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

Optimal. Leaf size=71 \[ \frac {\left ((d+e x)^3\right )^m F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{-3 m} \Gamma \left (3 m+1,-\frac {b c (d+e x) \log (F)}{e}\right )}{b c \log (F)} \]

[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))

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Rubi [A]  time = 0.06, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {2188, 2181} \[ \frac {\left ((d+e x)^3\right )^m F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{-3 m} \text {Gamma}\left (3 m+1,-\frac {b c \log (F) (d+e x)}{e}\right )}{b c \log (F)} \]

Antiderivative was successfully verified.

[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 2181

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

Rule 2188

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*Int[z*(a + b*(F^(
g*ExpandToSum[v, x]))^n)^p, x])/z, 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*} \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 &=(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*}

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Mathematica [A]  time = 0.01, size = 71, normalized size = 1.00 \[ \frac {\left ((d+e x)^3\right )^m F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{-3 m} \Gamma \left (3 m+1,-\frac {b c (d+e x) \log (F)}{e}\right )}{b c \log (F)} \]

Antiderivative was successfully verified.

[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))

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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )}^{m} F^{b c x + a c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )}^{m} F^{{\left (b x + a\right )} c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 0.12, size = 0, normalized size = 0.00 \[ \int F^{\left (b x +a \right ) c} \left (e^{3} x^{3}+3 d \,e^{2} x^{2}+3 d^{2} e x +d^{3}\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )}^{m} F^{{\left (b x + a\right )} c}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int F^{c\,\left (a+b\,x\right )}\,{\left (d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3\right )}^m \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int F^{c \left (a + b x\right )} \left (\left (d + e x\right )^{3}\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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