3.97 \(\int \frac{1}{x^4 \sqrt{a^2+2 a b x^3+b^2 x^6}} \, dx\)

Optimal. Leaf size=122 \[ -\frac{a+b x^3}{3 a x^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{b \log (x) \left (a+b x^3\right )}{a^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^2 \sqrt{a^2+2 a b x^3+b^2 x^6}} \]

[Out]

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

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Rubi [A]  time = 0.0505751, antiderivative size = 122, 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} \[ -\frac{a+b x^3}{3 a x^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{b \log (x) \left (a+b x^3\right )}{a^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^2 \sqrt{a^2+2 a b x^3+b^2 x^6}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]),x]

[Out]

-(a + b*x^3)/(3*a*x^3*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) - (b*(a + b*x^3)*Log[x])/(a^2*Sqrt[a^2 + 2*a*b*x^3 + b^
2*x^6]) + (b*(a + b*x^3)*Log[a + b*x^3])/(3*a^2*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^4 \sqrt{a^2+2 a b x^3+b^2 x^6}} \, dx &=\frac{\left (a b+b^2 x^3\right ) \int \frac{1}{x^4 \left (a b+b^2 x^3\right )} \, dx}{\sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=\frac{\left (a b+b^2 x^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a b+b^2 x\right )} \, dx,x,x^3\right )}{3 \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=\frac{\left (a b+b^2 x^3\right ) \operatorname{Subst}\left (\int \left (\frac{1}{a b x^2}-\frac{1}{a^2 x}+\frac{b}{a^2 (a+b x)}\right ) \, dx,x,x^3\right )}{3 \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ &=-\frac{a+b x^3}{3 a x^3 \sqrt{a^2+2 a b x^3+b^2 x^6}}-\frac{b \left (a+b x^3\right ) \log (x)}{a^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}+\frac{b \left (a+b x^3\right ) \log \left (a+b x^3\right )}{3 a^2 \sqrt{a^2+2 a b x^3+b^2 x^6}}\\ \end{align*}

Mathematica [A]  time = 0.0157451, size = 54, normalized size = 0.44 \[ -\frac{\left (a+b x^3\right ) \left (-b x^3 \log \left (a+b x^3\right )+a+3 b x^3 \log (x)\right )}{3 a^2 x^3 \sqrt{\left (a+b x^3\right )^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]),x]

[Out]

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

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Maple [A]  time = 0.01, size = 51, normalized size = 0.4 \begin{align*} -{\frac{ \left ( b{x}^{3}+a \right ) \left ( 3\,b\ln \left ( x \right ){x}^{3}-b\ln \left ( b{x}^{3}+a \right ){x}^{3}+a \right ) }{3\,{a}^{2}{x}^{3}}{\frac{1}{\sqrt{ \left ( b{x}^{3}+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.73329, size = 80, normalized size = 0.66 \begin{align*} \frac{b x^{3} \log \left (b x^{3} + a\right ) - 3 \, b x^{3} \log \left (x\right ) - a}{3 \, a^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.531288, size = 31, normalized size = 0.25 \begin{align*} - \frac{1}{3 a x^{3}} - \frac{b \log{\left (x \right )}}{a^{2}} + \frac{b \log{\left (\frac{a}{b} + x^{3} \right )}}{3 a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.1288, size = 68, normalized size = 0.56 \begin{align*} \frac{1}{3} \,{\left (\frac{b \log \left ({\left | b x^{3} + a \right |}\right )}{a^{2}} - \frac{3 \, b \log \left ({\left | x \right |}\right )}{a^{2}} + \frac{b x^{3} - a}{a^{2} x^{3}}\right )} \mathrm{sgn}\left (b x^{3} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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