3.11 \(\int F^x (a+b F^x)^n \, dx\)

Optimal. Leaf size=24 \[ \frac {\left (a+b F^x\right )^{n+1}}{b (n+1) \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 24, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2246, 32} \[ \frac {\left (a+b F^x\right )^{n+1}}{b (n+1) \log (F)} \]

Antiderivative was successfully verified.

[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 2246

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*} \int F^x \left (a+b F^x\right )^n \, dx &=\frac {\operatorname {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*}

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Mathematica [A]  time = 0.02, size = 24, normalized size = 1.00 \[ \frac {\left (a+b F^x\right )^{n+1}}{b n \log (F)+b \log (F)} \]

Antiderivative was successfully verified.

[In]

Integrate[F^x*(a + b*F^x)^n,x]

[Out]

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

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fricas [A]  time = 0.43, size = 28, normalized size = 1.17 \[ \frac {{\left (F^{x} b + a\right )} {\left (F^{x} b + a\right )}^{n}}{{\left (b n + b\right )} \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [A]  time = 0.43, size = 24, normalized size = 1.00 \[ \frac {{\left (F^{x} b + a\right )}^{n + 1}}{b {\left (n + 1\right )} \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 25, normalized size = 1.04 \[ \frac {\left (b \,F^{x}+a \right )^{n +1}}{\left (n +1\right ) b \ln \relax (F )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 24, normalized size = 1.00 \[ \frac {{\left (F^{x} b + a\right )}^{n + 1}}{b {\left (n + 1\right )} \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.48, size = 24, normalized size = 1.00 \[ \frac {{\left (a+F^x\,b\right )}^{n+1}}{b\,\ln \relax (F)\,\left (n+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^x*(a + F^x*b)^n,x)

[Out]

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

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sympy [A]  time = 1.26, size = 82, normalized size = 3.42 \[ \begin {cases} \frac {x}{a} & \text {for}\: F = 1 \wedge b = 0 \wedge n = -1 \\x \left (a + b\right )^{n} & \text {for}\: F = 1 \\\frac {F^{x} a^{n}}{\log {\relax (F )}} & \text {for}\: b = 0 \\\frac {\log {\left (F^{x} + \frac {a}{b} \right )}}{b \log {\relax (F )}} & \text {for}\: n = -1 \\\frac {F^{x} b \left (F^{x} b + a\right )^{n}}{b n \log {\relax (F )} + b \log {\relax (F )}} + \frac {a \left (F^{x} b + a\right )^{n}}{b n \log {\relax (F )} + b \log {\relax (F )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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