\(\int \frac {x^4}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\) [68]

Optimal result

Integrand size = 26, antiderivative size = 240 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {x^2 \left (a+b x^3\right )}{2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{2/3} \left (a+b x^3\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{2/3} \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a^{2/3} \left (a+b x^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}} \] Output:

1/2*x^2*(b*x^3+a)/b/((b*x^3+a)^2)^(1/2)+1/3*a^(2/3)*(b*x^3+a)*arctan(1/3*( 
a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))*3^(1/2)/b^(5/3)/((b*x^3+a)^2)^(1/2)+ 
1/3*a^(2/3)*(b*x^3+a)*ln(a^(1/3)+b^(1/3)*x)/b^(5/3)/((b*x^3+a)^2)^(1/2)-1/ 
6*a^(2/3)*(b*x^3+a)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(5/3)/((b* 
x^3+a)^2)^(1/2)
 

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.55 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\left (a+b x^3\right ) \left (3 b^{2/3} x^2+2 \sqrt {3} a^{2/3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+2 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )\right )}{6 b^{5/3} \sqrt {\left (a+b x^3\right )^2}} \] Input:

Integrate[x^4/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]
 

Output:

((a + b*x^3)*(3*b^(2/3)*x^2 + 2*Sqrt[3]*a^(2/3)*ArcTan[(1 - (2*b^(1/3)*x)/ 
a^(1/3))/Sqrt[3]] + 2*a^(2/3)*Log[a^(1/3) + b^(1/3)*x] - a^(2/3)*Log[a^(2/ 
3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]))/(6*b^(5/3)*Sqrt[(a + b*x^3)^2])
 

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.68, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {1384, 27, 843, 821, 16, 1142, 25, 27, 1082, 217, 1103}

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.

\(\displaystyle \int \frac {x^4}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\)

\(\Big \downarrow \) 1384

\(\displaystyle \frac {b \left (a+b x^3\right ) \int \frac {x^4}{b \left (b x^3+a\right )}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a+b x^3\right ) \int \frac {x^4}{b x^3+a}dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {x^2}{2 b}-\frac {a \int \frac {x}{b x^3+a}dx}{b}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 821

\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{b}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 1082

\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {\left (a+b x^3\right ) \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\)

Input:

Int[x^4/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]
 

Output:

((a + b*x^3)*(x^2/(2*b) - (a*(-1/3*Log[a^(1/3) + b^(1/3)*x]/(a^(1/3)*b^(2/ 
3)) + (-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3)) + 
Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/(2*b^(1/3)))/(3*a^(1/3)*b^( 
1/3))))/b))/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]
 

Defintions of rubi rules used

rule 16

Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + 
b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
 

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 217

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( 
-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & 
& (LtQ[a, 0] || LtQ[b, 0])
 

rule 821

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) 
 Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 
*x^2), x], x] /; FreeQ[{a, b}, x]
 

rule 843

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n 
 - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* 
p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1142

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 1384

Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S 
imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac 
Part[p]))   Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] 
&& EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n 
- 1)] && NeQ[u, x^(2*n - 1)] &&  !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
 

Maple [C] (warning: unable to verify)

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

Time = 2.16 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.32

method	result	size
risch	\(\frac {x^{2} \sqrt {\left (b \,x^{3}+a \right )^{2}}}{2 \left (b \,x^{3}+a \right ) b}-\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{3 \left (b \,x^{3}+a \right ) b^{2}}\)	\(77\)
default	\(\frac {\left (b \,x^{3}+a \right ) \left (3 x^{2} b \left (\frac {a}{b}\right )^{\frac {1}{3}}+2 \sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (-2 x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) a +2 \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right ) a -\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right ) a \right )}{6 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}}\)	\(113\)

Input:

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

Output:

1/2*x^2*((b*x^3+a)^2)^(1/2)/(b*x^3+a)/b-1/3*((b*x^3+a)^2)^(1/2)/(b*x^3+a)/ 
b^2*a*sum(1/_R*ln(x-_R),_R=RootOf(_Z^3*b+a))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.51 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {3 \, x^{2} - 2 \, \sqrt {3} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) - \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + 2 \, \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{6 \, b} \] Input:

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

Output:

1/6*(3*x^2 - 2*sqrt(3)*(a^2/b^2)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*(a^2/b^2) 
^(1/3) - sqrt(3)*a)/a) - (a^2/b^2)^(1/3)*log(a*x^2 - b*x*(a^2/b^2)^(2/3) + 
 a*(a^2/b^2)^(1/3)) + 2*(a^2/b^2)^(1/3)*log(a*x + b*(a^2/b^2)^(2/3)))/b
 

Sympy [F]

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

integrate(x**4/((b*x**3+a)**2)**(1/2),x)
 

Output:

Integral(x**4/sqrt((a + b*x**3)**2), x)
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.45 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {x^{2}}{2 \, b} - \frac {\sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {a \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {a \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \] Input:

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

Output:

1/2*x^2/b - 1/3*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/ 
3))/(b^2*(a/b)^(1/3)) - 1/6*a*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^2* 
(a/b)^(1/3)) + 1/3*a*log(x + (a/b)^(1/3))/(b^2*(a/b)^(1/3))
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.61 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {x^{2} \mathrm {sgn}\left (b x^{3} + a\right )}{2 \, b} + \frac {\left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, b} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right ) \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, b^{3}} - \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right ) \mathrm {sgn}\left (b x^{3} + a\right )}{6 \, b^{3}} \] Input:

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

Output:

1/2*x^2*sgn(b*x^3 + a)/b + 1/3*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))*sgn 
(b*x^3 + a)/b + 1/3*sqrt(3)*(-a*b^2)^(2/3)*arctan(1/3*sqrt(3)*(2*x + (-a/b 
)^(1/3))/(-a/b)^(1/3))*sgn(b*x^3 + a)/b^3 - 1/6*(-a*b^2)^(2/3)*log(x^2 + x 
*(-a/b)^(1/3) + (-a/b)^(2/3))*sgn(b*x^3 + a)/b^3
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\int \frac {x^4}{\sqrt {{\left (b\,x^3+a\right )}^2}} \,d x \] Input:

int(x^4/((a + b*x^3)^2)^(1/2),x)
 

Output:

int(x^4/((a + b*x^3)^2)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.34 \[ \int \frac {x^4}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) a +3 b^{\frac {2}{3}} a^{\frac {1}{3}} x^{2}-\mathrm {log}\left (a^{\frac {2}{3}}-b^{\frac {1}{3}} a^{\frac {1}{3}} x +b^{\frac {2}{3}} x^{2}\right ) a +2 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a}{6 b^{\frac {5}{3}} a^{\frac {1}{3}}} \] Input:

int(x^4/((b*x^3+a)^2)^(1/2),x)
 

Output:

(2*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a + 3*b**(2/ 
3)*a**(1/3)*x**2 - log(a**(2/3) - b**(1/3)*a**(1/3)*x + b**(2/3)*x**2)*a + 
 2*log(a**(1/3) + b**(1/3)*x)*a)/(6*b**(2/3)*a**(1/3)*b)