\(\int e^{n \cos (a+b x)} \sin (2 (a+b x)) \, dx\) [746]

Optimal result

Integrand size = 19, antiderivative size = 43 \[ \int e^{n \cos (a+b x)} \sin (2 (a+b x)) \, dx=\frac {2 e^{n \cos (a+b x)}}{b n^2}-\frac {2 e^{n \cos (a+b x)} \cos (a+b x)}{b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {12, 2207, 2225} \[ \int e^{n \cos (a+b x)} \sin (2 (a+b x)) \, dx=\frac {2 e^{n \cos (a+b x)}}{b n^2}-\frac {2 \cos (a+b x) e^{n \cos (a+b x)}}{b n} \]

[In]

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

[Out]

(2*E^(n*Cos[a + b*x]))/(b*n^2) - (2*E^(n*Cos[a + b*x])*Cos[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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.65 \[ \int e^{n \cos (a+b x)} \sin (2 (a+b x)) \, dx=-\frac {2 e^{n \cos (a+b x)} (-1+n \cos (a+b x))}{b n^2} \]

[In]

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

[Out]

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

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.02 (sec) , antiderivative size = 105, normalized size of antiderivative = 2.44

method	result	size
risch	\(-\frac {{\mathrm e}^{n \cos \left (x b \right ) \cos \left (a \right )-n \sin \left (x b \right ) \sin \left (a \right )} {\mathrm e}^{-i b x} {\mathrm e}^{-i a}}{b n}-\frac {{\mathrm e}^{n \cos \left (x b \right ) \cos \left (a \right )-n \sin \left (x b \right ) \sin \left (a \right )} {\mathrm e}^{i b x} {\mathrm e}^{i a}}{b n}+\frac {2 \,{\mathrm e}^{n \left (\cos \left (x b \right ) \cos \left (a \right )-\sin \left (x b \right ) \sin \left (a \right )\right )}}{b \,n^{2}}\)	\(105\)

[In]

int(exp(n*cos(b*x+a))*sin(2*b*x+2*a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.63 \[ \int e^{n \cos (a+b x)} \sin (2 (a+b x)) \, dx=-\frac {2 \, {\left (n \cos \left (b x + a\right ) - 1\right )} e^{\left (n \cos \left (b x + a\right )\right )}}{b n^{2}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int e^{n \cos (a+b x)} \sin (2 (a+b x)) \, dx=\int e^{n \cos {\left (a + b x \right )}} \sin {\left (2 a + 2 b x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.86 \[ \int e^{n \cos (a+b x)} \sin (2 (a+b x)) \, dx=-\frac {2 \, {\left (n \cos \left (b x + a\right ) e^{\left (n \cos \left (b x + a\right )\right )} - e^{\left (n \cos \left (b x + a\right )\right )}\right )}}{b n^{2}} \]

[In]

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

[Out]

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

Giac [F]

\[ \int e^{n \cos (a+b x)} \sin (2 (a+b x)) \, dx=\int { e^{\left (n \cos \left (b x + a\right )\right )} \sin \left (2 \, b x + 2 \, a\right ) \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.63 \[ \int e^{n \cos (a+b x)} \sin (2 (a+b x)) \, dx=-\frac {2\,{\mathrm {e}}^{n\,\cos \left (a+b\,x\right )}\,\left (n\,\cos \left (a+b\,x\right )-1\right )}{b\,n^2} \]

[In]

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

[Out]

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