrt {d x^{3} + c} x^{4}}{{\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^4 \sqrt {c+d x^3}}{\left (8 c-d x^3\right )^2} \, dx=\int \frac {x^4\,\sqrt {d\,x^3+c}}{{\left (8\,c-d\,x^3\right )}^2} \,d x \]