\(\int \frac {x}{(8 c-d x^3) (c+d x^3)^{3/2}} \, dx\) [508]

Optimal result

Integrand size = 25, antiderivative size = 632 \[ \int \frac {x}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {2 x^2}{27 c^2 \sqrt {c+d x^3}}-\frac {2 \sqrt {c+d x^3}}{27 c^2 d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}-\frac {\arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{54 \sqrt {3} c^{11/6} d^{2/3}}+\frac {\text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{162 c^{11/6} d^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{162 c^{11/6} d^{2/3}}+\frac {\sqrt {2-\sqrt {3}} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{9\ 3^{3/4} c^{5/3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {2 \sqrt {2} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} c^{5/3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:

2/27*x^2/c^2/(d*x^3+c)^(1/2)-2/27*(d*x^3+c)^(1/2)/c^2/d^(2/3)/((1+3^(1/2)) 
*c^(1/3)+d^(1/3)*x)-1/162*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)/d^(2/3)+1/162*arctanh(1/3*(c^(1/3)+d^(1/3)*x) 
^2/c^(1/6)/(d*x^3+c)^(1/2))/c^(11/6)/d^(2/3)-1/162*arctanh(1/3*(d*x^3+c)^( 
1/2)/c^(1/2))/c^(11/6)/d^(2/3)+1/27*(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^(5/3)/d^(2/3)/(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)-2/81*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^(5/3)/d^(2/3)/( 
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 = 10.10 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.20 \[ \int \frac {x}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\frac {x^2 \left (160 c-25 c \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+2 d x^3 \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )}{2160 c^3 \sqrt {c+d x^3}} \] Input:

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

Output:

(x^2*(160*c - 25*c*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3 
)/c), (d*x^3)/(8*c)] + 2*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)]))/(2160*c^3*Sqrt[c + d*x^3])
 

Rubi [A] (verified)

Time = 1.63 (sec) , antiderivative size = 631, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {972, 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.

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

\(\Big \downarrow \) 972

\(\displaystyle \frac {2 x^2}{27 c^2 \sqrt {c+d x^3}}-\frac {2 \int \frac {d x \left (5 c-d x^3\right )}{2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{27 c^2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 x^2}{27 c^2 \sqrt {c+d x^3}}-\frac {\int \frac {x \left (5 c-d x^3\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{27 c^2}\)

\(\Big \downarrow \) 1054

\(\displaystyle \frac {2 x^2}{27 c^2 \sqrt {c+d x^3}}-\frac {\int \left (\frac {x}{\sqrt {d x^3+c}}-\frac {3 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}\right )dx}{27 c^2}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 x^2}{27 c^2 \sqrt {c+d x^3}}-\frac {\frac {2 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{2 \sqrt {3} d^{2/3}}-\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{6 d^{2/3}}+\frac {\sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{6 d^{2/3}}+\frac {2 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}}{27 c^2}\)

Input:

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

Output:

(2*x^2)/(27*c^2*Sqrt[c + d*x^3]) - ((2*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)) - (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*Sqr 
t[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]*Sqr 
t[c + d*x^3]))/(27*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 972

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]
 

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 = 1.05 (sec) , antiderivative size = 875, normalized size of antiderivative = 1.38

method	result	size
default	\(\text {Expression too large to display}\)	\(875\)
elliptic	\(\text {Expression too large to display}\)	\(875\)

Input:

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

Output:

2/27*x^2/c^2/((x^3+c/d)*d)^(1/2)+2/81*I/c^2*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)))-1/243*I/c^2/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*(...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2525 vs. \(2 (444) = 888\).

Time = 0.58 (sec) , antiderivative size = 2525, normalized size of antiderivative = 4.00 \[ \int \frac {x}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

1/1944*(144*sqrt(d*x^3 + c)*d*x^2 + 144*(d*x^3 + c)*sqrt(d)*weierstrassZet 
a(0, -4*c/d, weierstrassPInverse(0, -4*c/d, x)) + (c^2*d^2*x^3 + c^3*d + s 
qrt(-3)*(c^2*d^2*x^3 + c^3*d))*(1/(c^11*d^4))^(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^8*d^5*x^7 + 64*c^9*d^4*x^4 + 32 
*c^10*d^3*x + sqrt(-3)*(5*c^8*d^5*x^7 + 64*c^9*d^4*x^4 + 32*c^10*d^3*x))*( 
1/(c^11*d^4))^(2/3) + 3*sqrt(d*x^3 + c)*(6*(5*c^10*d^5*x^5 + 32*c^11*d^4*x 
^2 - sqrt(-3)*(5*c^10*d^5*x^5 + 32*c^11*d^4*x^2))*(1/(c^11*d^4))^(5/6) - 2 
*(7*c^6*d^4*x^6 + 152*c^7*d^3*x^3 + 64*c^8*d^2)*sqrt(1/(c^11*d^4)) + (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 + 80*c^3*d^ 
2*x^4 + 160*c^4*d*x))*(1/(c^11*d^4))^(1/6)) - 9*(c^4*d^4*x^8 + 38*c^5*d^3* 
x^5 + 64*c^6*d^2*x^2 - sqrt(-3)*(c^4*d^4*x^8 + 38*c^5*d^3*x^5 + 64*c^6*d^2 
*x^2))*(1/(c^11*d^4))^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512 
*c^3)) - (c^2*d^2*x^3 + c^3*d + sqrt(-3)*(c^2*d^2*x^3 + c^3*d))*(1/(c^11*d 
^4))^(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^8*d^5*x^7 + 64*c^9*d^4*x^4 + 32*c^10*d^3*x + sqrt(-3)*(5*c^8*d^5*x^7 + 6 
4*c^9*d^4*x^4 + 32*c^10*d^3*x))*(1/(c^11*d^4))^(2/3) - 3*sqrt(d*x^3 + c)*( 
6*(5*c^10*d^5*x^5 + 32*c^11*d^4*x^2 - sqrt(-3)*(5*c^10*d^5*x^5 + 32*c^11*d 
^4*x^2))*(1/(c^11*d^4))^(5/6) - 2*(7*c^6*d^4*x^6 + 152*c^7*d^3*x^3 + 64*c^ 
8*d^2)*sqrt(1/(c^11*d^4)) + (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 + 80*c^3*d^2*x^4 + 160*c^4*d*x))*(1/(c^11*d^4))^(...
 

Sympy [F]

\[ \int \frac {x}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=- \int \frac {x}{- 8 c^{2} \sqrt {c + d x^{3}} - 7 c d x^{3} \sqrt {c + d x^{3}} + d^{2} x^{6} \sqrt {c + d x^{3}}}\, dx \] Input:

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

Output:

-Integral(x/(-8*c**2*sqrt(c + d*x**3) - 7*c*d*x**3*sqrt(c + d*x**3) + d**2 
*x**6*sqrt(c + d*x**3)), x)
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {x}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x}{-d^{3} x^{9}+6 c \,d^{2} x^{6}+15 c^{2} d \,x^{3}+8 c^{3}}d x \] Input:

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

Output:

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)