\(\int \frac {x^2}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\) [91]

Optimal result

Integrand size = 26, antiderivative size = 44 \[ \int \frac {x^2}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 b \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[Out]

1/3*(b*x^3+a)*ln(b*x^3+a)/b/((b*x^3+a)^2)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.115, Rules used = {1366, 622, 31} \[ \int \frac {x^2}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 b \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

[In]

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

[Out]

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

Rule 31

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

Rule 622

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[(b/2 + c*x)/Sqrt[a + b*x + c*x^2], Int[1/(b/2
+ c*x), x], x] /; FreeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 1366

Int[(x_)^(m_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*x +
c*x^2)^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[Simplify[m - n + 1], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3} \text {Subst}\left (\int \frac {1}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx,x,x^3\right ) \\ & = \frac {\left (a b+b^2 x^3\right ) \text {Subst}\left (\int \frac {1}{a b+b^2 x} \, dx,x,x^3\right )}{3 \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ & = \frac {\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 b \sqrt {a^2+2 a b x^3+b^2 x^6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.16 \[ \int \frac {x^2}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\frac {\sqrt {a^2}}{b}-\frac {\sqrt {a^2+2 a b x^3+b^2 x^6}}{b}}{x^3}\right )}{3 b} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 0.34 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.50

method	result	size
pseudoelliptic	\(\frac {\ln \left (b \,x^{3}+a \right ) \operatorname {csgn}\left (b \,x^{3}+a \right )}{3 b}\)	\(22\)
default	\(\frac {\left (b \,x^{3}+a \right ) \ln \left (b \,x^{3}+a \right )}{3 b \sqrt {\left (b \,x^{3}+a \right )^{2}}}\)	\(32\)
risch	\(\frac {\sqrt {\left (b \,x^{3}+a \right )^{2}}\, \ln \left (b \,x^{3}+a \right )}{3 \left (b \,x^{3}+a \right ) b}\)	\(34\)

[In]

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

[Out]

1/3*ln(b*x^3+a)/b*csgn(b*x^3+a)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.30 \[ \int \frac {x^2}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\log \left (b x^{3} + a\right )}{3 \, b} \]

[In]

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

[Out]

1/3*log(b*x^3 + a)/b

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.34 \[ \int \frac {x^2}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\log \left (x^{3} + \frac {a}{b}\right )}{3 \, b} \]

[In]

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

[Out]

1/3*log(x^3 + a/b)/b

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.50 \[ \int \frac {x^2}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\log \left ({\left | b x^{3} + a \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, b} \]

[In]

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

[Out]

1/3*log(abs(b*x^3 + a))*sgn(b*x^3 + a)/b

Mupad [B] (verification not implemented)

Time = 8.43 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.75 \[ \int \frac {x^2}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx=\frac {\ln \left (b^2\,x^3+a\,b\right )\,\mathrm {sign}\left (2\,b^2\,x^3+2\,a\,b\right )}{3\,\sqrt {b^2}} \]

[In]

int(x^2/((a + b*x^3)^2)^(1/2),x)

[Out]

(log(a*b + b^2*x^3)*sign(2*a*b + 2*b^2*x^3))/(3*(b^2)^(1/2))