\(\int \frac {x}{\sqrt {c+d x^3} (4 c+d x^3)} \, dx\) [451]

Optimal result

Integrand size = 24, antiderivative size = 206 \[ \int \frac {x}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=-\frac {\arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3\ 2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}+\frac {\arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3\ 2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{9\ 2^{2/3} c^{5/6} d^{2/3}} \] Output:

-1/18*arctan(3^(1/2)*c^(1/6)*(c^(1/3)+2^(1/3)*d^(1/3)*x)/(d*x^3+c)^(1/2))* 
2^(1/3)*3^(1/2)/c^(5/6)/d^(2/3)+1/18*arctan(1/3*(d*x^3+c)^(1/2)*3^(1/2)/c^ 
(1/2))*2^(1/3)*3^(1/2)/c^(5/6)/d^(2/3)-1/6*arctanh(c^(1/6)*(c^(1/3)-2^(1/3 
)*d^(1/3)*x)/(d*x^3+c)^(1/2))*2^(1/3)/c^(5/6)/d^(2/3)+1/18*arctanh((d*x^3+ 
c)^(1/2)/c^(1/2))*2^(1/3)/c^(5/6)/d^(2/3)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.

Time = 10.05 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.33 \[ \int \frac {x}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\frac {x^2 \sqrt {\frac {c+d x^3}{c}} \operatorname {AppellF1}\left (\frac {2}{3},\frac {1}{2},1,\frac {5}{3},-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )}{8 c \sqrt {c+d x^3}} \] Input:

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

Output:

(x^2*Sqrt[(c + d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), -1/4*(d* 
x^3)/c])/(8*c*Sqrt[c + d*x^3])
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {986}

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.

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

\(\Big \downarrow \) 986

\(\displaystyle -\frac {\arctan \left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3\ 2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}+\frac {\arctan \left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3\ 2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}-\frac {\text {arctanh}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}}+\frac {\text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{9\ 2^{2/3} c^{5/6} d^{2/3}}\)

Input:

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

Output:

-1/3*ArcTan[(Sqrt[3]*c^(1/6)*(c^(1/3) + 2^(1/3)*d^(1/3)*x))/Sqrt[c + d*x^3 
]]/(2^(2/3)*Sqrt[3]*c^(5/6)*d^(2/3)) + ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqr 
t[c])]/(3*2^(2/3)*Sqrt[3]*c^(5/6)*d^(2/3)) - ArcTanh[(c^(1/6)*(c^(1/3) - 2 
^(1/3)*d^(1/3)*x))/Sqrt[c + d*x^3]]/(3*2^(2/3)*c^(5/6)*d^(2/3)) + ArcTanh[ 
Sqrt[c + d*x^3]/Sqrt[c]]/(9*2^(2/3)*c^(5/6)*d^(2/3))
 

Defintions of rubi rules used

rule 986

Int[(x_)/(((a_) + (b_.)*(x_)^3)*Sqrt[(c_) + (d_.)*(x_)^3]), x_Symbol] :> Wi 
th[{q = Rt[d/c, 3]}, Simp[q*(ArcTanh[Sqrt[c + d*x^3]/Rt[c, 2]]/(9*2^(2/3)*b 
*Rt[c, 2])), x] + (-Simp[q*(ArcTanh[Rt[c, 2]*((1 - 2^(1/3)*q*x)/Sqrt[c + d* 
x^3])]/(3*2^(2/3)*b*Rt[c, 2])), x] + Simp[q*(ArcTan[Sqrt[c + d*x^3]/(Sqrt[3 
]*Rt[c, 2])]/(3*2^(2/3)*Sqrt[3]*b*Rt[c, 2])), x] - Simp[q*(ArcTan[Sqrt[3]*R 
t[c, 2]*((1 + 2^(1/3)*q*x)/Sqrt[c + d*x^3])]/(3*2^(2/3)*Sqrt[3]*b*Rt[c, 2]) 
), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[4*b*c - a*d, 
0] && PosQ[c]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 3.

