### 3.368 $$\int e^{\sin (x)} \sin (2 x) \, dx$$

Optimal. Leaf size=15 $2 e^{\sin (x)} \sin (x)-2 e^{\sin (x)}$

[Out]

-2*E^Sin[x] + 2*E^Sin[x]*Sin[x]

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Rubi [A]  time = 0.0165931, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 9, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.333, Rules used = {12, 2176, 2194} $2 e^{\sin (x)} \sin (x)-2 e^{\sin (x)}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^Sin[x]*Sin[2*x],x]

[Out]

-2*E^Sin[x] + 2*E^Sin[x]*Sin[x]

Rule 12

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

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]

Rubi steps

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

Mathematica [A]  time = 0.0153761, size = 11, normalized size = 0.73 $e^{\sin (x)} (2 \sin (x)-2)$

Antiderivative was successfully veriﬁed.

[In]

Integrate[E^Sin[x]*Sin[2*x],x]

[Out]

E^Sin[x]*(-2 + 2*Sin[x])

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Maple [A]  time = 0.009, size = 14, normalized size = 0.9 \begin{align*} -2\,{{\rm e}^{\sin \left ( x \right ) }}+2\,{{\rm e}^{\sin \left ( x \right ) }}\sin \left ( x \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(exp(sin(x))*sin(2*x),x)

[Out]

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

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Maxima [A]  time = 0.94368, size = 12, normalized size = 0.8 \begin{align*} 2 \,{\left (\sin \left (x\right ) - 1\right )} e^{\sin \left (x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(sin(x) - 1)*e^sin(x)

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Fricas [A]  time = 1.97079, size = 34, normalized size = 2.27 \begin{align*} 2 \,{\left (\sin \left (x\right ) - 1\right )} e^{\sin \left (x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(sin(x) - 1)*e^sin(x)

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Sympy [A]  time = 2.2503, size = 15, normalized size = 1. \begin{align*} 2 e^{\sin{\left (x \right )}} \sin{\left (x \right )} - 2 e^{\sin{\left (x \right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*exp(sin(x))*sin(x) - 2*exp(sin(x))

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Giac [B]  time = 1.09381, size = 721, normalized size = 48.07 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x)^6*tan(x)^2 - 2*e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x
)^5*tan(x)^2 - e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x)^6 + 8*e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/
2*x)^5*tan(x) - 5*e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x)^4*tan(x)^2 + 2*e^(2*tan(1/2*x)/(tan(1/2*x)^2
+ 1))*tan(1/2*x)^5 - 16*e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x)^4*tan(x) + 12*e^(2*tan(1/2*x)/(tan(1/2*
x)^2 + 1))*tan(1/2*x)^3*tan(x)^2 + 5*e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x)^4 - 5*e^(2*tan(1/2*x)/(tan
(1/2*x)^2 + 1))*tan(1/2*x)^2*tan(x)^2 - 12*e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x)^3 + 16*e^(2*tan(1/2*
x)/(tan(1/2*x)^2 + 1))*tan(1/2*x)^2*tan(x) - 2*e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x)*tan(x)^2 + 5*e^(
2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x)^2 - 8*e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x)*tan(x) + e^(2
*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(x)^2 + 2*e^(2*tan(1/2*x)/(tan(1/2*x)^2 + 1))*tan(1/2*x) - e^(2*tan(1/2*x)/
(tan(1/2*x)^2 + 1)))/(tan(1/2*x)^6*tan(x)^2 + tan(1/2*x)^6 + 3*tan(1/2*x)^4*tan(x)^2 + 3*tan(1/2*x)^4 + 3*tan(
1/2*x)^2*tan(x)^2 + 3*tan(1/2*x)^2 + tan(x)^2 + 1)