\(\int \frac {f^{a+b x^2}}{x^5} \, dx\) [78]

Optimal result

Integrand size = 13, antiderivative size = 58 \[ \int \frac {f^{a+b x^2}}{x^5} \, dx=-\frac {f^{a+b x^2}}{4 x^4}-\frac {b f^{a+b x^2} \log (f)}{4 x^2}+\frac {1}{4} b^2 f^a \operatorname {ExpIntegralEi}\left (b x^2 \log (f)\right ) \log ^2(f) \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2245, 2241} \[ \int \frac {f^{a+b x^2}}{x^5} \, dx=\frac {1}{4} b^2 f^a \log ^2(f) \operatorname {ExpIntegralEi}\left (b x^2 \log (f)\right )-\frac {b \log (f) f^{a+b x^2}}{4 x^2}-\frac {f^{a+b x^2}}{4 x^4} \]

[In]

Int[f^(a + b*x^2)/x^5,x]

[Out]

-1/4*f^(a + b*x^2)/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 2241

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]

Rule 2245

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]))

Rubi steps \begin{align*} \text {integral}& = -\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] (verified)

Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83 \[ \int \frac {f^{a+b x^2}}{x^5} \, dx=\frac {f^a \left (b^2 x^4 \operatorname {ExpIntegralEi}\left (b x^2 \log (f)\right ) \log ^2(f)-f^{b x^2} \left (1+b x^2 \log (f)\right )\right )}{4 x^4} \]

[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)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.90

method	result	size
risch	\(-\frac {f^{a} \left (\ln \left (f \right )^{2} \operatorname {Ei}_{1}\left (-b \,x^{2} \ln \left (f \right )\right ) b^{2} x^{4}+\ln \left (f \right ) f^{b \,x^{2}} b \,x^{2}+f^{b \,x^{2}}\right )}{4 x^{4}}\)	\(52\)
meijerg	\(\frac {f^{a} b^{2} \ln \left (f \right )^{2} \left (-\frac {1}{2 b^{2} x^{4} \ln \left (f \right )^{2}}-\frac {1}{b \,x^{2} \ln \left (f \right )}-\frac {3}{4}+\ln \left (x \right )+\frac {\ln \left (-b \right )}{2}+\frac {\ln \left (\ln \left (f \right )\right )}{2}+\frac {9 b^{2} x^{4} \ln \left (f \right )^{2}+12 b \,x^{2} \ln \left (f \right )+6}{12 b^{2} x^{4} \ln \left (f \right )^{2}}-\frac {\left (3+3 b \,x^{2} \ln \left (f \right )\right ) {\mathrm e}^{b \,x^{2} \ln \left (f \right )}}{6 b^{2} x^{4} \ln \left (f \right )^{2}}-\frac {\ln \left (-b \,x^{2} \ln \left (f \right )\right )}{2}-\frac {\operatorname {Ei}_{1}\left (-b \,x^{2} \ln \left (f \right )\right )}{2}\right )}{2}\)	\(139\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.83 \[ \int \frac {f^{a+b x^2}}{x^5} \, dx=\frac {b^{2} f^{a} x^{4} {\rm Ei}\left (b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{2} - {\left (b x^{2} \log \left (f\right ) + 1\right )} f^{b x^{2} + a}}{4 \, x^{4}} \]

[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

Sympy [F]

\[ \int \frac {f^{a+b x^2}}{x^5} \, dx=\int \frac {f^{a + b x^{2}}}{x^{5}}\, dx \]

[In]

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

[Out]

Integral(f**(a + b*x**2)/x**5, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.38 \[ \int \frac {f^{a+b x^2}}{x^5} \, dx=-\frac {1}{2} \, b^{2} f^{a} \Gamma \left (-2, -b x^{2} \log \left (f\right )\right ) \log \left (f\right )^{2} \]

[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

Giac [F]

\[ \int \frac {f^{a+b x^2}}{x^5} \, dx=\int { \frac {f^{b x^{2} + a}}{x^{5}} \,d x } \]

[In]

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

[Out]

integrate(f^(b*x^2 + a)/x^5, x)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98 \[ \int \frac {f^{a+b x^2}}{x^5} \, dx=-\frac {b^2\,f^a\,{\ln \left (f\right )}^2\,\left (f^{b\,x^2}\,\left (\frac {1}{2\,b\,x^2\,\ln \left (f\right )}+\frac {1}{2\,b^2\,x^4\,{\ln \left (f\right )}^2}\right )+\frac {\mathrm {expint}\left (-b\,x^2\,\ln \left (f\right )\right )}{2}\right )}{2} \]

[In]

int(f^(a + b*x^2)/x^5,x)

[Out]

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