Time = 1.11 (sec) , antiderivative size = 416, normalized size of antiderivative = 2.02

method	result	size
default	\(-\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}+4 c \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d}{6 d c}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right )}}\right )}{2 \underline {\hspace {1.25 ex}}\alpha \sqrt {d \,x^{3}+c}}\right )}{9 d^{3} c}\)	\(416\)
elliptic	\(-\frac {i \sqrt {2}\, \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (d \,\textit {\_Z}^{3}+4 c \right )}{\sum }\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {2}\, \sqrt {\frac {i d \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {d \left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i d \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right )}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}\, d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}+2 \underline {\hspace {1.25 ex}}\alpha ^{2} d^{2}-\left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha d -\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \operatorname {EllipticPi}\left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \underline {\hspace {1.25 ex}}\alpha ^{2} \sqrt {3}\, d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha \sqrt {3}+i \sqrt {3}\, c d -3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \underline {\hspace {1.25 ex}}\alpha -3 c d}{6 d c}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{d \left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right )}}\right )}{2 \underline {\hspace {1.25 ex}}\alpha \sqrt {d \,x^{3}+c}}\right )}{9 d^{3} c}\)	\(416\)

Input:

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

Output:

-1/9*I/d^3/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),1/6/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)),_alpha=RootOf(_Z^3*d+4*c))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2289 vs. \(2 (141) = 282\).

Time = 0.75 (sec) , antiderivative size = 2289, normalized size of antiderivative = 11.11 \[ \int \frac {x}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/36*(1/432)^(1/6)*(sqrt(-3) + 1)*(-1/(c^5*d^4))^(1/6)*log((d^3*x^9 - 66* 
c*d^2*x^6 - 72*c^2*d*x^3 - 32*c^3 - 24*(1/2)^(2/3)*(c^4*d^5*x^7 - c^5*d^4* 
x^4 - 2*c^6*d^3*x + sqrt(-3)*(c^4*d^5*x^7 - c^5*d^4*x^4 - 2*c^6*d^3*x))*(- 
1/(c^5*d^4))^(2/3) - 6*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^ 
2*x^2 - sqrt(-3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2))*(-1/(c^5*d 
^4))^(1/3) + 6*sqrt(d*x^3 + c)*(648*(1/432)^(5/6)*(sqrt(-3)*c^5*d^5*x^5 - 
c^5*d^5*x^5)*(-1/(c^5*d^4))^(5/6) + sqrt(1/3)*(5*c^3*d^4*x^6 - 20*c^4*d^3* 
x^3 - 16*c^5*d^2)*sqrt(-1/(c^5*d^4)) - (1/432)^(1/6)*(c*d^3*x^7 - 16*c^2*d 
^2*x^4 - 8*c^3*d*x + sqrt(-3)*(c*d^3*x^7 - 16*c^2*d^2*x^4 - 8*c^3*d*x))*(- 
1/(c^5*d^4))^(1/6)))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) + 1 
/36*(1/432)^(1/6)*(sqrt(-3) + 1)*(-1/(c^5*d^4))^(1/6)*log((d^3*x^9 - 66*c* 
d^2*x^6 - 72*c^2*d*x^3 - 32*c^3 - 24*(1/2)^(2/3)*(c^4*d^5*x^7 - c^5*d^4*x^ 
4 - 2*c^6*d^3*x + sqrt(-3)*(c^4*d^5*x^7 - c^5*d^4*x^4 - 2*c^6*d^3*x))*(-1/ 
(c^5*d^4))^(2/3) - 6*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2* 
x^2 - sqrt(-3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2))*(-1/(c^5*d^4 
))^(1/3) - 6*sqrt(d*x^3 + c)*(648*(1/432)^(5/6)*(sqrt(-3)*c^5*d^5*x^5 - c^ 
5*d^5*x^5)*(-1/(c^5*d^4))^(5/6) + sqrt(1/3)*(5*c^3*d^4*x^6 - 20*c^4*d^3*x^ 
3 - 16*c^5*d^2)*sqrt(-1/(c^5*d^4)) - (1/432)^(1/6)*(c*d^3*x^7 - 16*c^2*d^2 
*x^4 - 8*c^3*d*x + sqrt(-3)*(c*d^3*x^7 - 16*c^2*d^2*x^4 - 8*c^3*d*x))*(-1/ 
(c^5*d^4))^(1/6)))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) - ...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 20.07 (sec) , antiderivative size = 453, normalized size of antiderivative = 2.20 \[ \int \frac {x}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\frac {\sqrt {3}\,{314928}^{1/3}\,\ln \left (\frac {{\left (\sqrt {d\,x^3+c}+\sqrt {3}\,\sqrt {-c}-2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\right )}^3\,\left (54\,\sqrt {d\,x^3+c}-54\,\sqrt {3}\,\sqrt {-c}+54\,2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\right )}{{\left (d^{1/3}\,x-2^{2/3}\,{\left (-c\right )}^{1/3}\right )}^6}\right )}{2916\,{\left (-c\right )}^{5/6}\,d^{2/3}}+\frac {\sqrt {3}\,{314928}^{1/3}\,\ln \left (\frac {{\left (2\,\sqrt {3}\,\sqrt {-c}-2\,\sqrt {d\,x^3+c}+2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x+2^{1/3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\,3{}\mathrm {i}\right )}^3\,\left (108\,\sqrt {d\,x^3+c}+108\,\sqrt {3}\,\sqrt {-c}+54\,2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x+2^{1/3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\,162{}\mathrm {i}\right )}{{\left (2\,d^{1/3}\,x+2^{2/3}\,{\left (-c\right )}^{1/3}-2^{2/3}\,\sqrt {3}\,{\left (-c\right )}^{1/3}\,1{}\mathrm {i}\right )}^6}\right )\,\sqrt {-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}{2916\,{\left (-c\right )}^{5/6}\,d^{2/3}}+\frac {\sqrt {3}\,{314928}^{1/3}\,\ln \left (\frac {{\left (2\,\sqrt {d\,x^3+c}+2\,\sqrt {3}\,\sqrt {-c}+2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x-2^{1/3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\,3{}\mathrm {i}\right )}^3\,\left (108\,\sqrt {d\,x^3+c}-108\,\sqrt {3}\,\sqrt {-c}-54\,2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x+2^{1/3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\,162{}\mathrm {i}\right )}{{\left (2\,d^{1/3}\,x+2^{2/3}\,{\left (-c\right )}^{1/3}+2^{2/3}\,\sqrt {3}\,{\left (-c\right )}^{1/3}\,1{}\mathrm {i}\right )}^6}\right )\,\sqrt {\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\,1{}\mathrm {i}}{2916\,{\left (-c\right )}^{5/6}\,d^{2/3}} \] Input:

