\(\int \frac {f^{a+\frac {b}{x}}}{x^2} \, dx\) [122]

Optimal result

Integrand size = 13, antiderivative size = 18 \[ \int \frac {f^{a+\frac {b}{x}}}{x^2} \, dx=-\frac {f^{a+\frac {b}{x}}}{b \log (f)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2240} \[ \int \frac {f^{a+\frac {b}{x}}}{x^2} \, dx=-\frac {f^{a+\frac {b}{x}}}{b \log (f)} \]

[In]

Int[f^(a + b/x)/x^2,x]

[Out]

-(f^(a + b/x)/(b*Log[f]))

Rule 2240

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

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+\frac {b}{x}}}{x^2} \, dx=-\frac {f^{a+\frac {b}{x}}}{b \log (f)} \]

[In]

Integrate[f^(a + b/x)/x^2,x]

[Out]

-(f^(a + b/x)/(b*Log[f]))

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.06

method	result	size
derivativedivides	\(-\frac {f^{a +\frac {b}{x}}}{b \ln \left (f \right )}\)	\(19\)
default	\(-\frac {f^{a +\frac {b}{x}}}{b \ln \left (f \right )}\)	\(19\)
parallelrisch	\(-\frac {f^{a +\frac {b}{x}}}{b \ln \left (f \right )}\)	\(19\)
norman	\(-\frac {{\mathrm e}^{\left (a +\frac {b}{x}\right ) \ln \left (f \right )}}{b \ln \left (f \right )}\)	\(21\)
risch	\(-\frac {f^{\frac {a x +b}{x}}}{b \ln \left (f \right )}\)	\(21\)
meijerg	\(\frac {f^{a} \left (1-{\mathrm e}^{\frac {b \ln \left (f \right )}{x}}\right )}{\ln \left (f \right ) b}\)	\(24\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {f^{a+\frac {b}{x}}}{x^2} \, dx=-\frac {f^{\frac {a x + b}{x}}}{b \log \left (f\right )} \]

[In]

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

[Out]

-f^((a*x + b)/x)/(b*log(f))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {f^{a+\frac {b}{x}}}{x^2} \, dx=\begin {cases} - \frac {f^{a + \frac {b}{x}}}{b \log {\left (f \right )}} & \text {for}\: b \log {\left (f \right )} \neq 0 \\- \frac {1}{x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-f**(a + b/x)/(b*log(f)), Ne(b*log(f), 0)), (-1/x, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+\frac {b}{x}}}{x^2} \, dx=-\frac {f^{a + \frac {b}{x}}}{b \log \left (f\right )} \]

[In]

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

[Out]

-f^(a + b/x)/(b*log(f))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.11 \[ \int \frac {f^{a+\frac {b}{x}}}{x^2} \, dx=-\frac {f^{\frac {a x + b}{x}}}{b \log \left (f\right )} \]

[In]

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

[Out]

-f^((a*x + b)/x)/(b*log(f))

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int \frac {f^{a+\frac {b}{x}}}{x^2} \, dx=-\frac {f^{a+\frac {b}{x}}}{b\,\ln \left (f\right )} \]

[In]

int(f^(a + b/x)/x^2,x)

[Out]

-f^(a + b/x)/(b*log(f))