3.150 \(\int \frac{f^{a+\frac{b}{x^2}}}{x^8} \, dx\)

Optimal. Leaf size=109 \[ \frac{15 \sqrt{\pi } f^a \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (f)}}{x}\right )}{16 b^{7/2} \log ^{\frac{7}{2}}(f)}+\frac{5 f^{a+\frac{b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac{15 f^{a+\frac{b}{x^2}}}{8 b^3 x \log ^3(f)}-\frac{f^{a+\frac{b}{x^2}}}{2 b x^5 \log (f)} \]

[Out]

(15*f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x])/(16*b^(7/2)*Log[f]^(7/2)) - (15*f^(a + b/x^2))/(8*b^3*x*Log[f
]^3) + (5*f^(a + b/x^2))/(4*b^2*x^3*Log[f]^2) - f^(a + b/x^2)/(2*b*x^5*Log[f])

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Rubi [A]  time = 0.112402, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {2212, 2211, 2204} \[ \frac{15 \sqrt{\pi } f^a \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (f)}}{x}\right )}{16 b^{7/2} \log ^{\frac{7}{2}}(f)}+\frac{5 f^{a+\frac{b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac{15 f^{a+\frac{b}{x^2}}}{8 b^3 x \log ^3(f)}-\frac{f^{a+\frac{b}{x^2}}}{2 b x^5 \log (f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)/x^8,x]

[Out]

(15*f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x])/(16*b^(7/2)*Log[f]^(7/2)) - (15*f^(a + b/x^2))/(8*b^3*x*Log[f
]^3) + (5*f^(a + b/x^2))/(4*b^2*x^3*Log[f]^2) - f^(a + b/x^2)/(2*b*x^5*Log[f])

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])

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 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]

Rubi steps

\begin{align*} \int \frac{f^{a+\frac{b}{x^2}}}{x^8} \, dx &=-\frac{f^{a+\frac{b}{x^2}}}{2 b x^5 \log (f)}-\frac{5 \int \frac{f^{a+\frac{b}{x^2}}}{x^6} \, dx}{2 b \log (f)}\\ &=\frac{5 f^{a+\frac{b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac{f^{a+\frac{b}{x^2}}}{2 b x^5 \log (f)}+\frac{15 \int \frac{f^{a+\frac{b}{x^2}}}{x^4} \, dx}{4 b^2 \log ^2(f)}\\ &=-\frac{15 f^{a+\frac{b}{x^2}}}{8 b^3 x \log ^3(f)}+\frac{5 f^{a+\frac{b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac{f^{a+\frac{b}{x^2}}}{2 b x^5 \log (f)}-\frac{15 \int \frac{f^{a+\frac{b}{x^2}}}{x^2} \, dx}{8 b^3 \log ^3(f)}\\ &=-\frac{15 f^{a+\frac{b}{x^2}}}{8 b^3 x \log ^3(f)}+\frac{5 f^{a+\frac{b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac{f^{a+\frac{b}{x^2}}}{2 b x^5 \log (f)}+\frac{15 \operatorname{Subst}\left (\int f^{a+b x^2} \, dx,x,\frac{1}{x}\right )}{8 b^3 \log ^3(f)}\\ &=\frac{15 f^a \sqrt{\pi } \text{erfi}\left (\frac{\sqrt{b} \sqrt{\log (f)}}{x}\right )}{16 b^{7/2} \log ^{\frac{7}{2}}(f)}-\frac{15 f^{a+\frac{b}{x^2}}}{8 b^3 x \log ^3(f)}+\frac{5 f^{a+\frac{b}{x^2}}}{4 b^2 x^3 \log ^2(f)}-\frac{f^{a+\frac{b}{x^2}}}{2 b x^5 \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.0557144, size = 86, normalized size = 0.79 \[ \frac{15 \sqrt{\pi } f^a \text{Erfi}\left (\frac{\sqrt{b} \sqrt{\log (f)}}{x}\right )}{16 b^{7/2} \log ^{\frac{7}{2}}(f)}-\frac{f^{a+\frac{b}{x^2}} \left (4 b^2 \log ^2(f)-10 b x^2 \log (f)+15 x^4\right )}{8 b^3 x^5 \log ^3(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)/x^8,x]

