\(\int e^{a+b x} \cos ^3(c+d x) \sin (c+d x) \, dx\) [44]

Optimal result

Integrand size = 22, antiderivative size = 129 \[ \int e^{a+b x} \cos ^3(c+d x) \sin (c+d x) \, dx=-\frac {d e^{a+b x} \cos (2 c+2 d x)}{2 \left (b^2+4 d^2\right )}-\frac {d e^{a+b x} \cos (4 c+4 d x)}{2 \left (b^2+16 d^2\right )}+\frac {b e^{a+b x} \sin (2 c+2 d x)}{4 \left (b^2+4 d^2\right )}+\frac {b e^{a+b x} \sin (4 c+4 d x)}{8 \left (b^2+16 d^2\right )} \]

[Out]

-1/2*d*exp(b*x+a)*cos(2*d*x+2*c)/(b^2+4*d^2)-1/2*d*exp(b*x+a)*cos(4*d*x+4*c)/(b^2+16*d^2)+1/4*b*exp(b*x+a)*sin
(2*d*x+2*c)/(b^2+4*d^2)+1/8*b*exp(b*x+a)*sin(4*d*x+4*c)/(b^2+16*d^2)

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {4557, 4517} \[ \int e^{a+b x} \cos ^3(c+d x) \sin (c+d x) \, dx=\frac {b e^{a+b x} \sin (2 c+2 d x)}{4 \left (b^2+4 d^2\right )}+\frac {b e^{a+b x} \sin (4 c+4 d x)}{8 \left (b^2+16 d^2\right )}-\frac {d e^{a+b x} \cos (2 c+2 d x)}{2 \left (b^2+4 d^2\right )}-\frac {d e^{a+b x} \cos (4 c+4 d x)}{2 \left (b^2+16 d^2\right )} \]

[In]

Int[E^(a + b*x)*Cos[c + d*x]^3*Sin[c + d*x],x]

[Out]

-1/2*(d*E^(a + b*x)*Cos[2*c + 2*d*x])/(b^2 + 4*d^2) - (d*E^(a + b*x)*Cos[4*c + 4*d*x])/(2*(b^2 + 16*d^2)) + (b
*E^(a + b*x)*Sin[2*c + 2*d*x])/(4*(b^2 + 4*d^2)) + (b*E^(a + b*x)*Sin[4*c + 4*d*x])/(8*(b^2 + 16*d^2))

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 \left (\frac {1}{4} e^{a+b x} \sin (2 c+2 d x)+\frac {1}{8} e^{a+b x} \sin (4 c+4 d x)\right ) \, dx \\ & = \frac {1}{8} \int e^{a+b x} \sin (4 c+4 d x) \, dx+\frac {1}{4} \int e^{a+b x} \sin (2 c+2 d x) \, dx \\ & = -\frac {d e^{a+b x} \cos (2 c+2 d x)}{2 \left (b^2+4 d^2\right )}-\frac {d e^{a+b x} \cos (4 c+4 d x)}{2 \left (b^2+16 d^2\right )}+\frac {b e^{a+b x} \sin (2 c+2 d x)}{4 \left (b^2+4 d^2\right )}+\frac {b e^{a+b x} \sin (4 c+4 d x)}{8 \left (b^2+16 d^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.63 \[ \int e^{a+b x} \cos ^3(c+d x) \sin (c+d x) \, dx=\frac {1}{8} e^{a+b x} \left (\frac {2 (-2 d \cos (2 (c+d x))+b \sin (2 (c+d x)))}{b^2+4 d^2}+\frac {-4 d \cos (4 (c+d x))+b \sin (4 (c+d x))}{b^2+16 d^2}\right ) \]

[In]

Integrate[E^(a + b*x)*Cos[c + d*x]^3*Sin[c + d*x],x]

[Out]

(E^(a + b*x)*((2*(-2*d*Cos[2*(c + d*x)] + b*Sin[2*(c + d*x)]))/(b^2 + 4*d^2) + (-4*d*Cos[4*(c + d*x)] + b*Sin[
4*(c + d*x)])/(b^2 + 16*d^2)))/8

