3.565 \(\int a^x b^x \, dx\)

Optimal. Leaf size=14 \[ \frac {a^x b^x}{\log (a)+\log (b)} \]

[Out]

a^x*b^x/(ln(a)+ln(b))

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Rubi [A]  time = 0.01, antiderivative size = 14, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2287, 2194} \[ \frac {a^x b^x}{\log (a)+\log (b)} \]

Antiderivative was successfully verified.

[In]

Int[a^x*b^x,x]

[Out]

(a^x*b^x)/(Log[a] + Log[b])

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 2287

Int[(u_.)*(F_)^(v_)*(G_)^(w_), x_Symbol] :> With[{z = v*Log[F] + w*Log[G]}, Int[u*NormalizeIntegrand[E^z, x],
x] /; BinomialQ[z, x] || (PolynomialQ[z, x] && LeQ[Exponent[z, x], 2])] /; FreeQ[{F, G}, x]

Rubi steps

\begin {align*} \int a^x b^x \, dx &=\int e^{x (\log (a)+\log (b))} \, dx\\ &=\frac {a^x b^x}{\log (a)+\log (b)}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 14, normalized size = 1.00 \[ \frac {a^x b^x}{\log (a)+\log (b)} \]

Antiderivative was successfully verified.

[In]

Integrate[a^x*b^x,x]

[Out]

(a^x*b^x)/(Log[a] + Log[b])

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fricas [A]  time = 0.40, size = 14, normalized size = 1.00 \[ \frac {a^{x} b^{x}}{\log \relax (a) + \log \relax (b)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a^x*b^x,x, algorithm="fricas")

[Out]

a^x*b^x/(log(a) + log(b))

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giac [B]  time = 0.25, size = 242, normalized size = 17.29 \[ 2 \, {\left (\frac {2 \, {\left (\log \left ({\left | a \right |}\right ) + \log \left ({\left | b \right |}\right )\right )} \cos \left (-\frac {1}{2} \, \pi x \mathrm {sgn}\relax (a) - \frac {1}{2} \, \pi x \mathrm {sgn}\relax (b) + \pi x\right )}{{\left (2 \, \pi - \pi \mathrm {sgn}\relax (a) - \pi \mathrm {sgn}\relax (b)\right )}^{2} + 4 \, {\left (\log \left ({\left | a \right |}\right ) + \log \left ({\left | b \right |}\right )\right )}^{2}} + \frac {{\left (2 \, \pi - \pi \mathrm {sgn}\relax (a) - \pi \mathrm {sgn}\relax (b)\right )} \sin \left (-\frac {1}{2} \, \pi x \mathrm {sgn}\relax (a) - \frac {1}{2} \, \pi x \mathrm {sgn}\relax (b) + \pi x\right )}{{\left (2 \, \pi - \pi \mathrm {sgn}\relax (a) - \pi \mathrm {sgn}\relax (b)\right )}^{2} + 4 \, {\left (\log \left ({\left | a \right |}\right ) + \log \left ({\left | b \right |}\right )\right )}^{2}}\right )} e^{\left (x {\left (\log \left ({\left | a \right |}\right ) + \log \left ({\left | b \right |}\right )\right )}\right )} - \frac {{\left (\frac {i e^{\left (\frac {1}{2} \, {\left (\pi {\left (\mathrm {sgn}\relax (a) - 1\right )} + \pi {\left (\mathrm {sgn}\relax (b) - 1\right )}\right )} i x\right )}}{\pi i \mathrm {sgn}\relax (a) + \pi i \mathrm {sgn}\relax (b) - 2 \, \pi i + 2 \, \log \left ({\left | a \right |}\right ) + 2 \, \log \left ({\left | b \right |}\right )} + \frac {i e^{\left (-\frac {1}{2} \, {\left (\pi {\left (\mathrm {sgn}\relax (a) - 1\right )} + \pi {\left (\mathrm {sgn}\relax (b) - 1\right )}\right )} i x\right )}}{\pi i \mathrm {sgn}\relax (a) + \pi i \mathrm {sgn}\relax (b) - 2 \, \pi i - 2 \, \log \left ({\left | a \right |}\right ) - 2 \, \log \left ({\left | b \right |}\right )}\right )} e^{\left (x {\left (\log \left ({\left | a \right |}\right ) + \log \left ({\left | b \right |}\right )\right )}\right )}}{i} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a^x*b^x,x, algorithm="giac")

[Out]

2*(2*(log(abs(a)) + log(abs(b)))*cos(-1/2*pi*x*sgn(a) - 1/2*pi*x*sgn(b) + pi*x)/((2*pi - pi*sgn(a) - pi*sgn(b)
)^2 + 4*(log(abs(a)) + log(abs(b)))^2) + (2*pi - pi*sgn(a) - pi*sgn(b))*sin(-1/2*pi*x*sgn(a) - 1/2*pi*x*sgn(b)
+ pi*x)/((2*pi - pi*sgn(a) - pi*sgn(b))^2 + 4*(log(abs(a)) + log(abs(b)))^2))*e^(x*(log(abs(a)) + log(abs(b))
)) - (i*e^(1/2*(pi*(sgn(a) - 1) + pi*(sgn(b) - 1))*i*x)/(pi*i*sgn(a) + pi*i*sgn(b) - 2*pi*i + 2*log(abs(a)) +
2*log(abs(b))) + i*e^(-1/2*(pi*(sgn(a) - 1) + pi*(sgn(b) - 1))*i*x)/(pi*i*sgn(a) + pi*i*sgn(b) - 2*pi*i - 2*lo
g(abs(a)) - 2*log(abs(b))))*e^(x*(log(abs(a)) + log(abs(b))))/i

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maple [A]  time = 0.00, size = 15, normalized size = 1.07 \[ \frac {a^{x} b^{x}}{\ln \relax (a )+\ln \relax (b )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a^x*b^x,x)

[Out]

a^x*b^x/(ln(a)+ln(b))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a^x*b^x,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(log(b)/log(a)>0)', see `assume
?` for more details)Is log(b)/log(a) equal to -1?

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mupad [B]  time = 3.60, size = 14, normalized size = 1.00 \[ \frac {a^x\,b^x}{\ln \relax (a)+\ln \relax (b)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a^x*b^x,x)

[Out]

(a^x*b^x)/(log(a) + log(b))

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sympy [A]  time = 0.57, size = 24, normalized size = 1.71 \[ \begin {cases} \frac {a^{x} b^{x}}{\log {\relax (a )} + \log {\relax (b )}} & \text {for}\: a \neq \frac {1}{b} \\\tilde {\infty } b^{x} \left (\frac {1}{b}\right )^{x} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a**x*b**x,x)

[Out]

Piecewise((a**x*b**x/(log(a) + log(b)), Ne(a, 1/b)), (zoo*b**x*(1/b)**x, True))

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