Integrand size = 25, antiderivative size = 665 \[ \int \frac {x}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {5 x^2}{648 c^3 \sqrt {c+d x^3}}+\frac {x^2}{216 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {5 \sqrt {c+d x^3}}{648 c^3 d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}-\frac {5 \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{1296 \sqrt {3} c^{17/6} d^{2/3}}+\frac {5 \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{3888 c^{17/6} d^{2/3}}-\frac {5 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3888 c^{17/6} d^{2/3}}+\frac {5 \sqrt {2-\sqrt {3}} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{432\ 3^{3/4} c^{8/3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {5 \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{324 \sqrt {2} \sqrt [4]{3} c^{8/3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
5/648*x^2/c^3/(d*x^3+c)^(1/2)+1/216*x^2/c^2/(-d*x^3+8*c)/(d*x^3+c)^(1/2)-5 /648*(d*x^3+c)^(1/2)/c^3/d^(2/3)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)-5/3888*ar ctan(3^(1/2)*c^(1/6)*(c^(1/3)+d^(1/3)*x)/(d*x^3+c)^(1/2))*3^(1/2)/c^(17/6) /d^(2/3)+5/3888*arctanh(1/3*(c^(1/3)+d^(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2)) /c^(17/6)/d^(2/3)-5/3888*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(17/6)/d^( 2/3)+5/1296*(1/2*6^(1/2)-1/2*2^(1/2))*(c^(1/3)+d^(1/3)*x)*((c^(2/3)-c^(1/3 )*d^(1/3)*x+d^(2/3)*x^2)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)^2)^(1/2)*Elliptic E(((1-3^(1/2))*c^(1/3)+d^(1/3)*x)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x),I*3^(1/2 )+2*I)*3^(1/4)/c^(8/3)/d^(2/3)/(c^(1/3)*(c^(1/3)+d^(1/3)*x)/((1+3^(1/2))*c ^(1/3)+d^(1/3)*x)^2)^(1/2)/(d*x^3+c)^(1/2)-5/1944*(c^(1/3)+d^(1/3)*x)*((c^ (2/3)-c^(1/3)*d^(1/3)*x+d^(2/3)*x^2)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)^2)^(1 /2)*EllipticF(((1-3^(1/2))*c^(1/3)+d^(1/3)*x)/((1+3^(1/2))*c^(1/3)+d^(1/3) *x),I*3^(1/2)+2*I)*2^(1/2)*3^(3/4)/c^(8/3)/d^(2/3)/(c^(1/3)*(c^(1/3)+d^(1/ 3)*x)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)^2)^(1/2)/(d*x^3+c)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 10.16 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.25 \[ \int \frac {x}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {16 c x^2 \left (43 c-5 d x^3\right )+5 c x^2 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+d x^5 \left (8 c-d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{10368 c^4 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \] Input:
Integrate[x/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
Output:
(16*c*x^2*(43*c - 5*d*x^3) + 5*c*x^2*(-8*c + d*x^3)*Sqrt[1 + (d*x^3)/c]*Ap pellF1[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)])/(10368*c^4*(8*c - d*x^3)*Sqrt[c + d*x^3])
Time = 1.77 (sec) , antiderivative size = 669, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {972, 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.
| \(\displaystyle \int \frac {x}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 972 |
| \(\displaystyle \frac {\int \frac {5 d x \left (d x^3+10 c\right )}{2 \left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx}{216 c^2 d}+\frac {x^2}{216 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5 \int \frac {x \left (d x^3+10 c\right )}{\left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx}{432 c^2}+\frac {x^2}{216 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 1049 |
| \(\displaystyle \frac {5 \left (\frac {2 x^2}{3 c \sqrt {c+d x^3}}-\frac {2 \int \frac {9 c d x \left (2 c-d x^3\right )}{2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{27 c^2 d}\right )}{432 c^2}+\frac {x^2}{216 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {5 \left (\frac {2 x^2}{3 c \sqrt {c+d x^3}}-\frac {\int \frac {x \left (2 c-d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{3 c}\right )}{432 c^2}+\frac {x^2}{216 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {5 \left (\frac {2 x^2}{3 c \sqrt {c+d x^3}}-\frac {\int \left (\frac {x}{\sqrt {d x^3+c}}-\frac {6 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}\right )dx}{3 c}\right )}{432 c^2}+\frac {x^2}{216 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {5 \left (\frac {2 x^2}{3 c \sqrt {c+d x^3}}-\frac {\frac {2 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{\sqrt {3} d^{2/3}}-\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{3 d^{2/3}}+\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^{2/3}}+\frac {2 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}}{3 c}\right )}{432 c^2}+\frac {x^2}{216 c^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}\) |
Input:
Int[x/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
Output:
x^2/(216*c^2*(8*c - d*x^3)*Sqrt[c + d*x^3]) + (5*((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)) + (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)) - (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)) + (c^(1/6)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt [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[A rcSin[((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)))/(432*c ^2)
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[(-b)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x ^n)^(q + 1)/(a*e*n*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*n*(b*c - a*d)*(p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*b*(m + 1) + n*( b*c - a*d)*(p + 1) + d*b*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{ a, b, c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] & & 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 = 1.11 (sec) , antiderivative size = 904, normalized size of antiderivative = 1.36
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(904\) |
| elliptic | \(\text {Expression too large to display}\) | \(904\) |
Input:
int(x/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/243*x^2/c^3/((x^3+c/d)*d)^(1/2)+1/1944/c^3*x^2*(d*x^3+c)^(1/2)/(-d*x^3+8 *c)+5/1944*I/c^3*3^(1/2)/d*(-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)))-5/5 832*I/c^3/d^3*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)*_alpha^2*3^(1/2)*d-I...