Maple [A] (verified)

Time = 0.83 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.83

method	result	size
parallelrisch	\(\frac {\left (\left (b^{3}+4 b \,d^{2}\right ) \sin \left (4 d x +4 c \right )+\left (-4 b^{2} d -16 d^{3}\right ) \cos \left (4 d x +4 c \right )+2 \left (b^{2}+16 d^{2}\right ) \left (b \sin \left (2 d x +2 c \right )-2 d \cos \left (2 d x +2 c \right )\right )\right ) {\mathrm e}^{x b +a}}{8 b^{4}+160 b^{2} d^{2}+512 d^{4}}\)	\(107\)
default	\(-\frac {d \,{\mathrm e}^{x b +a} \cos \left (2 d x +2 c \right )}{2 \left (b^{2}+4 d^{2}\right )}-\frac {d \,{\mathrm e}^{x b +a} \cos \left (4 d x +4 c \right )}{2 \left (b^{2}+16 d^{2}\right )}+\frac {b \,{\mathrm e}^{x b +a} \sin \left (2 d x +2 c \right )}{4 b^{2}+16 d^{2}}+\frac {b \,{\mathrm e}^{x b +a} \sin \left (4 d x +4 c \right )}{8 b^{2}+128 d^{2}}\)	\(118\)
risch	\(-\frac {i {\mathrm e}^{x b +a} \left (-8 i d \left (b^{2}+4 d^{2}\right ) \cos \left (4 d x +4 c \right )-i \left (-2 b^{3}-8 b \,d^{2}\right ) \sin \left (4 d x +4 c \right )-8 i d \left (b^{2}+16 d^{2}\right ) \cos \left (2 d x +2 c \right )-i \left (-4 b^{3}-64 b \,d^{2}\right ) \sin \left (2 d x +2 c \right )\right )}{16 \left (4 i d +b \right ) \left (2 i d +b \right ) \left (2 i d -b \right ) \left (4 i d -b \right )}\)	\(139\)
norman	\(\frac {-\frac {d \left (b^{2}+10 d^{2}\right ) {\mathrm e}^{x b +a}}{b^{4}+20 b^{2} d^{2}+64 d^{4}}-\frac {6 b \left (b^{2}+2 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{b^{4}+20 b^{2} d^{2}+64 d^{4}}+\frac {6 b \left (b^{2}+2 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{b^{4}+20 b^{2} d^{2}+64 d^{4}}+\frac {2 b \left (b^{2}+10 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{b^{4}+20 b^{2} d^{2}+64 d^{4}}-\frac {2 b \left (b^{2}+10 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{b^{4}+20 b^{2} d^{2}+64 d^{4}}-\frac {30 d \left (b^{2}+2 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{b^{4}+20 b^{2} d^{2}+64 d^{4}}-\frac {d \left (b^{2}+10 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{b^{4}+20 b^{2} d^{2}+64 d^{4}}+\frac {8 d \left (2 b^{2}+11 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{b^{4}+20 b^{2} d^{2}+64 d^{4}}+\frac {8 d \left (2 b^{2}+11 d^{2}\right ) {\mathrm e}^{x b +a} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{b^{4}+20 b^{2} d^{2}+64 d^{4}}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(441\)

[In]

int(exp(b*x+a)*cos(d*x+c)^3*sin(d*x+c),x,method=_RETURNVERBOSE)

[Out]

((b^3+4*b*d^2)*sin(4*d*x+4*c)+(-4*b^2*d-16*d^3)*cos(4*d*x+4*c)+2*(b^2+16*d^2)*(b*sin(2*d*x+2*c)-2*d*cos(2*d*x+
2*c)))*exp(b*x+a)/(8*b^4+160*b^2*d^2+512*d^4)

