Integrand size = 27, antiderivative size = 675 \[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2}{27 c^2 x^4 \sqrt {c+d x^3}}-\frac {91 \sqrt {c+d x^3}}{864 c^3 x^4}+\frac {113 d \sqrt {c+d x^3}}{432 c^4 x}-\frac {113 d^{4/3} \sqrt {c+d x^3}}{432 c^4 \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}-\frac {d^{4/3} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3456 \sqrt {3} c^{23/6}}+\frac {d^{4/3} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{10368 c^{23/6}}-\frac {d^{4/3} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{10368 c^{23/6}}+\frac {113 \sqrt {2-\sqrt {3}} d^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{288\ 3^{3/4} c^{11/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {113 d^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{216 \sqrt {2} \sqrt [4]{3} c^{11/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
2/27/c^2/x^4/(d*x^3+c)^(1/2)-91/864*(d*x^3+c)^(1/2)/c^3/x^4+113/432*d*(d*x ^3+c)^(1/2)/c^4/x-113/432*d^(4/3)*(d*x^3+c)^(1/2)/c^4/((1+3^(1/2))*c^(1/3) +d^(1/3)*x)-1/10368*d^(4/3)*arctan(3^(1/2)*c^(1/6)*(c^(1/3)+d^(1/3)*x)/(d* x^3+c)^(1/2))*3^(1/2)/c^(23/6)+1/10368*d^(4/3)*arctanh(1/3*(c^(1/3)+d^(1/3 )*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/c^(23/6)-1/10368*d^(4/3)*arctanh(1/3*(d*x^ 3+c)^(1/2)/c^(1/2))/c^(23/6)+113/864*(1/2*6^(1/2)-1/2*2^(1/2))*d^(4/3)*(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^(11/3)/(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)-113/1 296*d^(4/3)*(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^( 11/3)/(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 = 11.12 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.23 \[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {160 c \left (-27 c^2+135 c d x^3+226 d^2 x^6\right )-9025 c d^2 x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+452 d^3 x^9 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{138240 c^5 x^4 \sqrt {c+d x^3}} \] Input:
Integrate[1/(x^5*(8*c - d*x^3)*(c + d*x^3)^(3/2)),x]
Output:
(160*c*(-27*c^2 + 135*c*d*x^3 + 226*d^2*x^6) - 9025*c*d^2*x^6*Sqrt[1 + (d* x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), (d*x^3)/(8*c)] + 452*d^3* x^9*Sqrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), (d*x^3)/( 8*c)])/(138240*c^5*x^4*Sqrt[c + d*x^3])
Time = 1.88 (sec) , antiderivative size = 686, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {972, 27, 1053, 27, 1053, 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.
| \(\displaystyle \int \frac {1}{x^5 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 972 |
| \(\displaystyle \frac {2}{27 c^2 x^4 \sqrt {c+d x^3}}-\frac {2 \int -\frac {d \left (91 c-11 d x^3\right )}{2 x^5 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{27 c^2 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {91 c-11 d x^3}{x^5 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{27 c^2}+\frac {2}{27 c^2 x^4 \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 1053 |
| \(\displaystyle \frac {-\frac {\int \frac {c d \left (3616 c-455 d x^3\right )}{2 x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{32 c^2}-\frac {91 \sqrt {c+d x^3}}{32 c x^4}}{27 c^2}+\frac {2}{27 c^2 x^4 \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {d \int \frac {3616 c-455 d x^3}{x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{64 c}-\frac {91 \sqrt {c+d x^3}}{32 c x^4}}{27 c^2}+\frac {2}{27 c^2 x^4 \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 1053 |
