\(\int \frac {f^{a+\frac {b}{x^2}}}{x^5} \, dx\) [136]

Optimal result

Integrand size = 13, antiderivative size = 44 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^5} \, dx=\frac {f^{a+\frac {b}{x^2}}}{2 b^2 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^2 \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2243, 2240} \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^5} \, dx=\frac {f^{a+\frac {b}{x^2}}}{2 b^2 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^2 \log (f)} \]

[In]

Int[f^(a + b/x^2)/x^5,x]

[Out]

f^(a + b/x^2)/(2*b^2*Log[f]^2) - f^(a + b/x^2)/(2*b*x^2*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^2 \log (f)}-\frac {\int \frac {f^{a+\frac {b}{x^2}}}{x^3} \, dx}{b \log (f)} \\ & = \frac {f^{a+\frac {b}{x^2}}}{2 b^2 \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^2 \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.73 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^5} \, dx=\frac {f^{a+\frac {b}{x^2}} \left (x^2-b \log (f)\right )}{2 b^2 x^2 \log ^2(f)} \]

[In]

Integrate[f^(a + b/x^2)/x^5,x]

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.80

method	result	size
meijerg	\(-\frac {f^{a} \left (1-\frac {\left (2-\frac {2 b \ln \left (f \right )}{x^{2}}\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{2}\right )}{2 b^{2} \ln \left (f \right )^{2}}\)	\(35\)
risch	\(-\frac {\left (b \ln \left (f \right )-x^{2}\right ) f^{\frac {a \,x^{2}+b}{x^{2}}}}{2 \ln \left (f \right )^{2} b^{2} x^{2}}\)	\(36\)
parallelrisch	\(\frac {-f^{a +\frac {b}{x^{2}}} b \ln \left (f \right )+f^{a +\frac {b}{x^{2}}} x^{2}}{2 x^{2} \ln \left (f \right )^{2} b^{2}}\)	\(41\)
norman	\(\frac {-\frac {x^{2} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{2 b \ln \left (f \right )}+\frac {x^{4} {\mathrm e}^{\left (a +\frac {b}{x^{2}}\right ) \ln \left (f \right )}}{2 b^{2} \ln \left (f \right )^{2}}}{x^{4}}\)	\(52\)

[In]

int(f^(a+b/x^2)/x^5,x,method=_RETURNVERBOSE)

[Out]

-1/2*f^a/b^2/ln(f)^2*(1-1/2*(2-2*b*ln(f)/x^2)*exp(b*ln(f)/x^2))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.77 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^5} \, dx=\frac {{\left (x^{2} - b \log \left (f\right )\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{2 \, b^{2} x^{2} \log \left (f\right )^{2}} \]

[In]

integrate(f^(a+b/x^2)/x^5,x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.66 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^5} \, dx=\frac {f^{a + \frac {b}{x^{2}}} \left (- b \log {\left (f \right )} + x^{2}\right )}{2 b^{2} x^{2} \log {\left (f \right )}^{2}} \]

[In]

integrate(f**(a+b/x**2)/x**5,x)

[Out]

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

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 = 0.50 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^5} \, dx=\frac {f^{a} \Gamma \left (2, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, b^{2} \log \left (f\right )^{2}} \]

[In]

integrate(f^(a+b/x^2)/x^5,x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^5} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{5}} \,d x } \]

[In]

integrate(f^(a+b/x^2)/x^5,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)/x^5, x)

Mupad [B] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.82 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^5} \, dx=-\frac {f^{a+\frac {b}{x^2}}\,\left (\frac {1}{2\,b\,\ln \left (f\right )}-\frac {x^2}{2\,b^2\,{\ln \left (f\right )}^2}\right )}{x^2} \]

[In]

int(f^(a + b/x^2)/x^5,x)

[Out]

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