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.88 \[ \int e^{a+b x} \cos ^3(c+d x) \sin (c+d x) \, dx=\frac {{\left (6 \, b d^{2} \cos \left (d x + c\right ) + {\left (b^{3} + 4 \, b d^{2}\right )} \cos \left (d x + c\right )^{3}\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) + {\left (3 \, b^{2} d \cos \left (d x + c\right )^{2} - 4 \, {\left (b^{2} d + 4 \, d^{3}\right )} \cos \left (d x + c\right )^{4} + 6 \, d^{3}\right )} e^{\left (b x + a\right )}}{b^{4} + 20 \, b^{2} d^{2} + 64 \, d^{4}} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)^3*sin(d*x+c),x, algorithm="fricas")

[Out]

((6*b*d^2*cos(d*x + c) + (b^3 + 4*b*d^2)*cos(d*x + c)^3)*e^(b*x + a)*sin(d*x + c) + (3*b^2*d*cos(d*x + c)^2 -
4*(b^2*d + 4*d^3)*cos(d*x + c)^4 + 6*d^3)*e^(b*x + a))/(b^4 + 20*b^2*d^2 + 64*d^4)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 7.38 (sec) , antiderivative size = 1357, normalized size of antiderivative = 10.52 \[ \int e^{a+b x} \cos ^3(c+d x) \sin (c+d x) \, dx=\text {Too large to display} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)**3*sin(d*x+c),x)

[Out]

Piecewise((x*exp(a)*sin(c)*cos(c)**3, Eq(b, 0) & Eq(d, 0)), (-I*x*exp(a)*exp(-4*I*d*x)*sin(c + d*x)**4/16 - x*
exp(a)*exp(-4*I*d*x)*sin(c + d*x)**3*cos(c + d*x)/4 + 3*I*x*exp(a)*exp(-4*I*d*x)*sin(c + d*x)**2*cos(c + d*x)*
*2/8 + x*exp(a)*exp(-4*I*d*x)*sin(c + d*x)*cos(c + d*x)**3/4 - I*x*exp(a)*exp(-4*I*d*x)*cos(c + d*x)**4/16 - e
xp(a)*exp(-4*I*d*x)*sin(c + d*x)**4/(24*d) + 5*I*exp(a)*exp(-4*I*d*x)*sin(c + d*x)**3*cos(c + d*x)/(48*d) + 11
*I*exp(a)*exp(-4*I*d*x)*sin(c + d*x)*cos(c + d*x)**3/(48*d) + exp(a)*exp(-4*I*d*x)*cos(c + d*x)**4/(24*d), Eq(
b, -4*I*d)), (I*x*exp(a)*exp(-2*I*d*x)*sin(c + d*x)**4/8 + x*exp(a)*exp(-2*I*d*x)*sin(c + d*x)**3*cos(c + d*x)
/4 + x*exp(a)*exp(-2*I*d*x)*sin(c + d*x)*cos(c + d*x)**3/4 - I*x*exp(a)*exp(-2*I*d*x)*cos(c + d*x)**4/8 + exp(
a)*exp(-2*I*d*x)*sin(c + d*x)**4/(16*d) + exp(a)*exp(-2*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**2/(4*d) - I*exp(a
)*exp(-2*I*d*x)*sin(c + d*x)*cos(c + d*x)**3/(6*d) - 7*exp(a)*exp(-2*I*d*x)*cos(c + d*x)**4/(48*d), Eq(b, -2*I
*d)), (-I*x*exp(a)*exp(2*I*d*x)*sin(c + d*x)**4/8 + x*exp(a)*exp(2*I*d*x)*sin(c + d*x)**3*cos(c + d*x)/4 + x*e
xp(a)*exp(2*I*d*x)*sin(c + d*x)*cos(c + d*x)**3/4 + I*x*exp(a)*exp(2*I*d*x)*cos(c + d*x)**4/8 + exp(a)*exp(2*I
*d*x)*sin(c + d*x)**4/(16*d) + exp(a)*exp(2*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**2/(4*d) + I*exp(a)*exp(2*I*d*
x)*sin(c + d*x)*cos(c + d*x)**3/(6*d) - 7*exp(a)*exp(2*I*d*x)*cos(c + d*x)**4/(48*d), Eq(b, 2*I*d)), (I*x*exp(
a)*exp(4*I*d*x)*sin(c + d*x)**4/16 - x*exp(a)*exp(4*I*d*x)*sin(c + d*x)**3*cos(c + d*x)/4 - 3*I*x*exp(a)*exp(4
*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**2/8 + x*exp(a)*exp(4*I*d*x)*sin(c + d*x)*cos(c + d*x)**3/4 + I*x*exp(a)*
exp(4*I*d*x)*cos(c + d*x)**4/16 - exp(a)*exp(4*I*d*x)*sin(c + d*x)**4/(24*d) - 5*I*exp(a)*exp(4*I*d*x)*sin(c +
d*x)**3*cos(c + d*x)/(48*d) - 11*I*exp(a)*exp(4*I*d*x)*sin(c + d*x)*cos(c + d*x)**3/(48*d) + exp(a)*exp(4*I*d
*x)*cos(c + d*x)**4/(24*d), Eq(b, 4*I*d)), (b**3*exp(a)*exp(b*x)*sin(c + d*x)*cos(c + d*x)**3/(b**4 + 20*b**2*
d**2 + 64*d**4) + 3*b**2*d*exp(a)*exp(b*x)*sin(c + d*x)**2*cos(c + d*x)**2/(b**4 + 20*b**2*d**2 + 64*d**4) - b
**2*d*exp(a)*exp(b*x)*cos(c + d*x)**4/(b**4 + 20*b**2*d**2 + 64*d**4) + 6*b*d**2*exp(a)*exp(b*x)*sin(c + d*x)*
*3*cos(c + d*x)/(b**4 + 20*b**2*d**2 + 64*d**4) + 10*b*d**2*exp(a)*exp(b*x)*sin(c + d*x)*cos(c + d*x)**3/(b**4
+ 20*b**2*d**2 + 64*d**4) + 6*d**3*exp(a)*exp(b*x)*sin(c + d*x)**4/(b**4 + 20*b**2*d**2 + 64*d**4) + 12*d**3*
exp(a)*exp(b*x)*sin(c + d*x)**2*cos(c + d*x)**2/(b**4 + 20*b**2*d**2 + 64*d**4) - 10*d**3*exp(a)*exp(b*x)*cos(
c + d*x)**4/(b**4 + 20*b**2*d**2 + 64*d**4), True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 550 vs. \(2 (117) = 234\).

