3.6.64 \(\int e^x x \cos (x) \, dx\) [564]

Optimal. Leaf size=30 \[ \frac {1}{2} e^x x \cos (x)-\frac {1}{2} e^x \sin (x)+\frac {1}{2} e^x x \sin (x) \]

[Out]

1/2*exp(x)*x*cos(x)-1/2*exp(x)*sin(x)+1/2*exp(x)*x*sin(x)

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Rubi [A]
time = 0.03, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4518, 4554, 4517} \begin {gather*} -\frac {1}{2} e^x \sin (x)+\frac {1}{2} e^x x \sin (x)+\frac {1}{2} e^x x \cos (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^x*x*Cos[x])/2 - (E^x*Sin[x])/2 + (E^x*x*Sin[x])/2

Rule 4517

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(S
in[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] - Simp[e*F^(c*(a + b*x))*(Cos[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]

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]

Rule 4554

Int[Cos[(d_.) + (e_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_.), x_Symbol] :> Module[{u
= IntHide[F^(c*(a + b*x))*Cos[d + e*x]^n, x]}, Dist[(f*x)^m, u, x] - Dist[f*m, Int[(f*x)^(m - 1)*u, x], x]] /
; FreeQ[{F, a, b, c, d, e, f}, x] && IGtQ[n, 0] && GtQ[m, 0]

Rubi steps

\begin {align*} \int e^x x \cos (x) \, dx &=\frac {1}{2} e^x x \cos (x)+\frac {1}{2} e^x x \sin (x)-\int \left (\frac {1}{2} e^x \cos (x)+\frac {1}{2} e^x \sin (x)\right ) \, dx\\ &=\frac {1}{2} e^x x \cos (x)+\frac {1}{2} e^x x \sin (x)-\frac {1}{2} \int e^x \cos (x) \, dx-\frac {1}{2} \int e^x \sin (x) \, dx\\ &=\frac {1}{2} e^x x \cos (x)-\frac {1}{2} e^x \sin (x)+\frac {1}{2} e^x x \sin (x)\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 18, normalized size = 0.60 \begin {gather*} \frac {1}{2} e^x (x \cos (x)+(-1+x) \sin (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(E^x*(x*Cos[x] + (-1 + x)*Sin[x]))/2

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Maple [A]
time = 0.02, size = 20, normalized size = 0.67

method	result	size
default	\(\frac {{\mathrm e}^{x} x \cos \left (x \right )}{2}-\left (-\frac {x}{2}+\frac {1}{2}\right ) {\mathrm e}^{x} \sin \left (x \right )\)	\(20\)
risch	\(\left (\frac {1}{8}-\frac {i}{8}\right ) \left (-1+i+2 x \right ) {\mathrm e}^{\left (1+i\right ) x}+\left (\frac {1}{8}+\frac {i}{8}\right ) \left (-1-i+2 x \right ) {\mathrm e}^{\left (1-i\right ) x}\)	\(36\)
norman	\(\frac {{\mathrm e}^{x} x \tan \left (\frac {x}{2}\right )+\frac {{\mathrm e}^{x} x}{2}-{\mathrm e}^{x} \tan \left (\frac {x}{2}\right )-\frac {{\mathrm e}^{x} x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{2}}{1+\tan ^{2}\left (\frac {x}{2}\right )}\)	\(45\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x)*x*cos(x),x,method=_RETURNVERBOSE)

[Out]

1/2*exp(x)*x*cos(x)-(-1/2*x+1/2)*exp(x)*sin(x)

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Maxima [A]
time = 3.27, size = 17, normalized size = 0.57 \begin {gather*} \frac {1}{2} \, x \cos \left (x\right ) e^{x} + \frac {1}{2} \, {\left (x - 1\right )} e^{x} \sin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x*cos(x),x, algorithm="maxima")

[Out]

1/2*x*cos(x)*e^x + 1/2*(x - 1)*e^x*sin(x)

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Fricas [A]
time = 0.85, size = 17, normalized size = 0.57 \begin {gather*} \frac {1}{2} \, x \cos \left (x\right ) e^{x} + \frac {1}{2} \, {\left (x - 1\right )} e^{x} \sin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x*cos(x),x, algorithm="fricas")

[Out]

1/2*x*cos(x)*e^x + 1/2*(x - 1)*e^x*sin(x)

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Sympy [A]
time = 0.17, size = 27, normalized size = 0.90 \begin {gather*} \frac {x e^{x} \sin {\left (x \right )}}{2} + \frac {x e^{x} \cos {\left (x \right )}}{2} - \frac {e^{x} \sin {\left (x \right )}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x*cos(x),x)

[Out]

x*exp(x)*sin(x)/2 + x*exp(x)*cos(x)/2 - exp(x)*sin(x)/2

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Giac [A]
time = 1.26, size = 15, normalized size = 0.50 \begin {gather*} \frac {1}{2} \, {\left (x \cos \left (x\right ) + {\left (x - 1\right )} \sin \left (x\right )\right )} e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(x)*x*cos(x),x, algorithm="giac")

[Out]

1/2*(x*cos(x) + (x - 1)*sin(x))*e^x

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Mupad [B]
time = 0.07, size = 17, normalized size = 0.57 \begin {gather*} \frac {{\mathrm {e}}^x\,\left (x\,\cos \left (x\right )-\sin \left (x\right )+x\,\sin \left (x\right )\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*exp(x)*cos(x),x)

[Out]

(exp(x)*(x*cos(x) - sin(x) + x*sin(x)))/2

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