\(\int F^{d x} (a+b F^{c+d x})^n \, dx\) [12]

Optimal result

Integrand size = 19, antiderivative size = 36 \[ \int F^{d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {F^{-c} \left (a+b F^{c+d x}\right )^{1+n}}{b d (1+n) \log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2279, 2278, 32} \[ \int F^{d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {F^{-c} \left (a+b F^{c+d x}\right )^{n+1}}{b d (n+1) \log (F)} \]

[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 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]

Rule 2279

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*} \text {integral}& = F^{-c} \int F^{c+d x} \left (a+b F^{c+d x}\right )^n \, dx \\ & = \frac {F^{-c} \text {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*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.97 \[ \int F^{d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {F^{-c} \left (a+b F^{c+d x}\right )^{1+n}}{b d \log (F)+b d n \log (F)} \]

[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]))

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.33

method	result	size
risch	\(\frac {\left (F^{c} F^{d x} b +a \right ) F^{-c} \left (F^{c} F^{d x} b +a \right )^{n}}{b \left (1+n \right ) \ln \left (F \right ) d}\)	\(48\)
norman	\(\frac {{\mathrm e}^{d \ln \left (F \right ) x} {\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{c \ln \left (F \right )} {\mathrm e}^{d \ln \left (F \right ) x}\right )}}{\ln \left (F \right ) d \left (1+n \right )}+\frac {F^{-c} a \,{\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{c \ln \left (F \right )} {\mathrm e}^{d \ln \left (F \right ) x}\right )}}{b d \ln \left (F \right ) \left (1+n \right )}\)	\(81\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.39 \[ \int F^{d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {{\left (F^{d x + c} b + a\right )}^{n} {\left (\frac {F^{d x + c} b}{F^{c}} + \frac {a}{F^{c}}\right )}}{{\left (b d n + b d\right )} \log \left (F\right )} \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (26) = 52\).

Time = 0.96 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.97 \[ \int F^{d x} \left (a+b F^{c+d x}\right )^n \, dx=\begin {cases} x \left (F^{c} b + a\right )^{n} & \text {for}\: d = 0 \\x \left (a + b\right )^{n} & \text {for}\: \log {\left (F \right )} = 0 \\\frac {\begin {cases} F^{d x} a^{n} & \text {for}\: F^{c} = 0 \vee b = 0 \\\frac {F^{- c} \left (\begin {cases} \frac {\left (F^{c} F^{d x} b + a\right )^{n + 1}}{n + 1} & \text {for}\: n \neq -1 \\\log {\left (F^{c} F^{d x} b + a \right )} & \text {otherwise} \end {cases}\right )}{b} & \text {otherwise} \end {cases}}{d \log {\left (F \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int F^{d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {{\left (F^{d x + c} b + a\right )}^{n + 1}}{F^{c} b d {\left (n + 1\right )} \log \left (F\right )} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.00 \[ \int F^{d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {{\left (F^{d x + c} b + a\right )}^{n + 1}}{F^{c} b d {\left (n + 1\right )} \log \left (F\right )} \]

[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))

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.53 \[ \int F^{d x} \left (a+b F^{c+d x}\right )^n \, dx={\left (a+F^{c+d\,x}\,b\right )}^n\,\left (\frac {F^{d\,x}}{d\,\ln \left (F\right )\,\left (n+1\right )}+\frac {a}{F^c\,b\,d\,\ln \left (F\right )\,\left (n+1\right )}\right ) \]

[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)))