Time = 0.23 (sec) , antiderivative size = 550, normalized size of antiderivative = 4.26 \[ \int e^{a+b x} \cos ^3(c+d x) \sin (c+d x) \, dx=-\frac {{\left (4 \, b^{2} d \cos \left (4 \, c\right ) e^{a} + 16 \, d^{3} \cos \left (4 \, c\right ) e^{a} - b^{3} e^{a} \sin \left (4 \, c\right ) - 4 \, b d^{2} e^{a} \sin \left (4 \, c\right )\right )} \cos \left (4 \, d x\right ) e^{\left (b x\right )} + {\left (4 \, b^{2} d \cos \left (4 \, c\right ) e^{a} + 16 \, d^{3} \cos \left (4 \, c\right ) e^{a} + b^{3} e^{a} \sin \left (4 \, c\right ) + 4 \, b d^{2} e^{a} \sin \left (4 \, c\right )\right )} \cos \left (4 \, d x + 8 \, c\right ) e^{\left (b x\right )} + 2 \, {\left (2 \, b^{2} d \cos \left (4 \, c\right ) e^{a} + 32 \, d^{3} \cos \left (4 \, c\right ) e^{a} + b^{3} e^{a} \sin \left (4 \, c\right ) + 16 \, b d^{2} e^{a} \sin \left (4 \, c\right )\right )} \cos \left (2 \, d x + 6 \, c\right ) e^{\left (b x\right )} + 2 \, {\left (2 \, b^{2} d \cos \left (4 \, c\right ) e^{a} + 32 \, d^{3} \cos \left (4 \, c\right ) e^{a} - b^{3} e^{a} \sin \left (4 \, c\right ) - 16 \, b d^{2} e^{a} \sin \left (4 \, c\right )\right )} \cos \left (2 \, d x - 2 \, c\right ) e^{\left (b x\right )} - {\left (b^{3} \cos \left (4 \, c\right ) e^{a} + 4 \, b d^{2} \cos \left (4 \, c\right ) e^{a} + 4 \, b^{2} d e^{a} \sin \left (4 \, c\right ) + 16 \, d^{3} e^{a} \sin \left (4 \, c\right )\right )} e^{\left (b x\right )} \sin \left (4 \, d x\right ) - {\left (b^{3} \cos \left (4 \, c\right ) e^{a} + 4 \, b d^{2} \cos \left (4 \, c\right ) e^{a} - 4 \, b^{2} d e^{a} \sin \left (4 \, c\right ) - 16 \, d^{3} e^{a} \sin \left (4 \, c\right )\right )} e^{\left (b x\right )} \sin \left (4 \, d x + 8 \, c\right ) - 2 \, {\left (b^{3} \cos \left (4 \, c\right ) e^{a} + 16 \, b d^{2} \cos \left (4 \, c\right ) e^{a} - 2 \, b^{2} d e^{a} \sin \left (4 \, c\right ) - 32 \, d^{3} e^{a} \sin \left (4 \, c\right )\right )} e^{\left (b x\right )} \sin \left (2 \, d x + 6 \, c\right ) - 2 \, {\left (b^{3} \cos \left (4 \, c\right ) e^{a} + 16 \, b d^{2} \cos \left (4 \, c\right ) e^{a} + 2 \, b^{2} d e^{a} \sin \left (4 \, c\right ) + 32 \, d^{3} e^{a} \sin \left (4 \, c\right )\right )} e^{\left (b x\right )} \sin \left (2 \, d x - 2 \, c\right )}{16 \, {\left (b^{4} \cos \left (4 \, c\right )^{2} + b^{4} \sin \left (4 \, c\right )^{2} + 64 \, {\left (\cos \left (4 \, c\right )^{2} + \sin \left (4 \, c\right )^{2}\right )} d^{4} + 20 \, {\left (b^{2} \cos \left (4 \, c\right )^{2} + b^{2} \sin \left (4 \, c\right )^{2}\right )} d^{2}\right )}} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)^3*sin(d*x+c),x, algorithm="maxima")

