Integrand size = 21, antiderivative size = 64 \[ \int e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \sin (a+b x) \, dx=\frac {4 e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )}}{b n^2}-\frac {4 e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \cos \left (\frac {a}{2}+\frac {b x}{2}\right )}{b n} \] Output:
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
Time = 0.01 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.56 \[ \int e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \sin (a+b x) \, dx=-\frac {4 e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \left (-1+n \cos \left (\frac {1}{2} (a+b x)\right )\right )}{b n^2} \] Input:
Integrate[E^(n*Cos[(a + b*x)/2])*Sin[a + b*x],x]
Output:
(-4*E^(n*Cos[(a + b*x)/2])*(-1 + n*Cos[(a + b*x)/2]))/(b*n^2)
Time = 0.23 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {4879, 27, 2607, 2624}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sin (a+b x) e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \, dx\) |
\(\Big \downarrow \) 4879 |
\(\displaystyle -\frac {2 \int 2 e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \cos \left (\frac {a}{2}+\frac {b x}{2}\right )d\cos \left (\frac {a}{2}+\frac {b x}{2}\right )}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {4 \int e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )} \cos \left (\frac {a}{2}+\frac {b x}{2}\right )d\cos \left (\frac {a}{2}+\frac {b x}{2}\right )}{b}\) |
\(\Big \downarrow \) 2607 |
\(\displaystyle -\frac {4 \left (\frac {\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )}}{n}-\frac {\int e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )}d\cos \left (\frac {a}{2}+\frac {b x}{2}\right )}{n}\right )}{b}\) |
\(\Big \downarrow \) 2624 |
\(\displaystyle -\frac {4 \left (\frac {\cos \left (\frac {a}{2}+\frac {b x}{2}\right ) e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )}}{n}-\frac {e^{n \cos \left (\frac {a}{2}+\frac {b x}{2}\right )}}{n^2}\right )}{b}\) |
Input:
Int[E^(n*Cos[(a + b*x)/2])*Sin[a + b*x],x]
Output:
(-4*(-(E^(n*Cos[a/2 + (b*x)/2])/n^2) + (E^(n*Cos[a/2 + (b*x)/2])*Cos[a/2 + (b*x)/2])/n))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] - Simp[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]
Int[((F_)^(v_))^(n_.), x_Symbol] :> Simp[(F^v)^n/(n*Log[F]*D[v, x]), x] /; FreeQ[{F, n}, x] && LinearQ[v, x]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa ctors[Cos[v], x]}, -d/Coefficient[v, x, 1] Subst[Int[SubstFor[1, Cos[v]/d , u/Sin[v], x], x], x, Cos[v]/d]], x] /; !FalseQ[v] && FunctionOfQ[Nonfree Factors[Cos[v], x], u/Sin[v], x]]
Result contains complex when optimal does not.
Time = 0.93 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.92
method | result | size |
risch | \(-\frac {2 \,{\mathrm e}^{n \cos \left (\frac {a}{2}\right ) \cos \left (\frac {b x}{2}\right )-n \sin \left (\frac {a}{2}\right ) \sin \left (\frac {b x}{2}\right )} {\mathrm e}^{-\frac {i b x}{2}} {\mathrm e}^{-\frac {i a}{2}}}{b n}-\frac {2 \,{\mathrm e}^{n \cos \left (\frac {a}{2}\right ) \cos \left (\frac {b x}{2}\right )-n \sin \left (\frac {a}{2}\right ) \sin \left (\frac {b x}{2}\right )} {\mathrm e}^{\frac {i b x}{2}} {\mathrm e}^{\frac {i a}{2}}}{b n}+\frac {4 \,{\mathrm e}^{n \left (\cos \left (\frac {a}{2}\right ) \cos \left (\frac {b x}{2}\right )-\sin \left (\frac {a}{2}\right ) \sin \left (\frac {b x}{2}\right )\right )}}{b \,n^{2}}\) | \(123\) |
Input:
int(exp(n*cos(1/2*a+1/2*b*x))*sin(b*x+a),x,method=_RETURNVERBOSE)
Output:
-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*(cos(1/2*a)*cos(1/2*b*x )-sin(1/2*a)*sin(1/2*b*x)))
Time = 0.07 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.52 \[ \int e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \sin (a+b x) \, dx=-\frac {4 \, {\left (n \cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right ) - 1\right )} e^{\left (n \cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )}}{b n^{2}} \] Input:
integrate(exp(n*cos(1/2*b*x+1/2*a))*sin(b*x+a),x, algorithm="fricas")
Output:
-4*(n*cos(1/2*b*x + 1/2*a) - 1)*e^(n*cos(1/2*b*x + 1/2*a))/(b*n^2)
\[ \int e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \sin (a+b x) \, dx=\int e^{n \cos {\left (\frac {a}{2} + \frac {b x}{2} \right )}} \sin {\left (a + b x \right )}\, dx \] Input:
integrate(exp(n*cos(1/2*b*x+1/2*a))*sin(b*x+a),x)
Output:
Integral(exp(n*cos(a/2 + b*x/2))*sin(a + b*x), x)
\[ \int e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \sin (a+b x) \, dx=\int { e^{\left (n \cos \left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )\right )} \sin \left (b x + a\right ) \,d x } \] Input:
integrate(exp(n*cos(1/2*b*x+1/2*a))*sin(b*x+a),x, algorithm="maxima")
Output:
integrate(e^(n*cos(1/2*b*x + 1/2*a))*sin(b*x + a), x)
Leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (50) = 100\).
Time = 0.20 (sec) , antiderivative size = 195, normalized size of antiderivative = 3.05 \[ \int e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \sin (a+b x) \, dx=\frac {4 \, {\left (n e^{\left (-\frac {n \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} - n}{\tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + 1}\right )} \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + e^{\left (-\frac {n \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} - n}{\tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + 1}\right )} \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} - n e^{\left (-\frac {n \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} - n}{\tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + 1}\right )} + e^{\left (-\frac {n \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} - n}{\tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + 1}\right )}\right )}}{b n^{2} \tan \left (\frac {1}{4} \, b x + \frac {1}{4} \, a\right )^{2} + b n^{2}} \] Input:
integrate(exp(n*cos(1/2*b*x+1/2*a))*sin(b*x+a),x, algorithm="giac")
Output:
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)
Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.52 \[ \int e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \sin (a+b x) \, dx=-\frac {4\,{\mathrm {e}}^{n\,\cos \left (\frac {a}{2}+\frac {b\,x}{2}\right )}\,\left (n\,\cos \left (\frac {a}{2}+\frac {b\,x}{2}\right )-1\right )}{b\,n^2} \] Input:
int(exp(n*cos(a/2 + (b*x)/2))*sin(a + b*x),x)
Output:
-(4*exp(n*cos(a/2 + (b*x)/2))*(n*cos(a/2 + (b*x)/2) - 1))/(b*n^2)
\[ \int e^{n \cos \left (\frac {1}{2} (a+b x)\right )} \sin (a+b x) \, dx=\int e^{\cos \left (\frac {b x}{2}+\frac {a}{2}\right ) n} \sin \left (b x +a \right )d x \] Input:
int(exp(n*cos(1/2*b*x+1/2*a))*sin(b*x+a),x)
Output:
int(e**(cos((a + b*x)/2)*n)*sin(a + b*x),x)