3.266 \(\int \frac{F^{a+b (c+d x)^2}}{(c+d x)^{11}} \, dx\)

Optimal. Leaf size=31 \[ \frac{b^5 F^a \log ^5(F) \text{Gamma}\left (-5,-b \log (F) (c+d x)^2\right )}{2 d} \]

[Out]

(b^5*F^a*Gamma[-5, -(b*(c + d*x)^2*Log[F])]*Log[F]^5)/(2*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0642407, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {2218} \[ \frac{b^5 F^a \log ^5(F) \text{Gamma}\left (-5,-b \log (F) (c+d x)^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^2)/(c + d*x)^11,x]

[Out]

(b^5*F^a*Gamma[-5, -(b*(c + d*x)^2*Log[F])]*Log[F]^5)/(2*d)

Rule 2218

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Simp[(F^a*(e + f*
x)^(m + 1)*Gamma[(m + 1)/n, -(b*(c + d*x)^n*Log[F])])/(f*n*(-(b*(c + d*x)^n*Log[F]))^((m + 1)/n)), x] /; FreeQ
[{F, a, b, c, d, e, f, m, n}, x] && EqQ[d*e - c*f, 0]

Rubi steps

\begin{align*} \int \frac{F^{a+b (c+d x)^2}}{(c+d x)^{11}} \, dx &=\frac{b^5 F^a \Gamma \left (-5,-b (c+d x)^2 \log (F)\right ) \log ^5(F)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0068067, size = 31, normalized size = 1. \[ \frac{b^5 F^a \log ^5(F) \text{Gamma}\left (-5,-b \log (F) (c+d x)^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^11,x]

[Out]

(b^5*F^a*Gamma[-5, -(b*(c + d*x)^2*Log[F])]*Log[F]^5)/(2*d)

________________________________________________________________________________________

Maple [B]  time = 0.137, size = 185, normalized size = 6. \begin{align*} -{\frac{{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{10\,d \left ( dx+c \right ) ^{10}}}-{\frac{b\ln \left ( F \right ){F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{40\,d \left ( dx+c \right ) ^{8}}}-{\frac{{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{120\,d \left ( dx+c \right ) ^{6}}}-{\frac{{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{240\,d \left ( dx+c \right ) ^{4}}}-{\frac{{b}^{4} \left ( \ln \left ( F \right ) \right ) ^{4}{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{240\,d \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{5} \left ( \ln \left ( F \right ) \right ) ^{5}{F}^{a}{\it Ei} \left ( 1,-b \left ( dx+c \right ) ^{2}\ln \left ( F \right ) \right ) }{240\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b*(d*x+c)^2)/(d*x+c)^11,x)

[Out]

-1/10/d/(d*x+c)^10*F^(b*(d*x+c)^2)*F^a-1/40/d*b*ln(F)/(d*x+c)^8*F^(b*(d*x+c)^2)*F^a-1/120/d*b^2*ln(F)^2/(d*x+c
)^6*F^(b*(d*x+c)^2)*F^a-1/240/d*b^3*ln(F)^3/(d*x+c)^4*F^(b*(d*x+c)^2)*F^a-1/240/d*b^4*ln(F)^4/(d*x+c)^2*F^(b*(
d*x+c)^2)*F^a-1/240/d*b^5*ln(F)^5*F^a*Ei(1,-b*(d*x+c)^2*ln(F))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{11}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^11,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^11, x)

________________________________________________________________________________________

Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^11,x, algorithm="fricas")

[Out]

Exception raised: TypeError

________________________________________________________________________________________

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**(a+b*(d*x+c)**2)/(d*x+c)**11,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{{\left (d x + c\right )}^{2} b + a}}{{\left (d x + c\right )}^{11}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b*(d*x+c)^2)/(d*x+c)^11,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^11, x)