Integrand size = 27, antiderivative size = 654 \[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {\sqrt {c+d x^3}}{32 c^2 x^4}+\frac {d \sqrt {c+d x^3}}{16 c^3 x}-\frac {d^{4/3} \sqrt {c+d x^3}}{16 c^3 \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}-\frac {d^{4/3} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{384 \sqrt {3} c^{17/6}}+\frac {d^{4/3} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{1152 c^{17/6}}-\frac {d^{4/3} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{1152 c^{17/6}}+\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} d^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{32 c^{8/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {d^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{8 \sqrt {2} \sqrt [4]{3} c^{8/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
-1/32*(d*x^3+c)^(1/2)/c^2/x^4+1/16*d*(d*x^3+c)^(1/2)/c^3/x-1/16*d^(4/3)*(d *x^3+c)^(1/2)/c^3/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)-1/1152*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^(17/6)+1/1152 *d^(4/3)*arctanh(1/3*(c^(1/3)+d^(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/c^(17/ 6)-1/1152*d^(4/3)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(17/6)+1/32*3^(1/ 4)*(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)/c^(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)-1/48*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)*Ellipti cF(((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)/(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 = 11.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.23 \[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {160 c \left (-c^2+c d x^3+2 d^2 x^6\right )-75 c d^2 x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+4 d^3 x^9 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{5120 c^4 x^4 \sqrt {c+d x^3}} \] Input:
Integrate[1/(x^5*(8*c - d*x^3)*Sqrt[c + d*x^3]),x]
Output:
(160*c*(-c^2 + c*d*x^3 + 2*d^2*x^6) - 75*c*d^2*x^6*Sqrt[1 + (d*x^3)/c]*App ellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), (d*x^3)/(8*c)] + 4*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)])/(5120* c^4*x^4*Sqrt[c + d*x^3])
Time = 1.79 (sec) , antiderivative size = 657, 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 = {980, 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.
| \(\displaystyle \int \frac {1}{x^5 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx\) |
| \(\Big \downarrow \) 980 |
| \(\displaystyle \frac {\int -\frac {d \left (32 c-5 d x^3\right )}{2 x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{32 c^2}-\frac {\sqrt {c+d x^3}}{32 c^2 x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {d \int \frac {32 c-5 d x^3}{x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{64 c^2}-\frac {\sqrt {c+d x^3}}{32 c^2 x^4}\) |
| \(\Big \downarrow \) 1053 |
| \(\displaystyle -\frac {d \left (-\frac {\int -\frac {8 c d x \left (15 c-2 d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{8 c^2}-\frac {4 \sqrt {c+d x^3}}{c x}\right )}{64 c^2}-\frac {\sqrt {c+d x^3}}{32 c^2 x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {d \left (\frac {d \int \frac {x \left (15 c-2 d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{c}-\frac {4 \sqrt {c+d x^3}}{c x}\right )}{64 c^2}-\frac {\sqrt {c+d x^3}}{32 c^2 x^4}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle -\frac {d \left (\frac {d \int \left (\frac {2 x}{\sqrt {d x^3+c}}-\frac {c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}\right )dx}{c}-\frac {4 \sqrt {c+d x^3}}{c x}\right )}{64 c^2}-\frac {\sqrt {c+d x^3}}{32 c^2 x^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d \left (\frac {d \left (\frac {4 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {2 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{6 \sqrt {3} d^{2/3}}-\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{18 d^{2/3}}+\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{18 d^{2/3}}+\frac {4 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )}{c}-\frac {4 \sqrt {c+d x^3}}{c x}\right )}{64 c^2}-\frac {\sqrt {c+d x^3}}{32 c^2 x^4}\) |
Input:
Int[1/(x^5*(8*c - d*x^3)*Sqrt[c + d*x^3]),x]
Output:
-1/32*Sqrt[c + d*x^3]/(c^2*x^4) - (d*((-4*Sqrt[c + d*x^3])/(c*x) + (d*((4* 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]])/(6*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])])/(18*d^(2/3)) + (c^(1/6)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(1 8*d^(2/3)) - (2*3^(1/4)*Sqrt[2 - Sqrt[3]]*c^(1/3)*(c^(1/3) + d^(1/3)*x)*Sq rt[(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 + Sqr t[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]) + (4*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^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[(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*c*e*(m + 1))), x] - Simp[1/(a*c*e^n*(m + 1)) Int[(e*x)^(m + n)*( a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -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 = 2.42 (sec) , antiderivative size = 882, normalized size of antiderivative = 1.35
| method | result | size |
| risch | \(\text {Expression too large to display}\) | \(882\) |
| elliptic | \(\text {Expression too large to display}\) | \(887\) |
| default | \(\text {Expression too large to display}\) | \(1351\) |
Input:
int(1/x^5/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/32*(d*x^3+c)^(1/2)*(-2*d*x^3+c)/c^3/x^4-1/64/c^3*d^2*(1/27*I/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*(-c*d^2)^(2/3)*_alpha* 3^(1/2)+I*3^(1/2)*c*d-3*(-c*d^2)^(2/3)*_alpha-3*c*d)/c,(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)),_a lpha=RootOf(_Z^3*d-8*c))-4/3*I*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)^...
