Integrand size = 20, antiderivative size = 63 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=-\frac {d e^{a+b x} \cos (2 c+2 d x)}{b^2+4 d^2}+\frac {b e^{a+b x} \sin (2 c+2 d x)}{2 \left (b^2+4 d^2\right )} \] Output:
-d*exp(b*x+a)*cos(2*d*x+2*c)/(b^2+4*d^2)+b*exp(b*x+a)*sin(2*d*x+2*c)/(2*b^ 2+8*d^2)
Time = 0.09 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=\frac {e^{a+b x} (-2 d \cos (2 (c+d x))+b \sin (2 (c+d x)))}{2 \left (b^2+4 d^2\right )} \] Input:
Integrate[E^(a + b*x)*Cos[c + d*x]*Sin[c + d*x],x]
Output:
(E^(a + b*x)*(-2*d*Cos[2*(c + d*x)] + b*Sin[2*(c + d*x)]))/(2*(b^2 + 4*d^2 ))
Time = 0.25 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4972, 27, 4932}
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.
\(\displaystyle \int e^{a+b x} \sin (c+d x) \cos (c+d x) \, dx\) |
\(\Big \downarrow \) 4972 |
\(\displaystyle \int \frac {1}{2} e^{a+b x} \sin (2 c+2 d x)dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} \int e^{a+b x} \sin (2 c+2 d x)dx\) |
\(\Big \downarrow \) 4932 |
\(\displaystyle \frac {1}{2} \left (\frac {b e^{a+b x} \sin (2 c+2 d x)}{b^2+4 d^2}-\frac {2 d e^{a+b x} \cos (2 c+2 d x)}{b^2+4 d^2}\right )\) |
Input:
Int[E^(a + b*x)*Cos[c + d*x]*Sin[c + d*x],x]
Output:
((-2*d*E^(a + b*x)*Cos[2*c + 2*d*x])/(b^2 + 4*d^2) + (b*E^(a + b*x)*Sin[2* c + 2*d*x])/(b^2 + 4*d^2))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(Sin[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x ] - Simp[e*F^(c*(a + b*x))*(Cos[d + e*x]/(e^2 + b^2*c^2*Log[F]^2)), x] /; F reeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]
Int[Cos[(f_.) + (g_.)*(x_)]^(n_.)*(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_ .) + (e_.)*(x_)]^(m_.), x_Symbol] :> Int[ExpandTrigReduce[F^(c*(a + b*x)), Sin[d + e*x]^m*Cos[f + g*x]^n, x], x] /; FreeQ[{F, a, b, c, d, e, f, g}, x] && IGtQ[m, 0] && IGtQ[n, 0]
Time = 0.77 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.71
method | result | size |
parallelrisch | \(\frac {{\mathrm e}^{b x +a} \left (-2 d \cos \left (2 d x +2 c \right )+b \sin \left (2 d x +2 c \right )\right )}{2 b^{2}+8 d^{2}}\) | \(45\) |
risch | \(-\frac {i {\mathrm e}^{b x +a} \left (4 i d \cos \left (2 d x +2 c \right )-2 i b \sin \left (2 d x +2 c \right )\right )}{4 \left (2 i d +b \right ) \left (2 i d -b \right )}\) | \(55\) |
default | \(-\frac {d \,{\mathrm e}^{b x +a} \cos \left (2 d x +2 c \right )}{b^{2}+4 d^{2}}+\frac {b \,{\mathrm e}^{b x +a} \sin \left (2 d x +2 c \right )}{2 b^{2}+8 d^{2}}\) | \(60\) |
orering | \(\frac {2 b \,{\mathrm e}^{b x +a} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{b^{2}+4 d^{2}}-\frac {b \,{\mathrm e}^{b x +a} \cos \left (d x +c \right ) \sin \left (d x +c \right )-{\mathrm e}^{b x +a} d \sin \left (d x +c \right )^{2}+{\mathrm e}^{b x +a} \cos \left (d x +c \right )^{2} d}{b^{2}+4 d^{2}}\) | \(101\) |
norman | \(\frac {-\frac {d \,{\mathrm e}^{b x +a}}{b^{2}+4 d^{2}}+\frac {2 b \,{\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2}+4 d^{2}}-\frac {2 b \,{\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{b^{2}+4 d^{2}}+\frac {6 d \,{\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{2}+4 d^{2}}-\frac {d \,{\mathrm e}^{b x +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{b^{2}+4 d^{2}}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\) | \(160\) |
Input:
int(exp(b*x+a)*cos(d*x+c)*sin(d*x+c),x,method=_RETURNVERBOSE)
Output:
exp(b*x+a)*(-2*d*cos(2*d*x+2*c)+b*sin(2*d*x+2*c))/(2*b^2+8*d^2)
Time = 0.07 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=\frac {b \cos \left (d x + c\right ) e^{\left (b x + a\right )} \sin \left (d x + c\right ) - {\left (2 \, d \cos \left (d x + c\right )^{2} - d\right )} e^{\left (b x + a\right )}}{b^{2} + 4 \, d^{2}} \] Input:
integrate(exp(b*x+a)*cos(d*x+c)*sin(d*x+c),x, algorithm="fricas")
Output:
(b*cos(d*x + c)*e^(b*x + a)*sin(d*x + c) - (2*d*cos(d*x + c)^2 - d)*e^(b*x + a))/(b^2 + 4*d^2)
Result contains complex when optimal does not.
