3.766 \(\int (a F^{c+d x})^m (b F^{e+f x})^n \, dx\)

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

[Out]

(a*F^(d*x+c))^m*(b*F^(f*x+e))^n/(d*m+f*n)/ln(F)

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Rubi [A]  time = 0.10, antiderivative size = 36, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2281, 2227, 2194} \[ \frac {\left (a F^{c+d x}\right )^m \left (b F^{e+f x}\right )^n}{\log (F) (d m+f n)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2227

Int[(u_.)*(F_)^((a_.) + (b_.)*(v_)), x_Symbol] :> Int[u*F^(a + b*NormalizePowerOfLinear[v, x]), x] /; FreeQ[{F
, a, b}, x] && PolynomialQ[u, x] && PowerOfLinearQ[v, x] &&  !PowerOfLinearMatchQ[v, x]

Rule 2281

Int[(u_.)*((a_.)*(F_)^(v_))^(n_), x_Symbol] :> Dist[(a*F^v)^n/F^(n*v), Int[u*F^(n*v), x], x] /; FreeQ[{F, a, n
}, x] &&  !IntegerQ[n]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 36, normalized size = 1.00 \[ \frac {\left (a F^{c+d x}\right )^m \left (b F^{e+f x}\right )^n}{d m \log (F)+f n \log (F)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 46, normalized size = 1.28 \[ \frac {e^{\left ({\left (d m x + c m\right )} \log \relax (F) + {\left (f n x + e n\right )} \log \relax (F) + m \log \relax (a) + n \log \relax (b)\right )}}{{\left (d m + f n\right )} \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.73, size = 47, normalized size = 1.31 \[ \frac {e^{\left (d m x \log \relax (F) + f n x \log \relax (F) + c m \log \relax (F) + n e \log \relax (F) + m \log \relax (a) + n \log \relax (b)\right )}}{d m \log \relax (F) + f n \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 37, normalized size = 1.03 \[ \frac {\left (a \,F^{d x +c}\right )^{m} \left (b \,F^{f x +e}\right )^{n}}{\left (m d +f n \right ) \ln \relax (F )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*F^(d*x+c))^m*(b*F^(f*x+e))^n/(d*m+f*n)/ln(F)

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maxima [A]  time = 0.88, size = 65, normalized size = 1.81 \[ \frac {{\left (F^{e}\right )}^{n} a^{m} b^{n} e^{\left (m \log \left (F^{d x + c}\right ) + n \log \left ({\left (F^{d x + c}\right )}^{\frac {f}{d}}\right )\right )}}{{\left (d m + f n\right )} {\left (F^{\frac {c f}{d}}\right )}^{n} \log \relax (F)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int {\left (F^{c+d\,x}\,a\right )}^m\,{\left (F^{e+f\,x}\,b\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 43.31, size = 143, normalized size = 3.97 \[ \begin {cases} a^{m} b^{n} x & \text {for}\: F = 1 \wedge \left (F = 1 \vee d = - \frac {f n}{m}\right ) \\a^{m} b^{n} x \left (F^{c}\right )^{m} \left (F^{e}\right )^{n} \left (F^{f x}\right )^{n} \left (F^{- \frac {f n x}{m}}\right )^{m} + \frac {a^{m} b^{n} \left (F^{c}\right )^{m} \left (F^{e}\right )^{n} \left (F^{f x}\right )^{n} \left (F^{- \frac {f n x}{m}}\right )^{m}}{f n \log {\relax (F )}} & \text {for}\: d = - \frac {f n}{m} \\\frac {a^{m} b^{n} \left (F^{c}\right )^{m} \left (F^{e}\right )^{n} \left (F^{d x}\right )^{m} \left (F^{f x}\right )^{n}}{d m \log {\relax (F )} + f n \log {\relax (F )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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