\(\int \frac {1}{x^2 (8 c-d x^3) \sqrt {c+d x^3}} \, dx\) [493]

Optimal result
Mathematica [C] (warning: unable to verify)
Rubi [A] (verified)
Maple [C] (warning: unable to verify)
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 27, antiderivative size = 632 \[ \int \frac {1}{x^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=-\frac {\sqrt {c+d x^3}}{8 c^2 x}+\frac {\sqrt [3]{d} \sqrt {c+d x^3}}{8 c^2 \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}-\frac {\sqrt [3]{d} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{48 \sqrt {3} c^{11/6}}+\frac {\sqrt [3]{d} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{144 c^{11/6}}-\frac {\sqrt [3]{d} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{144 c^{11/6}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{d} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{16 c^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [3]{d} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{4 \sqrt {2} \sqrt [4]{3} c^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:

-1/8*(d*x^3+c)^(1/2)/c^2/x+1/8*d^(1/3)*(d*x^3+c)^(1/2)/c^2/((1+3^(1/2))*c^ 
(1/3)+d^(1/3)*x)-1/144*d^(1/3)*arctan(3^(1/2)*c^(1/6)*(c^(1/3)+d^(1/3)*x)/ 
(d*x^3+c)^(1/2))*3^(1/2)/c^(11/6)+1/144*d^(1/3)*arctanh(1/3*(c^(1/3)+d^(1/ 
3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/c^(11/6)-1/144*d^(1/3)*arctanh(1/3*(d*x^3 
+c)^(1/2)/c^(1/2))/c^(11/6)-1/16*3^(1/4)*(1/2*6^(1/2)-1/2*2^(1/2))*d^(1/3) 
*(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)+1/24*d^(1/ 
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)
 

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 11.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.22 \[ \int \frac {1}{x^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\frac {-80 c \left (c+d x^3\right )+25 c d x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )-d^2 x^6 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{640 c^3 x \sqrt {c+d x^3}} \] Input:

Integrate[1/(x^2*(8*c - d*x^3)*Sqrt[c + d*x^3]),x]
 

Output:

(-80*c*(c + d*x^3) + 25*c*d*x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 1/2, 1, 
5/3, -((d*x^3)/c), (d*x^3)/(8*c)] - d^2*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[5 
/3, 1/2, 1, 8/3, -((d*x^3)/c), (d*x^3)/(8*c)])/(640*c^3*x*Sqrt[c + d*x^3])
 

Rubi [A] (verified)

Time = 1.66 (sec) , antiderivative size = 632, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {980, 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.

\(\displaystyle \int \frac {1}{x^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx\)

\(\Big \downarrow \) 980

\(\displaystyle \frac {\int \frac {d x \left (10 c-d x^3\right )}{2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{8 c^2}-\frac {\sqrt {c+d x^3}}{8 c^2 x}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {d \int \frac {x \left (10 c-d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{16 c^2}-\frac {\sqrt {c+d x^3}}{8 c^2 x}\)

\(\Big \downarrow \) 1054

\(\displaystyle \frac {d \int \left (\frac {2 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}+\frac {x}{\sqrt {d x^3+c}}\right )dx}{16 c^2}-\frac {\sqrt {c+d x^3}}{8 c^2 x}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d \left (\frac {2 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3 \sqrt {3} d^{2/3}}+\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{9 d^{2/3}}-\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{9 d^{2/3}}+\frac {2 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}\right )}{16 c^2}-\frac {\sqrt {c+d x^3}}{8 c^2 x}\)

Input:

Int[1/(x^2*(8*c - d*x^3)*Sqrt[c + d*x^3]),x]
 

Output:

-1/8*Sqrt[c + d*x^3]/(c^2*x) + (d*((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]])/(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])])/(9*d^(2/3)) - (c^(1/6)*Arc 
Tanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(9*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[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])))/(16*c^2)
 

Defintions of rubi rules used

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 980
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]
 

rule 1054
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]
 

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 
Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.85 (sec) , antiderivative size = 868, normalized size of antiderivative = 1.37

method result size
elliptic \(\text {Expression too large to display}\) \(868\)
risch \(\text {Expression too large to display}\) \(872\)
default \(\text {Expression too large to display}\) \(874\)

Input:

int(1/x^2/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/8*(d*x^3+c)^(1/2)/c^2/x-1/24*I/c^2*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)))-1/216*I/c^2/d^2*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)*_alph 
a^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)^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2259 vs. \(2 (444) = 888\).

