3.138 \(\int \frac{f^{a+\frac{b}{x^2}}}{x^9} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0957998, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2212, 2209} \[ \frac{3 f^{a+\frac{b}{x^2}}}{2 b^2 x^4 \log ^2(f)}-\frac{3 f^{a+\frac{b}{x^2}}}{b^3 x^2 \log ^3(f)}+\frac{3 f^{a+\frac{b}{x^2}}}{b^4 \log ^4(f)}-\frac{f^{a+\frac{b}{x^2}}}{2 b x^6 \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)/x^9,x]

[Out]

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

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rubi steps

\begin{align*} \int \frac{f^{a+\frac{b}{x^2}}}{x^9} \, dx &=-\frac{f^{a+\frac{b}{x^2}}}{2 b x^6 \log (f)}-\frac{3 \int \frac{f^{a+\frac{b}{x^2}}}{x^7} \, dx}{b \log (f)}\\ &=\frac{3 f^{a+\frac{b}{x^2}}}{2 b^2 x^4 \log ^2(f)}-\frac{f^{a+\frac{b}{x^2}}}{2 b x^6 \log (f)}+\frac{6 \int \frac{f^{a+\frac{b}{x^2}}}{x^5} \, dx}{b^2 \log ^2(f)}\\ &=-\frac{3 f^{a+\frac{b}{x^2}}}{b^3 x^2 \log ^3(f)}+\frac{3 f^{a+\frac{b}{x^2}}}{2 b^2 x^4 \log ^2(f)}-\frac{f^{a+\frac{b}{x^2}}}{2 b x^6 \log (f)}-\frac{6 \int \frac{f^{a+\frac{b}{x^2}}}{x^3} \, dx}{b^3 \log ^3(f)}\\ &=\frac{3 f^{a+\frac{b}{x^2}}}{b^4 \log ^4(f)}-\frac{3 f^{a+\frac{b}{x^2}}}{b^3 x^2 \log ^3(f)}+\frac{3 f^{a+\frac{b}{x^2}}}{2 b^2 x^4 \log ^2(f)}-\frac{f^{a+\frac{b}{x^2}}}{2 b x^6 \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.0103556, size = 58, normalized size = 0.67 \[ \frac{f^{a+\frac{b}{x^2}} \left (3 b^2 x^2 \log ^2(f)-b^3 \log ^3(f)-6 b x^4 \log (f)+6 x^6\right )}{2 b^4 x^6 \log ^4(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)/x^9,x]

[Out]

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

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Maple [A]  time = 0.017, size = 98, normalized size = 1.1 \begin{align*}{\frac{1}{{x}^{8}} \left ( 3\,{\frac{{x}^{8}}{{b}^{4} \left ( \ln \left ( f \right ) \right ) ^{4}}{{\rm e}^{ \left ( a+{\frac{b}{{x}^{2}}} \right ) \ln \left ( f \right ) }}}-3\,{\frac{{x}^{6}}{ \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}}{{\rm e}^{ \left ( a+{\frac{b}{{x}^{2}}} \right ) \ln \left ( f \right ) }}}+{\frac{3\,{x}^{4}}{2\, \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}}{{\rm e}^{ \left ( a+{\frac{b}{{x}^{2}}} \right ) \ln \left ( f \right ) }}}-{\frac{{x}^{2}}{2\,b\ln \left ( f \right ) }{{\rm e}^{ \left ( a+{\frac{b}{{x}^{2}}} \right ) \ln \left ( f \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^2)/x^9,x)

[Out]

(3/b^4/ln(f)^4*x^8*exp((a+b/x^2)*ln(f))-3/b^3/ln(f)^3*x^6*exp((a+b/x^2)*ln(f))+3/2/b^2/ln(f)^2*x^4*exp((a+b/x^
2)*ln(f))-1/2/b/ln(f)*x^2*exp((a+b/x^2)*ln(f)))/x^8

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Maxima [C]  time = 1.2621, size = 30, normalized size = 0.35 \begin{align*} \frac{f^{a} \Gamma \left (4, -\frac{b \log \left (f\right )}{x^{2}}\right )}{2 \, b^{4} \log \left (f\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)/x^9,x, algorithm="maxima")

[Out]

1/2*f^a*gamma(4, -b*log(f)/x^2)/(b^4*log(f)^4)

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Fricas [A]  time = 1.72871, size = 142, normalized size = 1.65 \begin{align*} \frac{{\left (6 \, x^{6} - 6 \, b x^{4} \log \left (f\right ) + 3 \, b^{2} x^{2} \log \left (f\right )^{2} - b^{3} \log \left (f\right )^{3}\right )} f^{\frac{a x^{2} + b}{x^{2}}}}{2 \, b^{4} x^{6} \log \left (f\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)/x^9,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.142659, size = 58, normalized size = 0.67 \begin{align*} \frac{f^{a + \frac{b}{x^{2}}} \left (- b^{3} \log{\left (f \right )}^{3} + 3 b^{2} x^{2} \log{\left (f \right )}^{2} - 6 b x^{4} \log{\left (f \right )} + 6 x^{6}\right )}{2 b^{4} x^{6} \log{\left (f \right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b/x**2)/x**9,x)

[Out]

f**(a + b/x**2)*(-b**3*log(f)**3 + 3*b**2*x**2*log(f)**2 - 6*b*x**4*log(f) + 6*x**6)/(2*b**4*x**6*log(f)**4)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{a + \frac{b}{x^{2}}}}{x^{9}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)/x^9,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)/x^9, x)