\(\int F^{c+d x} (a+b F^{c+d x})^n \, dx\) [13]

Optimal result

Integrand size = 21, antiderivative size = 31 \[ \int F^{c+d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {\left (a+b F^{c+d x}\right )^{1+n}}{b d (1+n) \log (F)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {2278, 32} \[ \int F^{c+d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {\left (a+b F^{c+d x}\right )^{n+1}}{b d (n+1) \log (F)} \]

[In]

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

[Out]

(a + b*F^(c + d*x))^(1 + n)/(b*d*(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^{c+d x}\right )}{d \log (F)} \\ & = \frac {\left (a+b F^{c+d x}\right )^{1+n}}{b d (1+n) \log (F)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.97 \[ \int F^{c+d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {\left (a+b F^{c+d x}\right )^{1+n}}{b d \log (F)+b d n \log (F)} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.03

method	result	size
derivativedivides	\(\frac {\left (a +b \,F^{d x +c}\right )^{1+n}}{b d \left (1+n \right ) \ln \left (F \right )}\)	\(32\)
default	\(\frac {\left (a +b \,F^{d x +c}\right )^{1+n}}{b d \left (1+n \right ) \ln \left (F \right )}\)	\(32\)
risch	\(\frac {\left (a +b \,F^{d x +c}\right ) \left (a +b \,F^{d x +c}\right )^{n}}{b d \ln \left (F \right ) \left (1+n \right )}\)	\(41\)
parallelrisch	\(\frac {F^{d x +c} \left (a +b \,F^{d x +c}\right )^{n} b +\left (a +b \,F^{d x +c}\right )^{n} a}{b d \ln \left (F \right ) \left (1+n \right )}\)	\(55\)
norman	\(\frac {{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )} {\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right ) d \left (1+n \right )}+\frac {a \,{\mathrm e}^{n \ln \left (a +b \,{\mathrm e}^{\left (d x +c \right ) \ln \left (F \right )}\right )}}{\ln \left (F \right ) d b \left (1+n \right )}\)	\(75\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.26 \[ \int F^{c+d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {{\left (F^{d x + c} b + a\right )} {\left (F^{d x + c} b + a\right )}^{n}}{{\left (b d n + b d\right )} \log \left (F\right )} \]

[In]

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

[Out]

(F^(d*x + c)*b + a)*(F^(d*x + c)*b + a)^n/((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. 299 vs. \(2 (22) = 44\).

Time = 1.09 (sec) , antiderivative size = 299, normalized size of antiderivative = 9.65 \[ \int F^{c+d x} \left (a+b F^{c+d x}\right )^n \, dx=\begin {cases} \tilde {\infty } x & \text {for}\: F = 1 \wedge a = 0 \wedge b = 0 \wedge d = 0 \wedge n = -1 \\x \left (a + b\right )^{n} & \text {for}\: F = 1 \\\frac {0^{n} F^{c + d x}}{d \log {\left (F \right )}} & \text {for}\: a = - F^{c + d x} b \\\frac {F^{c + d x} a^{n}}{d \log {\left (F \right )}} & \text {for}\: b = 0 \\F^{c} x \left (F^{c} b + a\right )^{n} & \text {for}\: d = 0 \\\frac {\log {\left (F^{c + d x} + \frac {a}{b} \right )}}{b d \log {\left (F \right )}} & \text {for}\: n = -1 \\\frac {2 F^{c + d x} a b \left (F^{c + d x} b + a\right )^{n}}{F^{c + d x} b^{2} d n \log {\left (F \right )} + F^{c + d x} b^{2} d \log {\left (F \right )} + a b d n \log {\left (F \right )} + a b d \log {\left (F \right )}} + \frac {F^{2 c + 2 d x} b^{2} \left (F^{c + d x} b + a\right )^{n}}{F^{c + d x} b^{2} d n \log {\left (F \right )} + F^{c + d x} b^{2} d \log {\left (F \right )} + a b d n \log {\left (F \right )} + a b d \log {\left (F \right )}} + \frac {a^{2} \left (F^{c + d x} b + a\right )^{n}}{F^{c + d x} b^{2} d n \log {\left (F \right )} + F^{c + d x} b^{2} d \log {\left (F \right )} + a b d n \log {\left (F \right )} + a b d \log {\left (F \right )}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((zoo*x, Eq(F, 1) & Eq(a, 0) & Eq(b, 0) & Eq(d, 0) & Eq(n, -1)), (x*(a + b)**n, Eq(F, 1)), (0**n*F**(
c + d*x)/(d*log(F)), Eq(a, -F**(c + d*x)*b)), (F**(c + d*x)*a**n/(d*log(F)), Eq(b, 0)), (F**c*x*(F**c*b + a)**
n, Eq(d, 0)), (log(F**(c + d*x) + a/b)/(b*d*log(F)), Eq(n, -1)), (2*F**(c + d*x)*a*b*(F**(c + d*x)*b + a)**n/(
F**(c + d*x)*b**2*d*n*log(F) + F**(c + d*x)*b**2*d*log(F) + a*b*d*n*log(F) + a*b*d*log(F)) + F**(2*c + 2*d*x)*
b**2*(F**(c + d*x)*b + a)**n/(F**(c + d*x)*b**2*d*n*log(F) + F**(c + d*x)*b**2*d*log(F) + a*b*d*n*log(F) + a*b
*d*log(F)) + a**2*(F**(c + d*x)*b + a)**n/(F**(c + d*x)*b**2*d*n*log(F) + F**(c + d*x)*b**2*d*log(F) + a*b*d*n
*log(F) + a*b*d*log(F)), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int F^{c+d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {{\left (F^{d x + c} b + a\right )}^{n + 1}}{b d {\left (n + 1\right )} \log \left (F\right )} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.00 \[ \int F^{c+d x} \left (a+b F^{c+d x}\right )^n \, dx=\frac {{\left (F^{d x + c} b + a\right )}^{n + 1}}{b d {\left (n + 1\right )} \log \left (F\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.68 \[ \int F^{c+d x} \left (a+b F^{c+d x}\right )^n \, dx={\left (a+F^{c+d\,x}\,b\right )}^n\,\left (\frac {F^{c+d\,x}}{d\,\ln \left (F\right )\,\left (n+1\right )}+\frac {a}{b\,d\,\ln \left (F\right )\,\left (n+1\right )}\right ) \]

[In]

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

[Out]

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