3.252 \(\int x (1+e^x \sin (x))^2 \, dx\)

Optimal. Leaf size=128 \[ \frac{x^2}{2}+\frac{1}{8} e^{2 x} x-\frac{3 e^{2 x}}{32}+\frac{1}{4} e^{2 x} x \sin ^2(x)-\frac{1}{16} e^{2 x} \sin ^2(x)+e^x x \sin (x)+\frac{1}{32} e^{2 x} \sin (2 x)-e^x x \cos (x)+e^x \cos (x)-\frac{1}{32} e^{2 x} \cos (2 x)-\frac{1}{4} e^{2 x} x \sin (x) \cos (x)+\frac{1}{16} e^{2 x} \sin (x) \cos (x) \]

[Out]

(-3*E^(2*x))/32 + (E^(2*x)*x)/8 + x^2/2 + E^x*Cos[x] - E^x*x*Cos[x] - (E^(2*x)*Cos[2*x])/32 + E^x*x*Sin[x] + (
E^(2*x)*Cos[x]*Sin[x])/16 - (E^(2*x)*x*Cos[x]*Sin[x])/4 - (E^(2*x)*Sin[x]^2)/16 + (E^(2*x)*x*Sin[x]^2)/4 + (E^
(2*x)*Sin[2*x])/32

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Rubi [A]  time = 0.189774, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {6742, 4432, 4465, 4433, 4434, 2194, 4469, 12} \[ \frac{x^2}{2}+\frac{1}{8} e^{2 x} x-\frac{3 e^{2 x}}{32}+\frac{1}{4} e^{2 x} x \sin ^2(x)-\frac{1}{16} e^{2 x} \sin ^2(x)+e^x x \sin (x)+\frac{1}{32} e^{2 x} \sin (2 x)-e^x x \cos (x)+e^x \cos (x)-\frac{1}{32} e^{2 x} \cos (2 x)-\frac{1}{4} e^{2 x} x \sin (x) \cos (x)+\frac{1}{16} e^{2 x} \sin (x) \cos (x) \]

Antiderivative was successfully verified.

[In]

Int[x*(1 + E^x*Sin[x])^2,x]

[Out]

(-3*E^(2*x))/32 + (E^(2*x)*x)/8 + x^2/2 + E^x*Cos[x] - E^x*x*Cos[x] - (E^(2*x)*Cos[2*x])/32 + E^x*x*Sin[x] + (
E^(2*x)*Cos[x]*Sin[x])/16 - (E^(2*x)*x*Cos[x]*Sin[x])/4 - (E^(2*x)*Sin[x]^2)/16 + (E^(2*x)*x*Sin[x]^2)/4 + (E^
(2*x)*Sin[2*x])/32

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rule 4432

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 4465

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*((f_.)*(x_))^(m_.)*Sin[(d_.) + (e_.)*(x_)]^(n_.), x_Symbol] :> Module[{u
= IntHide[F^(c*(a + b*x))*Sin[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]

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]

Rule 4434

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(n_), x_Symbol] :> Simp[(b*c*Log[F]*F^(c*(a + b*
x))*Sin[d + e*x]^n)/(e^2*n^2 + b^2*c^2*Log[F]^2), x] + (Dist[(n*(n - 1)*e^2)/(e^2*n^2 + b^2*c^2*Log[F]^2), Int
[F^(c*(a + b*x))*Sin[d + e*x]^(n - 2), x], x] - Simp[(e*n*F^(c*(a + b*x))*Cos[d + e*x]*Sin[d + e*x]^(n - 1))/(
e^2*n^2 + b^2*c^2*Log[F]^2), x]) /; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2*n^2 + b^2*c^2*Log[F]^2, 0] && GtQ[
n, 1]