Time = 0.52 (sec) , antiderivative size = 2259, normalized size of antiderivative = 3.57 \[ \int \frac {1}{x^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\text {Too large to display} \] Input:

integrate(1/x^2/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")
 

Output:

1/1728*(2*c^2*x*(d^2/c^11)^(1/6)*log((d^4*x^9 + 318*c*d^3*x^6 + 1200*c^2*d 
^2*x^3 + 640*c^3*d + 18*(5*c^8*d^2*x^7 + 64*c^9*d*x^4 + 32*c^10*x)*(d^2/c^ 
11)^(2/3) + 6*sqrt(d*x^3 + c)*(6*(5*c^10*d*x^5 + 32*c^11*x^2)*(d^2/c^11)^( 
5/6) + (7*c^6*d^2*x^6 + 152*c^7*d*x^3 + 64*c^8)*sqrt(d^2/c^11) + (c^2*d^3* 
x^7 + 80*c^3*d^2*x^4 + 160*c^4*d*x)*(d^2/c^11)^(1/6)) + 18*(c^4*d^3*x^8 + 
38*c^5*d^2*x^5 + 64*c^6*d*x^2)*(d^2/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*(d^2/c^11)^(1/6)*log((d^4*x^9 + 318*c 
*d^3*x^6 + 1200*c^2*d^2*x^3 + 640*c^3*d + 18*(5*c^8*d^2*x^7 + 64*c^9*d*x^4 
 + 32*c^10*x)*(d^2/c^11)^(2/3) - 6*sqrt(d*x^3 + c)*(6*(5*c^10*d*x^5 + 32*c 
^11*x^2)*(d^2/c^11)^(5/6) + (7*c^6*d^2*x^6 + 152*c^7*d*x^3 + 64*c^8)*sqrt( 
d^2/c^11) + (c^2*d^3*x^7 + 80*c^3*d^2*x^4 + 160*c^4*d*x)*(d^2/c^11)^(1/6)) 
 + 18*(c^4*d^3*x^8 + 38*c^5*d^2*x^5 + 64*c^6*d*x^2)*(d^2/c^11)^(1/3))/(d^3 
*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512*c^3)) - 216*sqrt(d)*x*weierstras 
sZeta(0, -4*c/d, weierstrassPInverse(0, -4*c/d, x)) + (sqrt(-3)*c^2*x + c^ 
2*x)*(d^2/c^11)^(1/6)*log((d^4*x^9 + 318*c*d^3*x^6 + 1200*c^2*d^2*x^3 + 64 
0*c^3*d - 9*(5*c^8*d^2*x^7 + 64*c^9*d*x^4 + 32*c^10*x + sqrt(-3)*(5*c^8*d^ 
2*x^7 + 64*c^9*d*x^4 + 32*c^10*x))*(d^2/c^11)^(2/3) + 3*sqrt(d*x^3 + c)*(6 
*(5*c^10*d*x^5 + 32*c^11*x^2 - sqrt(-3)*(5*c^10*d*x^5 + 32*c^11*x^2))*(d^2 
/c^11)^(5/6) - 2*(7*c^6*d^2*x^6 + 152*c^7*d*x^3 + 64*c^8)*sqrt(d^2/c^11) + 
 (c^2*d^3*x^7 + 80*c^3*d^2*x^4 + 160*c^4*d*x + sqrt(-3)*(c^2*d^3*x^7 + ...
 

Sympy [F]

\[ \int \frac {1}{x^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=- \int \frac {1}{- 8 c x^{2} \sqrt {c + d x^{3}} + d x^{5} \sqrt {c + d x^{3}}}\, dx \] Input:

integrate(1/x**2/(-d*x**3+8*c)/(d*x**3+c)**(1/2),x)
 

Output:

-Integral(1/(-8*c*x**2*sqrt(c + d*x**3) + d*x**5*sqrt(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {1}{x^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\int { -\frac {1}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )} x^{2}} \,d x } \] Input:

integrate(1/x^2/(-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^2), x)
 

Giac [F]

\[ \int \frac {1}{x^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\int { -\frac {1}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )} x^{2}} \,d x } \] Input:

integrate(1/x^2/(-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^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\int \frac {1}{x^2\,\sqrt {d\,x^3+c}\,\left (8\,c-d\,x^3\right )} \,d x \] Input:

int(1/(x^2*(c + d*x^3)^(1/2)*(8*c - d*x^3)),x)
 

Output:

int(1/(x^2*(c + d*x^3)^(1/2)*(8*c - d*x^3)), x)
 

Reduce [F]

\[ \int \frac {1}{x^2 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}}{-d^{2} x^{8}+7 c d \,x^{5}+8 c^{2} x^{2}}d x \] Input:

int(1/x^2/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x)
 

Output:

int(sqrt(c + d*x**3)/(8*c**2*x**2 + 7*c*d*x**5 - d**2*x**8),x)