\(\int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx\) [38]

Optimal result

Integrand size = 20, antiderivative size = 63 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=-\frac {d e^{a+b x} \cos (2 c+2 d x)}{b^2+4 d^2}+\frac {b e^{a+b x} \sin (2 c+2 d x)}{2 \left (b^2+4 d^2\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4557, 12, 4517} \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=\frac {b e^{a+b x} \sin (2 c+2 d x)}{2 \left (b^2+4 d^2\right )}-\frac {d e^{a+b x} \cos (2 c+2 d x)}{b^2+4 d^2} \]

[In]

Int[E^(a + b*x)*Cos[c + d*x]*Sin[c + d*x],x]

[Out]

-((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])/(2*(b^2 + 4*d^2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4517

Int[(F_)^((c_.)*((a_.) + (b_.)*(x_)))*Sin[(d_.) + (e_.)*(x_)], x_Symbol] :> Simp[b*c*Log[F]*F^(c*(a + b*x))*(S
in[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]
/; FreeQ[{F, a, b, c, d, e}, x] && NeQ[e^2 + b^2*c^2*Log[F]^2, 0]

Rule 4557

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]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{2} e^{a+b x} \sin (2 c+2 d x) \, dx \\ & = \frac {1}{2} \int e^{a+b x} \sin (2 c+2 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 )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=\frac {e^{a+b x} (-2 d \cos (2 (c+d x))+b \sin (2 (c+d x)))}{2 \left (b^2+4 d^2\right )} \]

[In]

Integrate[E^(a + b*x)*Cos[c + d*x]*Sin[c + d*x],x]

[Out]

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

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(\frac {{\mathrm e}^{x b +a} \left (b \sin \left (2 d x +2 c \right )-2 d \cos \left (2 d x +2 c \right )\right )}{2 b^{2}+8 d^{2}}\)	\(45\)
risch	\(-\frac {i {\mathrm e}^{x b +a} \left (4 i d \cos \left (2 d x +2 c \right )-2 i b \sin \left (2 d x +2 c \right )\right )}{4 \left (2 i d +b \right ) \left (2 i d -b \right )}\)	\(55\)
default	\(-\frac {d \,{\mathrm e}^{x b +a} \cos \left (2 d x +2 c \right )}{b^{2}+4 d^{2}}+\frac {b \,{\mathrm e}^{x b +a} \sin \left (2 d x +2 c \right )}{2 b^{2}+8 d^{2}}\)	\(60\)
norman	\(\frac {-\frac {d \,{\mathrm e}^{x b +a}}{b^{2}+4 d^{2}}+\frac {2 b \,{\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{2}+4 d^{2}}-\frac {2 b \,{\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{b^{2}+4 d^{2}}+\frac {6 d \,{\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{2}+4 d^{2}}-\frac {d \,{\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{b^{2}+4 d^{2}}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}\)	\(160\)

[In]

int(exp(b*x+a)*cos(d*x+c)*sin(d*x+c),x,method=_RETURNVERBOSE)

[Out]

exp(b*x+a)*(b*sin(2*d*x+2*c)-2*d*cos(2*d*x+2*c))/(2*b^2+8*d^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.89 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=\frac {b \cos \left (d x + c\right ) e^{\left (b x + a\right )} \sin \left (d x + c\right ) - {\left (2 \, d \cos \left (d x + c\right )^{2} - d\right )} e^{\left (b x + a\right )}}{b^{2} + 4 \, d^{2}} \]

[In]

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

[Out]

(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)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.77 (sec) , antiderivative size = 325, normalized size of antiderivative = 5.16 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=\begin {cases} x e^{a} \sin {\left (c \right )} \cos {\left (c \right )} & \text {for}\: b = 0 \wedge d = 0 \\\frac {i x e^{a} e^{- 2 i d x} \sin ^{2}{\left (c + d x \right )}}{4} + \frac {x e^{a} e^{- 2 i d x} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2} - \frac {i x e^{a} e^{- 2 i d x} \cos ^{2}{\left (c + d x \right )}}{4} + \frac {i e^{a} e^{- 2 i d x} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} & \text {for}\: b = - 2 i d \\- \frac {i x e^{a} e^{2 i d x} \sin ^{2}{\left (c + d x \right )}}{4} + \frac {x e^{a} e^{2 i d x} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{2} + \frac {i x e^{a} e^{2 i d x} \cos ^{2}{\left (c + d x \right )}}{4} - \frac {i e^{a} e^{2 i d x} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{4 d} & \text {for}\: b = 2 i d \\\frac {b e^{a} e^{b x} \sin {\left (c + d x \right )} \cos {\left (c + d x \right )}}{b^{2} + 4 d^{2}} + \frac {d e^{a} e^{b x} \sin ^{2}{\left (c + d x \right )}}{b^{2} + 4 d^{2}} - \frac {d e^{a} e^{b x} \cos ^{2}{\left (c + d x \right )}}{b^{2} + 4 d^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)*sin(d*x+c),x)

[Out]

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)*ex
p(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)*cos(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))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.70 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=-\frac {{\left (2 \, d \cos \left (2 \, d x + 2 \, c\right ) - b \sin \left (2 \, d x + 2 \, c\right )\right )} e^{\left (b x + a\right )}}{2 \, {\left (b^{2} + 4 \, d^{2}\right )}} \]

[In]

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

[Out]

-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)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.87 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=-\frac {1}{2} \, {\left (\frac {2 \, d \cos \left (2 \, d x + 2 \, c\right )}{b^{2} + 4 \, d^{2}} - \frac {b \sin \left (2 \, d x + 2 \, c\right )}{b^{2} + 4 \, d^{2}}\right )} e^{\left (b x + a\right )} \]

[In]

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

[Out]

-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)

Mupad [B] (verification not implemented)

Time = 27.15 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.73 \[ \int e^{a+b x} \cos (c+d x) \sin (c+d x) \, dx=-\frac {{\mathrm {e}}^{a+b\,x}\,\left (2\,d\,\cos \left (2\,c+2\,d\,x\right )-b\,\sin \left (2\,c+2\,d\,x\right )\right )}{2\,\left (b^2+4\,d^2\right )} \]

[In]

int(cos(c + d*x)*exp(a + b*x)*sin(c + d*x),x)

[Out]

-(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))