3.78 \(\int \frac{f^{a+b x^2}}{x^5} \, dx\)

Optimal. Leaf size=58 \[ \frac{1}{4} b^2 f^a \log ^2(f) \text{Ei}\left (b x^2 \log (f)\right )-\frac{f^{a+b x^2}}{4 x^4}-\frac{b \log (f) f^{a+b x^2}}{4 x^2} \]

[Out]

-f^(a + b*x^2)/(4*x^4) - (b*f^(a + b*x^2)*Log[f])/(4*x^2) + (b^2*f^a*ExpIntegralEi[b*x^2*Log[f]]*Log[f]^2)/4

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Rubi [A]  time = 0.0659541, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2214, 2210} \[ \frac{1}{4} b^2 f^a \log ^2(f) \text{Ei}\left (b x^2 \log (f)\right )-\frac{f^{a+b x^2}}{4 x^4}-\frac{b \log (f) f^{a+b x^2}}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)/x^5,x]

[Out]

-f^(a + b*x^2)/(4*x^4) - (b*f^(a + b*x^2)*Log[f])/(4*x^2) + (b^2*f^a*ExpIntegralEi[b*x^2*Log[f]]*Log[f]^2)/4

Rule 2214

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
+ 1)*F^(a + b*(c + d*x)^n))/(d*(m + 1)), x] - Dist[(b*n*Log[F])/(m + 1), 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[-4, (m + 1)/n, 5] && IntegerQ[n
] && ((GtQ[n, 0] && LtQ[m, -1]) || (GtQ[-n, 0] && LeQ[-n, m + 1]))

Rule 2210

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

Rubi steps

\begin{align*} \int \frac{f^{a+b x^2}}{x^5} \, dx &=-\frac{f^{a+b x^2}}{4 x^4}+\frac{1}{2} (b \log (f)) \int \frac{f^{a+b x^2}}{x^3} \, dx\\ &=-\frac{f^{a+b x^2}}{4 x^4}-\frac{b f^{a+b x^2} \log (f)}{4 x^2}+\frac{1}{2} \left (b^2 \log ^2(f)\right ) \int \frac{f^{a+b x^2}}{x} \, dx\\ &=-\frac{f^{a+b x^2}}{4 x^4}-\frac{b f^{a+b x^2} \log (f)}{4 x^2}+\frac{1}{4} b^2 f^a \text{Ei}\left (b x^2 \log (f)\right ) \log ^2(f)\\ \end{align*}

Mathematica [A]  time = 0.0184995, size = 48, normalized size = 0.83 \[ \frac{f^a \left (b^2 x^4 \log ^2(f) \text{Ei}\left (b x^2 \log (f)\right )-f^{b x^2} \left (b x^2 \log (f)+1\right )\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)/x^5,x]

[Out]

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

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Maple [A]  time = 0.026, size = 57, normalized size = 1. \begin{align*} -{\frac{{f}^{a}{f}^{b{x}^{2}}}{4\,{x}^{4}}}-{\frac{{f}^{a}\ln \left ( f \right ) b{f}^{b{x}^{2}}}{4\,{x}^{2}}}-{\frac{{f}^{a} \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}{\it Ei} \left ( 1,-b{x}^{2}\ln \left ( f \right ) \right ) }{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^2+a)/x^5,x)

[Out]

-1/4*f^a/x^4*f^(b*x^2)-1/4*f^a*ln(f)*b/x^2*f^(b*x^2)-1/4*f^a*ln(f)^2*b^2*Ei(1,-b*x^2*ln(f))

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Maxima [A]  time = 1.22192, size = 30, normalized size = 0.52 \begin{align*} -\frac{1}{2} \, b^{2} f^{a} \Gamma \left (-2, -b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^5,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.75448, size = 113, normalized size = 1.95 \begin{align*} \frac{b^{2} f^{a} x^{4}{\rm Ei}\left (b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{2} -{\left (b x^{2} \log \left (f\right ) + 1\right )} f^{b x^{2} + a}}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^5,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{a + b x^{2}}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**2+a)/x**5,x)

[Out]

Integral(f**(a + b*x**2)/x**5, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{f^{b x^{2} + a}}{x^{5}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)/x^5,x, algorithm="giac")

[Out]

integrate(f^(b*x^2 + a)/x^5, x)