Integrand size = 27, antiderivative size = 671 \[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {x^2}{81 c^2 d \sqrt {c+d x^3}}+\frac {x^2}{27 c d \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {\sqrt {c+d x^3}}{81 c^2 d^{5/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}-\frac {\arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{81 \sqrt {3} c^{11/6} d^{5/3}}+\frac {\text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{243 c^{11/6} d^{5/3}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{243 c^{11/6} d^{5/3}}-\frac {\sqrt {2-\sqrt {3}} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{54\ 3^{3/4} c^{5/3} d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt {2} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{81 \sqrt [4]{3} c^{5/3} d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \]
1/243*arctanh(1/3*(c^(1/3)+d^(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/c^(11/6)/ d^(5/3)-1/243*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(11/6)/d^(5/3)-1/243* arctan(c^(1/6)*(c^(1/3)+d^(1/3)*x)*3^(1/2)/(d*x^3+c)^(1/2))/c^(11/6)/d^(5/ 3)*3^(1/2)-1/81*x^2/c^2/d/(d*x^3+c)^(1/2)+1/27*x^2/c/d/(-d*x^3+8*c)/(d*x^3 +c)^(1/2)+1/81*(d*x^3+c)^(1/2)/c^2/d^(5/3)/(d^(1/3)*x+c^(1/3)*(1+3^(1/2))) +1/243*(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)/c^(5/3)/ 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)-1/162*(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)*3^(1/4)/c^(5/3)/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 = 10.12 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.25 \[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {80 c x^2 \left (-5 c+d x^3\right )+50 c x^2 \left (8 c-d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+d x^5 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{6480 c^3 d \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \]
(80*c*x^2*(-5*c + d*x^3) + 50*c*x^2*(8*c - d*x^3)*Sqrt[1 + (d*x^3)/c]*Appe llF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), (d*x^3)/(8*c)] + 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 )])/(6480*c^3*d*(8*c - d*x^3)*Sqrt[c + d*x^3])
Time = 1.08 (sec) , antiderivative size = 676, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {971, 27, 1049, 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.
\(\displaystyle \int \frac {x^4}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 971 |
\(\displaystyle \frac {x^2}{27 c d \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\int \frac {x \left (4 c-5 d x^3\right )}{2 \left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx}{27 c d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^2}{27 c d \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\int \frac {x \left (4 c-5 d x^3\right )}{\left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx}{54 c d}\) |
\(\Big \downarrow \) 1049 |
\(\displaystyle \frac {x^2}{27 c d \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\frac {2 x^2}{3 c \sqrt {c+d x^3}}-\frac {2 \int \frac {9 c d x \left (20 c-d x^3\right )}{2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{27 c^2 d}}{54 c d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {x^2}{27 c d \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\frac {2 x^2}{3 c \sqrt {c+d x^3}}-\frac {\int \frac {x \left (20 c-d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{3 c}}{54 c d}\) |
\(\Big \downarrow \) 1054 |
\(\displaystyle \frac {x^2}{27 c d \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\frac {2 x^2}{3 c \sqrt {c+d x^3}}-\frac {\int \left (\frac {12 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}+\frac {x}{\sqrt {d x^3+c}}\right )dx}{3 c}}{54 c d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^2}{27 c d \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\frac {2 x^2}{3 c \sqrt {c+d x^3}}-\frac {\frac {2 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {2 \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{\sqrt {3} d^{2/3}}+\frac {2 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{3 d^{2/3}}-\frac {2 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^{2/3}}+\frac {2 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}}{3 c}}{54 c d}\) |
x^2/(27*c*d*(8*c - d*x^3)*Sqrt[c + d*x^3]) - ((2*x^2)/(3*c*Sqrt[c + d*x^3] ) - ((2*Sqrt[c + d*x^3])/(d^(2/3)*((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)) - ( 2*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)) + (2*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)) - (2*c^(1/6)*ArcTanh[Sqrt[c + d*x^3]/(3*Sq rt[c])])/(3*d^(2/3)) - (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]*EllipticE[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]) + (2*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]))/(3*c))/(54*c *d)
3.5.50.3.1 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 + 1)/(n*(b*c - a*d)*(p + 1))), x] - Simp[e^n/(n*(b*c - a*d) *(p + 1)) Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e , q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*g*n*(b*c - a*d)*(p + 1))) , x] + Simp[1/(a*n*(b*c - a*d)*(p + 1)) Int[(g*x)^m*(a + b*x^n)^(p + 1)*( c + d*x^n)^q*Simp[c*(b*e - a*f)*(m + 1) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, q}, x] && IGtQ[n, 0] && LtQ[p, -1]
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 = 4.62 (sec) , antiderivative size = 910, normalized size of antiderivative = 1.36
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(910\) |
default | \(\text {Expression too large to display}\) | \(1789\) |
1/243*x^2/c^2/d*(d*x^3+c)^(1/2)/(-d*x^3+8*c)-2/243/d*x^2/c^2/((x^3+c/d)*d) ^(1/2)-1/243*I/c^2/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))) -2/729*I/c^2/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*...
