3.338 \(\int \frac{F^{a+\frac{b}{(c+d x)^2}}}{(c+d x)^{12}} \, dx\)

Optimal. Leaf size=49 \[ \frac{F^a \text{Gamma}\left (\frac{11}{2},-\frac{b \log (F)}{(c+d x)^2}\right )}{2 d (c+d x)^{11} \left (-\frac{b \log (F)}{(c+d x)^2}\right )^{11/2}} \]

[Out]

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

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Rubi [A]  time = 0.0453887, antiderivative size = 49, 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} \[ \frac{F^a \text{Gamma}\left (\frac{11}{2},-\frac{b \log (F)}{(c+d x)^2}\right )}{2 d (c+d x)^{11} \left (-\frac{b \log (F)}{(c+d x)^2}\right )^{11/2}} \]

Antiderivative was successfully verified.

[In]

Int[F^(a + b/(c + d*x)^2)/(c + d*x)^12,x]

[Out]

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

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+\frac{b}{(c+d x)^2}}}{(c+d x)^{12}} \, dx &=\frac{F^a \Gamma \left (\frac{11}{2},-\frac{b \log (F)}{(c+d x)^2}\right )}{2 d (c+d x)^{11} \left (-\frac{b \log (F)}{(c+d x)^2}\right )^{11/2}}\\ \end{align*}

Mathematica [A]  time = 0.0406678, size = 49, normalized size = 1. \[ \frac{F^a \text{Gamma}\left (\frac{11}{2},-\frac{b \log (F)}{(c+d x)^2}\right )}{2 d (c+d x)^{11} \left (-\frac{b \log (F)}{(c+d x)^2}\right )^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[F^(a + b/(c + d*x)^2)/(c + d*x)^12,x]

[Out]

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

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Maple [A]  time = 0.254, size = 208, normalized size = 4.2 \begin{align*} -{\frac{{F}^{a}}{2\,d \left ( dx+c \right ) ^{9}b\ln \left ( F \right ) }{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{9\,{F}^{a}}{4\, \left ( \ln \left ( F \right ) \right ) ^{2}{b}^{2}d \left ( dx+c \right ) ^{7}}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}-{\frac{63\,{F}^{a}}{8\,d{b}^{3} \left ( \ln \left ( F \right ) \right ) ^{3} \left ( dx+c \right ) ^{5}}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{315\,{F}^{a}}{16\,d{b}^{4} \left ( \ln \left ( F \right ) \right ) ^{4} \left ( dx+c \right ) ^{3}}{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}-{\frac{945\,{F}^{a}}{32\,d \left ( \ln \left ( F \right ) \right ) ^{5}{b}^{5} \left ( dx+c \right ) }{F}^{{\frac{b}{ \left ( dx+c \right ) ^{2}}}}}+{\frac{945\,{F}^{a}\sqrt{\pi }}{64\,d \left ( \ln \left ( F \right ) \right ) ^{5}{b}^{5}}{\it Erf} \left ({\frac{1}{dx+c}\sqrt{-b\ln \left ( F \right ) }} \right ){\frac{1}{\sqrt{-b\ln \left ( F \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(a+b/(d*x+c)^2)/(d*x+c)^12,x)

[Out]

-1/2/d*F^a*F^(b/(d*x+c)^2)/(d*x+c)^9/b/ln(F)+9/4/d*F^a/b^2/ln(F)^2*F^(b/(d*x+c)^2)/(d*x+c)^7-63/8/d*F^a/b^3/ln
(F)^3*F^(b/(d*x+c)^2)/(d*x+c)^5+315/16/d*F^a/b^4/ln(F)^4*F^(b/(d*x+c)^2)/(d*x+c)^3-945/32/d*F^a/b^5/ln(F)^5*F^
(b/(d*x+c)^2)/(d*x+c)+945/64/d*F^a/b^5/ln(F)^5*Pi^(1/2)/(-b*ln(F))^(1/2)*erf((-b*ln(F))^(1/2)/(d*x+c))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{a + \frac{b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{12}}\,{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)^12,x, algorithm="maxima")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^12, x)