[Out]

(15*f^a*Sqrt[Pi]*Erfi[(Sqrt[b]*Sqrt[Log[f]])/x])/(16*b^(7/2)*Log[f]^(7/2)) - (f^(a + b/x^2)*(15*x^4 - 10*b*x^2
*Log[f] + 4*b^2*Log[f]^2))/(8*b^3*x^5*Log[f]^3)

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Maple [A]  time = 0.044, size = 102, normalized size = 0.9 \begin{align*} -{\frac{{f}^{a}}{2\,{x}^{5}b\ln \left ( f \right ) }{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{5\,{f}^{a}}{4\,{b}^{2}{x}^{3} \left ( \ln \left ( f \right ) \right ) ^{2}}{f}^{{\frac{b}{{x}^{2}}}}}-{\frac{15\,{f}^{a}}{8\,{b}^{3}x \left ( \ln \left ( f \right ) \right ) ^{3}}{f}^{{\frac{b}{{x}^{2}}}}}+{\frac{15\,{f}^{a}\sqrt{\pi }}{16\, \left ( \ln \left ( f \right ) \right ) ^{3}{b}^{3}}{\it Erf} \left ({\frac{1}{x}\sqrt{-b\ln \left ( f \right ) }} \right ){\frac{1}{\sqrt{-b\ln \left ( f \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^2)/x^8,x)

[Out]

-1/2*f^a*f^(b/x^2)/x^5/b/ln(f)+5/4*f^a/ln(f)^2/b^2*f^(b/x^2)/x^3-15/8*f^a/ln(f)^3/b^3*f^(b/x^2)/x+15/16*f^a/ln
(f)^3/b^3*Pi^(1/2)/(-b*ln(f))^(1/2)*erf((-b*ln(f))^(1/2)/x)

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Maxima [A]  time = 1.24372, size = 38, normalized size = 0.35 \begin{align*} \frac{f^{a} \Gamma \left (\frac{7}{2}, -\frac{b \log \left (f\right )}{x^{2}}\right )}{2 \, x^{7} \left (-\frac{b \log \left (f\right )}{x^{2}}\right )^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)/x^8,x, algorithm="maxima")

[Out]

1/2*f^a*gamma(7/2, -b*log(f)/x^2)/(x^7*(-b*log(f)/x^2)^(7/2))

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Fricas [A]  time = 2.11687, size = 227, normalized size = 2.08 \begin{align*} -\frac{15 \, \sqrt{\pi } \sqrt{-b \log \left (f\right )} f^{a} x^{5} \operatorname{erf}\left (\frac{\sqrt{-b \log \left (f\right )}}{x}\right ) + 2 \,{\left (15 \, b x^{4} \log \left (f\right ) - 10 \, b^{2} x^{2} \log \left (f\right )^{2} + 4 \, b^{3} \log \left (f\right )^{3}\right )} f^{\frac{a x^{2} + b}{x^{2}}}}{16 \, b^{4} x^{5} \log \left (f\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)/x^8,x, algorithm="fricas")

[Out]

-1/16*(15*sqrt(pi)*sqrt(-b*log(f))*f^a*x^5*erf(sqrt(-b*log(f))/x) + 2*(15*b*x^4*log(f) - 10*b^2*x^2*log(f)^2 +
4*b^3*log(f)^3)*f^((a*x^2 + b)/x^2))/(b^4*x^5*log(f)^4)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(a+b/x**2)/x**8,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{a + \frac{b}{x^{2}}}}{x^{8}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(a+b/x^2)/x^8,x, algorithm="giac")

[Out]

integrate(f^(a + b/x^2)/x^8, x)