3.1.36 \(\int \frac {(a+b (F^{g (e+f x)})^n)^2}{c+d x} \, dx\) [36]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.18, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2214, 2213, 2209} \begin {gather*} \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} \end {gather*}

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 2209

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

Rule 2213

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 2214

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]

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 = 0.38, size = 108, normalized size = 0.81 \begin {gather*} \frac {2 a b F^{-\frac {f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^n \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right )+b^2 F^{-\frac {2 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{2 n} \text {Ei}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right )+a^2 \log (c+d x)}{d} \end {gather*}

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

________________________________________________________________________________________

Maple [F]
time = 0.04, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )^{2}}{d x +c}\, dx\]

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)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

________________________________________________________________________________________

Fricas [A]
time = 0.43, size = 105, normalized size = 0.78 \begin {gather*} \frac {a^{2} \log \left (d x + c\right ) + \frac {b^{2} {\rm Ei}\left (\frac {2 \, {\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right )}{F^{\frac {2 \, {\left (c f g n - d g n e\right )}}{d}}} + \frac {2 \, a b {\rm Ei}\left (\frac {{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right )}{F^{\frac {c f g n - d g n e}{d}}}}{d} \end {gather*}

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]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \left (F^{e g} F^{f g x}\right )^{n}\right )^{2}}{c + d x}\, dx \end {gather*}

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)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,{\left (F^{g\,\left (e+f\,x\right )}\right )}^n\right )}^2}{c+d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________