3.744 \(\int e^{n \sin (\frac{1}{2} (a+b x))} \sin (a+b x) \, dx\)

Optimal. Leaf size=64 \[ \frac{4 \sin \left (\frac{a}{2}+\frac{b x}{2}\right ) e^{n \sin \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n}-\frac{4 e^{n \sin \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n^2} \]

[Out]

(-4*E^(n*Sin[a/2 + (b*x)/2]))/(b*n^2) + (4*E^(n*Sin[a/2 + (b*x)/2])*Sin[a/2 + (b*x)/2])/(b*n)

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Rubi [A]  time = 0.0384814, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {12, 2176, 2194} \[ \frac{4 \sin \left (\frac{a}{2}+\frac{b x}{2}\right ) e^{n \sin \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n}-\frac{4 e^{n \sin \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*E^(n*Sin[a/2 + (b*x)/2]))/(b*n^2) + (4*E^(n*Sin[a/2 + (b*x)/2])*Sin[a/2 + (b*x)/2])/(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 \left (\frac{1}{2} (a+b x)\right )} \sin (a+b x) \, dx &=\frac{2 \operatorname{Subst}\left (\int 2 e^{n x} x \, dx,x,\sin \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b}\\ &=\frac{4 \operatorname{Subst}\left (\int e^{n x} x \, dx,x,\sin \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b}\\ &=\frac{4 e^{n \sin \left (\frac{a}{2}+\frac{b x}{2}\right )} \sin \left (\frac{a}{2}+\frac{b x}{2}\right )}{b n}-\frac{4 \operatorname{Subst}\left (\int e^{n x} \, dx,x,\sin \left (\frac{a}{2}+\frac{b x}{2}\right )\right )}{b n}\\ &=-\frac{4 e^{n \sin \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n^2}+\frac{4 e^{n \sin \left (\frac{a}{2}+\frac{b x}{2}\right )} \sin \left (\frac{a}{2}+\frac{b x}{2}\right )}{b n}\\ \end{align*}

Mathematica [A]  time = 0.0302343, size = 36, normalized size = 0.56 \[ \frac{4 e^{n \sin \left (\frac{1}{2} (a+b x)\right )} \left (n \sin \left (\frac{1}{2} (a+b x)\right )-1\right )}{b n^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0., size = 122, normalized size = 1.9 \begin{align*}{\frac{2\,i{{\rm e}^{-{\frac{i}{2}}bx}}{{\rm e}^{-{\frac{i}{2}}a}}}{nb}{{\rm e}^{n\sin \left ({\frac{a}{2}} \right ) \cos \left ({\frac{bx}{2}} \right ) +n\cos \left ({\frac{a}{2}} \right ) \sin \left ({\frac{bx}{2}} \right ) }}}-{\frac{2\,i{{\rm e}^{{\frac{i}{2}}bx}}{{\rm e}^{{\frac{i}{2}}a}}}{nb}{{\rm e}^{n\sin \left ({\frac{a}{2}} \right ) \cos \left ({\frac{bx}{2}} \right ) +n\cos \left ({\frac{a}{2}} \right ) \sin \left ({\frac{bx}{2}} \right ) }}}-4\,{\frac{{{\rm e}^{n \left ( \sin \left ( a/2 \right ) \cos \left ( 1/2\,bx \right ) +\cos \left ( a/2 \right ) \sin \left ( 1/2\,bx \right ) \right ) }}}{{n}^{2}b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(n*sin(1/2*a+1/2*b*x))*sin(b*x+a),x)

[Out]

2*I/n/b*exp(n*sin(1/2*a)*cos(1/2*b*x)+n*cos(1/2*a)*sin(1/2*b*x))*exp(-1/2*I*b*x)*exp(-1/2*I*a)-2*I/n/b*exp(n*s
in(1/2*a)*cos(1/2*b*x)+n*cos(1/2*a)*sin(1/2*b*x))*exp(1/2*I*b*x)*exp(1/2*I*a)-4/n^2/b*exp(n*(sin(1/2*a)*cos(1/
2*b*x)+cos(1/2*a)*sin(1/2*b*x)))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{\left (n \sin \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )\right )} \sin \left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sin(1/2*a+1/2*b*x))*sin(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.16646, size = 90, normalized size = 1.41 \begin{align*} \frac{4 \,{\left (n \sin \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) - 1\right )} e^{\left (n \sin \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )\right )}}{b n^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sin(1/2*a+1/2*b*x))*sin(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{n \sin{\left (\frac{a}{2} + \frac{b x}{2} \right )}} \sin{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sin(1/2*a+1/2*b*x))*sin(b*x+a),x)

[Out]

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

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Giac [B]  time = 1.26535, size = 186, normalized size = 2.91 \begin{align*} \frac{4 \,{\left (2 \, n e^{\left (\frac{2 \, n \tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )}{\tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )^{2} + 1}\right )} \tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right ) - e^{\left (\frac{2 \, n \tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )}{\tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )^{2} + 1}\right )} \tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )^{2} - e^{\left (\frac{2 \, n \tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )}{\tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )^{2} + 1}\right )}\right )}}{b n^{2} \tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )^{2} + b n^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(exp(n*sin(1/2*a+1/2*b*x))*sin(b*x+a),x, algorithm="giac")

[Out]

4*(2*n*e^(2*n*tan(1/4*b*x + 1/4*a)/(tan(1/4*b*x + 1/4*a)^2 + 1))*tan(1/4*b*x + 1/4*a) - e^(2*n*tan(1/4*b*x + 1
/4*a)/(tan(1/4*b*x + 1/4*a)^2 + 1))*tan(1/4*b*x + 1/4*a)^2 - e^(2*n*tan(1/4*b*x + 1/4*a)/(tan(1/4*b*x + 1/4*a)
^2 + 1)))/(b*n^2*tan(1/4*b*x + 1/4*a)^2 + b*n^2)