3.103 \(\int \frac{1}{x (a^2+2 a b x^3+b^2 x^6)^{3/2}} \, dx\)

Optimal. Leaf size=147 \[ \frac{1}{6 a \left (a+b x^3\right ) \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{1}{3 a^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{\log (x) \left (a+b x^3\right )}{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}} \]

[Out]

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

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Rubi [A]  time = 0.0827271, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {1355, 266, 44} \[ \frac{1}{6 a \left (a+b x^3\right ) \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{1}{3 a^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{\log (x) \left (a+b x^3\right )}{a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{\left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^3 \sqrt{a^2+2 a b x^3+b^2 x^6}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)),x]

[Out]

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

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 266

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0279135, size = 74, normalized size = 0.5 \[ \frac{a \left (3 a+2 b x^3\right )+6 \log (x) \left (a+b x^3\right )^2-2 \left (a+b x^3\right )^2 \log \left (a+b x^3\right )}{6 a^3 \left (a+b x^3\right ) \sqrt{\left (a+b x^3\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x*(a^2 + 2*a*b*x^3 + b^2*x^6)^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.02, size = 107, normalized size = 0.7 \begin{align*}{\frac{ \left ( 6\,\ln \left ( x \right ){x}^{6}{b}^{2}-2\,\ln \left ( b{x}^{3}+a \right ){x}^{6}{b}^{2}+12\,\ln \left ( x \right ){x}^{3}ab-4\,\ln \left ( b{x}^{3}+a \right ){x}^{3}ab+2\,ab{x}^{3}+6\,\ln \left ( x \right ){a}^{2}-2\,\ln \left ( b{x}^{3}+a \right ){a}^{2}+3\,{a}^{2} \right ) \left ( b{x}^{3}+a \right ) }{6\,{a}^{3}} \left ( \left ( b{x}^{3}+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x)

[Out]

1/6*(6*ln(x)*x^6*b^2-2*ln(b*x^3+a)*x^6*b^2+12*ln(x)*x^3*a*b-4*ln(b*x^3+a)*x^3*a*b+2*a*b*x^3+6*ln(x)*a^2-2*ln(b
*x^3+a)*a^2+3*a^2)*(b*x^3+a)/a^3/((b*x^3+a)^2)^(3/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.82563, size = 196, normalized size = 1.33 \begin{align*} \frac{2 \, a b x^{3} + 3 \, a^{2} - 2 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )} \log \left (b x^{3} + a\right ) + 6 \,{\left (b^{2} x^{6} + 2 \, a b x^{3} + a^{2}\right )} \log \left (x\right )}{6 \,{\left (a^{3} b^{2} x^{6} + 2 \, a^{4} b x^{3} + a^{5}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/6*(2*a*b*x^3 + 3*a^2 - 2*(b^2*x^6 + 2*a*b*x^3 + a^2)*log(b*x^3 + a) + 6*(b^2*x^6 + 2*a*b*x^3 + a^2)*log(x))/
(a^3*b^2*x^6 + 2*a^4*b*x^3 + a^5)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x \left (\left (a + b x^{3}\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b**2*x**6+2*a*b*x**3+a**2)**(3/2),x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x/(b^2*x^6+2*a*b*x^3+a^2)^(3/2),x, algorithm="giac")

[Out]

sage0*x