Integrand size = 27, antiderivative size = 668 \[ \int \frac {x^7}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=-\frac {2 x^2}{81 c d^2 \sqrt {c+d x^3}}+\frac {8 x^2}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}+\frac {2 \sqrt {c+d x^3}}{81 c d^{8/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}+\frac {4 \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{81 \sqrt {3} c^{5/6} d^{8/3}}-\frac {4 \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{243 c^{5/6} d^{8/3}}+\frac {4 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{243 c^{5/6} d^{8/3}}-\frac {\sqrt {2-\sqrt {3}} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{27\ 3^{3/4} c^{2/3} d^{8/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {2 \sqrt {2} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{81 \sqrt [4]{3} c^{2/3} d^{8/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
-2/81*x^2/c/d^2/(d*x^3+c)^(1/2)+8/27*x^2/d^2/(-d*x^3+8*c)/(d*x^3+c)^(1/2)+ 2/81*(d*x^3+c)^(1/2)/c/d^(8/3)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)+4/243*arcta n(3^(1/2)*c^(1/6)*(c^(1/3)+d^(1/3)*x)/(d*x^3+c)^(1/2))*3^(1/2)/c^(5/6)/d^( 8/3)-4/243*arctanh(1/3*(c^(1/3)+d^(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/c^(5 /6)/d^(8/3)+4/243*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(5/6)/d^(8/3)-1/8 1*(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)*EllipticE(((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^(2/3)/d^(8/3)/(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)+2/243*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) *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)*3^(3/4)/c^(2/3)/d^(8/3)/(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 = 10.13 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.25 \[ \int \frac {x^7}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\frac {80 c x^2 \left (4 c+d x^3\right )+40 c x^2 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+d x^5 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{3240 c^2 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \] Input:
Integrate[x^7/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
Output:
(80*c*x^2*(4*c + d*x^3) + 40*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 )])/(3240*c^2*d^2*(8*c - d*x^3)*Sqrt[c + d*x^3])
Time = 1.75 (sec) , antiderivative size = 669, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {970, 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.
| \(\displaystyle \int \frac {x^7}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 970 |
| \(\displaystyle \frac {8 x^2}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\int \frac {c x \left (7 d x^3+16 c\right )}{\left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx}{27 c d^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {8 x^2}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\int \frac {x \left (7 d x^3+16 c\right )}{\left (8 c-d x^3\right ) \left (d x^3+c\right )^{3/2}}dx}{27 d^2}\) |
| \(\Big \downarrow \) 1049 |
| \(\displaystyle \frac {8 x^2}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\frac {2 x^2}{3 c \sqrt {c+d x^3}}-\frac {2 \int -\frac {9 c d x \left (d x^3+16 c\right )}{2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{27 c^2 d}}{27 d^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {8 x^2}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\frac {\int \frac {x \left (d x^3+16 c\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{3 c}+\frac {2 x^2}{3 c \sqrt {c+d x^3}}}{27 d^2}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {8 x^2}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\frac {\int \left (\frac {24 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}-\frac {x}{\sqrt {d x^3+c}}\right )dx}{3 c}+\frac {2 x^2}{3 c \sqrt {c+d x^3}}}{27 d^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {8 x^2}{27 d^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}}-\frac {\frac {-\frac {2 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {4 \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{\sqrt {3} d^{2/3}}+\frac {4 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{3 d^{2/3}}-\frac {4 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^{2/3}}-\frac {2 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}}{3 c}+\frac {2 x^2}{3 c \sqrt {c+d x^3}}}{27 d^2}\) |
Input:
Int[x^7/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]
Output:
(8*x^2)/(27*d^2*(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) ) - (4*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)) + (4*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)) - (4*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]*Elli pticF[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))/ (27*d^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[(-a)*e^(2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n) ^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Simp[e^(2*n) /(b*n*(b*c - a*d)*(p + 1)) Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d *x^n)^q*Simp[a*c*(m - 2*n + 1) + (a*d*(m - n + n*q + 1) + b*c*n*(p + 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] && GtQ[m - n + 1, n] && 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 = 1.99 (sec) , antiderivative size = 910, normalized size of antiderivative = 1.36
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(910\) |
| default | \(\text {Expression too large to display}\) | \(2256\) |
Input:
int(x^7/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
2/243/d^2*x^2/c/((x^3+c/d)*d)^(1/2)+8/243*x^2/c/d^2*(d*x^3+c)^(1/2)/(-d*x^ 3+8*c)-2/243*I/d^3/c*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)))+8 /729*I/d^5/c*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. 2681 vs. \(2 (475) = 950\).
