Integrand size = 27, antiderivative size = 654 \[ \int \frac {\sqrt {c+d x^3}}{x^5 \left (8 c-d x^3\right )} \, dx=-\frac {\sqrt {c+d x^3}}{32 c x^4}-\frac {d \sqrt {c+d x^3}}{16 c^2 x}+\frac {d^{4/3} \sqrt {c+d x^3}}{16 c^2 \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}-\frac {\sqrt {3} d^{4/3} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{128 c^{11/6}}+\frac {d^{4/3} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{128 c^{11/6}}-\frac {d^{4/3} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{128 c^{11/6}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} d^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{32 c^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {d^{4/3} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{8 \sqrt {2} \sqrt [4]{3} c^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:
-1/32*(d*x^3+c)^(1/2)/c/x^4-1/16*d*(d*x^3+c)^(1/2)/c^2/x+1/16*d^(4/3)*(d*x ^3+c)^(1/2)/c^2/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)-1/128*3^(1/2)*d^(4/3)*arct an(3^(1/2)*c^(1/6)*(c^(1/3)+d^(1/3)*x)/(d*x^3+c)^(1/2))/c^(11/6)+1/128*d^( 4/3)*arctanh(1/3*(c^(1/3)+d^(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/c^(11/6)-1 /128*d^(4/3)*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(11/6)-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^(5/3)/(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)*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^(5/3)/(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 \[ \int \frac {\sqrt {c+d x^3}}{x^5 \left (8 c-d x^3\right )} \, dx=\frac {125 c d^2 x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )-4 \left (40 c \left (c^2+3 c d x^3+2 d^2 x^6\right )+d^3 x^9 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )}{5120 c^3 x^4 \sqrt {c+d x^3}} \] Input:
Integrate[Sqrt[c + d*x^3]/(x^5*(8*c - d*x^3)),x]
Output:
(125*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)] - 4*(40*c*(c^2 + 3*c*d*x^3 + 2*d^2*x^6) + 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)]))/(51 20*c^3*x^4*Sqrt[c + d*x^3])
Time = 1.78 (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 = {975, 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.
| \(\displaystyle \int \frac {\sqrt {c+d x^3}}{x^5 \left (8 c-d x^3\right )} \, dx\) |
| \(\Big \downarrow \) 975 |
| \(\displaystyle \frac {\int \frac {d \left (5 d x^3+32 c\right )}{2 x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{32 c}-\frac {\sqrt {c+d x^3}}{32 c x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d \int \frac {5 d x^3+32 c}{x^2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{64 c}-\frac {\sqrt {c+d x^3}}{32 c x^4}\) |
| \(\Big \downarrow \) 1053 |
| \(\displaystyle \frac {d \left (-\frac {\int -\frac {8 c d x \left (25 c-2 d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{8 c^2}-\frac {4 \sqrt {c+d x^3}}{c x}\right )}{64 c}-\frac {\sqrt {c+d x^3}}{32 c x^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d \left (\frac {d \int \frac {x \left (25 c-2 d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{c}-\frac {4 \sqrt {c+d x^3}}{c x}\right )}{64 c}-\frac {\sqrt {c+d x^3}}{32 c x^4}\) |
| \(\Big \downarrow \) 1054 |
| \(\displaystyle \frac {d \left (\frac {d \int \left (\frac {9 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}+\frac {2 x}{\sqrt {d x^3+c}}\right )dx}{c}-\frac {4 \sqrt {c+d x^3}}{c x}\right )}{64 c}-\frac {\sqrt {c+d x^3}}{32 c x^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d \left (\frac {d \left (\frac {4 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {2 \sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt {3} \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{2 d^{2/3}}+\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{2 d^{2/3}}-\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 d^{2/3}}+\frac {4 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )}{c}-\frac {4 \sqrt {c+d x^3}}{c x}\right )}{64 c}-\frac {\sqrt {c+d x^3}}{32 c x^4}\) |
Input:
Int[Sqrt[c + d*x^3]/(x^5*(8*c - d*x^3)),x]
Output:
-1/32*Sqrt[c + d*x^3]/(c*x^4) + (d*((-4*Sqrt[c + d*x^3])/(c*x) + (d*((4*Sq rt[c + d*x^3])/(d^(2/3)*((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)) - (Sqrt[3]*c^ (1/6)*ArcTan[(Sqrt[3]*c^(1/6)*(c^(1/3) + d^(1/3)*x))/Sqrt[c + d*x^3]])/(2* d^(2/3)) + (c^(1/6)*ArcTanh[(c^(1/3) + d^(1/3)*x)^2/(3*c^(1/6)*Sqrt[c + d* x^3])])/(2*d^(2/3)) - (c^(1/6)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2*d^ (2/3)) - (2*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]) + (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 + S qrt[3])*c^(1/3) + d^(1/3)*x)^2]*Sqrt[c + d*x^3])))/c))/(64*c)
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/ (a*e*(m + 1))), x] - Simp[1/(a*e^n*(m + 1)) Int[(e*x)^(m + n)*(a + b*x^n) ^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] && IntBinomi alQ[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 = 2.40 (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 | \(\text {Expression too large to display}\) | \(1782\) |
Input:
int((d*x^3+c)^(1/2)/x^5/(-d*x^3+8*c),x,method=_RETURNVERBOSE)
Output:
-1/32*(d*x^3+c)^(1/2)*(2*d*x^3+c)/c^2/x^4+1/64/c^2*d^2*(-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)^(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)))-1/3*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)*_al pha*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^...
