3.742 \(\int e^{n \sin (a+b x)} \sin (2 (a+b x)) \, dx\)

Optimal. Leaf size=43 \[ \frac {2 \sin (a+b x) e^{n \sin (a+b x)}}{b n}-\frac {2 e^{n \sin (a+b x)}}{b n^2} \]

[Out]

-2*exp(n*sin(b*x+a))/b/n^2+2*exp(n*sin(b*x+a))*sin(b*x+a)/b/n

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Rubi [A]  time = 0.03, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2176, 2194} \[ \frac {2 \sin (a+b x) e^{n \sin (a+b x)}}{b n}-\frac {2 e^{n \sin (a+b x)}}{b n^2} \]

Antiderivative was successfully verified.

[In]

Int[E^(n*Sin[a + b*x])*Sin[2*(a + b*x)],x]

[Out]

(-2*E^(n*Sin[a + b*x]))/(b*n^2) + (2*E^(n*Sin[a + b*x])*Sin[a + b*x])/(b*n)

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

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Mathematica [A]  time = 0.03, size = 28, normalized size = 0.65 \[ \frac {2 e^{n \sin (a+b x)} (n \sin (a+b x)-1)}{b n^2} \]

Antiderivative was successfully verified.

[In]

Integrate[E^(n*Sin[a + b*x])*Sin[2*(a + b*x)],x]

[Out]

(2*E^(n*Sin[a + b*x])*(-1 + n*Sin[a + b*x]))/(b*n^2)

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fricas [A]  time = 0.65, size = 27, normalized size = 0.63 \[ \frac {2 \, {\left (n \sin \left (b x + a\right ) - 1\right )} e^{\left (n \sin \left (b x + a\right )\right )}}{b n^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sin(b*x+a))*sin(2*b*x+2*a),x, algorithm="fricas")

[Out]

2*(n*sin(b*x + a) - 1)*e^(n*sin(b*x + a))/(b*n^2)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{\left (n \sin \left (b x + a\right )\right )} \sin \left (2 \, b x + 2 \, a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sin(b*x+a))*sin(2*b*x+2*a),x, algorithm="giac")

[Out]

integrate(e^(n*sin(b*x + a))*sin(2*b*x + 2*a), x)

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maple [C]  time = 0.00, size = 104, normalized size = 2.42 \[ -\frac {i {\mathrm e}^{n \sin \left (b x \right ) \cos \relax (a )+n \cos \left (b x \right ) \sin \relax (a )} {\mathrm e}^{i b x} {\mathrm e}^{i a}}{n b}+\frac {i {\mathrm e}^{n \sin \left (b x \right ) \cos \relax (a )+n \cos \left (b x \right ) \sin \relax (a )} {\mathrm e}^{-i b x} {\mathrm e}^{-i a}}{n b}-\frac {2 \,{\mathrm e}^{n \left (\sin \left (b x \right ) \cos \relax (a )+\cos \left (b x \right ) \sin \relax (a )\right )}}{n^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*sin(b*x+a))*sin(2*b*x+2*a),x)

[Out]

-I/n/b*exp(n*sin(b*x)*cos(a)+n*cos(b*x)*sin(a))*exp(I*b*x)*exp(I*a)+I/n/b*exp(n*sin(b*x)*cos(a)+n*cos(b*x)*sin
(a))*exp(-I*b*x)*exp(-I*a)-2/n^2/b*exp(n*(sin(b*x)*cos(a)+cos(b*x)*sin(a)))

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maxima [A]  time = 0.35, size = 37, normalized size = 0.86 \[ \frac {2 \, {\left (n e^{\left (n \sin \left (b x + a\right )\right )} \sin \left (b x + a\right ) - e^{\left (n \sin \left (b x + a\right )\right )}\right )}}{b n^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sin(b*x+a))*sin(2*b*x+2*a),x, algorithm="maxima")

[Out]

2*(n*e^(n*sin(b*x + a))*sin(b*x + a) - e^(n*sin(b*x + a)))/(b*n^2)

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mupad [B]  time = 0.00, size = 27, normalized size = 0.63 \[ \frac {2\,{\mathrm {e}}^{n\,\sin \left (a+b\,x\right )}\,\left (n\,\sin \left (a+b\,x\right )-1\right )}{b\,n^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*sin(a + b*x))*sin(2*a + 2*b*x),x)

[Out]

(2*exp(n*sin(a + b*x))*(n*sin(a + b*x) - 1))/(b*n^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int e^{n \sin {\left (a + b x \right )}} \sin {\left (2 a + 2 b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sin(b*x+a))*sin(2*b*x+2*a),x)

[Out]

Integral(exp(n*sin(a + b*x))*sin(2*a + 2*b*x), x)

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