3.1.37 \(\int \frac {(a+b (F^{g (e+f x)})^n)^2}{(c+d x)^2} \, dx\) [37]

Optimal. Leaf size=202 \[ -\frac {a^2}{d (c+d x)}-\frac {2 a b \left (F^{e g+f g x}\right )^n}{d (c+d x)}-\frac {b^2 \left (F^{e g+f g x}\right )^{2 n}}{d (c+d x)}+\frac {2 a b f F^{\left (e-\frac {c f}{d}\right ) g n-g n (e+f x)} \left (F^{e g+f g x}\right )^n g n \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log (F)}{d^2}+\frac {2 b^2 f F^{2 \left (e-\frac {c f}{d}\right ) g n-2 g n (e+f x)} \left (F^{e g+f g x}\right )^{2 n} g n \text {Ei}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right ) \log (F)}{d^2} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2214, 2208, 2213, 2209} \begin {gather*} -\frac {a^2}{d (c+d x)}+\frac {2 a b f g n \log (F) \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^2}-\frac {2 a b \left (F^{e g+f g x}\right )^n}{d (c+d x)}+\frac {2 b^2 f g n \log (F) \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^2}-\frac {b^2 \left (F^{e g+f g x}\right )^{2 n}}{d (c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2208

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(c + d*x)^(m
+ 1)*((b*F^(g*(e + f*x)))^n/(d*(m + 1))), x] - Dist[f*g*n*(Log[F]/(d*(m + 1))), Int[(c + d*x)^(m + 1)*(b*F^(g*
(e + f*x)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && LtQ[m, -1] && IntegerQ[2*m] &&  !TrueQ[$UseGamm
a]

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

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Mathematica [A]
time = 0.60, size = 136, normalized size = 0.67 \begin {gather*} \frac {-\frac {d \left (a+b \left (F^{g (e+f x)}\right )^n\right )^2}{c+d x}+2 a b f F^{-\frac {f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^n g n \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right ) \log (F)+2 b^2 f F^{-\frac {2 f g n (c+d x)}{d}} \left (F^{g (e+f x)}\right )^{2 n} g n \text {Ei}\left (\frac {2 f g n (c+d x) \log (F)}{d}\right ) \log (F)}{d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \left (F^{g \left (f x +e \right )}\right )^{n}\right )^{2}}{\left (d x +c \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.39, size = 183, normalized size = 0.91 \begin {gather*} -\frac {2 \, F^{f g n x + g n e} a b d + F^{2 \, f g n x + 2 \, g n e} b^{2} d + a^{2} d - \frac {2 \, {\left (b^{2} d f g n x + b^{2} c f g n\right )} {\rm Ei}\left (\frac {2 \, {\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right )}{F^{\frac {2 \, {\left (c f g n - d g n e\right )}}{d}}} - \frac {2 \, {\left (a b d f g n x + a b c f g n\right )} {\rm Ei}\left (\frac {{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) \log \left (F\right )}{F^{\frac {c f g n - d g n e}{d}}}}{d^{3} x + c d^{2}} \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)^2,x, algorithm="fricas")

[Out]

-(2*F^(f*g*n*x + g*n*e)*a*b*d + F^(2*f*g*n*x + 2*g*n*e)*b^2*d + a^2*d - 2*(b^2*d*f*g*n*x + b^2*c*f*g*n)*Ei(2*(
d*f*g*n*x + c*f*g*n)*log(F)/d)*log(F)/F^(2*(c*f*g*n - d*g*n*e)/d) - 2*(a*b*d*f*g*n*x + a*b*c*f*g*n)*Ei((d*f*g*
n*x + c*f*g*n)*log(F)/d)*log(F)/F^((c*f*g*n - d*g*n*e)/d))/(d^3*x + c*d^2)

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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}}{\left (c + d x\right )^{2}}\, 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)**2,x)

[Out]

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

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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)^2,x, algorithm="giac")

[Out]

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

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

[Out]

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

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