3.74 \(\int f^{a+b x^2} x^3 \, dx\)

Optimal. Leaf size=44 \[ \frac{x^2 f^{a+b x^2}}{2 b \log (f)}-\frac{f^{a+b x^2}}{2 b^2 \log ^2(f)} \]

[Out]

-f^(a + b*x^2)/(2*b^2*Log[f]^2) + (f^(a + b*x^2)*x^2)/(2*b*Log[f])

________________________________________________________________________________________

Rubi [A]  time = 0.0380601, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2212, 2209} \[ \frac{x^2 f^{a+b x^2}}{2 b \log (f)}-\frac{f^{a+b x^2}}{2 b^2 \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

Int[f^(a + b*x^2)*x^3,x]

[Out]

-f^(a + b*x^2)/(2*b^2*Log[f]^2) + (f^(a + b*x^2)*x^2)/(2*b*Log[f])

Rule 2212

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^(m
- n + 1)*F^(a + b*(c + d*x)^n))/(b*d*n*Log[F]), x] - Dist[(m - n + 1)/(b*n*Log[F]), Int[(c + d*x)^(m - n)*F^(
a + b*(c + d*x)^n), x], x] /; FreeQ[{F, a, b, c, d}, x] && IntegerQ[(2*(m + 1))/n] && LtQ[0, (m + 1)/n, 5] &&
IntegerQ[n] && (LtQ[0, n, m + 1] || LtQ[m, n, 0])

Rule 2209

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*} \int f^{a+b x^2} x^3 \, dx &=\frac{f^{a+b x^2} x^2}{2 b \log (f)}-\frac{\int f^{a+b x^2} x \, dx}{b \log (f)}\\ &=-\frac{f^{a+b x^2}}{2 b^2 \log ^2(f)}+\frac{f^{a+b x^2} x^2}{2 b \log (f)}\\ \end{align*}

Mathematica [A]  time = 0.0065671, size = 29, normalized size = 0.66 \[ \frac{f^{a+b x^2} \left (b x^2 \log (f)-1\right )}{2 b^2 \log ^2(f)} \]

Antiderivative was successfully verified.

[In]

Integrate[f^(a + b*x^2)*x^3,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.002, size = 28, normalized size = 0.6 \begin{align*}{\frac{ \left ( b{x}^{2}\ln \left ( f \right ) -1 \right ){f}^{b{x}^{2}+a}}{2\, \left ( \ln \left ( f \right ) \right ) ^{2}{b}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(b*x^2+a)*x^3,x)

[Out]

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

________________________________________________________________________________________

Maxima [A]  time = 1.13082, size = 43, normalized size = 0.98 \begin{align*} \frac{{\left (b f^{a} x^{2} \log \left (f\right ) - f^{a}\right )} f^{b x^{2}}}{2 \, b^{2} \log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^3,x, algorithm="maxima")

[Out]

1/2*(b*f^a*x^2*log(f) - f^a)*f^(b*x^2)/(b^2*log(f)^2)

________________________________________________________________________________________

Fricas [A]  time = 1.52543, size = 72, normalized size = 1.64 \begin{align*} \frac{{\left (b x^{2} \log \left (f\right ) - 1\right )} f^{b x^{2} + a}}{2 \, b^{2} \log \left (f\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.114831, size = 41, normalized size = 0.93 \begin{align*} \begin{cases} \frac{f^{a + b x^{2}} \left (b x^{2} \log{\left (f \right )} - 1\right )}{2 b^{2} \log{\left (f \right )}^{2}} & \text{for}\: 2 b^{2} \log{\left (f \right )}^{2} \neq 0 \\\frac{x^{4}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f**(b*x**2+a)*x**3,x)

[Out]

Piecewise((f**(a + b*x**2)*(b*x**2*log(f) - 1)/(2*b**2*log(f)**2), Ne(2*b**2*log(f)**2, 0)), (x**4/4, True))

________________________________________________________________________________________

Giac [B]  time = 1.29085, size = 932, normalized size = 21.18 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(f^(b*x^2+a)*x^3,x, algorithm="giac")

[Out]

1/2*(2*((pi*b*x^2*sgn(f) - pi*b*x^2)*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))/((pi^2*b^2*sgn(f) - pi^2
*b^2 + 2*b^2*log(abs(f))^2)^2 + 4*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2) + (pi^2*b^2*sgn(f) - pi^
2*b^2 + 2*b^2*log(abs(f))^2)*(b*x^2*log(abs(f)) - 1)/((pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)^2 + 4
*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2))*cos(-1/2*pi*b*x^2*sgn(f) + 1/2*pi*b*x^2 - 1/2*pi*a*sgn(f
) + 1/2*pi*a) + ((pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)*(pi*b*x^2*sgn(f) - pi*b*x^2)/((pi^2*b^2*sg
n(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2)^2 + 4*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2) - 4*(pi*b^2*l
og(abs(f))*sgn(f) - pi*b^2*log(abs(f)))*(b*x^2*log(abs(f)) - 1)/((pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f
))^2)^2 + 4*(pi*b^2*log(abs(f))*sgn(f) - pi*b^2*log(abs(f)))^2))*sin(-1/2*pi*b*x^2*sgn(f) + 1/2*pi*b*x^2 - 1/2
*pi*a*sgn(f) + 1/2*pi*a))*e^(b*x^2*log(abs(f)) + a*log(abs(f))) - 1/4*((2*b*i*x^2*log(abs(f)) - pi*b*x^2*sgn(f
) + pi*b*x^2 - 2*i)*e^(1/2*(pi*b*x^2*(sgn(f) - 1) + pi*a*(sgn(f) - 1))*i)/(2*pi*b^2*i*log(abs(f))*sgn(f) - 2*p
i*b^2*i*log(abs(f)) + pi^2*b^2*sgn(f) - pi^2*b^2 + 2*b^2*log(abs(f))^2) + (2*b*i*x^2*log(abs(f)) + pi*b*x^2*sg
n(f) - pi*b*x^2 - 2*i)*e^(-1/2*(pi*b*x^2*(sgn(f) - 1) + pi*a*(sgn(f) - 1))*i)/(2*pi*b^2*i*log(abs(f))*sgn(f) -
2*pi*b^2*i*log(abs(f)) - pi^2*b^2*sgn(f) + pi^2*b^2 - 2*b^2*log(abs(f))^2))*e^(b*x^2*log(abs(f)) + a*log(abs(
f)))/i