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]

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

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Rubi [A]  time = 0.04, antiderivative size = 44, normalized size of antiderivative = 1.00, 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 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]

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])

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*}

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Mathematica [A]  time = 0.01, 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)

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fricas [A]  time = 0.43, size = 27, normalized size = 0.61 \[ \frac {{\left (b x^{2} \log \relax (f) - 1\right )} f^{b x^{2} + a}}{2 \, b^{2} \log \relax (f)^{2}} \]

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)

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giac [B]  time = 0.52, size = 690, normalized size = 15.68 \[ \text {result too large to display} \]

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

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maple [A]  time = 0.01, size = 28, normalized size = 0.64 \[ \frac {\left (b \,x^{2} \ln \relax (f )-1\right ) f^{b \,x^{2}+a}}{2 b^{2} \ln \relax (f )^{2}} \]

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

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maxima [A]  time = 0.45, size = 32, normalized size = 0.73 \[ \frac {{\left (b f^{a} x^{2} \log \relax (f) - f^{a}\right )} f^{b x^{2}}}{2 \, b^{2} \log \relax (f)^{2}} \]

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)

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mupad [B]  time = 3.46, size = 27, normalized size = 0.61 \[ \frac {f^{b\,x^2+a}\,\left (\frac {b\,x^2\,\ln \relax (f)}{2}-\frac {1}{2}\right )}{b^2\,{\ln \relax (f)}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(f^(a + b*x^2)*x^3,x)

[Out]

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

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sympy [A]  time = 0.12, size = 41, normalized size = 0.93 \[ \begin {cases} \frac {f^{a + b x^{2}} \left (b x^{2} \log {\relax (f )} - 1\right )}{2 b^{2} \log {\relax (f )}^{2}} & \text {for}\: 2 b^{2} \log {\relax (f )}^{2} \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases} \]

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))

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