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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0239831, antiderivative size = 46, normalized size of antiderivative = 1., 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*}

Mathematica [A]  time = 0.0121115, size = 46, normalized size = 1. \[ -\frac{1}{2} f^a x^{m+1} \left (-b x^2 \log (f)\right )^{\frac{1}{2} (-m-1)} \text{Gamma}\left (\frac{m+1}{2},-b x^2 \log (f)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[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

________________________________________________________________________________________

Maple [B]  time = 0.035, size = 140, normalized size = 3. \begin{align*}{\frac{{f}^{a}}{2} \left ( -b \right ) ^{-{\frac{m}{2}}-{\frac{1}{2}}} \left ( \ln \left ( f \right ) \right ) ^{-{\frac{m}{2}}-{\frac{1}{2}}} \left ( 2\,{\frac{{x}^{1+m} \left ( -b \right ) ^{m/2+1/2} \left ( \ln \left ( f \right ) \right ) ^{m/2+1/2} \left ( m/2+1/2 \right ) \left ( -b{x}^{2}\ln \left ( f \right ) \right ) ^{-m/2-1/2}\Gamma \left ( m/2+1/2 \right ) }{1+m}}+2\,{\frac{{x}^{1+m} \left ( -b \right ) ^{m/2+1/2} \left ( \ln \left ( f \right ) \right ) ^{m/2+1/2} \left ( -m/2-1/2 \right ) \left ( -b{x}^{2}\ln \left ( f \right ) \right ) ^{-m/2-1/2}\Gamma \left ( m/2+1/2,-b{x}^{2}\ln \left ( f \right ) \right ) }{1+m}} \right ) } \end{align*}

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/(1+m)*x^(1+m)*(-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/(1+m)*x^(1+m)*(-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)))

________________________________________________________________________________________

Maxima [A]  time = 1.22648, size = 51, normalized size = 1.11 \begin{align*} -\frac{1}{2} \, \left (-b x^{2} \log \left (f\right )\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 \left (f\right )\right ) \end{align*}

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))

________________________________________________________________________________________

Fricas [A]  time = 1.56654, size = 126, normalized size = 2.74 \begin{align*} \frac{e^{\left (-\frac{1}{2} \,{\left (m - 1\right )} \log \left (-b \log \left (f\right )\right ) + a \log \left (f\right )\right )} \Gamma \left (\frac{1}{2} \, m + \frac{1}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, b \log \left (f\right )} \end{align*}

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))

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{a + b x^{2}} x^{m}\, dx \end{align*}

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)

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int f^{b x^{2} + a} x^{m}\,{d x} \end{align*}

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)