3.1890 \(\int \frac{1}{(a+\frac{b}{x^2})^3 x^{11}} \, dx\)

Optimal. Leaf size=86 \[ \frac{3 a^2}{2 b^4 \left (a x^2+b\right )}+\frac{a^2}{4 b^3 \left (a x^2+b\right )^2}-\frac{3 a^2 \log \left (a x^2+b\right )}{b^5}+\frac{6 a^2 \log (x)}{b^5}+\frac{3 a}{2 b^4 x^2}-\frac{1}{4 b^3 x^4} \]

[Out]

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

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Rubi [A]  time = 0.0543483, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {263, 266, 44} \[ \frac{3 a^2}{2 b^4 \left (a x^2+b\right )}+\frac{a^2}{4 b^3 \left (a x^2+b\right )^2}-\frac{3 a^2 \log \left (a x^2+b\right )}{b^5}+\frac{6 a^2 \log (x)}{b^5}+\frac{3 a}{2 b^4 x^2}-\frac{1}{4 b^3 x^4} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b/x^2)^3*x^11),x]

[Out]

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

Rule 263

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

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}{\left (a+\frac{b}{x^2}\right )^3 x^{11}} \, dx &=\int \frac{1}{x^5 \left (b+a x^2\right )^3} \, dx\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x^3 (b+a x)^3} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{1}{b^3 x^3}-\frac{3 a}{b^4 x^2}+\frac{6 a^2}{b^5 x}-\frac{a^3}{b^3 (b+a x)^3}-\frac{3 a^3}{b^4 (b+a x)^2}-\frac{6 a^3}{b^5 (b+a x)}\right ) \, dx,x,x^2\right )\\ &=-\frac{1}{4 b^3 x^4}+\frac{3 a}{2 b^4 x^2}+\frac{a^2}{4 b^3 \left (b+a x^2\right )^2}+\frac{3 a^2}{2 b^4 \left (b+a x^2\right )}+\frac{6 a^2 \log (x)}{b^5}-\frac{3 a^2 \log \left (b+a x^2\right )}{b^5}\\ \end{align*}

Mathematica [A]  time = 0.0441437, size = 74, normalized size = 0.86 \[ \frac{\frac{b \left (18 a^2 b x^4+12 a^3 x^6+4 a b^2 x^2-b^3\right )}{x^4 \left (a x^2+b\right )^2}-12 a^2 \log \left (a x^2+b\right )+24 a^2 \log (x)}{4 b^5} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b/x^2)^3*x^11),x]

[Out]

((b*(-b^3 + 4*a*b^2*x^2 + 18*a^2*b*x^4 + 12*a^3*x^6))/(x^4*(b + a*x^2)^2) + 24*a^2*Log[x] - 12*a^2*Log[b + a*x
^2])/(4*b^5)

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Maple [A]  time = 0.013, size = 79, normalized size = 0.9 \begin{align*} -{\frac{1}{4\,{b}^{3}{x}^{4}}}+{\frac{3\,a}{2\,{b}^{4}{x}^{2}}}+{\frac{{a}^{2}}{4\,{b}^{3} \left ( a{x}^{2}+b \right ) ^{2}}}+{\frac{3\,{a}^{2}}{2\,{b}^{4} \left ( a{x}^{2}+b \right ) }}+6\,{\frac{{a}^{2}\ln \left ( x \right ) }{{b}^{5}}}-3\,{\frac{{a}^{2}\ln \left ( a{x}^{2}+b \right ) }{{b}^{5}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+1/x^2*b)^3/x^11,x)

[Out]

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

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Maxima [A]  time = 1.00575, size = 124, normalized size = 1.44 \begin{align*} \frac{12 \, a^{3} x^{6} + 18 \, a^{2} b x^{4} + 4 \, a b^{2} x^{2} - b^{3}}{4 \,{\left (a^{2} b^{4} x^{8} + 2 \, a b^{5} x^{6} + b^{6} x^{4}\right )}} - \frac{3 \, a^{2} \log \left (a x^{2} + b\right )}{b^{5}} + \frac{3 \, a^{2} \log \left (x^{2}\right )}{b^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^11,x, algorithm="maxima")

[Out]

1/4*(12*a^3*x^6 + 18*a^2*b*x^4 + 4*a*b^2*x^2 - b^3)/(a^2*b^4*x^8 + 2*a*b^5*x^6 + b^6*x^4) - 3*a^2*log(a*x^2 +
b)/b^5 + 3*a^2*log(x^2)/b^5

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Fricas [A]  time = 1.42516, size = 274, normalized size = 3.19 \begin{align*} \frac{12 \, a^{3} b x^{6} + 18 \, a^{2} b^{2} x^{4} + 4 \, a b^{3} x^{2} - b^{4} - 12 \,{\left (a^{4} x^{8} + 2 \, a^{3} b x^{6} + a^{2} b^{2} x^{4}\right )} \log \left (a x^{2} + b\right ) + 24 \,{\left (a^{4} x^{8} + 2 \, a^{3} b x^{6} + a^{2} b^{2} x^{4}\right )} \log \left (x\right )}{4 \,{\left (a^{2} b^{5} x^{8} + 2 \, a b^{6} x^{6} + b^{7} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^11,x, algorithm="fricas")

[Out]

1/4*(12*a^3*b*x^6 + 18*a^2*b^2*x^4 + 4*a*b^3*x^2 - b^4 - 12*(a^4*x^8 + 2*a^3*b*x^6 + a^2*b^2*x^4)*log(a*x^2 +
b) + 24*(a^4*x^8 + 2*a^3*b*x^6 + a^2*b^2*x^4)*log(x))/(a^2*b^5*x^8 + 2*a*b^6*x^6 + b^7*x^4)

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Sympy [A]  time = 1.35903, size = 90, normalized size = 1.05 \begin{align*} \frac{6 a^{2} \log{\left (x \right )}}{b^{5}} - \frac{3 a^{2} \log{\left (x^{2} + \frac{b}{a} \right )}}{b^{5}} + \frac{12 a^{3} x^{6} + 18 a^{2} b x^{4} + 4 a b^{2} x^{2} - b^{3}}{4 a^{2} b^{4} x^{8} + 8 a b^{5} x^{6} + 4 b^{6} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x**2)**3/x**11,x)

[Out]

6*a**2*log(x)/b**5 - 3*a**2*log(x**2 + b/a)/b**5 + (12*a**3*x**6 + 18*a**2*b*x**4 + 4*a*b**2*x**2 - b**3)/(4*a
**2*b**4*x**8 + 8*a*b**5*x**6 + 4*b**6*x**4)

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Giac [A]  time = 1.19774, size = 108, normalized size = 1.26 \begin{align*} \frac{3 \, a^{2} \log \left (x^{2}\right )}{b^{5}} - \frac{3 \, a^{2} \log \left ({\left | a x^{2} + b \right |}\right )}{b^{5}} + \frac{12 \, a^{3} x^{6} + 18 \, a^{2} b x^{4} + 4 \, a b^{2} x^{2} - b^{3}}{4 \,{\left (a x^{4} + b x^{2}\right )}^{2} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b/x^2)^3/x^11,x, algorithm="giac")

[Out]

3*a^2*log(x^2)/b^5 - 3*a^2*log(abs(a*x^2 + b))/b^5 + 1/4*(12*a^3*x^6 + 18*a^2*b*x^4 + 4*a*b^2*x^2 - b^3)/((a*x
^4 + b*x^2)^2*b^4)