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 4469

Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)]^(m_.), x_Symbol] :
> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g
}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin{align*} \int x \left (1+e^x \sin (x)\right )^2 \, dx &=\int \left (x+2 e^x x \sin (x)+e^{2 x} x \sin ^2(x)\right ) \, dx\\ &=\frac{x^2}{2}+2 \int e^x x \sin (x) \, dx+\int e^{2 x} x \sin ^2(x) \, dx\\ &=\frac{1}{8} e^{2 x} x+\frac{x^2}{2}-e^x x \cos (x)+e^x x \sin (x)-\frac{1}{4} e^{2 x} x \cos (x) \sin (x)+\frac{1}{4} e^{2 x} x \sin ^2(x)-2 \int \left (-\frac{1}{2} e^x \cos (x)+\frac{1}{2} e^x \sin (x)\right ) \, dx-\int \left (\frac{e^{2 x}}{8}-\frac{1}{4} e^{2 x} \cos (x) \sin (x)+\frac{1}{4} e^{2 x} \sin ^2(x)\right ) \, dx\\ &=\frac{1}{8} e^{2 x} x+\frac{x^2}{2}-e^x x \cos (x)+e^x x \sin (x)-\frac{1}{4} e^{2 x} x \cos (x) \sin (x)+\frac{1}{4} e^{2 x} x \sin ^2(x)-\frac{1}{8} \int e^{2 x} \, dx+\frac{1}{4} \int e^{2 x} \cos (x) \sin (x) \, dx-\frac{1}{4} \int e^{2 x} \sin ^2(x) \, dx+\int e^x \cos (x) \, dx-\int e^x \sin (x) \, dx\\ &=-\frac{e^{2 x}}{16}+\frac{1}{8} e^{2 x} x+\frac{x^2}{2}+e^x \cos (x)-e^x x \cos (x)+e^x x \sin (x)+\frac{1}{16} e^{2 x} \cos (x) \sin (x)-\frac{1}{4} e^{2 x} x \cos (x) \sin (x)-\frac{1}{16} e^{2 x} \sin ^2(x)+\frac{1}{4} e^{2 x} x \sin ^2(x)-\frac{1}{16} \int e^{2 x} \, dx+\frac{1}{4} \int \frac{1}{2} e^{2 x} \sin (2 x) \, dx\\ &=-\frac{3 e^{2 x}}{32}+\frac{1}{8} e^{2 x} x+\frac{x^2}{2}+e^x \cos (x)-e^x x \cos (x)+e^x x \sin (x)+\frac{1}{16} e^{2 x} \cos (x) \sin (x)-\frac{1}{4} e^{2 x} x \cos (x) \sin (x)-\frac{1}{16} e^{2 x} \sin ^2(x)+\frac{1}{4} e^{2 x} x \sin ^2(x)+\frac{1}{8} \int e^{2 x} \sin (2 x) \, dx\\ &=-\frac{3 e^{2 x}}{32}+\frac{1}{8} e^{2 x} x+\frac{x^2}{2}+e^x \cos (x)-e^x x \cos (x)-\frac{1}{32} e^{2 x} \cos (2 x)+e^x x \sin (x)+\frac{1}{16} e^{2 x} \cos (x) \sin (x)-\frac{1}{4} e^{2 x} x \cos (x) \sin (x)-\frac{1}{16} e^{2 x} \sin ^2(x)+\frac{1}{4} e^{2 x} x \sin ^2(x)+\frac{1}{32} e^{2 x} \sin (2 x)\\ \end{align*}

Mathematica [A]  time = 0.164996, size = 67, normalized size = 0.52 \[ \frac{1}{8} \left (4 x^2+e^{2 x} (2 x-1)+8 e^x x \sin (x)-e^{2 x} x \cos (2 x)-8 e^x (x-1) \cos (x)-e^{2 x} (2 x-1) \sin (x) \cos (x)\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*(1 + E^x*Sin[x])^2,x]

[Out]

(4*x^2 + E^(2*x)*(-1 + 2*x) - 8*E^x*(-1 + x)*Cos[x] - E^(2*x)*x*Cos[2*x] + 8*E^x*x*Sin[x] - E^(2*x)*(-1 + 2*x)
*Cos[x]*Sin[x])/8

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Maple [A]  time = 0.009, size = 63, normalized size = 0.5 \begin{align*}{\frac{{x}^{2}}{2}}+2\, \left ( -x/2+1/2 \right ){{\rm e}^{x}}\cos \left ( x \right ) +{{\rm e}^{x}}x\sin \left ( x \right ) +{\frac{ \left ({{\rm e}^{x}} \right ) ^{2}x}{4}}-{\frac{ \left ({{\rm e}^{x}} \right ) ^{2}}{8}}-{\frac{x{{\rm e}^{2\,x}}\cos \left ( 2\,x \right ) }{8}}+{\frac{{{\rm e}^{2\,x}}\sin \left ( 2\,x \right ) }{2} \left ( -{\frac{x}{4}}+{\frac{1}{8}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(1+exp(x)*sin(x))^2,x)

