3.1.72 \(\int (a+b (F^{g (e+f x)})^n) (c+d x)^m \, dx\) [72]

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

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2214, 2213, 2212} \begin {gather*} \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)} \end {gather*}

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 2212

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

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 \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*}

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Mathematica [F]
time = 0.15, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a+b \left (F^{g (e+f x)}\right )^n\right ) (c+d x)^m \, dx \end {gather*}

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

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]
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)*(d*x+c)^m,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.10, size = 114, normalized size = 0.98 \begin {gather*} \frac {{\left (b d m + b d\right )} e^{\left (-\frac {d m \log \left (-\frac {f g n \log \left (F\right )}{d}\right ) + {\left (c f g n - d g n e\right )} \log \left (F\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 {gather*}

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*m*log(-f*g*n*log(F)/d) + (c*f*g*n - d*g*n*e)*log(F))/d)*gamma(m + 1, -(d*f*g*n*x + c*f*g
*n)*log(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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, 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)*(d*x+c)^m,x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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