3.265 \(\int \frac{F^{a+b (c+d x)^2}}{(c+d x)^9} \, dx\)

Optimal. Leaf size=31 \[ -\frac{b^4 F^a \log ^4(F) \text{Gamma}\left (-4,-b \log (F) (c+d x)^2\right )}{2 d} \]

[Out]

-(b^4*F^a*Gamma[-4, -(b*(c + d*x)^2*Log[F])]*Log[F]^4)/(2*d)

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Rubi [A]  time = 0.062995, 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} \[ -\frac{b^4 F^a \log ^4(F) \text{Gamma}\left (-4,-b \log (F) (c+d x)^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b*(c + d*x)^2)/(c + d*x)^9,x]

[Out]

-(b^4*F^a*Gamma[-4, -(b*(c + d*x)^2*Log[F])]*Log[F]^4)/(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)^9} \, dx &=-\frac{b^4 F^a \Gamma \left (-4,-b (c+d x)^2 \log (F)\right ) \log ^4(F)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0067544, size = 31, normalized size = 1. \[ -\frac{b^4 F^a \log ^4(F) \text{Gamma}\left (-4,-b \log (F) (c+d x)^2\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b*(c + d*x)^2)/(c + d*x)^9,x]

[Out]

-(b^4*F^a*Gamma[-4, -(b*(c + d*x)^2*Log[F])]*Log[F]^4)/(2*d)

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Maple [B]  time = 0.091, size = 152, normalized size = 4.9 \begin{align*} -{\frac{{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{8\,d \left ( dx+c \right ) ^{8}}}-{\frac{b\ln \left ( F \right ){F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{24\,d \left ( dx+c \right ) ^{6}}}-{\frac{{b}^{2} \left ( \ln \left ( F \right ) \right ) ^{2}{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{48\,d \left ( dx+c \right ) ^{4}}}-{\frac{{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3}{F}^{b \left ( dx+c \right ) ^{2}}{F}^{a}}{48\,d \left ( dx+c \right ) ^{2}}}-{\frac{{b}^{4} \left ( \ln \left ( F \right ) \right ) ^{4}{F}^{a}{\it Ei} \left ( 1,-b \left ( dx+c \right ) ^{2}\ln \left ( F \right ) \right ) }{48\,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)^9,x)

[Out]

-1/8/d/(d*x+c)^8*F^(b*(d*x+c)^2)*F^a-1/24/d*b*ln(F)/(d*x+c)^6*F^(b*(d*x+c)^2)*F^a-1/48/d*b^2*ln(F)^2/(d*x+c)^4
*F^(b*(d*x+c)^2)*F^a-1/48/d*b^3*ln(F)^3/(d*x+c)^2*F^(b*(d*x+c)^2)*F^a-1/48/d*b^4*ln(F)^4*F^a*Ei(1,-b*(d*x+c)^2
*ln(F))

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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 )}^{9}}\,{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)^9,x, algorithm="maxima")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^9, x)

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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)^9,x, algorithm="fricas")

[Out]

Exception raised: TypeError

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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**(a+b*(d*x+c)**2)/(d*x+c)**9,x)

[Out]

Timed out

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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 )}^{9}}\,{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)^9,x, algorithm="giac")

[Out]

integrate(F^((d*x + c)^2*b + a)/(d*x + c)^9, x)