3.1.28 \(\int (a+b (F^{g (e+f x)})^n) \, dx\) [28]

Optimal. Leaf size=30 \[ a x+\frac {b \left (F^{g (e+f x)}\right )^n}{f g n \log (F)} \]

[Out]

a*x+b*(F^(g*(f*x+e)))^n/f/g/n/ln(F)

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Rubi [A]
time = 0.01, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {2225} \begin {gather*} a x+\frac {b \left (F^{g (e+f x)}\right )^n}{f g n \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[a + b*(F^(g*(e + f*x)))^n,x]

[Out]

a*x + (b*(F^(g*(e + f*x)))^n)/(f*g*n*Log[F])

Rule 2225

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 30, normalized size = 1.00 \begin {gather*} a x+\frac {b \left (F^{g (e+f x)}\right )^n}{f g n \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

a*x + (b*(F^(g*(e + f*x)))^n)/(f*g*n*Log[F])

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Maple [A]
time = 0.05, size = 31, normalized size = 1.03

method	result	size
default	\(a x +\frac {b \left (F^{g \left (f x +e \right )}\right )^{n}}{f g n \ln \left (F \right )}\)	\(31\)
norman	\(a x +\frac {b \,{\mathrm e}^{n \ln \left ({\mathrm e}^{g \left (f x +e \right ) \ln \left (F \right )}\right )}}{n g f \ln \left (F \right )}\)	\(34\)
derivativedivides	\(\frac {b \left (F^{g \left (f x +e \right )}\right )^{n}+a \ln \left (\left (F^{g \left (f x +e \right )}\right )^{n}\right )}{g f \ln \left (F \right ) n}\)	\(43\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a+b*(F^(g*(f*x+e)))^n,x,method=_RETURNVERBOSE)

[Out]

a*x+b*(F^(g*(f*x+e)))^n/f/g/n/ln(F)

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Maxima [A]
time = 0.28, size = 30, normalized size = 1.00 \begin {gather*} a x + \frac {F^{{\left (f x + e\right )} g n} b}{f g 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,x, algorithm="maxima")

[Out]

a*x + F^((f*x + e)*g*n)*b/(f*g*n*log(F))

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Fricas [A]
time = 0.39, size = 38, normalized size = 1.27 \begin {gather*} \frac {a f g n x \log \left (F\right ) + F^{f g n x + g n e} b}{f g 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,x, algorithm="fricas")

[Out]

(a*f*g*n*x*log(F) + F^(f*g*n*x + g*n*e)*b)/(f*g*n*log(F))

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Sympy [A]
time = 0.04, size = 32, normalized size = 1.07 \begin {gather*} a x + \begin {cases} \frac {b \left (F^{g \left (e + f x\right )}\right )^{n}}{f g n \log {\left (F \right )}} & \text {for}\: f g n \log {\left (F \right )} \neq 0 \\b x & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a+b*(F**(g*(f*x+e)))**n,x)

[Out]

a*x + Piecewise((b*(F**(g*(e + f*x)))**n/(f*g*n*log(F)), Ne(f*g*n*log(F), 0)), (b*x, True))

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Giac [A]
time = 2.60, size = 32, normalized size = 1.07 \begin {gather*} a x + \frac {F^{f g n x + g n e} b}{f g 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,x, algorithm="giac")

[Out]

a*x + F^(f*g*n*x + g*n*e)*b/(f*g*n*log(F))

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Mupad [B]
time = 3.60, size = 31, normalized size = 1.03 \begin {gather*} a\,x+\frac {b\,{\left (F^{e\,g+f\,g\,x}\right )}^n}{f\,g\,n\,\ln \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a + b*(F^(g*(e + f*x)))^n,x)

[Out]

a*x + (b*(F^(e*g + f*g*x))^n)/(f*g*n*log(F))

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