3.80 \(\int \frac {f^{a+b x^2}}{x^9} \, dx\)

Optimal. Leaf size=24 \[ -\frac {1}{2} b^4 f^a \log ^4(f) \Gamma \left (-4,-b x^2 \log (f)\right ) \]

[Out]

-1/2*f^a/x^8*Ei(5,-b*x^2*ln(f))

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2218} \[ -\frac {1}{2} b^4 f^a \log ^4(f) \text {Gamma}\left (-4,-b x^2 \log (f)\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)/x^9,x]

[Out]

-(b^4*f^a*Gamma[-4, -(b*x^2*Log[f])]*Log[f]^4)/2

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin {align*} \int \frac {f^{a+b x^2}}{x^9} \, dx &=-\frac {1}{2} b^4 f^a \Gamma \left (-4,-b x^2 \log (f)\right ) \log ^4(f)\\ \end {align*}

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Mathematica [A]  time = 0.00, size = 24, normalized size = 1.00 \[ -\frac {1}{2} b^4 f^a \log ^4(f) \Gamma \left (-4,-b x^2 \log (f)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)/x^9,x]

[Out]

-1/2*(b^4*f^a*Gamma[-4, -(b*x^2*Log[f])]*Log[f]^4)

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fricas [B]  time = 0.41, size = 71, normalized size = 2.96 \[ \frac {b^{4} f^{a} x^{8} {\rm Ei}\left (b x^{2} \log \relax (f)\right ) \log \relax (f)^{4} - {\left (b^{3} x^{6} \log \relax (f)^{3} + b^{2} x^{4} \log \relax (f)^{2} + 2 \, b x^{2} \log \relax (f) + 6\right )} f^{b x^{2} + a}}{48 \, x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^9,x, algorithm="fricas")

[Out]

1/48*(b^4*f^a*x^8*Ei(b*x^2*log(f))*log(f)^4 - (b^3*x^6*log(f)^3 + b^2*x^4*log(f)^2 + 2*b*x^2*log(f) + 6)*f^(b*
x^2 + a))/x^8

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{b x^{2} + a}}{x^{9}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^9,x, algorithm="giac")

[Out]

integrate(f^(b*x^2 + a)/x^9, x)

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maple [B]  time = 0.07, size = 101, normalized size = 4.21 \[ -\frac {b^{4} f^{a} \Ei \left (1, -b \,x^{2} \ln \relax (f )\right ) \ln \relax (f )^{4}}{48}-\frac {b^{3} f^{a} f^{b \,x^{2}} \ln \relax (f )^{3}}{48 x^{2}}-\frac {b^{2} f^{a} f^{b \,x^{2}} \ln \relax (f )^{2}}{48 x^{4}}-\frac {b \,f^{a} f^{b \,x^{2}} \ln \relax (f )}{24 x^{6}}-\frac {f^{a} f^{b \,x^{2}}}{8 x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^2+a)/x^9,x)

[Out]

-1/8*f^a/x^8*f^(b*x^2)-1/24*f^a*ln(f)*b/x^6*f^(b*x^2)-1/48*f^a*ln(f)^2*b^2/x^4*f^(b*x^2)-1/48*f^a*ln(f)^3*b^3/
x^2*f^(b*x^2)-1/48*f^a*ln(f)^4*b^4*Ei(1,-b*x^2*ln(f))

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maxima [B]  time = 0.62, size = 22, normalized size = 0.92 \[ -\frac {1}{2} \, b^{4} f^{a} \Gamma \left (-4, -b x^{2} \log \relax (f)\right ) \log \relax (f)^{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^9,x, algorithm="maxima")

[Out]

-1/2*b^4*f^a*gamma(-4, -b*x^2*log(f))*log(f)^4

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mupad [B]  time = 3.57, size = 90, normalized size = 3.75 \[ -\frac {b^4\,f^a\,{\ln \relax (f)}^4\,\mathrm {expint}\left (-b\,x^2\,\ln \relax (f)\right )}{48}-\frac {b^4\,f^a\,f^{b\,x^2}\,{\ln \relax (f)}^4\,\left (\frac {1}{24\,b\,x^2\,\ln \relax (f)}+\frac {1}{24\,b^2\,x^4\,{\ln \relax (f)}^2}+\frac {1}{12\,b^3\,x^6\,{\ln \relax (f)}^3}+\frac {1}{4\,b^4\,x^8\,{\ln \relax (f)}^4}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^2)/x^9,x)

[Out]

- (b^4*f^a*log(f)^4*expint(-b*x^2*log(f)))/48 - (b^4*f^a*f^(b*x^2)*log(f)^4*(1/(24*b*x^2*log(f)) + 1/(24*b^2*x
^4*log(f)^2) + 1/(12*b^3*x^6*log(f)^3) + 1/(4*b^4*x^8*log(f)^4)))/2

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {f^{a + b x^{2}}}{x^{9}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**2+a)/x**9,x)

[Out]

Integral(f**(a + b*x**2)/x**9, x)

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