\(\int f^{a+\frac {b}{x^2}} \, dx\) [146]

Optimal result

Integrand size = 9, antiderivative size = 49 \[ \int f^{a+\frac {b}{x^2}} \, dx=f^{a+\frac {b}{x^2}} x-\sqrt {b} f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \sqrt {\log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2237, 2242, 2235} \[ \int f^{a+\frac {b}{x^2}} \, dx=x f^{a+\frac {b}{x^2}}-\sqrt {\pi } \sqrt {b} f^a \sqrt {\log (f)} \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \]

[In]

Int[f^(a + b/x^2),x]

[Out]

f^(a + b/x^2)*x - Sqrt[b]*f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x]*Sqrt[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 2237

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(c + d*x)*(F^(a + b*(c + d*x)^n)/d), x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

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
)]

Rubi steps \begin{align*} \text {integral}& = f^{a+\frac {b}{x^2}} x+(2 b \log (f)) \int \frac {f^{a+\frac {b}{x^2}}}{x^2} \, dx \\ & = f^{a+\frac {b}{x^2}} x-(2 b \log (f)) \text {Subst}\left (\int f^{a+b x^2} \, dx,x,\frac {1}{x}\right ) \\ & = f^{a+\frac {b}{x^2}} x-\sqrt {b} f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \sqrt {\log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x^2}} \, dx=f^{a+\frac {b}{x^2}} x-\sqrt {b} f^a \sqrt {\pi } \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right ) \sqrt {\log (f)} \]

[In]

Integrate[f^(a + b/x^2),x]

[Out]

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

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90

method	result	size
risch	\(f^{a} x \,f^{\frac {b}{x^{2}}}-\frac {f^{a} \ln \left (f \right ) b \sqrt {\pi }\, \operatorname {erf}\left (\frac {\sqrt {-b \ln \left (f \right )}}{x}\right )}{\sqrt {-b \ln \left (f \right )}}\)	\(44\)
meijerg	\(-\frac {f^{a} \sqrt {-b}\, \sqrt {\ln \left (f \right )}\, \left (-\frac {2 x \,{\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{\sqrt {-b}\, \sqrt {\ln \left (f \right )}}+\frac {2 \sqrt {b}\, \sqrt {\pi }\, \operatorname {erfi}\left (\frac {\sqrt {b}\, \sqrt {\ln \left (f \right )}}{x}\right )}{\sqrt {-b}}\right )}{2}\)	\(61\)

[In]

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

[Out]

f^a*x*f^(b/x^2)-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 = 42, normalized size of antiderivative = 0.86 \[ \int f^{a+\frac {b}{x^2}} \, dx=\sqrt {\pi } \sqrt {-b \log \left (f\right )} f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \left (f\right )}}{x}\right ) + f^{\frac {a x^{2} + b}{x^{2}}} x \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b/x**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.53 \[ \int f^{a+\frac {b}{x^2}} \, dx=\frac {1}{2} \, f^{a} x \sqrt {-\frac {b \log \left (f\right )}{x^{2}}} \Gamma \left (-\frac {1}{2}, -\frac {b \log \left (f\right )}{x^{2}}\right ) \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(a + b/x^2), x)

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.90 \[ \int f^{a+\frac {b}{x^2}} \, dx=f^a\,f^{\frac {b}{x^2}}\,x-\frac {b\,f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \left (f\right )}{x\,\sqrt {b\,\ln \left (f\right )}}\right )\,\ln \left (f\right )}{\sqrt {b\,\ln \left (f\right )}} \]

[In]

int(f^(a + b/x^2),x)

[Out]

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