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

Optimal. Leaf size=116 \[ \frac{b (c+d x)^m \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac{c f}{d}\right )-g n (e+f x)} \left (-\frac{f g n \log (F) (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{f g n \log (F) (c+d x)}{d}\right )}{f g n \log (F)}+\frac{a (c+d x)^{m+1}}{d (m+1)} \]

[Out]

(a*(c + d*x)^(1 + m))/(d*(1 + m)) + (b*F^((e - (c*f)/d)*g*n - g*n*(e + f*x))*(F^(e*g + f*g*x))^n*(c + d*x)^m*G
amma[1 + m, -((f*g*n*(c + d*x)*Log[F])/d)])/(f*g*n*Log[F]*(-((f*g*n*(c + d*x)*Log[F])/d))^m)

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Rubi [A]  time = 0.134753, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2183, 2182, 2181} \[ \frac{b (c+d x)^m \left (F^{e g+f g x}\right )^n F^{g n \left (e-\frac{c f}{d}\right )-g n (e+f x)} \left (-\frac{f g n \log (F) (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,-\frac{f g n \log (F) (c+d x)}{d}\right )}{f g n \log (F)}+\frac{a (c+d x)^{m+1}}{d (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(c + d*x)^(1 + m))/(d*(1 + m)) + (b*F^((e - (c*f)/d)*g*n - g*n*(e + f*x))*(F^(e*g + f*g*x))^n*(c + d*x)^m*G
amma[1 + m, -((f*g*n*(c + d*x)*Log[F])/d)])/(f*g*n*Log[F]*(-((f*g*n*(c + d*x)*Log[F])/d))^m)

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 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

\begin{align*} \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx &=\int \left (a (c+d x)^m+b \left (F^{e g+f g x}\right )^n (c+d x)^m\right ) \, dx\\ &=\frac{a (c+d x)^{1+m}}{d (1+m)}+b \int \left (F^{e g+f g x}\right )^n (c+d x)^m \, dx\\ &=\frac{a (c+d x)^{1+m}}{d (1+m)}+\left (b F^{-n (e g+f g x)} \left (F^{e g+f g x}\right )^n\right ) \int F^{n (e g+f g x)} (c+d x)^m \, dx\\ &=\frac{a (c+d x)^{1+m}}{d (1+m)}+\frac{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 (c+d x)^m \Gamma \left (1+m,-\frac{f g n (c+d x) \log (F)}{d}\right ) \left (-\frac{f g n (c+d x) \log (F)}{d}\right )^{-m}}{f g n \log (F)}\\ \end{align*}

Mathematica [F]  time = 0.119028, size = 0, normalized size = 0. \[ \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.039, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ({F}^{g \left ( fx+e \right ) } \right ) ^{n} \right ) \left ( dx+c \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55604, size = 266, normalized size = 2.29 \begin{align*} \frac{{\left (b d m + b d\right )} e^{\left (\frac{{\left (d e - c f\right )} g n \log \left (F\right ) - d m \log \left (-\frac{f g n \log \left (F\right )}{d}\right )}{d}\right )} \Gamma \left (m + 1, -\frac{{\left (d f g n x + c f g n\right )} \log \left (F\right )}{d}\right ) +{\left (a d f g n x + a c f g n\right )}{\left (d x + c\right )}^{m} \log \left (F\right )}{{\left (d f g m + d f g\right )} n \log \left (F\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*d*m + b*d)*e^(((d*e - c*f)*g*n*log(F) - d*m*log(-f*g*n*log(F)/d))/d)*gamma(m + 1, -(d*f*g*n*x + c*f*g*n)*l
og(F)/d) + (a*d*f*g*n*x + a*c*f*g*n)*(d*x + c)^m*log(F))/((d*f*g*m + d*f*g)*n*log(F))

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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((a+b*(F**(g*(f*x+e)))**n)*(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 ({\left (F^{{\left (f x + e\right )} g}\right )}^{n} b + a\right )}{\left (d x + c\right )}^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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