3.79 \(\int a^x \cos (x) \, dx\)

Optimal. Leaf size=31 \[ \frac{a^x \sin (x)}{\log ^2(a)+1}+\frac{a^x \log (a) \cos (x)}{\log ^2(a)+1} \]

[Out]

(a^x*Cos[x]*Log[a])/(1 + Log[a]^2) + (a^x*Sin[x])/(1 + Log[a]^2)

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Rubi [A]  time = 0.0099713, antiderivative size = 31, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 6, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {4433} \[ \frac{a^x \sin (x)}{\log ^2(a)+1}+\frac{a^x \log (a) \cos (x)}{\log ^2(a)+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^x*Cos[x]*Log[a])/(1 + Log[a]^2) + (a^x*Sin[x])/(1 + Log[a]^2)

Rule 4433

Int[Cos[(d_.) + (e_.)*(x_)]*(F_)^((c_.)*((a_.) + (b_.)*(x_))), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*x))*C
os[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x] + Simp[(e*F^(c*(a + b*x))*Sin[d + e*x])/(e^2 + b^2*c^2*Log[F]^2), x]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rubi steps

\begin{align*} \int a^x \cos (x) \, dx &=\frac{a^x \cos (x) \log (a)}{1+\log ^2(a)}+\frac{a^x \sin (x)}{1+\log ^2(a)}\\ \end{align*}

Mathematica [A]  time = 0.0176623, size = 20, normalized size = 0.65 \[ \frac{a^x (\log (a) \cos (x)+\sin (x))}{\log ^2(a)+1} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^x*(Cos[x]*Log[a] + Sin[x]))/(1 + Log[a]^2)

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Maple [B]  time = 0.017, size = 71, normalized size = 2.3 \begin{align*}{ \left ({\frac{\ln \left ( a \right ){{\rm e}^{x\ln \left ( a \right ) }}}{1+ \left ( \ln \left ( a \right ) \right ) ^{2}}}+2\,{\frac{{{\rm e}^{x\ln \left ( a \right ) }}\tan \left ( x/2 \right ) }{1+ \left ( \ln \left ( a \right ) \right ) ^{2}}}-{\frac{\ln \left ( a \right ){{\rm e}^{x\ln \left ( a \right ) }}}{1+ \left ( \ln \left ( a \right ) \right ) ^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}} \right ) \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/(1+ln(a)^2)*ln(a)*exp(x*ln(a))+2/(1+ln(a)^2)*exp(x*ln(a))*tan(1/2*x)-1/(1+ln(a)^2)*ln(a)*exp(x*ln(a))*tan(1
/2*x)^2)/(tan(1/2*x)^2+1)

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Maxima [A]  time = 0.948647, size = 32, normalized size = 1.03 \begin{align*} \frac{a^{x} \cos \left (x\right ) \log \left (a\right ) + a^{x} \sin \left (x\right )}{\log \left (a\right )^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^x*cos(x)*log(a) + a^x*sin(x))/(log(a)^2 + 1)

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Fricas [A]  time = 2.21215, size = 61, normalized size = 1.97 \begin{align*} \frac{{\left (\cos \left (x\right ) \log \left (a\right ) + \sin \left (x\right )\right )} a^{x}}{\log \left (a\right )^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(cos(x)*log(a) + sin(x))*a^x/(log(a)^2 + 1)

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Sympy [A]  time = 1.06125, size = 107, normalized size = 3.45 \begin{align*} \begin{cases} \frac{i x e^{- i x} \sin{\left (x \right )}}{2} + \frac{x e^{- i x} \cos{\left (x \right )}}{2} + \frac{i e^{- i x} \cos{\left (x \right )}}{2} & \text{for}\: a = e^{- i} \\- \frac{i x e^{i x} \sin{\left (x \right )}}{2} + \frac{x e^{i x} \cos{\left (x \right )}}{2} - \frac{i e^{i x} \cos{\left (x \right )}}{2} & \text{for}\: a = e^{i} \\\frac{a^{x} \log{\left (a \right )} \cos{\left (x \right )}}{\log{\left (a \right )}^{2} + 1} + \frac{a^{x} \sin{\left (x \right )}}{\log{\left (a \right )}^{2} + 1} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(a**x*cos(x),x)

[Out]

Piecewise((I*x*exp(-I*x)*sin(x)/2 + x*exp(-I*x)*cos(x)/2 + I*exp(-I*x)*cos(x)/2, Eq(a, exp(-I))), (-I*x*exp(I*
x)*sin(x)/2 + x*exp(I*x)*cos(x)/2 - I*exp(I*x)*cos(x)/2, Eq(a, exp(I))), (a**x*log(a)*cos(x)/(log(a)**2 + 1) +
a**x*sin(x)/(log(a)**2 + 1), True))

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Giac [C]  time = 1.09182, size = 444, normalized size = 14.32 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

abs(a)^x*(2*cos(1/2*pi*x*sgn(a) - 1/2*pi*x + x)*log(abs(a))/((pi - pi*sgn(a) - 2)^2 + 4*log(abs(a))^2) - (pi -
pi*sgn(a) - 2)*sin(1/2*pi*x*sgn(a) - 1/2*pi*x + x)/((pi - pi*sgn(a) - 2)^2 + 4*log(abs(a))^2)) + abs(a)^x*(2*
cos(1/2*pi*x*sgn(a) - 1/2*pi*x - x)*log(abs(a))/((pi - pi*sgn(a) + 2)^2 + 4*log(abs(a))^2) - (pi - pi*sgn(a) +
2)*sin(1/2*pi*x*sgn(a) - 1/2*pi*x - x)/((pi - pi*sgn(a) + 2)^2 + 4*log(abs(a))^2)) - 1/2*I*abs(a)^x*(-2*I*e^(
1/2*I*pi*x*sgn(a) - 1/2*I*pi*x + I*x)/(-2*I*pi + 2*I*pi*sgn(a) + 4*log(abs(a)) + 4*I) + 2*I*e^(-1/2*I*pi*x*sgn
(a) + 1/2*I*pi*x - I*x)/(2*I*pi - 2*I*pi*sgn(a) + 4*log(abs(a)) - 4*I)) - 1/2*I*abs(a)^x*(-2*I*e^(1/2*I*pi*x*s
gn(a) - 1/2*I*pi*x - I*x)/(-2*I*pi + 2*I*pi*sgn(a) + 4*log(abs(a)) - 4*I) + 2*I*e^(-1/2*I*pi*x*sgn(a) + 1/2*I*
pi*x + I*x)/(2*I*pi - 2*I*pi*sgn(a) + 4*log(abs(a)) + 4*I))