3.1.79 \(\int a^x \cos (x) \, dx\) [79]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 31, normalized size of antiderivative = 1.00, 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 = {4518} \begin {gather*} \frac {a^x \sin (x)}{\log ^2(a)+1}+\frac {a^x \log (a) \cos (x)}{\log ^2(a)+1} \end {gather*}

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 4518

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*}

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Mathematica [A]
time = 0.01, size = 20, normalized size = 0.65 \begin {gather*} \frac {a^x (\cos (x) \log (a)+\sin (x))}{1+\log ^2(a)} \end {gather*}

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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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 2.47, size = 83, normalized size = 2.68 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {\left (I x \text {Sin}\left [x\right ]+x \text {Cos}\left [x\right ]+\text {Sin}\left [x\right ]\right ) E^{-I x}}{2},a\text {==}E^{-I}\right \},\left \{\frac {\left (-I x \text {Sin}\left [x\right ]+x \text {Cos}\left [x\right ]+\text {Sin}\left [x\right ]\right ) E^{I x}}{2},a\text {==}E^I\right \}\right \},\frac {\text {Cos}\left [x\right ] \text {Log}\left [a\right ] a^x}{1+\text {Log}\left [a\right ]^2}+\frac {a^x \text {Sin}\left [x\right ]}{1+\text {Log}\left [a\right ]^2}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[a^x*Cos[x],x]')

[Out]

Piecewise[{{(I x Sin[x] + x Cos[x] + Sin[x]) E ^ (-I x) / 2, a == E ^ (-I)}, {(-I x Sin[x] + x Cos[x] + Sin[x]
) E ^ (I x) / 2, a == E ^ I}}, Cos[x] Log[a] a ^ x / (1 + Log[a] ^ 2) + a ^ x Sin[x] / (1 + Log[a] ^ 2)]

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Maple [A]
time = 0.03, size = 32, normalized size = 1.03

method	result	size
risch	\(\frac {a^{x} \cos \left (x \right ) \ln \left (a \right )}{1+\ln \left (a \right )^{2}}+\frac {a^{x} \sin \left (x \right )}{1+\ln \left (a \right )^{2}}\)	\(32\)
norman	\(\frac {\frac {\ln \left (a \right ) {\mathrm e}^{x \ln \left (a \right )}}{1+\ln \left (a \right )^{2}}+\frac {2 \,{\mathrm e}^{x \ln \left (a \right )} \tan \left (\frac {x}{2}\right )}{1+\ln \left (a \right )^{2}}-\frac {\ln \left (a \right ) {\mathrm e}^{x \ln \left (a \right )} \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{1+\ln \left (a \right )^{2}}}{1+\tan ^{2}\left (\frac {x}{2}\right )}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(a^x*cos(x),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 24, normalized size = 0.77 \begin {gather*} \frac {a^{x} \cos \left (x\right ) \log \left (a\right ) + a^{x} \sin \left (x\right )}{\log \left (a\right )^{2} + 1} \end {gather*}

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 = 0.37, size = 20, normalized size = 0.65 \begin {gather*} \frac {{\left (\cos \left (x\right ) \log \left (a\right ) + \sin \left (x\right )\right )} a^{x}}{\log \left (a\right )^{2} + 1} \end {gather*}

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 = 0.29, size = 107, normalized size = 3.45 \begin {gather*} \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 {gather*}

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] Result contains complex when optimal does not.
time = 0.00, size = 383, normalized size = 12.35 \begin {gather*} \mathrm {e}^{\ln \left |a\right |\cdot x} \left (\frac {4\cdot 2 \ln \left |a\right |\cdot \cos \left (\frac {\pi x \mathrm {sign}\left (a\right )-\pi x+2 x}{2}\right )}{\left (4 \ln \left |a\right |\right )^{2}+\left (2 \pi \mathrm {sign}\left (a\right )-2 \pi +4\right )^{2}}-\frac {2 \left (-2 \pi \mathrm {sign}\left (a\right )+2 \pi -4\right ) \sin \left (\frac {\pi x \mathrm {sign}\left (a\right )-\pi x+2 x}{2}\right )}{\left (4 \ln \left |a\right |\right )^{2}+\left (2 \pi \mathrm {sign}\left (a\right )-2 \pi +4\right )^{2}}\right )+\mathrm {e}^{\ln \left |a\right |\cdot x} \left (\frac {4\cdot 2 \ln \left |a\right |\cdot \cos \left (\frac {\pi x \mathrm {sign}\left (a\right )-\pi x-2 x}{2}\right )}{\left (4 \ln \left |a\right |\right )^{2}+\left (2 \pi \mathrm {sign}\left (a\right )-2 \pi -4\right )^{2}}-\frac {2 \left (-2 \pi \mathrm {sign}\left (a\right )+2 \pi +4\right ) \sin \left (\frac {\pi x \mathrm {sign}\left (a\right )-\pi x-2 x}{2}\right )}{\left (4 \ln \left |a\right |\right )^{2}+\left (2 \pi \mathrm {sign}\left (a\right )-2 \pi -4\right )^{2}}\right )+\frac {\mathrm {e}^{\ln \left |a\right |\cdot x} \left (-\frac {2 \mathrm {i} \mathrm {e}^{\frac {1}{2} \mathrm {i} \left (\pi x \mathrm {sign}\left (a\right )-\pi x-2 x\right )}}{4 \ln \left |a\right |+\pi \cdot 2 \mathrm {i} \mathrm {sign}\left (a\right )-\pi \cdot 2 \mathrm {i}-4 \mathrm {i}}+\frac {2 \mathrm {i} \mathrm {e}^{-\frac {1}{2} \mathrm {i} \left (\pi x \mathrm {sign}\left (a\right )-\pi x-2 x\right )}}{4 \ln \left |a\right |-\pi \cdot 2 \mathrm {i} \mathrm {sign}\left (a\right )+\pi \cdot 2 \mathrm {i}+4 \mathrm {i}}\right )}{2 \mathrm {i}}+\frac {\mathrm {e}^{\ln \left |a\right |\cdot x} \left (-\frac {2 \mathrm {i} \mathrm {e}^{\frac {1}{2} \mathrm {i} \left (\pi x \mathrm {sign}\left (a\right )-\pi x+2 x\right )}}{4 \ln \left |a\right |+\pi \cdot 2 \mathrm {i} \mathrm {sign}\left (a\right )-\pi \cdot 2 \mathrm {i}+4 \mathrm {i}}+\frac {2 \mathrm {i} \mathrm {e}^{-\frac {1}{2} \mathrm {i} \left (\pi x \mathrm {sign}\left (a\right )-\pi x+2 x\right )}}{4 \ln \left |a\right |-\pi \cdot 2 \mathrm {i} \mathrm {sign}\left (a\right )+\pi \cdot 2 \mathrm {i}-4 \mathrm {i}}\right )}{2 \mathrm {i}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)) + I*abs(a)^x*(I*e^(1/2*I*p
i*x*sgn(a) - 1/2*I*pi*x + I*x)/(-2*I*pi + 2*I*pi*sgn(a) + 4*log(abs(a)) + 4*I) - 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)) + I*abs(a)^x*(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) - I*e^(-1/2*I*pi*x*sgn(a) + 1/2*I*pi*x + I*x)/(2*I*p
i - 2*I*pi*sgn(a) + 4*log(abs(a)) + 4*I))

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Mupad [B]
time = 0.03, size = 20, normalized size = 0.65 \begin {gather*} \frac {a^x\,\left (\sin \left (x\right )+\ln \left (a\right )\,\cos \left (x\right )\right )}{{\ln \left (a\right )}^2+1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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