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Fricas [A]  time = 2.0105, size = 1300, normalized size = 26.53 \begin{align*} -\frac{945 \, \sqrt{\pi }{\left (d^{10} x^{9} + 9 \, c d^{9} x^{8} + 36 \, c^{2} d^{8} x^{7} + 84 \, c^{3} d^{7} x^{6} + 126 \, c^{4} d^{6} x^{5} + 126 \, c^{5} d^{5} x^{4} + 84 \, c^{6} d^{4} x^{3} + 36 \, c^{7} d^{3} x^{2} + 9 \, c^{8} d^{2} x + c^{9} d\right )} F^{a} \sqrt{-\frac{b \log \left (F\right )}{d^{2}}} \operatorname{erf}\left (\frac{d \sqrt{-\frac{b \log \left (F\right )}{d^{2}}}}{d x + c}\right ) + 2 \,{\left (16 \, b^{5} \log \left (F\right )^{5} - 72 \,{\left (b^{4} d^{2} x^{2} + 2 \, b^{4} c d x + b^{4} c^{2}\right )} \log \left (F\right )^{4} + 252 \,{\left (b^{3} d^{4} x^{4} + 4 \, b^{3} c d^{3} x^{3} + 6 \, b^{3} c^{2} d^{2} x^{2} + 4 \, b^{3} c^{3} d x + b^{3} c^{4}\right )} \log \left (F\right )^{3} - 630 \,{\left (b^{2} d^{6} x^{6} + 6 \, b^{2} c d^{5} x^{5} + 15 \, b^{2} c^{2} d^{4} x^{4} + 20 \, b^{2} c^{3} d^{3} x^{3} + 15 \, b^{2} c^{4} d^{2} x^{2} + 6 \, b^{2} c^{5} d x + b^{2} c^{6}\right )} \log \left (F\right )^{2} + 945 \,{\left (b d^{8} x^{8} + 8 \, b c d^{7} x^{7} + 28 \, b c^{2} d^{6} x^{6} + 56 \, b c^{3} d^{5} x^{5} + 70 \, b c^{4} d^{4} x^{4} + 56 \, b c^{5} d^{3} x^{3} + 28 \, b c^{6} d^{2} x^{2} + 8 \, b c^{7} d x + b c^{8}\right )} \log \left (F\right )\right )} F^{\frac{a d^{2} x^{2} + 2 \, a c d x + a c^{2} + b}{d^{2} x^{2} + 2 \, c d x + c^{2}}}}{64 \,{\left (b^{6} d^{10} x^{9} + 9 \, b^{6} c d^{9} x^{8} + 36 \, b^{6} c^{2} d^{8} x^{7} + 84 \, b^{6} c^{3} d^{7} x^{6} + 126 \, b^{6} c^{4} d^{6} x^{5} + 126 \, b^{6} c^{5} d^{5} x^{4} + 84 \, b^{6} c^{6} d^{4} x^{3} + 36 \, b^{6} c^{7} d^{3} x^{2} + 9 \, b^{6} c^{8} d^{2} x + b^{6} c^{9} d\right )} \log \left (F\right )^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(F^(a+b/(d*x+c)^2)/(d*x+c)^12,x, algorithm="fricas")

[Out]

-1/64*(945*sqrt(pi)*(d^10*x^9 + 9*c*d^9*x^8 + 36*c^2*d^8*x^7 + 84*c^3*d^7*x^6 + 126*c^4*d^6*x^5 + 126*c^5*d^5*
x^4 + 84*c^6*d^4*x^3 + 36*c^7*d^3*x^2 + 9*c^8*d^2*x + c^9*d)*F^a*sqrt(-b*log(F)/d^2)*erf(d*sqrt(-b*log(F)/d^2)
/(d*x + c)) + 2*(16*b^5*log(F)^5 - 72*(b^4*d^2*x^2 + 2*b^4*c*d*x + b^4*c^2)*log(F)^4 + 252*(b^3*d^4*x^4 + 4*b^
3*c*d^3*x^3 + 6*b^3*c^2*d^2*x^2 + 4*b^3*c^3*d*x + b^3*c^4)*log(F)^3 - 630*(b^2*d^6*x^6 + 6*b^2*c*d^5*x^5 + 15*
b^2*c^2*d^4*x^4 + 20*b^2*c^3*d^3*x^3 + 15*b^2*c^4*d^2*x^2 + 6*b^2*c^5*d*x + b^2*c^6)*log(F)^2 + 945*(b*d^8*x^8
+ 8*b*c*d^7*x^7 + 28*b*c^2*d^6*x^6 + 56*b*c^3*d^5*x^5 + 70*b*c^4*d^4*x^4 + 56*b*c^5*d^3*x^3 + 28*b*c^6*d^2*x^
2 + 8*b*c^7*d*x + b*c^8)*log(F))*F^((a*d^2*x^2 + 2*a*c*d*x + a*c^2 + b)/(d^2*x^2 + 2*c*d*x + c^2)))/((b^6*d^10
*x^9 + 9*b^6*c*d^9*x^8 + 36*b^6*c^2*d^8*x^7 + 84*b^6*c^3*d^7*x^6 + 126*b^6*c^4*d^6*x^5 + 126*b^6*c^5*d^5*x^4 +
84*b^6*c^6*d^4*x^3 + 36*b^6*c^7*d^3*x^2 + 9*b^6*c^8*d^2*x + b^6*c^9*d)*log(F)^6)

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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)**12,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{F^{a + \frac{b}{{\left (d x + c\right )}^{2}}}}{{\left (d x + c\right )}^{12}}\,{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)^12,x, algorithm="giac")

[Out]

integrate(F^(a + b/(d*x + c)^2)/(d*x + c)^12, x)