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

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

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 86, normalized size of antiderivative = 1.00, 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 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]

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])

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*}

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Mathematica [A]  time = 0.01, size = 58, normalized size = 0.67 \[ \frac {f^{a+\frac {b}{x^2}} \left (-b^3 \log ^3(f)+3 b^2 x^2 \log ^2(f)-6 b x^4 \log (f)+6 x^6\right )}{2 b^4 x^6 \log ^4(f)} \]

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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fricas [A]  time = 0.40, size = 60, normalized size = 0.70 \[ \frac {{\left (6 \, x^{6} - 6 \, b x^{4} \log \relax (f) + 3 \, b^{2} x^{2} \log \relax (f)^{2} - b^{3} \log \relax (f)^{3}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{2 \, b^{4} x^{6} \log \relax (f)^{4}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + \frac {b}{x^{2}}}}{x^{9}}\,{d x} \]

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)

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maple [A]  time = 0.03, size = 98, normalized size = 1.14 \[ \frac {-\frac {x^{2} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \relax (f )}}{2 b \ln \relax (f )}+\frac {3 x^{4} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \relax (f )}}{2 b^{2} \ln \relax (f )^{2}}-\frac {3 x^{6} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \relax (f )}}{b^{3} \ln \relax (f )^{3}}+\frac {3 x^{8} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \relax (f )}}{b^{4} \ln \relax (f )^{4}}}{x^{8}} \]

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

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maxima [C]  time = 1.28, size = 22, normalized size = 0.26 \[ \frac {f^{a} \Gamma \left (4, -\frac {b \log \relax (f)}{x^{2}}\right )}{2 \, b^{4} \log \relax (f)^{4}} \]

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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mupad [B]  time = 3.63, size = 60, normalized size = 0.70 \[ -\frac {f^{a+\frac {b}{x^2}}\,\left (\frac {1}{2\,b\,\ln \relax (f)}-\frac {3\,x^2}{2\,b^2\,{\ln \relax (f)}^2}+\frac {3\,x^4}{b^3\,{\ln \relax (f)}^3}-\frac {3\,x^6}{b^4\,{\ln \relax (f)}^4}\right )}{x^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^2)/x^9,x)

[Out]

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

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sympy [A]  time = 0.15, size = 58, normalized size = 0.67 \[ \frac {f^{a + \frac {b}{x^{2}}} \left (- b^{3} \log {\relax (f )}^{3} + 3 b^{2} x^{2} \log {\relax (f )}^{2} - 6 b x^{4} \log {\relax (f )} + 6 x^{6}\right )}{2 b^{4} x^{6} \log {\relax (f )}^{4}} \]

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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