\(\int e^{n \cos (\frac {a}{2}+\frac {b x}{2})} \sin (a+b x) \, dx\) [747]

Optimal result

Integrand size = 24, antiderivative size = 64 \[ \int e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \sin (a+b x) \, dx=\frac {4 e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )}}{b n^2}-\frac {4 e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \cos \left (\frac {a}{2}+\frac {b x}{2}\right )}{b n} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {12, 2207, 2225} \[ \int e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \sin (a+b x) \, dx=\frac {4 e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )}}{b n^2}-\frac {4 \cos \left (\frac {a}{2}+\frac {b x}{2}\right ) e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )}}{b n} \]

[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 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 {2 \text {Subst}\left (\int 2 e^{n x} x \, dx,x,\cos \left (\frac {a}{2}+\frac {b x}{2}\right )\right )}{b} \\ & = -\frac {4 \text {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 \text {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] (verified)

Time = 0.10 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.56 \[ \int e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \sin (a+b x) \, dx=-\frac {4 e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \left (-1+n \cos \left (\frac {1}{2} (a+b x)\right )\right )}{b n^2} \]

[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)

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 1.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.92

method	result	size
risch	\(-\frac {2 \,{\mathrm e}^{n \cos \left (\frac {a}{2}\right ) \cos \left (\frac {x b}{2}\right )-n \sin \left (\frac {a}{2}\right ) \sin \left (\frac {x b}{2}\right )} {\mathrm e}^{-\frac {i b x}{2}} {\mathrm e}^{-\frac {i a}{2}}}{b n}-\frac {2 \,{\mathrm e}^{n \cos \left (\frac {a}{2}\right ) \cos \left (\frac {x b}{2}\right )-n \sin \left (\frac {a}{2}\right ) \sin \left (\frac {x b}{2}\right )} {\mathrm e}^{\frac {i b x}{2}} {\mathrm e}^{\frac {i a}{2}}}{b n}+\frac {4 \,{\mathrm e}^{n \left (\cos \left (\frac {a}{2}\right ) \cos \left (\frac {x b}{2}\right )-\sin \left (\frac {a}{2}\right ) \sin \left (\frac {x b}{2}\right )\right )}}{b \,n^{2}}\)	\(123\)

[In]

int(exp(n*cos(1/2*a+1/2*x*b))*sin(b*x+a),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.52 \[ \int e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \sin (a+b x) \, dx=-\frac {4 \, {\left (n \cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 1\right )} e^{\left (n \cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )}}{b n^{2}} \]

[In]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

\[ \int e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \sin (a+b x) \, dx=\int { e^{\left (n \cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )} \sin \left (b x + a\right ) \,d x } \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (50) = 100\).

Time = 0.34 (sec) , antiderivative size = 195, normalized size of antiderivative = 3.05 \[ \int e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \sin (a+b x) \, dx=\frac {4 \, {\left (n e^{\left (-\frac {n \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} - n}{\tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + 1}\right )} \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + e^{\left (-\frac {n \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} - n}{\tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + 1}\right )} \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} - n e^{\left (-\frac {n \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} - n}{\tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + 1}\right )} + e^{\left (-\frac {n \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} - n}{\tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + 1}\right )}\right )}}{b n^{2} \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + b n^{2}} \]

[In]

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

Mupad [B] (verification not implemented)

Time = 26.77 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.52 \[ \int e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \sin (a+b x) \, dx=-\frac {4\,{\mathrm {e}}^{n\,\cos \left (\frac {a}{2}+\frac {b\,x}{2}\right )}\,\left (n\,\cos \left (\frac {a}{2}+\frac {b\,x}{2}\right )-1\right )}{b\,n^2} \]

[In]

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

[Out]

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