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

Optimal. Leaf size=61 \[ \frac{F^a (c+d x)^{m+1} \left (-\frac{b \log (F)}{(c+d x)^2}\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b \log (F)}{(c+d x)^2}\right )}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0536999, antiderivative size = 61, 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} \[ \frac{F^a (c+d x)^{m+1} \left (-\frac{b \log (F)}{(c+d x)^2}\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b \log (F)}{(c+d x)^2}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(F^a*(c + d*x)^(1 + m)*Gamma[(-1 - m)/2, -((b*Log[F])/(c + d*x)^2)]*(-((b*Log[F])/(c + d*x)^2))^((1 + m)/2))/(
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 F^{a+\frac{b}{(c+d x)^2}} (c+d x)^m \, dx &=\frac{F^a (c+d x)^{1+m} \Gamma \left (\frac{1}{2} (-1-m),-\frac{b \log (F)}{(c+d x)^2}\right ) \left (-\frac{b \log (F)}{(c+d x)^2}\right )^{\frac{1+m}{2}}}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.034179, size = 61, normalized size = 1. \[ \frac{F^a (c+d x)^{m+1} \left (-\frac{b \log (F)}{(c+d x)^2}\right )^{\frac{m+1}{2}} \text{Gamma}\left (\frac{1}{2} (-m-1),-\frac{b \log (F)}{(c+d x)^2}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.08, size = 0, normalized size = 0. \begin{align*} \int{F}^{a+{\frac{b}{ \left ( dx+c \right ) ^{2}}}} \left ( dx+c \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (d x + c\right )}^{m} 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}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((d*x + c)^m*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)), x)

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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)**m,x)

[Out]

Timed out

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

[Out]

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