\(\int f^{a+\frac {b}{x^2}} x \, dx\) [133]

Optimal result

Integrand size = 11, antiderivative size = 35 \[ \int f^{a+\frac {b}{x^2}} x \, dx=\frac {1}{2} f^{a+\frac {b}{x^2}} x^2-\frac {1}{2} b f^a \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^2}\right ) \log (f) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2245, 2241} \[ \int f^{a+\frac {b}{x^2}} x \, dx=\frac {1}{2} x^2 f^{a+\frac {b}{x^2}}-\frac {1}{2} b f^a \log (f) \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^2}\right ) \]

[In]

Int[f^(a + b/x^2)*x,x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.91 \[ \int f^{a+\frac {b}{x^2}} x \, dx=\frac {1}{2} f^a \left (f^{\frac {b}{x^2}} x^2-b \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x^2}\right ) \log (f)\right ) \]

[In]

Integrate[f^(a + b/x^2)*x,x]

[Out]

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

Maple [A] (verified)

Time = 0.04 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00

method	result	size
risch	\(\frac {f^{a} x^{2} f^{\frac {b}{x^{2}}}}{2}+\frac {f^{a} \ln \left (f \right ) b \,\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{2}}\right )}{2}\)	\(35\)
meijerg	\(\frac {f^{a} b \ln \left (f \right ) \left (\frac {x^{2}}{b \ln \left (f \right )}+1+2 \ln \left (x \right )-\ln \left (-b \right )-\ln \left (\ln \left (f \right )\right )-\frac {x^{2} \left (2+\frac {2 b \ln \left (f \right )}{x^{2}}\right )}{2 b \ln \left (f \right )}+\frac {x^{2} {\mathrm e}^{\frac {b \ln \left (f \right )}{x^{2}}}}{b \ln \left (f \right )}+\ln \left (-\frac {b \ln \left (f \right )}{x^{2}}\right )+\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x^{2}}\right )\right )}{2}\)	\(97\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x^2}} x \, dx=-\frac {1}{2} \, b f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x^{2}}\right ) \log \left (f\right ) + \frac {1}{2} \, f^{\frac {a x^{2} + b}{x^{2}}} x^{2} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b/x**2)*x, x)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.51 \[ \int f^{a+\frac {b}{x^2}} x \, dx=-\frac {1}{2} \, b f^{a} \Gamma \left (-1, -\frac {b \log \left (f\right )}{x^{2}}\right ) \log \left (f\right ) \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(a + b/x^2)*x, x)

Mupad [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.94 \[ \int f^{a+\frac {b}{x^2}} x \, dx=\frac {f^a\,f^{\frac {b}{x^2}}\,x^2}{2}+\frac {b\,f^a\,\ln \left (f\right )\,\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x^2}\right )}{2} \]

[In]

int(f^(a + b/x^2)*x,x)

[Out]

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