| \(\displaystyle \frac {-\frac {d \left (-\frac {\int -\frac {8 c d x \left (1805 c-226 d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{8 c^2}-\frac {452 \sqrt {c+d x^3}}{c x}\right )}{64 c}-\frac {91 \sqrt {c+d x^3}}{32 c x^4}}{27 c^2}+\frac {2}{27 c^2 x^4 \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {d \left (\frac {d \int \frac {x \left (1805 c-226 d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{c}-\frac {452 \sqrt {c+d x^3}}{c x}\right )}{64 c}-\frac {91 \sqrt {c+d x^3}}{32 c x^4}}{27 c^2}+\frac {2}{27 c^2 x^4 \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {-\frac {d \left (\frac {d \int \left (\frac {226 x}{\sqrt {d x^3+c}}-\frac {3 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}\right )dx}{c}-\frac {452 \sqrt {c+d x^3}}{c x}\right )}{64 c}-\frac {91 \sqrt {c+d x^3}}{32 c x^4}}{27 c^2}+\frac {2}{27 c^2 x^4 \sqrt {c+d x^3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {d \left (\frac {d \left (\frac {452 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {226 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{2 \sqrt {3} d^{2/3}}-\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{6 d^{2/3}}+\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 d^{2/3}}+\frac {452 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )}{c}-\frac {452 \sqrt {c+d x^3}}{c x}\right )}{64 c}-\frac {91 \sqrt {c+d x^3}}{32 c x^4}}{27 c^2}+\frac {2}{27 c^2 x^4 \sqrt {c+d x^3}}\) |
Input:
Int[1/(x^5*(8*c - d*x^3)*(c + d*x^3)^(3/2)),x]
Output:
2/(27*c^2*x^4*Sqrt[c + d*x^3]) + ((-91*Sqrt[c + d*x^3])/(32*c*x^4) - (d*(( -452*Sqrt[c + d*x^3])/(c*x) + (d*((452*Sqrt[c + d*x^3])/(d^(2/3)*((1 + Sqr t[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]])/(2*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])])/(6*d^(2/3)) + (c^(1/6)*Ar cTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(6*d^(2/3)) - (226*3^(1/4)*Sqrt[2 - Sq rt[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]) + (452*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])))/c))/(64*c))/(27*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[e*(g*x)^(m + 1)*(a + b *x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^n*( m + 1)) Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2 ) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && LtQ[m, -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 = 2.96 (sec) , antiderivative size = 911, normalized size of antiderivative = 1.35
| method | result | size |
| elliptic | \(\text {Expression too large to display}\) | \(911\) |
| risch | \(\text {Expression too large to display}\) | \(1344\) |
| default | \(\text {Expression too large to display}\) | \(1864\) |
Input:
int(1/x^5/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/32*(d*x^3+c)^(1/2)/c^3/x^4+3/16*d*(d*x^3+c)^(1/2)/c^4/x+2/27*d^2/c^4*x^ 2/((x^3+c/d)*d)^(1/2)+113/1296*I*d/c^4*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)))-1/15552*I/d/c^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)*_...
Leaf count of result is larger than twice the leaf count of optimal. 2529 vs. \(2 (479) = 958\).
Time = 2.54 (sec) , antiderivative size = 2529, normalized size of antiderivative = 3.75 \[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/x^5/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="fricas")
Output:
