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

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

[Out]

-1/2*f^a*x^(1+m)*GAMMA(1/2+1/2*m,-b*x^2*ln(f))*(-b*x^2*ln(f))^(-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} \[ -\frac {1}{2} f^a x^{m+1} \left (-b x^2 \log (f)\right )^{\frac {1}{2} (-m-1)} \text {Gamma}\left (\frac {m+1}{2},-b x^2 \log (f)\right ) \]

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

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

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)*x^m,x]

[Out]

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

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fricas [A]  time = 0.42, size = 40, normalized size = 0.87 \[ \frac {e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-b \log \relax (f)\right ) + a \log \relax (f)\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -b x^{2} \log \relax (f)\right )}{2 \, b \log \relax (f)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(f^(b*x^2 + a)*x^m, x)

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maple [B]  time = 0.06, size = 140, normalized size = 3.04 \[ \frac {\left (\frac {2 \left (\frac {m}{2}+\frac {1}{2}\right ) x^{m +1} \left (-b \right )^{\frac {m}{2}+\frac {1}{2}} \left (-b \,x^{2} \ln \relax (f )\right )^{-\frac {m}{2}-\frac {1}{2}} \ln \relax (f )^{\frac {m}{2}+\frac {1}{2}} \Gamma \left (\frac {m}{2}+\frac {1}{2}\right )}{m +1}+\frac {2 \left (-\frac {m}{2}-\frac {1}{2}\right ) x^{m +1} \left (-b \right )^{\frac {m}{2}+\frac {1}{2}} \left (-b \,x^{2} \ln \relax (f )\right )^{-\frac {m}{2}-\frac {1}{2}} \ln \relax (f )^{\frac {m}{2}+\frac {1}{2}} \Gamma \left (\frac {m}{2}+\frac {1}{2}, -b \,x^{2} \ln \relax (f )\right )}{m +1}\right ) f^{a} \left (-b \right )^{-\frac {m}{2}-\frac {1}{2}} \ln \relax (f )^{-\frac {m}{2}-\frac {1}{2}}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maxima [A]  time = 0.65, size = 38, normalized size = 0.83 \[ -\frac {1}{2} \, \left (-b x^{2} \log \relax (f)\right )^{-\frac {1}{2} \, m - \frac {1}{2}} f^{a} x^{m + 1} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -b x^{2} \log \relax (f)\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.63, size = 49, normalized size = 1.07 \[ \frac {f^a\,x^{m+1}\,\left (\Gamma \left (\frac {m}{2}+\frac {1}{2}\right )-\Gamma \left (\frac {m}{2}+\frac {1}{2},-b\,x^2\,\ln \relax (f)\right )\right )}{2\,{\left (-b\,x^2\,\ln \relax (f)\right )}^{\frac {m}{2}+\frac {1}{2}}} \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(f**(a + b*x**2)*x**m, x)

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