\(\int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx\) [148]

Optimal result

Integrand size = 13, antiderivative size = 63 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx=\frac {f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2243, 2242, 2235} \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx=\frac {\sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)} \]

[In]

Int[f^(a + b/x^2)/x^4,x]

[Out]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx=\frac {f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)} \]

[In]

Integrate[f^(a + b/x^2)/x^4,x]

[Out]

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

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92

method	result	size
risch	\(-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 x b \ln \left (f \right )}+\frac {f^{a} \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{4 \ln \left (f \right ) b \sqrt {-b \ln \left (f \right )}}\)	\(58\)
meijerg	\(-\frac {f^{a} \sqrt {-b}\, \left (\frac {\left (-b \right )^{\frac {3}{2}} \sqrt {\ln \left (f \right )}\, {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{x b}-\frac {\left (-b \right )^{\frac {3}{2}} \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{2 b^{\frac {3}{2}}}\right )}{2 \ln \left (f \right )^{\frac {3}{2}} b^{2}}\)	\(68\)

[In]

int(f^(a+b/x^2)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/2*f^a*f^(b/x^2)/x/b/ln(f)+1/4*f^a/ln(f)/b*Pi^(1/2)/(-b*ln(f))^(1/2)*erf((-b*ln(f))^(1/2)/x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx=-\frac {\sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} x \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) + 2 \, b f^{\frac {a x^{2} + b}{x^{2}}} \log \left (f\right )}{4 \, b^{2} x \log \left (f\right )^{2}} \]

[In]

integrate(f^(a+b/x^2)/x^4,x, algorithm="fricas")

[Out]

-1/4*(sqrt(pi)*sqrt(-b*log(f))*f^a*x*erf(sqrt(-b*log(f))/x) + 2*b*f^((a*x^2 + b)/x^2)*log(f))/(b^2*x*log(f)^2)

Sympy [F(-1)]

Timed out. \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx=\text {Timed out} \]

[In]

integrate(f**(a+b/x**2)/x**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.44 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx=\frac {f^{a} \Gamma \left (\frac {3}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right )}{2 \, x^{3} \left (-\frac {b \log \left (f\right )}{x^{2}}\right )^{\frac {3}{2}}} \]

[In]

integrate(f^(a+b/x^2)/x^4,x, algorithm="maxima")

[Out]

1/2*f^a*gamma(3/2, -b*log(f)/x^2)/(x^3*(-b*log(f)/x^2)^(3/2))

Giac [F]

\[ \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx=\int { \frac {f^{a + \frac {b}{x^{2}}}}{x^{4}} \,d x } \]

[In]

integrate(f^(a+b/x^2)/x^4,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)/x^4, x)

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.92 \[ \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx=\frac {f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (f\right )}{x\,\sqrt {b\,\ln \left (f\right )}}\right )}{4\,b\,\ln \left (f\right )\,\sqrt {b\,\ln \left (f\right )}}-\frac {f^a\,f^{\frac {b}{x^2}}}{2\,b\,x\,\ln \left (f\right )} \]

[In]

int(f^(a + b/x^2)/x^4,x)

[Out]

(f^a*pi^(1/2)*erfi((b*log(f))/(x*(b*log(f))^(1/2))))/(4*b*log(f)*(b*log(f))^(1/2)) - (f^a*f^(b/x^2))/(2*b*x*lo
g(f))