Leaf count of result is larger than twice the leaf count of optimal. 2684 vs. \(2 (472) = 944\).
Time = 0.76 (sec) , antiderivative size = 2684, normalized size of antiderivative = 4.04 \[ \int \frac {x}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(x/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
1/46656*(360*(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)) + 5*(c^3*d^3*x^6 - 7*c^4*d^2*x^3 - 8*c^5*d + sqrt(-3)*(c^3*d^3*x^6 - 7*c^4*d^2*x^3 - 8*c^5*d))*(1/(c^17*d^4)) ^(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^12 *d^5*x^7 + 64*c^13*d^4*x^4 + 32*c^14*d^3*x + sqrt(-3)*(5*c^12*d^5*x^7 + 64 *c^13*d^4*x^4 + 32*c^14*d^3*x))*(1/(c^17*d^4))^(2/3) + 3*sqrt(d*x^3 + c)*( 6*(5*c^15*d^5*x^5 + 32*c^16*d^4*x^2 - sqrt(-3)*(5*c^15*d^5*x^5 + 32*c^16*d ^4*x^2))*(1/(c^17*d^4))^(5/6) - 2*(7*c^9*d^4*x^6 + 152*c^10*d^3*x^3 + 64*c ^11*d^2)*sqrt(1/(c^17*d^4)) + (c^3*d^3*x^7 + 80*c^4*d^2*x^4 + 160*c^5*d*x + sqrt(-3)*(c^3*d^3*x^7 + 80*c^4*d^2*x^4 + 160*c^5*d*x))*(1/(c^17*d^4))^(1 /6)) - 9*(c^6*d^4*x^8 + 38*c^7*d^3*x^5 + 64*c^8*d^2*x^2 - sqrt(-3)*(c^6*d^ 4*x^8 + 38*c^7*d^3*x^5 + 64*c^8*d^2*x^2))*(1/(c^17*d^4))^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - 5*(c^3*d^3*x^6 - 7*c^4*d^2*x^3 - 8*c^5*d + sqrt(-3)*(c^3*d^3*x^6 - 7*c^4*d^2*x^3 - 8*c^5*d))*(1/(c^17*d^ 4))^(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 ^12*d^5*x^7 + 64*c^13*d^4*x^4 + 32*c^14*d^3*x + sqrt(-3)*(5*c^12*d^5*x^7 + 64*c^13*d^4*x^4 + 32*c^14*d^3*x))*(1/(c^17*d^4))^(2/3) - 3*sqrt(d*x^3 + c )*(6*(5*c^15*d^5*x^5 + 32*c^16*d^4*x^2 - sqrt(-3)*(5*c^15*d^5*x^5 + 32*c^1 6*d^4*x^2))*(1/(c^17*d^4))^(5/6) - 2*(7*c^9*d^4*x^6 + 152*c^10*d^3*x^3 + 6 4*c^11*d^2)*sqrt(1/(c^17*d^4)) + (c^3*d^3*x^7 + 80*c^4*d^2*x^4 + 160*c^...
\[ \int \frac {x}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x}{\left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x/(-d*x**3+8*c)**2/(d*x**3+c)**(3/2),x)
Output:
Integral(x/((-8*c + d*x**3)**2*(c + d*x**3)**(3/2)), x)
\[ \int \frac {x}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {x}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:
integrate(x/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
Output:
integrate(x/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2), x)
\[ \int \frac {x}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {x}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:
integrate(x/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
integrate(x/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2), x)
Timed out. \[ \int \frac {x}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x}{{\left (d\,x^3+c\right )}^{3/2}\,{\left (8\,c-d\,x^3\right )}^2} \,d x \] Input:
int(x/((c + d*x^3)^(3/2)*(8*c - d*x^3)^2),x)
Output:
int(x/((c + d*x^3)^(3/2)*(8*c - d*x^3)^2), x)
\[ \int \frac {x}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x}{d^{4} x^{12}-14 c \,d^{3} x^{9}+33 c^{2} d^{2} x^{6}+112 c^{3} d \,x^{3}+64 c^{4}}d x \] Input:
int(x/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x)
Output:
int((sqrt(c + d*x**3)*x)/(64*c**4 + 112*c**3*d*x**3 + 33*c**2*d**2*x**6 - 14*c*d**3*x**9 + d**4*x**12),x)