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

Optimal. Leaf size=73 \[ \frac{\left ((d+e x)^2\right )^{-m} F^{c \left (a-\frac{b d}{e}\right )} \left (-\frac{b c \log (F) (d+e x)}{e}\right )^{2 m} \text{Gamma}\left (1-2 m,-\frac{b c \log (F) (d+e x)}{e}\right )}{b c \log (F)} \]

[Out]

(F^(c*(a - (b*d)/e))*Gamma[1 - 2*m, -((b*c*(d + e*x)*Log[F])/e)]*(-((b*c*(d + e*x)*Log[F])/e))^(2*m))/(b*c*((d
+ e*x)^2)^m*Log[F])

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Rubi [A]  time = 0.0566247, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {2188, 2181} \[ \frac{\left ((d+e x)^2\right )^{-m} F^{c \left (a-\frac{b d}{e}\right )} \left (-\frac{b c \log (F) (d+e x)}{e}\right )^{2 m} \text{Gamma}\left (1-2 m,-\frac{b c \log (F) (d+e x)}{e}\right )}{b c \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^(c*(a - (b*d)/e))*Gamma[1 - 2*m, -((b*c*(d + e*x)*Log[F])/e)]*(-((b*c*(d + e*x)*Log[F])/e))^(2*m))/(b*c*((d
+ e*x)^2)^m*Log[F])

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]

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]

Rubi steps

\begin{align*} \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right )^{-m} \, dx &=(d+e x)^{2 m} \left ((d+e x)^2\right )^{-m} \int F^{c (a+b x)} (d+e x)^{-2 m} \, dx\\ &=\frac{F^{c \left (a-\frac{b d}{e}\right )} \left ((d+e x)^2\right )^{-m} \Gamma \left (1-2 m,-\frac{b c (d+e x) \log (F)}{e}\right ) \left (-\frac{b c (d+e x) \log (F)}{e}\right )^{2 m}}{b c \log (F)}\\ \end{align*}

Mathematica [A]  time = 0.0107577, size = 73, normalized size = 1. \[ \frac{\left ((d+e x)^2\right )^{-m} F^{c \left (a-\frac{b d}{e}\right )} \left (-\frac{b c \log (F) (d+e x)}{e}\right )^{2 m} \text{Gamma}\left (1-2 m,-\frac{b c \log (F) (d+e x)}{e}\right )}{b c \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^(c*(a - (b*d)/e))*Gamma[1 - 2*m, -((b*c*(d + e*x)*Log[F])/e)]*(-((b*c*(d + e*x)*Log[F])/e))^(2*m))/(b*c*((d
+ e*x)^2)^m*Log[F])

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Maple [F]  time = 0.099, size = 0, normalized size = 0. \begin{align*} \int{\frac{{F}^{c \left ( bx+a \right ) }}{ \left ({e}^{2}{x}^{2}+2\,dex+{d}^{2} \right ) ^{m}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )}^{m}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{F^{b c x + a c}}{{\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )}^{m}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (b x + a\right )} c}}{{\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )}^{m}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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