\(\int f^{a+\frac {b}{x}} \, dx\) [120]

Optimal result

Integrand size = 9, antiderivative size = 28 \[ \int f^{a+\frac {b}{x}} \, dx=f^{a+\frac {b}{x}} x-b f^a \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x}\right ) \log (f) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2237, 2241} \[ \int f^{a+\frac {b}{x}} \, dx=x f^{a+\frac {b}{x}}-b f^a \log (f) \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x}\right ) \]

[In]

Int[f^(a + b/x),x]

[Out]

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

Rule 2237

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_)), x_Symbol] :> Simp[(c + d*x)*(F^(a + b*(c + d*x)^n)/d), x]
- Dist[b*n*Log[F], Int[(c + d*x)^n*F^(a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[2/n]
&& ILtQ[n, 0]

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]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x}} \, dx=f^{a+\frac {b}{x}} x-b f^a \operatorname {ExpIntegralEi}\left (\frac {b \log (f)}{x}\right ) \log (f) \]

[In]

Integrate[f^(a + b/x),x]

[Out]

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

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.11

method	result	size
risch	\(f^{a} \operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x}\right ) b \ln \left (f \right )+f^{a} f^{\frac {b}{x}} x\)	\(31\)
meijerg	\(f^{a} b \ln \left (f \right ) \left (\frac {x}{b \ln \left (f \right )}+1+\ln \left (x \right )-\ln \left (-b \right )-\ln \left (\ln \left (f \right )\right )-\frac {x \left (2+\frac {2 b \ln \left (f \right )}{x}\right )}{2 b \ln \left (f \right )}+\frac {x \,{\mathrm e}^{\frac {b \ln \left (f \right )}{x}}}{b \ln \left (f \right )}+\ln \left (-\frac {b \ln \left (f \right )}{x}\right )+\operatorname {Ei}_{1}\left (-\frac {b \ln \left (f \right )}{x}\right )\right )\)	\(88\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.07 \[ \int f^{a+\frac {b}{x}} \, dx=-b f^{a} {\rm Ei}\left (\frac {b \log \left (f\right )}{x}\right ) \log \left (f\right ) + f^{\frac {a x + b}{x}} x \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b/x), x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.64 \[ \int f^{a+\frac {b}{x}} \, dx=-b f^{a} \Gamma \left (-1, -\frac {b \log \left (f\right )}{x}\right ) \log \left (f\right ) \]

[In]

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

[Out]

-b*f^a*gamma(-1, -b*log(f)/x)*log(f)

Giac [F]

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

[In]

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

[Out]

integrate(f^(a + b/x), x)

Mupad [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.96 \[ \int f^{a+\frac {b}{x}} \, dx=f^a\,\left (f^{b/x}\,x+b\,\ln \left (f\right )\,\mathrm {expint}\left (-\frac {b\,\ln \left (f\right )}{x}\right )\right ) \]

[In]

int(f^(a + b/x),x)

[Out]

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