3.1.16 \(\int (F^{e (c+d x)})^n (a+b (F^{e (c+d x)})^n)^p \, dx\) [16]

Optimal. Leaf size=41 \[ \frac {\left (a+b \left (F^{e (c+d x)}\right )^n\right )^{1+p}}{b d e n (1+p) \log (F)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {2278, 32} \begin {gather*} \frac {\left (a+b \left (F^{e (c+d x)}\right )^n\right )^{p+1}}{b d e n (p+1) \log (F)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [F]
time = 0.34, size = 0, normalized size = 0.00 \begin {gather*} \int \left (F^{e (c+d x)}\right )^n \left (a+b \left (F^{e (c+d x)}\right )^n\right )^p \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [A]
time = 0.04, size = 42, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {\left (a +b \left (F^{e \left (d x +c \right )}\right )^{n}\right )^{1+p}}{b d e n \left (1+p \right ) \ln \left (F \right )}\)	\(42\)
default	\(\frac {\left (a +b \left (F^{e \left (d x +c \right )}\right )^{n}\right )^{1+p}}{b d e n \left (1+p \right ) \ln \left (F \right )}\)	\(42\)
risch	\(\frac {\left (a +b \left (F^{e \left (d x +c \right )}\right )^{n}\right ) \left (a +b \left (F^{e \left (d x +c \right )}\right )^{n}\right )^{p}}{b \left (1+p \right ) \ln \left (F \right ) e d n}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 40, normalized size = 0.98 \begin {gather*} \frac {{\left (F^{{\left (d x + c\right )} e n} b + a\right )}^{p + 1}}{b d e n {\left (p + 1\right )} \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.35, size = 55, normalized size = 1.34 \begin {gather*} \frac {{\left (F^{{\left (d n x + c n\right )} e} b + a\right )} {\left (F^{{\left (d n x + c n\right )} e} b + a\right )}^{p} e^{\left (-1\right )}}{{\left (b d n p + b d n\right )} \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 90 vs. \(2 (29) = 58\).
time = 223.79, size = 90, normalized size = 2.20 \begin {gather*} \begin {cases} x \left (a + b \left (F^{c e}\right )^{n}\right )^{p} \left (F^{c e}\right )^{n} & \text {for}\: d = 0 \\x \left (a + b\right )^{p} & \text {for}\: e = 0 \vee n = 0 \vee \log {\left (F \right )} = 0 \\\frac {\begin {cases} a^{p} \left (F^{e \left (c + d x\right )}\right )^{n} & \text {for}\: b = 0 \\\frac {\begin {cases} \frac {\left (a + b \left (F^{e \left (c + d x\right )}\right )^{n}\right )^{p + 1}}{p + 1} & \text {for}\: p \neq -1 \\\log {\left (a + b \left (F^{e \left (c + d x\right )}\right )^{n} \right )} & \text {otherwise} \end {cases}}{b} & \text {otherwise} \end {cases}}{d e n \log {\left (F \right )}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 2.94, size = 43, normalized size = 1.05 \begin {gather*} \frac {{\left (F^{d n x e + c n e} b + a\right )}^{p + 1} e^{\left (-1\right )}}{b d n {\left (p + 1\right )} \log \left (F\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 3.51, size = 74, normalized size = 1.80 \begin {gather*} \left (\frac {{\left (F^{c\,e+d\,e\,x}\right )}^n}{d\,e\,n\,\ln \left (F\right )\,\left (p+1\right )}+\frac {a}{b\,d\,e\,n\,\ln \left (F\right )\,\left (p+1\right )}\right )\,{\left (a+b\,{\left (F^{c\,e+d\,e\,x}\right )}^n\right )}^p \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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