\(\int F^{c (a+b x)} (d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3)^m \, dx\) [21]

Optimal result

Integrand size = 39, antiderivative size = 71 \[ \int F^{c (a+b x)} \left (d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3\right )^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^3\right )^m \Gamma \left (1+3 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-3 m}}{b c \log (F)} \]

[Out]

F^(c*(a-b*d/e))*((e*x+d)^3)^m*GAMMA(1+3*m,-b*c*(e*x+d)*ln(F)/e)/b/c/ln(F)/((-b*c*(e*x+d)*ln(F)/e)^(3*m))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2219, 2212} \[ \int F^{c (a+b x)} \left (d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3\right )^m \, dx=\frac {\left ((d+e x)^3\right )^m F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{-3 m} \Gamma \left (3 m+1,-\frac {b c (d+e x) \log (F)}{e}\right )}{b c \log (F)} \]

[In]

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

[Out]

(F^(c*(a - (b*d)/e))*((d + e*x)^3)^m*Gamma[1 + 3*m, -((b*c*(d + e*x)*Log[F])/e)])/(b*c*Log[F]*(-((b*c*(d + e*x
)*Log[F])/e))^(3*m))

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 2219

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Module[{uu = NormalizePowerOfLine
ar[u, x], z}, Simp[z = If[PowerQ[uu] && FreeQ[uu[[2]], x], uu[[1]]^(m*uu[[2]]), uu^m]; (uu^m/z)*Int[z*(a + b*(
F^(g*ExpandToSum[v, x]))^n)^p, x], x]] /; FreeQ[{F, a, b, g, m, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ[u
, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) &&  !IntegerQ[m]

Rubi steps \begin{align*} \text {integral}& = (d+e x)^{-3 m} \left ((d+e x)^3\right )^m \int F^{c (a+b x)} (d+e x)^{3 m} \, dx \\ & = \frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^3\right )^m \Gamma \left (1+3 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-3 m}}{b c \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \left (d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3\right )^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^3\right )^m \Gamma \left (1+3 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-3 m}}{b c \log (F)} \]

[In]

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

[Out]

(F^(c*(a - (b*d)/e))*((d + e*x)^3)^m*Gamma[1 + 3*m, -((b*c*(d + e*x)*Log[F])/e)])/(b*c*Log[F]*(-((b*c*(d + e*x
)*Log[F])/e))^(3*m))

Maple [F]

\[\int F^{c \left (b x +a \right )} \left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}\right )^{m}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int F^{c (a+b x)} \left (d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3\right )^m \, dx=\int { {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \]

[In]

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

[Out]

integral((e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x + d^3)^m*F^(b*c*x + a*c), x)

Sympy [F]

\[ \int F^{c (a+b x)} \left (d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3\right )^m \, dx=\int F^{c \left (a + b x\right )} \left (\left (d + e x\right )^{3}\right )^{m}\, dx \]

[In]

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

[Out]

Integral(F**(c*(a + b*x))*((d + e*x)**3)**m, x)

Maxima [F]

\[ \int F^{c (a+b x)} \left (d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3\right )^m \, dx=\int { {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int F^{c (a+b x)} \left (d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3\right )^m \, dx=\int { {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )}^{m} F^{{\left (b x + a\right )} c} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int F^{c (a+b x)} \left (d^3+3 d^2 e x+3 d e^2 x^2+e^3 x^3\right )^m \, dx=\int F^{c\,\left (a+b\,x\right )}\,{\left (d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3\right )}^m \,d x \]

[In]

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

[Out]

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