\(\int F^x (a+b F^x)^n \, dx\) [11]

Optimal result

Integrand size = 13, antiderivative size = 24 \[ \int F^x \left (a+b F^x\right )^n \, dx=\frac {\left (a+b F^x\right )^{1+n}}{b (1+n) \log (F)} \]

[Out]

(a+b*F^x)^(1+n)/b/(1+n)/ln(F)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2278, 32} \[ \int F^x \left (a+b F^x\right )^n \, dx=\frac {\left (a+b F^x\right )^{n+1}}{b (n+1) \log (F)} \]

[In]

Int[F^x*(a + b*F^x)^n,x]

[Out]

(a + b*F^x)^(1 + n)/(b*(1 + n)*Log[F])

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rule 2278

Int[((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)*((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.),
x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int[(a + b*x)^p, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b,
c, d, e, n, p}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int (a+b x)^n \, dx,x,F^x\right )}{\log (F)} \\ & = \frac {\left (a+b F^x\right )^{1+n}}{b (1+n) \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int F^x \left (a+b F^x\right )^n \, dx=\frac {\left (a+b F^x\right )^{1+n}}{b \log (F)+b n \log (F)} \]

[In]

Integrate[F^x*(a + b*F^x)^n,x]

[Out]

(a + b*F^x)^(1 + n)/(b*Log[F] + b*n*Log[F])

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.04

method	result	size
derivativedivides	\(\frac {\left (a +b \,F^{x}\right )^{1+n}}{b \left (1+n \right ) \ln \left (F \right )}\)	\(25\)
default	\(\frac {\left (a +b \,F^{x}\right )^{1+n}}{b \left (1+n \right ) \ln \left (F \right )}\)	\(25\)
risch	\(\frac {\left (a +b \,F^{x}\right ) \left (a +b \,F^{x}\right )^{n}}{b \ln \left (F \right ) \left (1+n \right )}\)	\(30\)
parallelrisch	\(\frac {F^{x} \left (a +b \,F^{x}\right )^{n} b +\left (a +b \,F^{x}\right )^{n} a}{b \ln \left (F \right ) \left (1+n \right )}\)	\(40\)
norman	\(\frac {{\mathrm e}^{x \ln \left (F \right )} {\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{x \ln \left (F \right )}\right )}}{\ln \left (F \right ) \left (1+n \right )}+\frac {a \,{\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{x \ln \left (F \right )}\right )}}{\ln \left (F \right ) b \left (1+n \right )}\)	\(57\)

[In]

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

[Out]

(a+b*F^x)^(1+n)/b/(1+n)/ln(F)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 28, normalized size of antiderivative = 1.17 \[ \int F^x \left (a+b F^x\right )^n \, dx=\frac {{\left (F^{x} b + a\right )} {\left (F^{x} b + a\right )}^{n}}{{\left (b n + b\right )} \log \left (F\right )} \]

[In]

integrate(F^x*(a+b*F^x)^n,x, algorithm="fricas")

[Out]

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 82 vs. \(2 (17) = 34\).

Time = 0.43 (sec) , antiderivative size = 82, normalized size of antiderivative = 3.42 \[ \int F^x \left (a+b F^x\right )^n \, dx=\begin {cases} \frac {x}{a} & \text {for}\: F = 1 \wedge b = 0 \wedge n = -1 \\x \left (a + b\right )^{n} & \text {for}\: F = 1 \\\frac {F^{x} a^{n}}{\log {\left (F \right )}} & \text {for}\: b = 0 \\\frac {\log {\left (F^{x} + \frac {a}{b} \right )}}{b \log {\left (F \right )}} & \text {for}\: n = -1 \\\frac {F^{x} b \left (F^{x} b + a\right )^{n}}{b n \log {\left (F \right )} + b \log {\left (F \right )}} + \frac {a \left (F^{x} b + a\right )^{n}}{b n \log {\left (F \right )} + b \log {\left (F \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(F**x*(a+b*F**x)**n,x)

[Out]

Piecewise((x/a, Eq(F, 1) & Eq(b, 0) & Eq(n, -1)), (x*(a + b)**n, Eq(F, 1)), (F**x*a**n/log(F), Eq(b, 0)), (log
(F**x + a/b)/(b*log(F)), Eq(n, -1)), (F**x*b*(F**x*b + a)**n/(b*n*log(F) + b*log(F)) + a*(F**x*b + a)**n/(b*n*
log(F) + b*log(F)), True))

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int F^x \left (a+b F^x\right )^n \, dx=\frac {{\left (F^{x} b + a\right )}^{n + 1}}{b {\left (n + 1\right )} \log \left (F\right )} \]

[In]

integrate(F^x*(a+b*F^x)^n,x, algorithm="maxima")

[Out]

(F^x*b + a)^(n + 1)/(b*(n + 1)*log(F))

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int F^x \left (a+b F^x\right )^n \, dx=\frac {{\left (F^{x} b + a\right )}^{n + 1}}{b {\left (n + 1\right )} \log \left (F\right )} \]

[In]

integrate(F^x*(a+b*F^x)^n,x, algorithm="giac")

[Out]

(F^x*b + a)^(n + 1)/(b*(n + 1)*log(F))

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.00 \[ \int F^x \left (a+b F^x\right )^n \, dx=\frac {{\left (a+F^x\,b\right )}^{n+1}}{b\,\ln \left (F\right )\,\left (n+1\right )} \]

[In]

int(F^x*(a + F^x*b)^n,x)

[Out]

(a + F^x*b)^(n + 1)/(b*log(F)*(n + 1))