Time = 0.93 (sec) , antiderivative size = 325, normalized size of antiderivative = 5.16 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=\begin {cases} x e^{a} \sin {\left (c \right )} \cos {\left (c \right )} & \text {for}\: b = 0 \wedge d = 0 \\\frac {i x e^{a} e^{- 2 i d x} \sin ^{2}{\left (c + d x \right )}}{4} + \frac {x e^{a} e^{- 2 i d x} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2} - \frac {i x e^{a} e^{- 2 i d x} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {i e^{a} e^{- 2 i d x} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} & \text {for}\: b = - 2 i d \\- \frac {i x e^{a} e^{2 i d x} \sin ^{2}{\left (c + d x \right )}}{4} + \frac {x e^{a} e^{2 i d x} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2} + \frac {i x e^{a} e^{2 i d x} \cos ^{2}{\left (c + d x \right )}}{4} - \frac {i e^{a} e^{2 i d x} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} & \text {for}\: b = 2 i d \\\frac {b e^{a} e^{b x} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{b^{2} + 4 d^{2}} + \frac {d e^{a} e^{b x} \sin ^{2}{\left (c + d x \right )}}{b^{2} + 4 d^{2}} - \frac {d e^{a} e^{b x} \cos ^{2}{\left (c + d x \right )}}{b^{2} + 4 d^{2}} & \text {otherwise} \end {cases} \] Input:
integrate(exp(b*x+a)*cos(d*x+c)*sin(d*x+c),x)
Output:
Piecewise((x*exp(a)*sin(c)*cos(c), Eq(b, 0) & Eq(d, 0)), (I*x*exp(a)*exp(- 2*I*d*x)*sin(c + d*x)**2/4 + x*exp(a)*exp(-2*I*d*x)*sin(c + d*x)*cos(c + d *x)/2 - I*x*exp(a)*exp(-2*I*d*x)*cos(c + d*x)**2/4 + I*exp(a)*exp(-2*I*d*x )*sin(c + d*x)*cos(c + d*x)/(4*d), Eq(b, -2*I*d)), (-I*x*exp(a)*exp(2*I*d* x)*sin(c + d*x)**2/4 + x*exp(a)*exp(2*I*d*x)*sin(c + d*x)*cos(c + d*x)/2 + I*x*exp(a)*exp(2*I*d*x)*cos(c + d*x)**2/4 - I*exp(a)*exp(2*I*d*x)*sin(c + d*x)*cos(c + d*x)/(4*d), Eq(b, 2*I*d)), (b*exp(a)*exp(b*x)*sin(c + d*x)*c os(c + d*x)/(b**2 + 4*d**2) + d*exp(a)*exp(b*x)*sin(c + d*x)**2/(b**2 + 4* d**2) - d*exp(a)*exp(b*x)*cos(c + d*x)**2/(b**2 + 4*d**2), True))
Time = 0.06 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=-\frac {{\left (2 \, d \cos \left (2 \, d x + 2 \, c\right ) - b \sin \left (2 \, d x + 2 \, c\right )\right )} e^{\left (b x + a\right )}}{2 \, {\left (b^{2} + 4 \, d^{2}\right )}} \] Input:
integrate(exp(b*x+a)*cos(d*x+c)*sin(d*x+c),x, algorithm="maxima")
Output:
-1/2*(2*d*cos(2*d*x + 2*c) - b*sin(2*d*x + 2*c))*e^(b*x + a)/(b^2 + 4*d^2)
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=-\frac {1}{2} \, {\left (\frac {2 \, d \cos \left (2 \, d x + 2 \, c\right )}{b^{2} + 4 \, d^{2}} - \frac {b \sin \left (2 \, d x + 2 \, c\right )}{b^{2} + 4 \, d^{2}}\right )} e^{\left (b x + a\right )} \] Input:
integrate(exp(b*x+a)*cos(d*x+c)*sin(d*x+c),x, algorithm="giac")
Output:
-1/2*(2*d*cos(2*d*x + 2*c)/(b^2 + 4*d^2) - b*sin(2*d*x + 2*c)/(b^2 + 4*d^2 ))*e^(b*x + a)
Time = 0.55 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.73 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=-\frac {{\mathrm {e}}^{a+b\,x}\,\left (2\,d\,\cos \left (2\,c+2\,d\,x\right )-b\,\sin \left (2\,c+2\,d\,x\right )\right )}{2\,\left (b^2+4\,d^2\right )} \] Input:
int(cos(c + d*x)*exp(a + b*x)*sin(c + d*x),x)
Output:
-(exp(a + b*x)*(2*d*cos(2*c + 2*d*x) - b*sin(2*c + 2*d*x)))/(2*(b^2 + 4*d^ 2))
Time = 0.17 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=\frac {e^{b x +a} \left (-\cos \left (d x +c \right )^{2} d +\cos \left (d x +c \right ) \sin \left (d x +c \right ) b +\sin \left (d x +c \right )^{2} d \right )}{b^{2}+4 d^{2}} \] Input:
int(exp(b*x+a)*cos(d*x+c)*sin(d*x+c),x)
Output:
(e**(a + b*x)*( - cos(c + d*x)**2*d + cos(c + d*x)*sin(c + d*x)*b + sin(c + d*x)**2*d))/(b**2 + 4*d**2)