[Out]

1/2*x^2+2*(-1/2*x+1/2)*exp(x)*cos(x)+exp(x)*x*sin(x)+1/4*exp(x)^2*x-1/8*exp(x)^2-1/8*x*exp(2*x)*cos(2*x)+1/2*(
-1/4*x+1/8)*exp(2*x)*sin(2*x)

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Maxima [A]  time = 0.97353, size = 78, normalized size = 0.61 \begin{align*} -\frac{1}{8} \, x \cos \left (2 \, x\right ) e^{\left (2 \, x\right )} -{\left (x - 1\right )} \cos \left (x\right ) e^{x} - \frac{1}{16} \,{\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \sin \left (2 \, x\right ) + x e^{x} \sin \left (x\right ) + \frac{1}{2} \, x^{2} + \frac{1}{8} \,{\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(1+exp(x)*sin(x))^2,x, algorithm="maxima")

[Out]

-1/8*x*cos(2*x)*e^(2*x) - (x - 1)*cos(x)*e^x - 1/16*(2*x - 1)*e^(2*x)*sin(2*x) + x*e^x*sin(x) + 1/2*x^2 + 1/8*
(2*x - 1)*e^(2*x)

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Fricas [A]  time = 1.67328, size = 162, normalized size = 1.27 \begin{align*} -{\left (x - 1\right )} \cos \left (x\right ) e^{x} + \frac{1}{2} \, x^{2} - \frac{1}{8} \,{\left (2 \, x \cos \left (x\right )^{2} - 3 \, x + 1\right )} e^{\left (2 \, x\right )} - \frac{1}{8} \,{\left ({\left (2 \, x - 1\right )} \cos \left (x\right ) e^{\left (2 \, x\right )} - 8 \, x e^{x}\right )} \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(1+exp(x)*sin(x))^2,x, algorithm="fricas")

[Out]

-(x - 1)*cos(x)*e^x + 1/2*x^2 - 1/8*(2*x*cos(x)^2 - 3*x + 1)*e^(2*x) - 1/8*((2*x - 1)*cos(x)*e^(2*x) - 8*x*e^x
)*sin(x)

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Sympy [A]  time = 5.38626, size = 109, normalized size = 0.85 \begin{align*} \frac{x^{2}}{2} + \frac{3 x e^{2 x} \sin ^{2}{\left (x \right )}}{8} - \frac{x e^{2 x} \sin{\left (x \right )} \cos{\left (x \right )}}{4} + \frac{x e^{2 x} \cos ^{2}{\left (x \right )}}{8} + x e^{x} \sin{\left (x \right )} - x e^{x} \cos{\left (x \right )} - \frac{e^{2 x} \sin ^{2}{\left (x \right )}}{8} + \frac{e^{2 x} \sin{\left (x \right )} \cos{\left (x \right )}}{8} - \frac{e^{2 x} \cos ^{2}{\left (x \right )}}{8} + e^{x} \cos{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(1+exp(x)*sin(x))**2,x)

[Out]

x**2/2 + 3*x*exp(2*x)*sin(x)**2/8 - x*exp(2*x)*sin(x)*cos(x)/4 + x*exp(2*x)*cos(x)**2/8 + x*exp(x)*sin(x) - x*
exp(x)*cos(x) - exp(2*x)*sin(x)**2/8 + exp(2*x)*sin(x)*cos(x)/8 - exp(2*x)*cos(x)**2/8 + exp(x)*cos(x)

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Giac [A]  time = 1.10757, size = 77, normalized size = 0.6 \begin{align*} \frac{1}{2} \, x^{2} - \frac{1}{16} \,{\left (2 \, x \cos \left (2 \, x\right ) +{\left (2 \, x - 1\right )} \sin \left (2 \, x\right )\right )} e^{\left (2 \, x\right )} + \frac{1}{8} \,{\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} -{\left ({\left (x - 1\right )} \cos \left (x\right ) - x \sin \left (x\right )\right )} e^{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(1+exp(x)*sin(x))^2,x, algorithm="giac")

[Out]

1/2*x^2 - 1/16*(2*x*cos(2*x) + (2*x - 1)*sin(2*x))*e^(2*x) + 1/8*(2*x - 1)*e^(2*x) - ((x - 1)*cos(x) - x*sin(x
))*e^x