Leaf count of result is larger than twice the leaf count of optimal. 2403 vs. \(2 (462) = 924\).
Time = 1.12 (sec) , antiderivative size = 2403, normalized size of antiderivative = 3.67 \[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\text {Too large to display} \] Input:
integrate(1/x^5/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")
Output:
1/13824*(2*c^3*x^4*(d^8/c^17)^(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 + 18*(5*c^12*d^3*x^7 + 64*c^13*d^2*x^4 + 32*c^14*d *x)*(d^8/c^17)^(2/3) + 6*sqrt(d*x^3 + c)*(6*(5*c^15*d*x^5 + 32*c^16*x^2)*( d^8/c^17)^(5/6) + (7*c^9*d^4*x^6 + 152*c^10*d^3*x^3 + 64*c^11*d^2)*sqrt(d^ 8/c^17) + (c^3*d^7*x^7 + 80*c^4*d^6*x^4 + 160*c^5*d^5*x)*(d^8/c^17)^(1/6)) + 18*(c^6*d^6*x^8 + 38*c^7*d^5*x^5 + 64*c^8*d^4*x^2)*(d^8/c^17)^(1/3))/(d ^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - 2*c^3*x^4*(d^8/c^17)^( 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 + 18*(5 *c^12*d^3*x^7 + 64*c^13*d^2*x^4 + 32*c^14*d*x)*(d^8/c^17)^(2/3) - 6*sqrt(d *x^3 + c)*(6*(5*c^15*d*x^5 + 32*c^16*x^2)*(d^8/c^17)^(5/6) + (7*c^9*d^4*x^ 6 + 152*c^10*d^3*x^3 + 64*c^11*d^2)*sqrt(d^8/c^17) + (c^3*d^7*x^7 + 80*c^4 *d^6*x^4 + 160*c^5*d^5*x)*(d^8/c^17)^(1/6)) + 18*(c^6*d^6*x^8 + 38*c^7*d^5 *x^5 + 64*c^8*d^4*x^2)*(d^8/c^17)^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2 *d*x^3 - 512*c^3)) + 864*d^(3/2)*x^4*weierstrassZeta(0, -4*c/d, weierstras sPInverse(0, -4*c/d, x)) + (sqrt(-3)*c^3*x^4 + c^3*x^4)*(d^8/c^17)^(1/6)*l og((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^12*d ^3*x^7 + 64*c^13*d^2*x^4 + 32*c^14*d*x + sqrt(-3)*(5*c^12*d^3*x^7 + 64*c^1 3*d^2*x^4 + 32*c^14*d*x))*(d^8/c^17)^(2/3) + 3*sqrt(d*x^3 + c)*(6*(5*c^15* d*x^5 + 32*c^16*x^2 - sqrt(-3)*(5*c^15*d*x^5 + 32*c^16*x^2))*(d^8/c^17)^(5 /6) - 2*(7*c^9*d^4*x^6 + 152*c^10*d^3*x^3 + 64*c^11*d^2)*sqrt(d^8/c^17)...
\[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=- \int \frac {1}{- 8 c x^{5} \sqrt {c + d x^{3}} + d x^{8} \sqrt {c + d x^{3}}}\, dx \] Input:
integrate(1/x**5/(-d*x**3+8*c)/(d*x**3+c)**(1/2),x)
Output:
-Integral(1/(-8*c*x**5*sqrt(c + d*x**3) + d*x**8*sqrt(c + d*x**3)), x)
\[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\int { -\frac {1}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )} x^{5}} \,d x } \] Input:
integrate(1/x^5/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")
Output:
-integrate(1/(sqrt(d*x^3 + c)*(d*x^3 - 8*c)*x^5), x)
\[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\int { -\frac {1}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )} x^{5}} \,d x } \] Input:
integrate(1/x^5/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="giac")
Output:
integrate(-1/(sqrt(d*x^3 + c)*(d*x^3 - 8*c)*x^5), x)
Timed out. \[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\int \frac {1}{x^5\,\sqrt {d\,x^3+c}\,\left (8\,c-d\,x^3\right )} \,d x \] Input:
int(1/(x^5*(c + d*x^3)^(1/2)*(8*c - d*x^3)),x)
Output:
int(1/(x^5*(c + d*x^3)^(1/2)*(8*c - d*x^3)), x)
\[ \int \frac {1}{x^5 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {-2 \sqrt {d \,x^{3}+c}-32 \left (\int \frac {\sqrt {d \,x^{3}+c}}{-d^{2} x^{8}+7 c d \,x^{5}+8 c^{2} x^{2}}d x \right ) c d \,x^{4}+5 \left (\int \frac {\sqrt {d \,x^{3}+c}\, x}{-d^{2} x^{6}+7 c d \,x^{3}+8 c^{2}}d x \right ) d^{2} x^{4}}{64 c^{2} x^{4}} \] Input:
int(1/x^5/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x)
Output:
( - 2*sqrt(c + d*x**3) - 32*int(sqrt(c + d*x**3)/(8*c**2*x**2 + 7*c*d*x**5 - d**2*x**8),x)*c*d*x**4 + 5*int((sqrt(c + d*x**3)*x)/(8*c**2 + 7*c*d*x** 3 - d**2*x**6),x)*d**2*x**4)/(64*c**2*x**4)