\(\int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx\) [149]

Optimal result

Integrand size = 13, antiderivative size = 86 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=-\frac {3 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}+\frac {3 f^{a+\frac {b}{x^2}}}{4 b^2 x \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)} \]

[Out]

3/4*f^(a+b/x^2)/b^2/x/ln(f)^2-1/2*f^(a+b/x^2)/b/x^3/ln(f)-3/8*f^a*erfi(b^(1/2)*ln(f)^(1/2)/x)*Pi^(1/2)/b^(5/2)
/ln(f)^(5/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2243, 2242, 2235} \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=-\frac {3 \sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}+\frac {3 f^{a+\frac {b}{x^2}}}{4 b^2 x \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)} \]

[In]

Int[f^(a + b/x^2)/x^6,x]

[Out]

(-3*f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x])/(8*b^(5/2)*Log[f]^(5/2)) + (3*f^(a + b/x^2))/(4*b^2*x*Log[f]^
2) - f^(a + b/x^2)/(2*b*x^3*Log[f])

Rule 2235

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^2), x_Symbol] :> Simp[F^a*Sqrt[Pi]*(Erfi[(c + d*x)*Rt[b*Log[F], 2
]]/(2*d*Rt[b*Log[F], 2])), x] /; FreeQ[{F, a, b, c, d}, x] && PosQ[b]

Rule 2242

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[1/(d*(m + 1))
, Subst[Int[F^(a + b*x^2), x], x, (c + d*x)^(m + 1)], x] /; FreeQ[{F, a, b, c, d, m, n}, x] && EqQ[n, 2*(m + 1
)]

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^3 \log (f)}-\frac {3 \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx}{2 b \log (f)} \\ & = \frac {3 f^{a+\frac {b}{x^2}}}{4 b^2 x \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}+\frac {3 \int \frac {f^{a+\frac {b}{x^2}}}{x^2} \, dx}{4 b^2 \log ^2(f)} \\ & = \frac {3 f^{a+\frac {b}{x^2}}}{4 b^2 x \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)}-\frac {3 \text {Subst}\left (\int f^{a+b x^2} \, dx,x,\frac {1}{x}\right )}{4 b^2 \log ^2(f)} \\ & = -\frac {3 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}+\frac {3 f^{a+\frac {b}{x^2}}}{4 b^2 x \log ^2(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x^3 \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.86 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=-\frac {3 f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{8 b^{5/2} \log ^{\frac {5}{2}}(f)}+\frac {f^{a+\frac {b}{x^2}} \left (3 x^2-2 b \log (f)\right )}{4 b^2 x^3 \log ^2(f)} \]

[In]

Integrate[f^(a + b/x^2)/x^6,x]

[Out]

(-3*f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x])/(8*b^(5/2)*Log[f]^(5/2)) + (f^(a + b/x^2)*(3*x^2 - 2*b*Log[f]
))/(4*b^2*x^3*Log[f]^2)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92

method	result	size
meijerg	\(\frac {f^{a} \sqrt {-b}\, \left (-\frac {\left (-b \right )^{\frac {5}{2}} \sqrt {\ln \left (f \right )}\, \left (-\frac {10 b \ln \left (f \right )}{x^{2}}+15\right ) {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{10 x \,b^{2}}+\frac {3 \left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{4 b^{\frac {5}{2}}}\right )}{2 \ln \left (f \right )^{\frac {5}{2}} b^{3}}\)	\(79\)
risch	\(-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 x^{3} b \ln \left (f \right )}+\frac {3 f^{a} f^{\frac {b}{x^{2}}}}{4 \ln \left (f \right )^{2} b^{2} x}-\frac {3 f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{8 \ln \left (f \right )^{2} b^{2} \sqrt {-b \ln \left (f \right )}}\)	\(80\)

[In]

int(f^(a+b/x^2)/x^6,x,method=_RETURNVERBOSE)

[Out]

1/2*f^a/ln(f)^(5/2)/b^3*(-b)^(1/2)*(-1/10/x*(-b)^(5/2)*ln(f)^(1/2)*(-10*b*ln(f)/x^2+15)/b^2*exp(b*ln(f)/x^2)+3
/4*(-b)^(5/2)/b^(5/2)*Pi^(1/2)*erfi(b^(1/2)*ln(f)^(1/2)/x))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.88 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=\frac {3 \, \sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} x^{3} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) + 2 \, {\left (3 \, b x^{2} \log \left (f\right ) - 2 \, b^{2} \log \left (f\right )^{2}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{8 \, b^{3} x^{3} \log \left (f\right )^{3}} \]

[In]

integrate(f^(a+b/x^2)/x^6,x, algorithm="fricas")

[Out]

1/8*(3*sqrt(pi)*sqrt(-b*log(f))*f^a*x^3*erf(sqrt(-b*log(f))/x) + 2*(3*b*x^2*log(f) - 2*b^2*log(f)^2)*f^((a*x^2
+ b)/x^2))/(b^3*x^3*log(f)^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=\text {Timed out} \]

[In]

integrate(f**(a+b/x**2)/x**6,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.33 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=\frac {f^{a} \Gamma \left (\frac {5}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, x^{5} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {5}{2}}} \]

[In]

integrate(f^(a+b/x^2)/x^6,x, algorithm="maxima")

[Out]

1/2*f^a*gamma(5/2, -b*log(f)/x^2)/(x^5*(-b*log(f)/x^2)^(5/2))

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{6}} \,d x } \]

[In]

integrate(f^(a+b/x^2)/x^6,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)/x^6, x)

Mupad [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.92 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^6} \, dx=-\frac {f^a\,\left (3\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (f\right )}{x\,\sqrt {b\,\ln \left (f\right )}}\right )-\frac {6\,f^{\frac {b}{x^2}}\,\sqrt {b\,\ln \left (f\right )}}{x}\right )}{8\,b^2\,{\ln \left (f\right )}^2\,\sqrt {b\,\ln \left (f\right )}}-\frac {f^a\,f^{\frac {b}{x^2}}}{2\,b\,x^3\,\ln \left (f\right )} \]

[In]

int(f^(a + b/x^2)/x^6,x)

[Out]

- (f^a*(3*pi^(1/2)*erfi((b*log(f))/(x*(b*log(f))^(1/2))) - (6*f^(b/x^2)*(b*log(f))^(1/2))/x))/(8*b^2*log(f)^2*
(b*log(f))^(1/2)) - (f^a*f^(b/x^2))/(2*b*x^3*log(f))