\(\int \frac {x^4}{(8 c-d x^3)^2 \sqrt {c+d x^3}} \, dx\) [605]

Optimal result

Integrand size = 27, antiderivative size = 647 \[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {\sqrt {c+d x^3}}{27 c d^{5/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}+\frac {x^2 \sqrt {c+d x^3}}{27 c d \left (8 c-d x^3\right )}+\frac {\arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{27 \sqrt {3} c^{5/6} d^{5/3}}-\frac {\text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{81 c^{5/6} d^{5/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{81 c^{5/6} d^{5/3}}-\frac {\sqrt {2-\sqrt {3}} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right )|-7-4 \sqrt {3}\right )}{18\ 3^{3/4} c^{2/3} d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt {2} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x}\right ),-7-4 \sqrt {3}\right )}{27 \sqrt [4]{3} c^{2/3} d^{5/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}} \] Output:

1/27*(d*x^3+c)^(1/2)/c/d^(5/3)/((1+3^(1/2))*c^(1/3)+d^(1/3)*x)+1/27*x^2*(d 
*x^3+c)^(1/2)/c/d/(-d*x^3+8*c)+1/81*arctan(3^(1/2)*c^(1/6)*(c^(1/3)+d^(1/3 
)*x)/(d*x^3+c)^(1/2))*3^(1/2)/c^(5/6)/d^(5/3)-1/81*arctanh(1/3*(c^(1/3)+d^ 
(1/3)*x)^2/c^(1/6)/(d*x^3+c)^(1/2))/c^(5/6)/d^(5/3)+1/81*arctanh(1/3*(d*x^ 
3+c)^(1/2)/c^(1/2))/c^(5/6)/d^(5/3)-1/54*(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^(2/3)/d^(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/ 
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^(2/3)/d^(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 = 10.13 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.26 \[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\frac {80 c x^2 \left (c+d x^3\right )+10 c x^2 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )+d x^5 \left (-8 c+d x^3\right ) \sqrt {1+\frac {d x^3}{c}} \operatorname {AppellF1}\left (\frac {5}{3},\frac {1}{2},1,\frac {8}{3},-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{2160 c^2 d \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \] Input:

Integrate[x^4/((8*c - d*x^3)^2*Sqrt[c + d*x^3]),x]
 

Output:

(80*c*x^2*(c + d*x^3) + 10*c*x^2*(-8*c + d*x^3)*Sqrt[1 + (d*x^3)/c]*Appell 
F1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), (d*x^3)/(8*c)] + d*x^5*(-8*c + d*x^3)*S 
qrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), (d*x^3)/(8*c)] 
)/(2160*c^2*d*(8*c - d*x^3)*Sqrt[c + d*x^3])
 

Rubi [A] (verified)

Time = 1.65 (sec) , antiderivative size = 646, 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 = {971, 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.

\(\displaystyle \int \frac {x^4}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx\)

\(\Big \downarrow \) 971

\(\displaystyle \frac {x^2 \sqrt {c+d x^3}}{27 c d \left (8 c-d x^3\right )}-\frac {\int \frac {x \left (d x^3+4 c\right )}{2 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{27 c d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^2 \sqrt {c+d x^3}}{27 c d \left (8 c-d x^3\right )}-\frac {\int \frac {x \left (d x^3+4 c\right )}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx}{54 c d}\)

