3.564 \(\int a^x b^x x \, dx\)
Optimal. Leaf size=31 \[ \frac {x a^x b^x}{\log (a)+\log (b)}-\frac {a^x b^x}{(\log (a)+\log (b))^2} \]
[Out]
-a^x*b^x/(ln(a)+ln(b))^2+a^x*b^x*x/(ln(a)+ln(b))
________________________________________________________________________________________
Rubi [A] time = 0.03, antiderivative size = 31, normalized size of antiderivative = 1.00,
number of steps used = 3, number of rules used = 3, integrand size = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used =
{2287, 2176, 2194} \[ \frac {x a^x b^x}{\log (a)+\log (b)}-\frac {a^x b^x}{(\log (a)+\log (b))^2} \]
Antiderivative was successfully verified.
[In]
Int[a^x*b^x*x,x]
[Out]
-((a^x*b^x)/(Log[a] + Log[b])^2) + (a^x*b^x*x)/(Log[a] + Log[b])
Rule 2176
Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] && !$UseGamma === True
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 x \, dx &=\int e^{x (\log (a)+\log (b))} x \, dx\\ &=\frac {a^x b^x x}{\log (a)+\log (b)}-\frac {\int e^{x (\log (a)+\log (b))} \, dx}{\log (a)+\log (b)}\\ &=-\frac {a^x b^x}{(\log (a)+\log (b))^2}+\frac {a^x b^x x}{\log (a)+\log (b)}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.02, size = 26, normalized size = 0.84 \[ a^x b^x \left (\frac {x}{\log (a)+\log (b)}-\frac {1}{(\log (a)+\log (b))^2}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[a^x*b^x*x,x]
[Out]
a^x*b^x*(-(Log[a] + Log[b])^(-2) + x/(Log[a] + Log[b]))
________________________________________________________________________________________
fricas [A] time = 0.40, size = 34, normalized size = 1.10 \[ \frac {{\left (x \log \relax (a) + x \log \relax (b) - 1\right )} a^{x} b^{x}}{\log \relax (a)^{2} + 2 \, \log \relax (a) \log \relax (b) + \log \relax (b)^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(a^x*b^x*x,x, algorithm="fricas")
[Out]
(x*log(a) + x*log(b) - 1)*a^x*b^x/(log(a)^2 + 2*log(a)*log(b) + log(b)^2)
________________________________________________________________________________________
giac [B] time = 0.29, size = 1020, normalized size = 32.90 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(a^x*b^x*x,x, algorithm="giac")
[Out]
(2*((pi*x*sgn(a) + pi*x*sgn(b) - 2*pi*x)*(pi*log(abs(a))*sgn(a) + pi*log(abs(b))*sgn(a) + pi*log(abs(a))*sgn(b
) + pi*log(abs(b))*sgn(b) - 2*pi*log(abs(a)) - 2*pi*log(abs(b)))/((pi^2*sgn(a)*sgn(b) - 2*pi^2*sgn(a) - 2*pi^2
*sgn(b) + 3*pi^2 - 2*log(abs(a))^2 - 4*log(abs(a))*log(abs(b)) - 2*log(abs(b))^2)^2 + 4*(pi*log(abs(a))*sgn(a)
+ pi*log(abs(b))*sgn(a) + pi*log(abs(a))*sgn(b) + pi*log(abs(b))*sgn(b) - 2*pi*log(abs(a)) - 2*pi*log(abs(b))
)^2) - (pi^2*sgn(a)*sgn(b) - 2*pi^2*sgn(a) - 2*pi^2*sgn(b) + 3*pi^2 - 2*log(abs(a))^2 - 4*log(abs(a))*log(abs(
b)) - 2*log(abs(b))^2)*(x*log(abs(a)) + x*log(abs(b)) - 1)/((pi^2*sgn(a)*sgn(b) - 2*pi^2*sgn(a) - 2*pi^2*sgn(b
) + 3*pi^2 - 2*log(abs(a))^2 - 4*log(abs(a))*log(abs(b)) - 2*log(abs(b))^2)^2 + 4*(pi*log(abs(a))*sgn(a) + pi*
log(abs(b))*sgn(a) + pi*log(abs(a))*sgn(b) + pi*log(abs(b))*sgn(b) - 2*pi*log(abs(a)) - 2*pi*log(abs(b)))^2))*
cos(-1/2*pi*x*sgn(a) - 1/2*pi*x*sgn(b) + pi*x) - ((pi^2*sgn(a)*sgn(b) - 2*pi^2*sgn(a) - 2*pi^2*sgn(b) + 3*pi^2
- 2*log(abs(a))^2 - 4*log(abs(a))*log(abs(b)) - 2*log(abs(b))^2)*(pi*x*sgn(a) + pi*x*sgn(b) - 2*pi*x)/((pi^2*
sgn(a)*sgn(b) - 2*pi^2*sgn(a) - 2*pi^2*sgn(b) + 3*pi^2 - 2*log(abs(a))^2 - 4*log(abs(a))*log(abs(b)) - 2*log(a
bs(b))^2)^2 + 4*(pi*log(abs(a))*sgn(a) + pi*log(abs(b))*sgn(a) + pi*log(abs(a))*sgn(b) + pi*log(abs(b))*sgn(b)
- 2*pi*log(abs(a)) - 2*pi*log(abs(b)))^2) + 4*(pi*log(abs(a))*sgn(a) + pi*log(abs(b))*sgn(a) + pi*log(abs(a))
*sgn(b) + pi*log(abs(b))*sgn(b) - 2*pi*log(abs(a)) - 2*pi*log(abs(b)))*(x*log(abs(a)) + x*log(abs(b)) - 1)/((p
