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

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

[Out]

-15/8*f^(a+b/x^2)/b^3/x/ln(f)^3+5/4*f^(a+b/x^2)/b^2/x^3/ln(f)^2-1/2*f^(a+b/x^2)/b/x^5/ln(f)+15/16*f^a*erfi(b^(
1/2)*ln(f)^(1/2)/x)*Pi^(1/2)/b^(7/2)/ln(f)^(7/2)

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Rubi [A]  time = 0.11, antiderivative size = 109, normalized size of antiderivative = 1.00, 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 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^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*}

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Mathematica [A]  time = 0.06, 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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fricas [A]  time = 0.41, size = 88, normalized size = 0.81 \[ -\frac {15 \, \sqrt {\pi } \sqrt {-b \log \relax (f)} f^{a} x^{5} \operatorname {erf}\left (\frac {\sqrt {-b \log \relax (f)}}{x}\right ) + 2 \, {\left (15 \, b x^{4} \log \relax (f) - 10 \, b^{2} x^{2} \log \relax (f)^{2} + 4 \, b^{3} \log \relax (f)^{3}\right )} f^{\frac {a x^{2} + b}{x^{2}}}}{16 \, b^{4} x^{5} \log \relax (f)^{4}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + \frac {b}{x^{2}}}}{x^{8}}\,{d x} \]

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)

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maple [A]  time = 0.07, size = 102, normalized size = 0.94 \[ \frac {15 \sqrt {\pi }\, f^{a} \erf \left (\frac {\sqrt {-b \ln \relax (f )}}{x}\right )}{16 \sqrt {-b \ln \relax (f )}\, b^{3} \ln \relax (f )^{3}}-\frac {f^{a} f^{\frac {b}{x^{2}}}}{2 b \,x^{5} \ln \relax (f )}+\frac {5 f^{a} f^{\frac {b}{x^{2}}}}{4 b^{2} x^{3} \ln \relax (f )^{2}}-\frac {15 f^{a} f^{\frac {b}{x^{2}}}}{8 b^{3} x \ln \relax (f )^{3}} \]

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 = 1.27, size = 28, normalized size = 0.26 \[ \frac {f^{a} \Gamma \left (\frac {7}{2}, -\frac {b \log \relax (f)}{x^{2}}\right )}{2 \, x^{7} \left (-\frac {b \log \relax (f)}{x^{2}}\right )^{\frac {7}{2}}} \]

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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mupad [B]  time = 3.66, size = 102, normalized size = 0.94 \[ \frac {5\,f^a\,f^{\frac {b}{x^2}}}{4\,b^2\,x^3\,{\ln \relax (f)}^2}-\frac {f^a\,f^{\frac {b}{x^2}}}{2\,b\,x^5\,\ln \relax (f)}-\frac {15\,f^a\,f^{\frac {b}{x^2}}}{8\,b^3\,x\,{\ln \relax (f)}^3}+\frac {15\,f^a\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b\,\ln \relax (f)}{x\,\sqrt {b\,\ln \relax (f)}}\right )}{16\,b^3\,{\ln \relax (f)}^3\,\sqrt {b\,\ln \relax (f)}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^2)/x^8,x)

[Out]

(5*f^a*f^(b/x^2))/(4*b^2*x^3*log(f)^2) - (f^a*f^(b/x^2))/(2*b*x^5*log(f)) - (15*f^a*f^(b/x^2))/(8*b^3*x*log(f)
^3) + (15*f^a*pi^(1/2)*erfi((b*log(f))/(x*(b*log(f))^(1/2))))/(16*b^3*log(f)^3*(b*log(f))^(1/2))

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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**8,x)

[Out]

Timed out

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