\(\int f^{a+b x^2} x \, dx\) [75]

Optimal result

Integrand size = 11, antiderivative size = 20 \[ \int f^{a+b x^2} x \, dx=\frac {f^{a+b x^2}}{2 b \log (f)} \]

[Out]

1/2*f^(b*x^2+a)/b/ln(f)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2240} \[ \int f^{a+b x^2} x \, dx=\frac {f^{a+b x^2}}{2 b \log (f)} \]

[In]

Int[f^(a + b*x^2)*x,x]

[Out]

f^(a + b*x^2)/(2*b*Log[f])

Rule 2240

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

Rubi steps \begin{align*} \text {integral}& = \frac {f^{a+b x^2}}{2 b \log (f)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 20, normalized size of antiderivative = 1.00 \[ \int f^{a+b x^2} x \, dx=\frac {f^{a+b x^2}}{2 b \log (f)} \]

[In]

Integrate[f^(a + b*x^2)*x,x]

[Out]

f^(a + b*x^2)/(2*b*Log[f])

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.95

method	result	size
gosper	\(\frac {f^{b \,x^{2}+a}}{2 b \ln \left (f \right )}\)	\(19\)
derivativedivides	\(\frac {f^{b \,x^{2}+a}}{2 b \ln \left (f \right )}\)	\(19\)
default	\(\frac {f^{b \,x^{2}+a}}{2 b \ln \left (f \right )}\)	\(19\)
risch	\(\frac {f^{b \,x^{2}+a}}{2 b \ln \left (f \right )}\)	\(19\)
parallelrisch	\(\frac {f^{b \,x^{2}+a}}{2 b \ln \left (f \right )}\)	\(19\)
norman	\(\frac {{\mathrm e}^{\left (b \,x^{2}+a \right ) \ln \left (f \right )}}{2 b \ln \left (f \right )}\)	\(21\)
meijerg	\(-\frac {f^{a} \left (1-{\mathrm e}^{b \,x^{2} \ln \left (f \right )}\right )}{2 b \ln \left (f \right )}\)	\(25\)

[In]

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

[Out]

1/2*f^(b*x^2+a)/b/ln(f)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int f^{a+b x^2} x \, dx=\frac {f^{b x^{2} + a}}{2 \, b \log \left (f\right )} \]

[In]

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

[Out]

1/2*f^(b*x^2 + a)/(b*log(f))

Sympy [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 22, normalized size of antiderivative = 1.10 \[ \int f^{a+b x^2} x \, dx=\begin {cases} \frac {f^{a + b x^{2}}}{2 b \log {\left (f \right )}} & \text {for}\: b \log {\left (f \right )} \neq 0 \\\frac {x^{2}}{2} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((f**(a + b*x**2)/(2*b*log(f)), Ne(b*log(f), 0)), (x**2/2, True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int f^{a+b x^2} x \, dx=\frac {f^{b x^{2} + a}}{2 \, b \log \left (f\right )} \]

[In]

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

[Out]

1/2*f^(b*x^2 + a)/(b*log(f))

Giac [A] (verification not implemented)

none

Time = 0.50 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int f^{a+b x^2} x \, dx=\frac {f^{b x^{2} + a}}{2 \, b \log \left (f\right )} \]

[In]

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

[Out]

1/2*f^(b*x^2 + a)/(b*log(f))

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.90 \[ \int f^{a+b x^2} x \, dx=\frac {f^{b\,x^2+a}}{2\,b\,\ln \left (f\right )} \]

[In]

int(f^(a + b*x^2)*x,x)

[Out]

f^(a + b*x^2)/(2*b*log(f))