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

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

[Out]

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

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Rubi [A]  time = 0.0374232, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {12, 2176, 2194} \[ \frac{4 e^{n \cos \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n^2}-\frac{4 \cos \left (\frac{a}{2}+\frac{b x}{2}\right ) e^{n \cos \left (\frac{a}{2}+\frac{b x}{2}\right )}}{b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.176593, size = 36, normalized size = 0.56 \[ -\frac{4 e^{n \cos \left (\frac{1}{2} (a+b x)\right )} \left (n \cos \left (\frac{1}{2} (a+b x)\right )-1\right )}{b n^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.072, size = 124, normalized size = 1.9 \begin{align*} -2\,{\frac{{{\rm e}^{n\cos \left ( a/2 \right ) \cos \left ( 1/2\,bx \right ) -n\sin \left ( a/2 \right ) \sin \left ( 1/2\,bx \right ) }}{{\rm e}^{i/2bx}}{{\rm e}^{i/2a}}}{bn}}-2\,{\frac{{{\rm e}^{n\cos \left ( a/2 \right ) \cos \left ( 1/2\,bx \right ) -n\sin \left ( a/2 \right ) \sin \left ( 1/2\,bx \right ) }}{{\rm e}^{-i/2bx}}{{\rm e}^{-i/2a}}}{bn}}+4\,{\frac{{{\rm e}^{-n \left ( \sin \left ( a/2 \right ) \sin \left ( 1/2\,bx \right ) -\cos \left ( a/2 \right ) \cos \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*cos(1/2*a+1/2*b*x))*sin(b*x+a),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int e^{\left (n \cos \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*cos(1/2*a+1/2*b*x))*sin(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.09486, size = 92, normalized size = 1.44 \begin{align*} -\frac{4 \,{\left (n \cos \left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right ) - 1\right )} e^{\left (n \cos \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*cos(1/2*a+1/2*b*x))*sin(b*x+a),x, algorithm="fricas")

[Out]

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

[Out]

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

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Giac [B]  time = 1.2496, size = 263, normalized size = 4.11 \begin{align*} \frac{4 \,{\left (n e^{\left (-\frac{n \tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )^{2} - n}{\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{n \tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )^{2} - n}{\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} - n e^{\left (-\frac{n \tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )^{2} - n}{\tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )^{2} + 1}\right )} + e^{\left (-\frac{n \tan \left (\frac{1}{4} \, b x + \frac{1}{4} \, a\right )^{2} - n}{\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*cos(1/2*a+1/2*b*x))*sin(b*x+a),x, algorithm="giac")

[Out]

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