\(\Big \downarrow \) 1054

\(\displaystyle \frac {x^2 \sqrt {c+d x^3}}{27 c d \left (8 c-d x^3\right )}-\frac {\int \left (\frac {12 c x}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}-\frac {x}{\sqrt {d x^3+c}}\right )dx}{54 c d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {x^2 \sqrt {c+d x^3}}{27 c d \left (8 c-d x^3\right )}-\frac {-\frac {2 \sqrt {2} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right ),-7-4 \sqrt {3}\right )}{\sqrt [4]{3} d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}+\frac {\sqrt [4]{3} \sqrt {2-\sqrt {3}} \sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right ) \sqrt {\frac {c^{2/3}-\sqrt [3]{c} \sqrt [3]{d} x+d^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} E\left (\arcsin \left (\frac {\sqrt [3]{d} x+\left (1-\sqrt {3}\right ) \sqrt [3]{c}}{\sqrt [3]{d} x+\left (1+\sqrt {3}\right ) \sqrt [3]{c}}\right )|-7-4 \sqrt {3}\right )}{d^{2/3} \sqrt {\frac {\sqrt [3]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )^2}} \sqrt {c+d x^3}}-\frac {2 \sqrt [6]{c} \arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{\sqrt {3} d^{2/3}}+\frac {2 \sqrt [6]{c} \text {arctanh}\left (\frac {\left (\sqrt [3]{c}+\sqrt [3]{d} x\right )^2}{3 \sqrt [6]{c} \sqrt {c+d x^3}}\right )}{3 d^{2/3}}-\frac {2 \sqrt [6]{c} \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{3 d^{2/3}}-\frac {2 \sqrt {c+d x^3}}{d^{2/3} \left (\left (1+\sqrt {3}\right ) \sqrt [3]{c}+\sqrt [3]{d} x\right )}}{54 c d}\)

Input:

Int[x^4/((8*c - d*x^3)^2*Sqrt[c + d*x^3]),x]
 

Output:

(x^2*Sqrt[c + d*x^3])/(27*c*d*(8*c - d*x^3)) - ((-2*Sqrt[c + d*x^3])/(d^(2 
/3)*((1 + Sqrt[3])*c^(1/3) + d^(1/3)*x)) - (2*c^(1/6)*ArcTan[(Sqrt[3]*c^(1 
/6)*(c^(1/3) + d^(1/3)*x))/Sqrt[c + d*x^3]])/(Sqrt[3]*d^(2/3)) + (2*c^(1/6 
)*ArcTanh[(c^(1/3) + d^(1/3)*x)^2/(3*c^(1/6)*Sqrt[c + d*x^3])])/(3*d^(2/3) 
) - (2*c^(1/6)*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(3*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]))/(54*c*d)
 

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 971

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[e^(n - 1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)* 
((c + d*x^n)^(q + 1)/(n*(b*c - a*d)*(p + 1))), x] - Simp[e^n/(n*(b*c - a*d) 
*(p + 1))   Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - 
 n + 1) + d*(m + n*(p + q + 1) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e 
, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n, m - n + 
 1] && GtQ[m - n + 1, 0] && 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 = 1.59 (sec) , antiderivative size = 886, normalized size of antiderivative = 1.37

method	result	size
elliptic	\(\text {Expression too large to display}\)	\(886\)
default	\(\text {Expression too large to display}\)	\(1305\)

Input:

int(x^4/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

1/27*x^2*(d*x^3+c)^(1/2)/c/d/(-d*x^3+8*c)-1/81*I/c/d^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)))+2/243*I/d^4/c*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)*...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2538 vs. \(2 (458) = 916\).

Time = 0.39 (sec) , antiderivative size = 2538, normalized size of antiderivative = 3.92 \[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\text {Too large to display} \] Input:

integrate(x^4/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x, algorithm="fricas")
 

Output:

-1/972*(36*sqrt(d*x^3 + c)*d*x^2 + 36*(d*x^3 - 8*c)*sqrt(d)*weierstrassZet 
a(0, -4*c/d, weierstrassPInverse(0, -4*c/d, x)) + (c*d^3*x^3 - 8*c^2*d^2 + 
 sqrt(-3)*(c*d^3*x^3 - 8*c^2*d^2))*(1/(c^5*d^10))^(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^4*d^9*x^7 + 64*c^5*d^8*x^4 
+ 32*c^6*d^7*x + sqrt(-3)*(5*c^4*d^9*x^7 + 64*c^5*d^8*x^4 + 32*c^6*d^7*x)) 
*(1/(c^5*d^10))^(2/3) + 3*sqrt(d*x^3 + c)*(6*(5*c^5*d^10*x^5 + 32*c^6*d^9* 
x^2 - sqrt(-3)*(5*c^5*d^10*x^5 + 32*c^6*d^9*x^2))*(1/(c^5*d^10))^(5/6) - 2 
*(7*c^3*d^7*x^6 + 152*c^4*d^6*x^3 + 64*c^5*d^5)*sqrt(1/(c^5*d^10)) + (c*d^ 
4*x^7 + 80*c^2*d^3*x^4 + 160*c^3*d^2*x + sqrt(-3)*(c*d^4*x^7 + 80*c^2*d^3* 
x^4 + 160*c^3*d^2*x))*(1/(c^5*d^10))^(1/6)) - 9*(c^2*d^6*x^8 + 38*c^3*d^5* 
x^5 + 64*c^4*d^4*x^2 - sqrt(-3)*(c^2*d^6*x^8 + 38*c^3*d^5*x^5 + 64*c^4*d^4 
*x^2))*(1/(c^5*d^10))^(1/3))/(d^3*x^9 - 24*c*d^2*x^6 + 192*c^2*d*x^3 - 512 
*c^3)) - (c*d^3*x^3 - 8*c^2*d^2 + sqrt(-3)*(c*d^3*x^3 - 8*c^2*d^2))*(1/(c^ 
5*d^10))^(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^4*d^9*x^7 + 64*c^5*d^8*x^4 + 32*c^6*d^7*x + sqrt(-3)*(5*c^4*d^9*x^7 
+ 64*c^5*d^8*x^4 + 32*c^6*d^7*x))*(1/(c^5*d^10))^(2/3) - 3*sqrt(d*x^3 + c) 
*(6*(5*c^5*d^10*x^5 + 32*c^6*d^9*x^2 - sqrt(-3)*(5*c^5*d^10*x^5 + 32*c^6*d 
^9*x^2))*(1/(c^5*d^10))^(5/6) - 2*(7*c^3*d^7*x^6 + 152*c^4*d^6*x^3 + 64*c^ 
5*d^5)*sqrt(1/(c^5*d^10)) + (c*d^4*x^7 + 80*c^2*d^3*x^4 + 160*c^3*d^2*x + 
sqrt(-3)*(c*d^4*x^7 + 80*c^2*d^3*x^4 + 160*c^3*d^2*x))*(1/(c^5*d^10))^(...
 

Sympy [F]

\[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {x^{4}}{\left (- 8 c + d x^{3}\right )^{2} \sqrt {c + d x^{3}}}\, dx \] Input:

integrate(x**4/(-d*x**3+8*c)**2/(d*x**3+c)**(1/2),x)
 

Output:

Integral(x**4/((-8*c + d*x**3)**2*sqrt(c + d*x**3)), x)
 

Maxima [F]

\[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {x^{4}}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:

integrate(x^4/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x, algorithm="maxima")
 

Output:

integrate(x^4/(sqrt(d*x^3 + c)*(d*x^3 - 8*c)^2), x)
 

Giac [F]

\[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int { \frac {x^{4}}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )}^{2}} \,d x } \] Input:

integrate(x^4/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x, algorithm="giac")
 

Output:

integrate(x^4/(sqrt(d*x^3 + c)*(d*x^3 - 8*c)^2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {x^4}{\sqrt {d\,x^3+c}\,{\left (8\,c-d\,x^3\right )}^2} \,d x \] Input:

int(x^4/((c + d*x^3)^(1/2)*(8*c - d*x^3)^2),x)
 

Output:

int(x^4/((c + d*x^3)^(1/2)*(8*c - d*x^3)^2), x)
 

Reduce [F]

\[ \int \frac {x^4}{\left (8 c-d x^3\right )^2 \sqrt {c+d x^3}} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x^{4}}{d^{3} x^{9}-15 c \,d^{2} x^{6}+48 c^{2} d \,x^{3}+64 c^{3}}d x \] Input:

int(x^4/(-d*x^3+8*c)^2/(d*x^3+c)^(1/2),x)
 

Output:

int((sqrt(c + d*x**3)*x**4)/(64*c**3 + 48*c**2*d*x**3 - 15*c*d**2*x**6 + d 
**3*x**9),x)