Leaf count of result is larger than twice the leaf count of optimal. 2401 vs. \(2 (462) = 924\).
Time = 0.75 (sec) , antiderivative size = 2401, normalized size of antiderivative = 3.67 \[ \int \frac {\sqrt {c+d x^3}}{x^5 \left (8 c-d x^3\right )} \, dx=\text {Too large to display} \] Input:
integrate((d*x^3+c)^(1/2)/x^5/(-d*x^3+8*c),x, algorithm="fricas")
Output:
1/1536*(2*c^2*x^4*(d^8/c^11)^(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^8*d^3*x^7 + 64*c^9*d^2*x^4 + 32*c^10*d*x) *(d^8/c^11)^(2/3) + 6*sqrt(d*x^3 + c)*(6*(5*c^10*d*x^5 + 32*c^11*x^2)*(d^8 /c^11)^(5/6) + (7*c^6*d^4*x^6 + 152*c^7*d^3*x^3 + 64*c^8*d^2)*sqrt(d^8/c^1 1) + (c^2*d^7*x^7 + 80*c^3*d^6*x^4 + 160*c^4*d^5*x)*(d^8/c^11)^(1/6)) + 18 *(c^4*d^6*x^8 + 38*c^5*d^5*x^5 + 64*c^6*d^4*x^2)*(d^8/c^11)^(1/3))/(d^3*x^ 9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - 2*c^2*x^4*(d^8/c^11)^(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^8* d^3*x^7 + 64*c^9*d^2*x^4 + 32*c^10*d*x)*(d^8/c^11)^(2/3) - 6*sqrt(d*x^3 + c)*(6*(5*c^10*d*x^5 + 32*c^11*x^2)*(d^8/c^11)^(5/6) + (7*c^6*d^4*x^6 + 152 *c^7*d^3*x^3 + 64*c^8*d^2)*sqrt(d^8/c^11) + (c^2*d^7*x^7 + 80*c^3*d^6*x^4 + 160*c^4*d^5*x)*(d^8/c^11)^(1/6)) + 18*(c^4*d^6*x^8 + 38*c^5*d^5*x^5 + 64 *c^6*d^4*x^2)*(d^8/c^11)^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - 96*d^(3/2)*x^4*weierstrassZeta(0, -4*c/d, weierstrassPInverse( 0, -4*c/d, x)) + (sqrt(-3)*c^2*x^4 + c^2*x^4)*(d^8/c^11)^(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^8*d^3*x^7 + 64 *c^9*d^2*x^4 + 32*c^10*d*x + sqrt(-3)*(5*c^8*d^3*x^7 + 64*c^9*d^2*x^4 + 32 *c^10*d*x))*(d^8/c^11)^(2/3) + 3*sqrt(d*x^3 + c)*(6*(5*c^10*d*x^5 + 32*c^1 1*x^2 - sqrt(-3)*(5*c^10*d*x^5 + 32*c^11*x^2))*(d^8/c^11)^(5/6) - 2*(7*c^6 *d^4*x^6 + 152*c^7*d^3*x^3 + 64*c^8*d^2)*sqrt(d^8/c^11) + (c^2*d^7*x^7 ...
\[ \int \frac {\sqrt {c+d x^3}}{x^5 \left (8 c-d x^3\right )} \, dx=- \int \frac {\sqrt {c + d x^{3}}}{- 8 c x^{5} + d x^{8}}\, dx \] Input:
integrate((d*x**3+c)**(1/2)/x**5/(-d*x**3+8*c),x)
Output:
-Integral(sqrt(c + d*x**3)/(-8*c*x**5 + d*x**8), x)
\[ \int \frac {\sqrt {c+d x^3}}{x^5 \left (8 c-d x^3\right )} \, dx=\int { -\frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )} x^{5}} \,d x } \] Input:
integrate((d*x^3+c)^(1/2)/x^5/(-d*x^3+8*c),x, algorithm="maxima")
Output:
-integrate(sqrt(d*x^3 + c)/((d*x^3 - 8*c)*x^5), x)
\[ \int \frac {\sqrt {c+d x^3}}{x^5 \left (8 c-d x^3\right )} \, dx=\int { -\frac {\sqrt {d x^{3} + c}}{{\left (d x^{3} - 8 \, c\right )} x^{5}} \,d x } \] Input:
integrate((d*x^3+c)^(1/2)/x^5/(-d*x^3+8*c),x, algorithm="giac")
Output:
integrate(-sqrt(d*x^3 + c)/((d*x^3 - 8*c)*x^5), x)
Timed out. \[ \int \frac {\sqrt {c+d x^3}}{x^5 \left (8 c-d x^3\right )} \, dx=\int \frac {\sqrt {d\,x^3+c}}{x^5\,\left (8\,c-d\,x^3\right )} \,d x \] Input:
int((c + d*x^3)^(1/2)/(x^5*(8*c - d*x^3)),x)
Output:
int((c + d*x^3)^(1/2)/(x^5*(8*c - d*x^3)), x)
\[ \int \frac {\sqrt {c+d x^3}}{x^5 \left (8 c-d x^3\right )} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{-d \,x^{8}+8 c \,x^{5}}d x \] Input:
int((d*x^3+c)^(1/2)/x^5/(-d*x^3+8*c),x)
Output:
int(sqrt(c + d*x**3)/(8*c*x**5 - d*x**8),x)