\(\int F^{c (a+b x)} (d^2+2 d e x+e^2 x^2)^{-m} \, dx\) [25]

Optimal result

Integrand size = 30, antiderivative size = 73 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right )^{-m} \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^2\right )^{-m} \Gamma \left (1-2 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{2 m}}{b c \log (F)} \]

[Out]

F^(c*(a-b*d/e))*GAMMA(1-2*m,-b*c*(e*x+d)*ln(F)/e)*(-b*c*(e*x+d)*ln(F)/e)^(2*m)/b/c/(((e*x+d)^2)^m)/ln(F)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2219, 2212} \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right )^{-m} \, dx=\frac {\left ((d+e x)^2\right )^{-m} F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{2 m} \Gamma \left (1-2 m,-\frac {b c (d+e x) \log (F)}{e}\right )}{b c \log (F)} \]

[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 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)^{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] (verified)

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right )^{-m} \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^2\right )^{-m} \Gamma \left (1-2 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{2 m}}{b c \log (F)} \]

[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])

Maple [F]

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

[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)

Fricas [F]

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

[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)

Sympy [F(-2)]

Exception generated. \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right )^{-m} \, dx=\text {Exception raised: HeuristicGCDFailed} \]

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

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

[Out]

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