i^2*sgn(a)*sgn(b) - 2*pi^2*sgn(a) - 2*pi^2*sgn(b) + 3*pi^2 - 2*log(abs(a))^2 - 4*log(abs(a))*log(abs(b)) - 2*l
og(abs(b))^2)^2 + 4*(pi*log(abs(a))*sgn(a) + pi*log(abs(b))*sgn(a) + pi*log(abs(a))*sgn(b) + pi*log(abs(b))*sg
n(b) - 2*pi*log(abs(a)) - 2*pi*log(abs(b)))^2))*sin(-1/2*pi*x*sgn(a) - 1/2*pi*x*sgn(b) + pi*x))*e^(x*(log(abs(
a)) + log(abs(b)))) - 1/2*((2*i*x*log(abs(a)) + 2*i*x*log(abs(b)) - pi*x*sgn(a) - pi*x*sgn(b) + 2*pi*x - 2*i)*
e^(1/2*(pi*(sgn(a) - 1) + pi*(sgn(b) - 1))*i*x)/(2*pi*i*log(abs(a))*sgn(a) + 2*pi*i*log(abs(b))*sgn(a) + 2*pi*
i*log(abs(a))*sgn(b) + 2*pi*i*log(abs(b))*sgn(b) - pi^2*sgn(a)*sgn(b) - 4*pi*i*log(abs(a)) - 4*pi*i*log(abs(b)
) + 2*pi^2*sgn(a) + 2*pi^2*sgn(b) - 3*pi^2 + 2*log(abs(a))^2 + 4*log(abs(a))*log(abs(b)) + 2*log(abs(b))^2) +
(2*i*x*log(abs(a)) + 2*i*x*log(abs(b)) + pi*x*sgn(a) + pi*x*sgn(b) - 2*pi*x - 2*i)*e^(-1/2*(pi*(sgn(a) - 1) +
pi*(sgn(b) - 1))*i*x)/(2*pi*i*log(abs(a))*sgn(a) + 2*pi*i*log(abs(b))*sgn(a) + 2*pi*i*log(abs(a))*sgn(b) + 2*p
i*i*log(abs(b))*sgn(b) + pi^2*sgn(a)*sgn(b) - 4*pi*i*log(abs(a)) - 4*pi*i*log(abs(b)) - 2*pi^2*sgn(a) - 2*pi^2
*sgn(b) + 3*pi^2 - 2*log(abs(a))^2 - 4*log(abs(a))*log(abs(b)) - 2*log(abs(b))^2))*e^(x*(log(abs(a)) + log(abs
(b))))/i
________________________________________________________________________________________
maple [A] time = 0.01, size = 25, normalized size = 0.81 \[ \frac {\left (x \ln \relax (a )+x \ln \relax (b )-1\right ) a^{x} b^{x}}{\left (\ln \relax (a )+\ln \relax (b )\right )^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(a^x*b^x*x,x)
[Out]
(x*ln(b)+x*ln(a)-1)*a^x*b^x/(ln(a)+ln(b))^2
________________________________________________________________________________________
maxima [A] time = 0.79, size = 37, normalized size = 1.19 \[ \frac {{\left (x {\left (\log \relax (a) + \log \relax (b)\right )} - 1\right )} e^{\left (x \log \relax (a) + x \log \relax (b)\right )}}{\log \relax (a)^{2} + 2 \, \log \relax (a) \log \relax (b) + \log \relax (b)^{2}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(a^x*b^x*x,x, algorithm="maxima")
[Out]
(x*(log(a) + log(b)) - 1)*e^(x*log(a) + x*log(b))/(log(a)^2 + 2*log(a)*log(b) + log(b)^2)
________________________________________________________________________________________
mupad [B] time = 0.02, size = 23, normalized size = 0.74 \[ \frac {a^x\,b^x\,\left (x\,\left (\ln \relax (a)+\ln \relax (b)\right )-1\right )}{{\left (\ln \relax (a)+\ln \relax (b)\right )}^2} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(a^x*b^x*x,x)
[Out]
(a^x*b^x*(x*(log(a) + log(b)) - 1))/(log(a) + log(b))^2
________________________________________________________________________________________
sympy [A] time = 1.15, size = 97, normalized size = 3.13 \[ \begin {cases} \frac {a^{x} b^{x} x \log {\relax (a )}}{\log {\relax (a )}^{2} + 2 \log {\relax (a )} \log {\relax (b )} + \log {\relax (b )}^{2}} + \frac {a^{x} b^{x} x \log {\relax (b )}}{\log {\relax (a )}^{2} + 2 \log {\relax (a )} \log {\relax (b )} + \log {\relax (b )}^{2}} - \frac {a^{x} b^{x}}{\log {\relax (a )}^{2} + 2 \log {\relax (a )} \log {\relax (b )} + \log {\relax (b )}^{2}} & \text {for}\: a \neq \frac {1}{b} \\\tilde {\infty } b^{x} \left (\frac {1}{b}\right )^{x} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(a**x*b**x*x,x)
[Out]
Piecewise((a**x*b**x*x*log(a)/(log(a)**2 + 2*log(a)*log(b) + log(b)**2) + a**x*b**x*x*log(b)/(log(a)**2 + 2*lo
g(a)*log(b) + log(b)**2) - a**x*b**x/(log(a)**2 + 2*log(a)*log(b) + log(b)**2), Ne(a, 1/b)), (zoo*b**x*(1/b)**
x, True))
________________________________________________________________________________________