\(\int \frac {f^{a+\frac {b}{x^2}}}{x^9} \, dx\) [138]

Optimal result

Integrand size = 13, antiderivative size = 86 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^9} \, dx=\frac {3 f^{a+\frac {b}{x^2}}}{b^4 \log ^4(f)}-\frac {3 f^{a+\frac {b}{x^2}}}{b^3 x^2 \log ^3(f)}+\frac {3 f^{a+\frac {b}{x^2}}}{2 b^2 x^4 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^6 \log (f)} \]

[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)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2243, 2240} \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^9} \, dx=\frac {3 f^{a+\frac {b}{x^2}}}{b^4 \log ^4(f)}-\frac {3 f^{a+\frac {b}{x^2}}}{b^3 x^2 \log ^3(f)}+\frac {3 f^{a+\frac {b}{x^2}}}{2 b^2 x^4 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^6 \log (f)} \]

[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 2240

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 2243

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*} \text {integral}& = -\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] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^9} \, dx=\frac {f^{a+\frac {b}{x^2}} \left (6 x^6-6 b x^4 \log (f)+3 b^2 x^2 \log ^2(f)-b^3 \log ^3(f)\right )}{2 b^4 x^6 \log ^4(f)} \]

[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)

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69

method	result	size
meijerg	\(-\frac {f^{a} \left (6-\frac {\left (-\frac {4 b^{3} \ln \left (f \right )^{3}}{x^{6}}+\frac {12 b^{2} \ln \left (f \right )^{2}}{x^{4}}-\frac {24 b \ln \left (f \right )}{x^{2}}+24\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{4}\right )}{2 b^{4} \ln \left (f \right )^{4}}\)	\(59\)
risch	\(-\frac {\left (\ln \left (f \right )^{3} b^{3}-3 b^{2} x^{2} \ln \left (f \right )^{2}+6 b \,x^{4} \ln \left (f \right )-6 x^{6}\right ) f^{\frac {a \,x^{2}+b}{x^{2}}}}{2 \ln \left (f \right )^{4} b^{4} x^{6}}\)	\(60\)
parallelrisch	\(\frac {-f^{a +\frac {b}{x^{2}}} \ln \left (f \right )^{3} b^{3}+3 b^{2} f^{a +\frac {b}{x^{2}}} x^{2} \ln \left (f \right )^{2}-6 b \,f^{a +\frac {b}{x^{2}}} x^{4} \ln \left (f \right )+6 f^{a +\frac {b}{x^{2}}} x^{6}}{2 x^{6} \ln \left (f \right )^{4} b^{4}}\)	\(84\)
norman	\(\frac {-\frac {x^{2} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{2 b \ln \left (f \right )}+\frac {3 x^{4} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{2 b^{2} \ln \left (f \right )^{2}}-\frac {3 x^{6} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{b^{3} \ln \left (f \right )^{3}}+\frac {3 x^{8} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{b^{4} \ln \left (f \right )^{4}}}{x^{8}}\)	\(98\)

[In]

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

[Out]

-1/2*f^a/b^4/ln(f)^4*(6-1/4*(-4*b^3*ln(f)^3/x^6+12*b^2*ln(f)^2/x^4-24*b*ln(f)/x^2+24)*exp(b*ln(f)/x^2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.70 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^9} \, dx=\frac {{\left (6 \, x^{6} - 6 \, b x^{4} \log \left (f\right ) + 3 \, b^{2} x^{2} \log \left (f\right )^{2} - b^{3} \log \left (f\right )^{3}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{2 \, b^{4} x^{6} \log \left (f\right )^{4}} \]

[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)

Sympy [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.67 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^9} \, dx=\frac {f^{a + \frac {b}{x^{2}}} \left (- b^{3} \log {\left (f \right )}^{3} + 3 b^{2} x^{2} \log {\left (f \right )}^{2} - 6 b x^{4} \log {\left (f \right )} + 6 x^{6}\right )}{2 b^{4} x^{6} \log {\left (f \right )}^{4}} \]

[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)

Maxima [C] (verification not implemented)

Result contains higher order function than in optimal. Order 4 vs. order 3.

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.26 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^9} \, dx=\frac {f^{a} \Gamma \left (4, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, b^{4} \log \left (f\right )^{4}} \]

[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)

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^9} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{9}} \,d x } \]

[In]

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

[Out]

integrate(f^(a + b/x^2)/x^9, x)

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.70 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^9} \, dx=-\frac {f^{a+\frac {b}{x^2}}\,\left (\frac {1}{2\,b\,\ln \left (f\right )}-\frac {3\,x^2}{2\,b^2\,{\ln \left (f\right )}^2}+\frac {3\,x^4}{b^3\,{\ln \left (f\right )}^3}-\frac {3\,x^6}{b^4\,{\ln \left (f\right )}^4}\right )}{x^6} \]

[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