3.36 \(\int \frac{(a+b (F^{g (e+f x)})^n)^2}{c+d x} \, dx\)

Optimal. Leaf size=134 \[ \frac{a^2 \log (c+d x)}{d}+\frac{2 a b \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac{c f}{d}\right )-g n (e+f x)} \text{Ei}\left (\frac{f g n (c+d x) \log (F)}{d}\right )}{d}+\frac{b^2 \left (F^{e g+f g x}\right )^{2 n} F^{2 g n \left (e-\frac{c f}{d}\right )-2 g n (e+f x)} \text{Ei}\left (\frac{2 f g n (c+d x) \log (F)}{d}\right )}{d} \]

[Out]

(2*a*b*F^((e - (c*f)/d)*g*n - g*n*(e + f*x))*(F^(e*g + f*g*x))^n*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F])/d])/d
+ (b^2*F^(2*(e - (c*f)/d)*g*n - 2*g*n*(e + f*x))*(F^(e*g + f*g*x))^(2*n)*ExpIntegralEi[(2*f*g*n*(c + d*x)*Log[
F])/d])/d + (a^2*Log[c + d*x])/d

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Rubi [A]  time = 0.254095, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2183, 2182, 2178} \[ \frac{a^2 \log (c+d x)}{d}+\frac{2 a b \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac{c f}{d}\right )-g n (e+f x)} \text{Ei}\left (\frac{f g n (c+d x) \log (F)}{d}\right )}{d}+\frac{b^2 \left (F^{e g+f g x}\right )^{2 n} F^{2 g n \left (e-\frac{c f}{d}\right )-2 g n (e+f x)} \text{Ei}\left (\frac{2 f g n (c+d x) \log (F)}{d}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(F^(g*(e + f*x)))^n)^2/(c + d*x),x]

[Out]

(2*a*b*F^((e - (c*f)/d)*g*n - g*n*(e + f*x))*(F^(e*g + f*g*x))^n*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F])/d])/d
+ (b^2*F^(2*(e - (c*f)/d)*g*n - 2*g*n*(e + f*x))*(F^(e*g + f*g*x))^(2*n)*ExpIntegralEi[(2*f*g*n*(c + d*x)*Log[
F])/d])/d + (a^2*Log[c + d*x])/d

Rule 2183

Int[((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.))^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> In
t[ExpandIntegrand[(c + d*x)^m, (a + b*(F^(g*(e + f*x)))^n)^p, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, m, n},
x] && IGtQ[p, 0]

Rule 2182

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Dist[(b*F^(g*(e +
f*x)))^n/F^(g*n*(e + f*x)), Int[(c + d*x)^m*F^(g*n*(e + f*x)), x], x] /; FreeQ[{F, b, c, d, e, f, g, m, n}, x]

Rule 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rubi steps

\begin{align*} \int \frac{\left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{c+d x} \, dx &=\int \left (\frac{a^2}{c+d x}+\frac{2 a b \left (F^{e g+f g x}\right )^n}{c+d x}+\frac{b^2 \left (F^{e g+f g x}\right )^{2 n}}{c+d x}\right ) \, dx\\ &=\frac{a^2 \log (c+d x)}{d}+(2 a b) \int \frac{\left (F^{e g+f g x}\right )^n}{c+d x} \, dx+b^2 \int \frac{\left (F^{e g+f g x}\right )^{2 n}}{c+d x} \, dx\\ &=\frac{a^2 \log (c+d x)}{d}+\left (2 a b F^{-n (e g+f g x)} \left (F^{e g+f g x}\right )^n\right ) \int \frac{F^{n (e g+f g x)}}{c+d x} \, dx+\left (b^2 F^{-2 n (e g+f g x)} \left (F^{e g+f g x}\right )^{2 n}\right ) \int \frac{F^{2 n (e g+f g x)}}{c+d x} \, dx\\ &=\frac{2 a b F^{\left (e-\frac{c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n \text{Ei}\left (\frac{f g n (c+d x) \log (F)}{d}\right )}{d}+\frac{b^2 F^{2 \left (e-\frac{c f}{d}\right ) g n-2 g n (e+f x)} \left (F^{e g+f g x}\right )^{2 n} \text{Ei}\left (\frac{2 f g n (c+d x) \log (F)}{d}\right )}{d}+\frac{a^2 \log (c+d x)}{d}\\ \end{align*}

Mathematica [A]  time = 0.29943, size = 108, normalized size = 0.81 \[ \frac{a^2 \log (c+d x)+2 a b \left (F^{g (e+f x)}\right )^n F^{-\frac{f g n (c+d x)}{d}} \text{Ei}\left (\frac{f g n (c+d x) \log (F)}{d}\right )+b^2 \left (F^{g (e+f x)}\right )^{2 n} F^{-\frac{2 f g n (c+d x)}{d}} \text{Ei}\left (\frac{2 f g n (c+d x) \log (F)}{d}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(F^(g*(e + f*x)))^n)^2/(c + d*x),x]

[Out]

((2*a*b*(F^(g*(e + f*x)))^n*ExpIntegralEi[(f*g*n*(c + d*x)*Log[F])/d])/F^((f*g*n*(c + d*x))/d) + (b^2*(F^(g*(e
+ f*x)))^(2*n)*ExpIntegralEi[(2*f*g*n*(c + d*x)*Log[F])/d])/F^((2*f*g*n*(c + d*x))/d) + a^2*Log[c + d*x])/d

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( a+b \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) ^{2}}{dx+c}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*(F^(g*(f*x+e)))^n)^2/(d*x+c),x)

[Out]

int((a+b*(F^(g*(f*x+e)))^n)^2/(d*x+c),x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^2/(d*x+c),x, algorithm="maxima")

[Out]

(F^(e*g))^(2*n)*b^2*integrate((F^(f*g*x))^(2*n)/(d*x + c), x) + 2*(F^(e*g))^n*a*b*integrate((F^(f*g*x))^n/(d*x
+ c), x) + a^2*log(d*x + c)/d

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Fricas [A]  time = 1.52974, size = 208, normalized size = 1.55 \begin{align*} \frac{F^{\frac{2 \,{\left (d e - c f\right )} g n}{d}} b^{2}{\rm Ei}\left (\frac{2 \,{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) + 2 \, F^{\frac{{\left (d e - c f\right )} g n}{d}} a b{\rm Ei}\left (\frac{{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) + a^{2} \log \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^2/(d*x+c),x, algorithm="fricas")

[Out]

(F^(2*(d*e - c*f)*g*n/d)*b^2*Ei(2*(d*f*g*n*x + c*f*g*n)*log(F)/d) + 2*F^((d*e - c*f)*g*n/d)*a*b*Ei((d*f*g*n*x
+ c*f*g*n)*log(F)/d) + a^2*log(d*x + c))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \left (F^{e g} F^{f g x}\right )^{n}\right )^{2}}{c + d x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F**(g*(f*x+e)))**n)**2/(d*x+c),x)

[Out]

Integral((a + b*(F**(e*g)*F**(f*g*x))**n)**2/(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*(F^(g*(f*x+e)))^n)^2/(d*x+c),x, algorithm="giac")

[Out]

integrate(((F^((f*x + e)*g))^n*b + a)^2/(d*x + c), x)