\(\int f^{a+b x^2} x^m \, dx\) [69]

Optimal result

Integrand size = 13, antiderivative size = 46 \[ \int f^{a+b x^2} x^m \, dx=-\frac {1}{2} f^a x^{1+m} \Gamma \left (\frac {1+m}{2},-b x^2 \log (f)\right ) \left (-b x^2 \log (f)\right )^{\frac {1}{2} (-1-m)} \]

[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)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, 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} \[ \int f^{a+b x^2} x^m \, dx=-\frac {1}{2} f^a x^{m+1} \left (-b x^2 \log (f)\right )^{\frac {1}{2} (-m-1)} \Gamma \left (\frac {m+1}{2},-b x^2 \log (f)\right ) \]

[In]

Int[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))

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}& = -\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] (verified)

Time = 0.03 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^2} x^m \, dx=-\frac {1}{2} f^a x^{1+m} \Gamma \left (\frac {1+m}{2},-b x^2 \log (f)\right ) \left (-b x^2 \log (f)\right )^{\frac {1}{2} (-1-m)} \]

[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))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(139\) vs. \(2(38)=76\).

Time = 0.06 (sec) , antiderivative size = 140, normalized size of antiderivative = 3.04

method	result	size
meijerg	\(\frac {f^{a} \left (-b \right )^{-\frac {1}{2}-\frac {m}{2}} \ln \left (f \right )^{-\frac {1}{2}-\frac {m}{2}} \left (\frac {2 x^{1+m} \left (-b \right )^{\frac {m}{2}+\frac {1}{2}} \ln \left (f \right )^{\frac {m}{2}+\frac {1}{2}} \left (\frac {m}{2}+\frac {1}{2}\right ) \left (-b \,x^{2} \ln \left (f \right )\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {m}{2}+\frac {1}{2}\right )}{1+m}+\frac {2 x^{1+m} \left (-b \right )^{\frac {m}{2}+\frac {1}{2}} \ln \left (f \right )^{\frac {m}{2}+\frac {1}{2}} \left (-\frac {1}{2}-\frac {m}{2}\right ) \left (-b \,x^{2} \ln \left (f \right )\right )^{-\frac {1}{2}-\frac {m}{2}} \Gamma \left (\frac {m}{2}+\frac {1}{2}, -b \,x^{2} \ln \left (f \right )\right )}{1+m}\right )}{2}\)	\(140\)

[In]

int(f^(b*x^2+a)*x^m,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int f^{a+b x^2} x^m \, dx=\frac {e^{\left (-\frac {1}{2} \, {\left (m - 1\right )} \log \left (-b \log \left (f\right )\right ) + a \log \left (f\right )\right )} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -b x^{2} \log \left (f\right )\right )}{2 \, b \log \left (f\right )} \]

[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]

\[ \int f^{a+b x^2} x^m \, dx=\int f^{a + b x^{2}} x^{m}\, dx \]

[In]

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

[Out]

Integral(f**(a + b*x**2)*x**m, x)

Maxima [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int f^{a+b x^2} x^m \, dx=-\frac {1}{2} \, \left (-b x^{2} \log \left (f\right )\right )^{-\frac {1}{2} \, m - \frac {1}{2}} f^{a} x^{m + 1} \Gamma \left (\frac {1}{2} \, m + \frac {1}{2}, -b x^{2} \log \left (f\right )\right ) \]

[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))

Giac [F]

\[ \int f^{a+b x^2} x^m \, dx=\int { f^{b x^{2} + a} x^{m} \,d x } \]

[In]

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

[Out]

integrate(f^(b*x^2 + a)*x^m, x)

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.07 \[ \int f^{a+b x^2} x^m \, dx=\frac {f^a\,x^{m+1}\,\left (\Gamma \left (\frac {m}{2}+\frac {1}{2}\right )-\Gamma \left (\frac {m}{2}+\frac {1}{2},-b\,x^2\,\ln \left (f\right )\right )\right )}{2\,{\left (-b\,x^2\,\ln \left (f\right )\right )}^{\frac {m}{2}+\frac {1}{2}}} \]

[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))