1/124416*(32544*(d^2*x^7 + c*d*x^4)*sqrt(d)*weierstrassZeta(0, -4*c/d, wei erstrassPInverse(0, -4*c/d, x)) + (c^4*d*x^7 + c^5*x^4 + sqrt(-3)*(c^4*d*x ^7 + c^5*x^4))*(d^8/c^23)^(1/6)*log((d^9*x^9 + 318*c*d^8*x^6 + 1200*c^2*d^ 7*x^3 + 640*c^3*d^6 - 9*(5*c^16*d^3*x^7 + 64*c^17*d^2*x^4 + 32*c^18*d*x + sqrt(-3)*(5*c^16*d^3*x^7 + 64*c^17*d^2*x^4 + 32*c^18*d*x))*(d^8/c^23)^(2/3 ) + 3*sqrt(d*x^3 + c)*(6*(5*c^20*d*x^5 + 32*c^21*x^2 - sqrt(-3)*(5*c^20*d* x^5 + 32*c^21*x^2))*(d^8/c^23)^(5/6) - 2*(7*c^12*d^4*x^6 + 152*c^13*d^3*x^ 3 + 64*c^14*d^2)*sqrt(d^8/c^23) + (c^4*d^7*x^7 + 80*c^5*d^6*x^4 + 160*c^6* d^5*x + sqrt(-3)*(c^4*d^7*x^7 + 80*c^5*d^6*x^4 + 160*c^6*d^5*x))*(d^8/c^23 )^(1/6)) - 9*(c^8*d^6*x^8 + 38*c^9*d^5*x^5 + 64*c^10*d^4*x^2 - sqrt(-3)*(c ^8*d^6*x^8 + 38*c^9*d^5*x^5 + 64*c^10*d^4*x^2))*(d^8/c^23)^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - (c^4*d*x^7 + c^5*x^4 + sqrt( -3)*(c^4*d*x^7 + c^5*x^4))*(d^8/c^23)^(1/6)*log((d^9*x^9 + 318*c*d^8*x^6 + 1200*c^2*d^7*x^3 + 640*c^3*d^6 - 9*(5*c^16*d^3*x^7 + 64*c^17*d^2*x^4 + 32 *c^18*d*x + sqrt(-3)*(5*c^16*d^3*x^7 + 64*c^17*d^2*x^4 + 32*c^18*d*x))*(d^ 8/c^23)^(2/3) - 3*sqrt(d*x^3 + c)*(6*(5*c^20*d*x^5 + 32*c^21*x^2 - sqrt(-3 )*(5*c^20*d*x^5 + 32*c^21*x^2))*(d^8/c^23)^(5/6) - 2*(7*c^12*d^4*x^6 + 152 *c^13*d^3*x^3 + 64*c^14*d^2)*sqrt(d^8/c^23) + (c^4*d^7*x^7 + 80*c^5*d^6*x^ 4 + 160*c^6*d^5*x + sqrt(-3)*(c^4*d^7*x^7 + 80*c^5*d^6*x^4 + 160*c^6*d^5*x ))*(d^8/c^23)^(1/6)) - 9*(c^8*d^6*x^8 + 38*c^9*d^5*x^5 + 64*c^10*d^4*x^...
\[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=- \int \frac {1}{- 8 c^{2} x^{5} \sqrt {c + d x^{3}} - 7 c d x^{8} \sqrt {c + d x^{3}} + d^{2} x^{11} \sqrt {c + d x^{3}}}\, dx \] Input:
integrate(1/x**5/(-d*x**3+8*c)/(d*x**3+c)**(3/2),x)
Output:
-Integral(1/(-8*c**2*x**5*sqrt(c + d*x**3) - 7*c*d*x**8*sqrt(c + d*x**3) + d**2*x**11*sqrt(c + d*x**3)), x)
\[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { -\frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )} x^{5}} \,d x } \] Input:
integrate(1/x^5/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="maxima")
Output:
-integrate(1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)*x^5), x)
\[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int { -\frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )} x^{5}} \,d x } \] Input:
integrate(1/x^5/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="giac")
Output:
integrate(-1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)*x^5), x)
Timed out. \[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {1}{x^5\,{\left (d\,x^3+c\right )}^{3/2}\,\left (8\,c-d\,x^3\right )} \,d x \] Input:
int(1/(x^5*(c + d*x^3)^(3/2)*(8*c - d*x^3)),x)
Output:
int(1/(x^5*(c + d*x^3)^(3/2)*(8*c - d*x^3)), x)
\[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}\, c +10 \sqrt {d \,x^{3}+c}\, d \,x^{3}-25 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{4}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c \,d^{3} x^{4}-25 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x^{4}}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) d^{4} x^{7}+201 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c^{2} d^{2} x^{4}+201 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \right ) c \,d^{3} x^{7}}{64 c^{3} x^{4} \left (d \,x^{3}+c \right )} \] Input:
int(1/x^5/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x)
Output:
( - 2*sqrt(c + d*x**3)*c + 10*sqrt(c + d*x**3)*d*x**3 - 25*int((sqrt(c + d *x**3)*x**4)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d**2*x**6 - d**3*x**9),x)*c*d* *3*x**4 - 25*int((sqrt(c + d*x**3)*x**4)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d* *2*x**6 - d**3*x**9),x)*d**4*x**7 + 201*int((sqrt(c + d*x**3)*x)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d**2*x**6 - d**3*x**9),x)*c**2*d**2*x**4 + 201*int(( sqrt(c + d*x**3)*x)/(8*c**3 + 15*c**2*d*x**3 + 6*c*d**2*x**6 - d**3*x**9), x)*c*d**3*x**7)/(64*c**3*x**4*(c + d*x**3))