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

Output:

(3^(1/2)*314928^(1/3)*log((((c + d*x^3)^(1/2) + 3^(1/2)*(-c)^(1/2) - 2^(1/ 
3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x)^3*(54*(c + d*x^3)^(1/2) - 54*3^(1/2)*(-c) 
^(1/2) + 54*2^(1/3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x))/(d^(1/3)*x - 2^(2/3)*(- 
c)^(1/3))^6))/(2916*(-c)^(5/6)*d^(2/3)) + (3^(1/2)*314928^(1/3)*log(((2*3^ 
(1/2)*(-c)^(1/2) - 2*(c + d*x^3)^(1/2) + 2^(1/3)*(-c)^(1/6)*d^(1/3)*x*3i + 
 2^(1/3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x)^3*(108*(c + d*x^3)^(1/2) + 108*3^(1 
/2)*(-c)^(1/2) + 2^(1/3)*(-c)^(1/6)*d^(1/3)*x*162i + 54*2^(1/3)*3^(1/2)*(- 
c)^(1/6)*d^(1/3)*x))/(2*d^(1/3)*x + 2^(2/3)*(-c)^(1/3) - 2^(2/3)*3^(1/2)*( 
-c)^(1/3)*1i)^6)*((3^(1/2)*1i)/2 - 1/2)^(1/2))/(2916*(-c)^(5/6)*d^(2/3)) + 
 (3^(1/2)*314928^(1/3)*log(((2*(c + d*x^3)^(1/2) + 2*3^(1/2)*(-c)^(1/2) - 
2^(1/3)*(-c)^(1/6)*d^(1/3)*x*3i + 2^(1/3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x)^3* 
(108*(c + d*x^3)^(1/2) - 108*3^(1/2)*(-c)^(1/2) + 2^(1/3)*(-c)^(1/6)*d^(1/ 
3)*x*162i - 54*2^(1/3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x))/(2*d^(1/3)*x + 2^(2/ 
3)*(-c)^(1/3) + 2^(2/3)*3^(1/2)*(-c)^(1/3)*1i)^6)*((3^(1/2)*1i)/2 + 1/2)^( 
1/2)*1i)/(2916*(-c)^(5/6)*d^(2/3))
 

Reduce [F]

\[ \int \frac {x}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx=\int \frac {\sqrt {d \,x^{3}+c}\, x}{d^{2} x^{6}+5 c d \,x^{3}+4 c^{2}}d x \] Input:

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

Output:

int((sqrt(c + d*x**3)*x)/(4*c**2 + 5*c*d*x**3 + d**2*x**6),x)