Time = 0.67 (sec) , antiderivative size = 2681, normalized size of antiderivative = 4.01 \[ \int \frac {x^7}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(x^7/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
-1/729*(18*(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*d^5*x^6 - 7*c^2*d^4*x^3 - 8*c^3* d^3 + sqrt(-3)*(c*d^5*x^6 - 7*c^2*d^4*x^3 - 8*c^3*d^3))*(1/(c^5*d^16))^(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^4*d^13 *x^7 + 64*c^5*d^12*x^4 + 32*c^6*d^11*x + sqrt(-3)*(5*c^4*d^13*x^7 + 64*c^5 *d^12*x^4 + 32*c^6*d^11*x))*(1/(c^5*d^16))^(2/3) + 3*sqrt(d*x^3 + c)*(6*(5 *c^5*d^15*x^5 + 32*c^6*d^14*x^2 - sqrt(-3)*(5*c^5*d^15*x^5 + 32*c^6*d^14*x ^2))*(1/(c^5*d^16))^(5/6) - 2*(7*c^3*d^10*x^6 + 152*c^4*d^9*x^3 + 64*c^5*d ^8)*sqrt(1/(c^5*d^16)) + (c*d^5*x^7 + 80*c^2*d^4*x^4 + 160*c^3*d^3*x + sqr t(-3)*(c*d^5*x^7 + 80*c^2*d^4*x^4 + 160*c^3*d^3*x))*(1/(c^5*d^16))^(1/6)) - 9*(c^2*d^8*x^8 + 38*c^3*d^7*x^5 + 64*c^4*d^6*x^2 - sqrt(-3)*(c^2*d^8*x^8 + 38*c^3*d^7*x^5 + 64*c^4*d^6*x^2))*(1/(c^5*d^16))^(1/3))/(d^3*x^9 - 24*c *d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - (c*d^5*x^6 - 7*c^2*d^4*x^3 - 8*c^3* d^3 + sqrt(-3)*(c*d^5*x^6 - 7*c^2*d^4*x^3 - 8*c^3*d^3))*(1/(c^5*d^16))^(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^4*d^13 *x^7 + 64*c^5*d^12*x^4 + 32*c^6*d^11*x + sqrt(-3)*(5*c^4*d^13*x^7 + 64*c^5 *d^12*x^4 + 32*c^6*d^11*x))*(1/(c^5*d^16))^(2/3) - 3*sqrt(d*x^3 + c)*(6*(5 *c^5*d^15*x^5 + 32*c^6*d^14*x^2 - sqrt(-3)*(5*c^5*d^15*x^5 + 32*c^6*d^14*x ^2))*(1/(c^5*d^16))^(5/6) - 2*(7*c^3*d^10*x^6 + 152*c^4*d^9*x^3 + 64*c^5*d ^8)*sqrt(1/(c^5*d^16)) + (c*d^5*x^7 + 80*c^2*d^4*x^4 + 160*c^3*d^3*x + ...
\[ \int \frac {x^7}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x^{7}}{\left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(x**7/(-d*x**3+8*c)**2/(d*x**3+c)**(3/2),x)
Output:
Integral(x**7/((-8*c + d*x**3)**2*(c + d*x**3)**(3/2)), x)
\[ \int \frac {x^7}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {x^{7}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:
integrate(x^7/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
Output:
integrate(x^7/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2), x)
\[ \int \frac {x^7}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int { \frac {x^{7}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:
integrate(x^7/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
integrate(x^7/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2), x)
Timed out. \[ \int \frac {x^7}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {x^7}{{\left (d\,x^3+c\right )}^{3/2}\,{\left (8\,c-d\,x^3\right )}^2} \,d x \] Input:
int(x^7/((c + d*x^3)^(3/2)*(8*c - d*x^3)^2),x)
Output:
int(x^7/((c + d*x^3)^(3/2)*(8*c - d*x^3)^2), x)
\[ \int \frac {x^7}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x^{7}}{d^{4} x^{12}-14 c \,d^{3} x^{9}+33 c^{2} d^{2} x^{6}+112 c^{3} d \,x^{3}+64 c^{4}}d x \] Input:
int(x^7/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x)
Output:
int((sqrt(c + d*x**3)*x**7)/(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)