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

Optimal. Leaf size=39 \[ -\frac {\sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{2 \sqrt {b} \sqrt {\log (f)}} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2211, 2204} \[ -\frac {\sqrt {\pi } f^a \text {Erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{2 \sqrt {b} \sqrt {\log (f)}} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)/x^2,x]

[Out]

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

Rule 2204

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 2211

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

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Mathematica [A]  time = 0.01, size = 39, normalized size = 1.00 \[ -\frac {\sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{2 \sqrt {b} \sqrt {\log (f)}} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)/x^2,x]

[Out]

-1/2*(f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x])/(Sqrt[b]*Sqrt[Log[f]])

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fricas [A]  time = 0.43, size = 34, normalized size = 0.87 \[ \frac {\sqrt {\pi } \sqrt {-b \log \relax (f)} f^{a} \operatorname {erf}\left (\frac {\sqrt {-b \log \relax (f)}}{x}\right )}{2 \, b \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + \frac {b}{x^{2}}}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(a + b/x^2)/x^2, x)

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maple [A]  time = 0.05, size = 28, normalized size = 0.72 \[ -\frac {\sqrt {\pi }\, f^{a} \erf \left (\frac {\sqrt {-b \ln \relax (f )}}{x}\right )}{2 \sqrt {-b \ln \relax (f )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.30, size = 34, normalized size = 0.87 \[ -\frac {\sqrt {\pi } f^{a} {\left (\operatorname {erf}\left (\sqrt {-\frac {b \log \relax (f)}{x^{2}}}\right ) - 1\right )}}{2 \, x \sqrt {-\frac {b \log \relax (f)}{x^{2}}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.52, size = 28, normalized size = 0.72 \[ -\frac {f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \relax (f)}{x\,\sqrt {b\,\ln \relax (f)}}\right )}{2\,\sqrt {b\,\ln \relax (f)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^2)/x^2,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + \frac {b}{x^{2}}}}{x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b/x**2)/x**2, x)

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