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.65 (sec) , antiderivative size = 2723, normalized size of antiderivative = 4.06 \[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \]
-1/2916*(36*(d^2*x^6 - 7*c*d*x^3 - 8*c^2)*sqrt(d)*weierstrassZeta(0, -4*c/ d, weierstrassPInverse(0, -4*c/d, x)) - (c^2*d^4*x^6 - 7*c^3*d^3*x^3 - 8*c ^4*d^2 + sqrt(-3)*(c^2*d^4*x^6 - 7*c^3*d^3*x^3 - 8*c^4*d^2))*(1/(c^11*d^10 ))^(1/6)*log((d^3*x^9 + 318*c*d^2*x^6 + 1200*c^2*d*x^3 + 640*c^3 - 9*(5*c^ 8*d^9*x^7 + 64*c^9*d^8*x^4 + 32*c^10*d^7*x + sqrt(-3)*(5*c^8*d^9*x^7 + 64* c^9*d^8*x^4 + 32*c^10*d^7*x))*(1/(c^11*d^10))^(2/3) + 3*sqrt(d*x^3 + c)*(6 *(5*c^10*d^10*x^5 + 32*c^11*d^9*x^2 - sqrt(-3)*(5*c^10*d^10*x^5 + 32*c^11* d^9*x^2))*(1/(c^11*d^10))^(5/6) - 2*(7*c^6*d^7*x^6 + 152*c^7*d^6*x^3 + 64* c^8*d^5)*sqrt(1/(c^11*d^10)) + (c^2*d^4*x^7 + 80*c^3*d^3*x^4 + 160*c^4*d^2 *x + sqrt(-3)*(c^2*d^4*x^7 + 80*c^3*d^3*x^4 + 160*c^4*d^2*x))*(1/(c^11*d^1 0))^(1/6)) - 9*(c^4*d^6*x^8 + 38*c^5*d^5*x^5 + 64*c^6*d^4*x^2 - sqrt(-3)*( c^4*d^6*x^8 + 38*c^5*d^5*x^5 + 64*c^6*d^4*x^2))*(1/(c^11*d^10))^(1/3))/(d^ 3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) + (c^2*d^4*x^6 - 7*c^3*d^ 3*x^3 - 8*c^4*d^2 + sqrt(-3)*(c^2*d^4*x^6 - 7*c^3*d^3*x^3 - 8*c^4*d^2))*(1 /(c^11*d^10))^(1/6)*log((d^3*x^9 + 318*c*d^2*x^6 + 1200*c^2*d*x^3 + 640*c^ 3 - 9*(5*c^8*d^9*x^7 + 64*c^9*d^8*x^4 + 32*c^10*d^7*x + sqrt(-3)*(5*c^8*d^ 9*x^7 + 64*c^9*d^8*x^4 + 32*c^10*d^7*x))*(1/(c^11*d^10))^(2/3) - 3*sqrt(d* x^3 + c)*(6*(5*c^10*d^10*x^5 + 32*c^11*d^9*x^2 - sqrt(-3)*(5*c^10*d^10*x^5 + 32*c^11*d^9*x^2))*(1/(c^11*d^10))^(5/6) - 2*(7*c^6*d^7*x^6 + 152*c^7*d^ 6*x^3 + 64*c^8*d^5)*sqrt(1/(c^11*d^10)) + (c^2*d^4*x^7 + 80*c^3*d^3*x^4...
\[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x^{4}}{\left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {x^{4}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \]
\[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {x^{4}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x^4}{{\left (d\,x^3+c\right )}^{3/2}\,{\left (8\,c-d\,x^3\right )}^2} \,d x \]