3.286 \(\int F^{a+b (c+d x)^3} (c+d x)^2 \, dx\)

Optimal. Leaf size=27 \[ \frac {F^{a+b (c+d x)^3}}{3 b d \log (F)} \]

[Out]

1/3*F^(a+b*(d*x+c)^3)/b/d/ln(F)

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Rubi [A]  time = 0.07, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2209} \[ \frac {F^{a+b (c+d x)^3}}{3 b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^3)*(c + d*x)^2,x]

[Out]

F^(a + b*(c + d*x)^3)/(3*b*d*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]

Rubi steps

\begin {align*} \int F^{a+b (c+d x)^3} (c+d x)^2 \, dx &=\frac {F^{a+b (c+d x)^3}}{3 b d \log (F)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 27, normalized size = 1.00 \[ \frac {F^{a+b (c+d x)^3}}{3 b d \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^3)*(c + d*x)^2,x]

[Out]

F^(a + b*(c + d*x)^3)/(3*b*d*Log[F])

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fricas [A]  time = 0.42, size = 47, normalized size = 1.74 \[ \frac {F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{3 \, b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*F^(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)/(b*d*log(F))

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giac [A]  time = 0.23, size = 47, normalized size = 1.74 \[ \frac {F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{3 \, b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*F^(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)/(b*d*log(F))

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maple [A]  time = 0.01, size = 48, normalized size = 1.78 \[ \frac {F^{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a}}{3 b d \ln \relax (F )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+(d*x+c)^3*b)*(d*x+c)^2,x)

[Out]

1/3*F^(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)/b/d/ln(F)

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maxima [A]  time = 0.88, size = 25, normalized size = 0.93 \[ \frac {F^{{\left (d x + c\right )}^{3} b + a}}{3 \, b d \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*F^((d*x + c)^3*b + a)/(b*d*log(F))

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mupad [B]  time = 3.53, size = 25, normalized size = 0.93 \[ \frac {F^{a+b\,{\left (c+d\,x\right )}^3}}{3\,b\,d\,\ln \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a + b*(c + d*x)^3)*(c + d*x)^2,x)

[Out]

F^(a + b*(c + d*x)^3)/(3*b*d*log(F))

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sympy [A]  time = 0.18, size = 46, normalized size = 1.70 \[ \begin {cases} \frac {F^{a + b \left (c + d x\right )^{3}}}{3 b d \log {\relax (F )}} & \text {for}\: 3 b d \log {\relax (F )} \neq 0 \\c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F**(a+b*(d*x+c)**3)*(d*x+c)**2,x)

[Out]

Piecewise((F**(a + b*(c + d*x)**3)/(3*b*d*log(F)), Ne(3*b*d*log(F), 0)), (c**2*x + c*d*x**2 + d**2*x**3/3, Tru
e))

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