3.128 \(\int f^{a+\frac {b}{x^2}} x^m \, dx\)

Optimal. Leaf size=46 \[ \frac {1}{2} f^a x^{m+1} \left (-\frac {b \log (f)}{x^2}\right )^{\frac {m+1}{2}} \Gamma \left (\frac {1}{2} (-m-1),-\frac {b \log (f)}{x^2}\right ) \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 46, 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} \[ \frac {1}{2} f^a x^{m+1} \left (-\frac {b \log (f)}{x^2}\right )^{\frac {m+1}{2}} \text {Gamma}\left (\frac {1}{2} (-m-1),-\frac {b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b/x^2)*x^m,x]

[Out]

(f^a*x^(1 + m)*Gamma[(-1 - m)/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^((1 + m)/2))/2

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

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Mathematica [A]  time = 0.01, size = 46, normalized size = 1.00 \[ \frac {1}{2} f^a x^{m+1} \left (-\frac {b \log (f)}{x^2}\right )^{\frac {m+1}{2}} \Gamma \left (\frac {1}{2} (-m-1),-\frac {b \log (f)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b/x^2)*x^m,x]

[Out]

(f^a*x^(1 + m)*Gamma[(-1 - m)/2, -((b*Log[f])/x^2)]*(-((b*Log[f])/x^2))^((1 + m)/2))/2

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (f^{\frac {a x^{2} + b}{x^{2}}} x^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(f^((a*x^2 + b)/x^2)*x^m, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(a + b/x^2)*x^m, x)

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maple [B]  time = 0.06, size = 169, normalized size = 3.67 \[ -\frac {\left (\frac {2 b \,x^{m -1} \left (-b \right )^{-\frac {m}{2}-\frac {1}{2}} \left (-\frac {b \ln \relax (f )}{x^{2}}\right )^{\frac {m}{2}-\frac {1}{2}} \ln \relax (f )^{-\frac {m}{2}+\frac {1}{2}} \Gamma \left (-\frac {m}{2}+\frac {1}{2}\right )}{m +1}-\frac {2 b \,x^{m -1} \left (-b \right )^{-\frac {m}{2}-\frac {1}{2}} \left (-\frac {b \ln \relax (f )}{x^{2}}\right )^{\frac {m}{2}-\frac {1}{2}} \ln \relax (f )^{-\frac {m}{2}+\frac {1}{2}} \Gamma \left (-\frac {m}{2}+\frac {1}{2}, -\frac {b \ln \relax (f )}{x^{2}}\right )}{m +1}-\frac {2 x^{m +1} \left (-b \right )^{-\frac {m}{2}-\frac {1}{2}} \ln \relax (f )^{-\frac {m}{2}-\frac {1}{2}} {\mathrm e}^{\frac {b \ln \relax (f )}{x^{2}}}}{m +1}\right ) f^{a} \left (-b \right )^{\frac {m}{2}+\frac {1}{2}} \ln \relax (f )^{\frac {m}{2}+\frac {1}{2}}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a+b/x^2)*x^m,x)

[Out]

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

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maxima [A]  time = 1.37, size = 38, normalized size = 0.83 \[ \frac {1}{2} \, f^{a} x^{m + 1} \left (-\frac {b \log \relax (f)}{x^{2}}\right )^{\frac {1}{2} \, m + \frac {1}{2}} \Gamma \left (-\frac {1}{2} \, m - \frac {1}{2}, -\frac {b \log \relax (f)}{x^{2}}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.51, size = 54, normalized size = 1.17 \[ \frac {f^a\,x^{m+1}\,{\mathrm {e}}^{\frac {b\,\ln \relax (f)}{2\,x^2}}\,{\mathrm {M}}_{\frac {m}{4}+\frac {3}{4},-\frac {m}{4}-\frac {1}{4}}\left (\frac {b\,\ln \relax (f)}{x^2}\right )\,{\left (\frac {b\,\ln \relax (f)}{x^2}\right )}^{\frac {m}{4}-\frac {1}{4}}}{m+1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b/x^2)*x^m,x)

[Out]

(f^a*x^(m + 1)*exp((b*log(f))/(2*x^2))*whittakerM(m/4 + 3/4, - m/4 - 1/4, (b*log(f))/x^2)*((b*log(f))/x^2)^(m/
4 - 1/4))/(m + 1)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int f^{a + \frac {b}{x^{2}}} x^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b/x**2)*x**m, x)

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