3.12 \(\int F^{d x} (a+b F^{c+d x})^n \, dx\)

Optimal. Leaf size=36 \[ \frac {F^{-c} \left (a+b F^{c+d x}\right )^{n+1}}{b d (n+1) \log (F)} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2247, 2246, 32} \[ \frac {F^{-c} \left (a+b F^{c+d x}\right )^{n+1}}{b d (n+1) \log (F)} \]

Antiderivative was successfully verified.

[In]

Int[F^(d*x)*(a + b*F^(c + d*x))^n,x]

[Out]

(a + b*F^(c + d*x))^(1 + n)/(b*d*F^c*(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]

Rule 2247

Int[((a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.))^(p_.)*((G_)^((h_.)*((f_.) + (g_.)*(x_))))^(m_.),
x_Symbol] :> Dist[(G^(h*(f + g*x)))^m/(F^(e*(c + d*x)))^n, Int[(F^(e*(c + d*x)))^n*(a + b*(F^(e*(c + d*x)))^n)
^p, x], x] /; FreeQ[{F, G, a, b, c, d, e, f, g, h, m, n, p}, x] && EqQ[d*e*n*Log[F], g*h*m*Log[G]]

Rubi steps

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

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Mathematica [A]  time = 0.04, size = 35, normalized size = 0.97 \[ \frac {F^{-c} \left (a+b F^{c+d x}\right )^{n+1}}{b d n \log (F)+b d \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 50, normalized size = 1.39 \[ \frac {{\left (F^{d x + c} b + a\right )}^{n} {\left (\frac {F^{d x + c} b}{F^{c}} + \frac {a}{F^{c}}\right )}}{{\left (b d n + b d\right )} \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.44, size = 36, normalized size = 1.00 \[ \frac {{\left (F^{d x + c} b + a\right )}^{n + 1}}{F^{c} b d {\left (n + 1\right )} \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 81, normalized size = 2.25 \[ \frac {a \,F^{-c} {\mathrm e}^{n \ln \left (b \,{\mathrm e}^{c \ln \relax (F )} {\mathrm e}^{d x \ln \relax (F )}+a \right )}}{\left (n +1\right ) b d \ln \relax (F )}+\frac {{\mathrm e}^{n \ln \left (b \,{\mathrm e}^{c \ln \relax (F )} {\mathrm e}^{d x \ln \relax (F )}+a \right )} {\mathrm e}^{d x \ln \relax (F )}}{\left (n +1\right ) d \ln \relax (F )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.44, size = 36, normalized size = 1.00 \[ \frac {{\left (F^{d x + c} b + a\right )}^{n + 1}}{F^{c} b d {\left (n + 1\right )} \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 3.53, size = 55, normalized size = 1.53 \[ {\left (a+F^{c+d\,x}\,b\right )}^n\,\left (\frac {F^{d\,x}}{d\,\ln \relax (F)\,\left (n+1\right )}+\frac {a}{F^c\,b\,d\,\ln \relax (F)\,\left (n+1\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(F^(d*x)*(a + F^(c + d*x)*b)^n,x)

[Out]

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

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sympy [A]  time = 145.67, size = 141, normalized size = 3.92 \[ \begin {cases} \frac {x}{a} & \text {for}\: F = 1 \wedge b = 0 \wedge d = 0 \wedge n = -1 \\x \left (a + b\right )^{n} & \text {for}\: F = 1 \\\frac {F^{d x} a^{n}}{d \log {\relax (F )}} & \text {for}\: b = 0 \\x \left (F^{c} b + a\right )^{n} & \text {for}\: d = 0 \\\frac {F^{- c} \log {\left (F^{c} F^{d x} + \frac {a}{b} \right )}}{b d \log {\relax (F )}} & \text {for}\: n = -1 \\\frac {F^{c} F^{d x} b \left (F^{c} F^{d x} b + a\right )^{n}}{F^{c} b d n \log {\relax (F )} + F^{c} b d \log {\relax (F )}} + \frac {a \left (F^{c} F^{d x} b + a\right )^{n}}{F^{c} b d n \log {\relax (F )} + F^{c} b d \log {\relax (F )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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