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

Optimal. Leaf size=35 \[ f^a x^{m+1} \left (-\frac {b \log (f)}{x}\right )^{m+1} \Gamma \left (-m-1,-\frac {b \log (f)}{x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2218} \[ f^a x^{m+1} \left (-\frac {b \log (f)}{x}\right )^{m+1} \text {Gamma}\left (-m-1,-\frac {b \log (f)}{x}\right ) \]

Antiderivative was successfully verified.

[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 2218

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

Rubi steps

\begin {align*} \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}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 35, normalized size = 1.00 \[ f^a x^{m+1} \left (-\frac {b \log (f)}{x}\right )^{m+1} \Gamma \left (-m-1,-\frac {b \log (f)}{x}\right ) \]

Antiderivative was successfully verified.

[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)

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fricas [F]  time = 0.40, size = 0, normalized size = 0.00 \[ {\rm integral}\left (f^{\frac {a x + b}{x}} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(f^((a*x + b)/x)*x^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x}} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(a + b/x)*x^m, x)

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maple [B]  time = 0.06, size = 136, normalized size = 3.89 \[ \left (-\frac {x^{m} \left (-b \right )^{-m} \left (-\frac {b \ln \relax (f )}{x}\right )^{m} \ln \relax (f )^{-m} \Gamma \left (-m \right )}{m +1}+\frac {x^{m} \left (-b \right )^{-m} \left (-\frac {b \ln \relax (f )}{x}\right )^{m} \ln \relax (f )^{-m} \Gamma \left (-m , -\frac {b \ln \relax (f )}{x}\right )}{m +1}+\frac {x^{m +1} \left (-b \right )^{-m} \ln \relax (f )^{-m -1} {\mathrm e}^{\frac {b \ln \relax (f )}{x}}}{\left (m +1\right ) b}\right ) b \,f^{a} \left (-b \right )^{m} \ln \relax (f )^{m +1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.61, size = 35, normalized size = 1.00 \[ f^{a} x^{m + 1} \left (-\frac {b \log \relax (f)}{x}\right )^{m + 1} \Gamma \left (-m - 1, -\frac {b \log \relax (f)}{x}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 3.50, size = 52, normalized size = 1.49 \[ \frac {f^a\,x^{m+1}\,{\mathrm {e}}^{\frac {b\,\ln \relax (f)}{2\,x}}\,{\mathrm {M}}_{\frac {m}{2}+1,-\frac {m}{2}-\frac {1}{2}}\left (\frac {b\,\ln \relax (f)}{x}\right )\,{\left (\frac {b\,\ln \relax (f)}{x}\right )}^{m/2}}{m+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x}} x^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b/x)*x**m, x)

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