\(\int f^{a+\frac {b}{x}} x^m \, dx\) [115]

Optimal result

Integrand size = 13, antiderivative size = 35 \[ \int f^{a+\frac {b}{x}} x^m \, dx=f^a x^{1+m} \Gamma \left (-1-m,-\frac {b \log (f)}{x}\right ) \left (-\frac {b \log (f)}{x}\right )^{1+m} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, 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 = {2250} \[ \int f^{a+\frac {b}{x}} x^m \, dx=f^a x^{m+1} \left (-\frac {b \log (f)}{x}\right )^{m+1} \Gamma \left (-m-1,-\frac {b \log (f)}{x}\right ) \]

[In]

Int[f^(a + b/x)*x^m,x]

[Out]

f^a*x^(1 + m)*Gamma[-1 - m, -((b*Log[f])/x)]*(-((b*Log[f])/x))^(1 + m)

Rule 2250

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

Rubi steps \begin{align*} \text {integral}& = f^a x^{1+m} \Gamma \left (-1-m,-\frac {b \log (f)}{x}\right ) \left (-\frac {b \log (f)}{x}\right )^{1+m} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x}} x^m \, dx=f^a x^{1+m} \Gamma \left (-1-m,-\frac {b \log (f)}{x}\right ) \left (-\frac {b \log (f)}{x}\right )^{1+m} \]

[In]

Integrate[f^(a + b/x)*x^m,x]

[Out]

f^a*x^(1 + m)*Gamma[-1 - m, -((b*Log[f])/x)]*(-((b*Log[f])/x))^(1 + m)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(135\) vs. \(2(35)=70\).

Time = 0.09 (sec) , antiderivative size = 136, normalized size of antiderivative = 3.89

method	result	size
meijerg	\(f^{a} \left (-b \right )^{m} \ln \left (f \right )^{1+m} b \left (-\frac {x^{m} \left (-b \right )^{-m} \ln \left (f \right )^{-m} \Gamma \left (-m \right ) \left (-\frac {b \ln \left (f \right )}{x}\right )^{m}}{1+m}+\frac {x^{1+m} \left (-b \right )^{-m} \ln \left (f \right )^{-m -1} {\mathrm e}^{\frac {b \ln \left (f \right )}{x}}}{\left (1+m \right ) b}+\frac {x^{m} \left (-b \right )^{-m} \ln \left (f \right )^{-m} \left (-\frac {b \ln \left (f \right )}{x}\right )^{m} \Gamma \left (-m , -\frac {b \ln \left (f \right )}{x}\right )}{1+m}\right )\)	\(136\)

[In]

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

[Out]

f^a*(-b)^m*ln(f)^(1+m)*b*(-1/(1+m)*x^m*(-b)^(-m)*ln(f)^(-m)*GAMMA(-m)*(-b*ln(f)/x)^m+1/(1+m)*x^(1+m)*(-b)^(-m)
*ln(f)^(-m-1)/b*exp(b*ln(f)/x)+1/(1+m)*x^m*(-b)^(-m)*ln(f)^(-m)*(-b*ln(f)/x)^m*GAMMA(-m,-b*ln(f)/x))

Fricas [F]

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

[In]

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

[Out]

integral(f^((a*x + b)/x)*x^m, x)

Sympy [F]

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

[In]

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

[Out]

Integral(f**(a + b/x)*x**m, x)

Maxima [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 35, normalized size of antiderivative = 1.00 \[ \int f^{a+\frac {b}{x}} x^m \, dx=f^{a} x^{m + 1} \left (-\frac {b \log \left (f\right )}{x}\right )^{m + 1} \Gamma \left (-m - 1, -\frac {b \log \left (f\right )}{x}\right ) \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(f^(a + b/x)*x^m, x)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.49 \[ \int f^{a+\frac {b}{x}} x^m \, dx=\frac {f^a\,x^{m+1}\,{\mathrm {e}}^{\frac {b\,\ln \left (f\right )}{2\,x}}\,{\mathrm {M}}_{\frac {m}{2}+1,-\frac {m}{2}-\frac {1}{2}}\left (\frac {b\,\ln \left (f\right )}{x}\right )\,{\left (\frac {b\,\ln \left (f\right )}{x}\right )}^{m/2}}{m+1} \]

[In]

int(f^(a + b/x)*x^m,x)

[Out]

(f^a*x^(m + 1)*exp((b*log(f))/(2*x))*whittakerM(m/2 + 1, - m/2 - 1/2, (b*log(f))/x)*((b*log(f))/x)^(m/2))/(m +
1)