\(\int F^{c (a+b x)} (d^2+2 d e x+e^2 x^2)^m \, dx\) [83]

Optimal result

Integrand size = 28, antiderivative size = 71 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right )^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^2\right )^m \Gamma \left (1+2 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-2 m}}{b c \log (F)} \] Output:

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00 \[ \int F^{c (a+b x)} \left (d^2+2 d e x+e^2 x^2\right )^m \, dx=\frac {F^{c \left (a-\frac {b d}{e}\right )} \left ((d+e x)^2\right )^m \Gamma \left (1+2 m,-\frac {b c (d+e x) \log (F)}{e}\right ) \left (-\frac {b c (d+e x) \log (F)}{e}\right )^{-2 m}}{b c \log (F)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2008, 2612}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2008

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

\(\Big \downarrow \) 2612

\(\displaystyle \frac {\left ((d+e x)^2\right )^m F^{c \left (a-\frac {b d}{e}\right )} \left (-\frac {b c \log (F) (d+e x)}{e}\right )^{-2 m} \Gamma \left (2 m+1,-\frac {b c (d+e x) \log (F)}{e}\right )}{b c \log (F)}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 2008

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{a = Rt[Coeff[Px, x, 0], Expon[Px, 
x]], b = Rt[Coeff[Px, x, Expon[Px, x]], Expon[Px, x]]}, Simp[((a + b*x)^Exp 
on[Px, x])^p/(a + b*x)^(Expon[Px, x]*p)   Int[u*(a + b*x)^(Expon[Px, x]*p), 
 x], x] /; EqQ[Px, (a + b*x)^Expon[Px, x]]] /;  !IntegerQ[p] && PolyQ[Px, x 
] && GtQ[Expon[Px, x], 1] && NeQ[Coeff[Px, x, 0], 0]
 

rule 2612

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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(f**(a*c)*(f**(b*c*x)*(d**2 + 2*d*e*x + e**2*x**2)**m - 2*int((f**(b*c*x)* 
(d**2 + 2*d*e*x + e**2*x**2)**m)/(d + e*x),x)*e*m))/(log(f)*b*c)