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

Optimal. Leaf size=63 \[ \frac {\sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)} \]

[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)

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Rubi [A]  time = 0.05, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {2212, 2211, 2204} \[ \frac {\sqrt {\pi } f^a \text {Erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)} \]

Antiderivative was successfully verified.

[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 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
)]

Rule 2212

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*} \int \frac {f^{a+\frac {b}{x^2}}}{x^4} \, dx &=-\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 {\operatorname {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*}

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Mathematica [A]  time = 0.02, size = 63, normalized size = 1.00 \[ \frac {\sqrt {\pi } f^a \text {erfi}\left (\frac {\sqrt {b} \sqrt {\log (f)}}{x}\right )}{4 b^{3/2} \log ^{\frac {3}{2}}(f)}-\frac {f^{a+\frac {b}{x^2}}}{2 b x \log (f)} \]

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.42, size = 58, normalized size = 0.92 \[ -\frac {\sqrt {\pi } \sqrt {-b \log \relax (f)} f^{a} x \operatorname {erf}\left (\frac {\sqrt {-b \log \relax (f)}}{x}\right ) + 2 \, b f^{\frac {a x^{2} + b}{x^{2}}} \log \relax (f)}{4 \, b^{2} x \log \relax (f)^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + \frac {b}{x^{2}}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(a + b/x^2)/x^4, x)

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maple [A]  time = 0.06, size = 58, normalized size = 0.92 \[ \frac {\sqrt {\pi }\, f^{a} \erf \left (\frac {\sqrt {-b \ln \relax (f )}}{x}\right )}{4 \sqrt {-b \ln \relax (f )}\, b \ln \relax (f )}-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 b x \ln \relax (f )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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maxima [A]  time = 1.49, size = 28, normalized size = 0.44 \[ \frac {f^{a} \Gamma \left (\frac {3}{2}, -\frac {b \log \relax (f)}{x^{2}}\right )}{2 \, x^{3} \left (-\frac {b \log \relax (f)}{x^{2}}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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mupad [B]  time = 3.56, size = 58, normalized size = 0.92 \[ \frac {f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \relax (f)}{x\,\sqrt {b\,\ln \relax (f)}}\right )}{4\,b\,\ln \relax (f)\,\sqrt {b\,\ln \relax (f)}}-\frac {f^a\,f^{\frac {b}{x^2}}}{2\,b\,x\,\ln \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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