\(\int F^{a+b (c+d x)^3} (c+d x)^2 \, dx\) [286]

Optimal result

Integrand size = 21, antiderivative size = 27 \[ \int F^{a+b (c+d x)^3} (c+d x)^2 \, dx=\frac {F^{a+b (c+d x)^3}}{3 b d \log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.048, Rules used = {2240} \[ \int F^{a+b (c+d x)^3} (c+d x)^2 \, dx=\frac {F^{a+b (c+d x)^3}}{3 b d \log (F)} \]

[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 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 (c+d x)^3}}{3 b d \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00 \[ \int F^{a+b (c+d x)^3} (c+d x)^2 \, dx=\frac {F^{a+b (c+d x)^3}}{3 b d \log (F)} \]

[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])

Maple [A] (verified)

Time = 0.26 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {F^{a +b \left (d x +c \right )^{3}}}{3 b d \ln \left (F \right )}\)	\(26\)
default	\(\frac {F^{a +b \left (d x +c \right )^{3}}}{3 b d \ln \left (F \right )}\)	\(26\)
parallelrisch	\(\frac {F^{a +b \left (d x +c \right )^{3}}}{3 b d \ln \left (F \right )}\)	\(26\)
norman	\(\frac {{\mathrm e}^{\left (a +b \left (d x +c \right )^{3}\right ) \ln \left (F \right )}}{3 b d \ln \left (F \right )}\)	\(28\)
gosper	\(\frac {F^{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a}}{3 b d \ln \left (F \right )}\)	\(48\)
risch	\(\frac {F^{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a}}{3 b d \ln \left (F \right )}\)	\(48\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int F^{a+b (c+d x)^3} (c+d x)^2 \, dx=\frac {F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{3 \, b d \log \left (F\right )} \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 44 vs. \(2 (19) = 38\).

Time = 0.08 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.63 \[ \int F^{a+b (c+d x)^3} (c+d x)^2 \, dx=\begin {cases} \frac {F^{a + b \left (c + d x\right )^{3}}}{3 b d \log {\left (F \right )}} & \text {for}\: b d \log {\left (F \right )} \neq 0 \\c^{2} x + c d x^{2} + \frac {d^{2} x^{3}}{3} & \text {otherwise} \end {cases} \]

[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(b*d*log(F), 0)), (c**2*x + c*d*x**2 + d**2*x**3/3, True)
)

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int F^{a+b (c+d x)^3} (c+d x)^2 \, dx=\frac {F^{{\left (d x + c\right )}^{3} b + a}}{3 \, b d \log \left (F\right )} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.74 \[ \int F^{a+b (c+d x)^3} (c+d x)^2 \, dx=\frac {F^{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}}{3 \, b d \log \left (F\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93 \[ \int F^{a+b (c+d x)^3} (c+d x)^2 \, dx=\frac {F^{a+b\,{\left (c+d\,x\right )}^3}}{3\,b\,d\,\ln \left (F\right )} \]

[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))