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

Optimal. Leaf size=100 \[ -\frac {a}{d (c+d x)}+\frac {b f g n \log (F) \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac {c f}{d}\right )-g n (e+f x)} \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right )}{d^2}-\frac {b \left (F^{e g+f g x}\right )^n}{d (c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2183, 2177, 2182, 2178} \[ -\frac {a}{d (c+d x)}+\frac {b f g n \log (F) \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac {c f}{d}\right )-g n (e+f x)} \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right )}{d^2}-\frac {b \left (F^{e g+f g x}\right )^n}{d (c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2177

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] &&  !$UseGamma ===
True

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

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 78, normalized size = 0.78 \[ \frac {b f g n \log (F) \left (F^{g (e+f x)}\right )^n F^{-\frac {f g n (c+d x)}{d}} \text {Ei}\left (\frac {f g n (c+d x) \log (F)}{d}\right )-\frac {d \left (a+b \left (F^{g (e+f x)}\right )^n\right )}{c+d x}}{d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 87, normalized size = 0.87 \[ \frac {{\left (b d f g n x + b c f g n\right )} F^{\frac {{\left (d e - c f\right )} g n}{d}} {\rm Ei}\left (\frac {{\left (d f g n x + c f g n\right )} \log \relax (F)}{d}\right ) \log \relax (F) - F^{f g n x + e g n} b d - a d}{d^{3} x + c d^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {b \left (F^{\left (f x +e \right ) g}\right )^{n}+a}{\left (d x +c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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