[Out]

-1/16*((4*b^2*d*cos(4*c)*e^a + 16*d^3*cos(4*c)*e^a - b^3*e^a*sin(4*c) - 4*b*d^2*e^a*sin(4*c))*cos(4*d*x)*e^(b*
x) + (4*b^2*d*cos(4*c)*e^a + 16*d^3*cos(4*c)*e^a + b^3*e^a*sin(4*c) + 4*b*d^2*e^a*sin(4*c))*cos(4*d*x + 8*c)*e
^(b*x) + 2*(2*b^2*d*cos(4*c)*e^a + 32*d^3*cos(4*c)*e^a + b^3*e^a*sin(4*c) + 16*b*d^2*e^a*sin(4*c))*cos(2*d*x +
6*c)*e^(b*x) + 2*(2*b^2*d*cos(4*c)*e^a + 32*d^3*cos(4*c)*e^a - b^3*e^a*sin(4*c) - 16*b*d^2*e^a*sin(4*c))*cos(
2*d*x - 2*c)*e^(b*x) - (b^3*cos(4*c)*e^a + 4*b*d^2*cos(4*c)*e^a + 4*b^2*d*e^a*sin(4*c) + 16*d^3*e^a*sin(4*c))*
e^(b*x)*sin(4*d*x) - (b^3*cos(4*c)*e^a + 4*b*d^2*cos(4*c)*e^a - 4*b^2*d*e^a*sin(4*c) - 16*d^3*e^a*sin(4*c))*e^
(b*x)*sin(4*d*x + 8*c) - 2*(b^3*cos(4*c)*e^a + 16*b*d^2*cos(4*c)*e^a - 2*b^2*d*e^a*sin(4*c) - 32*d^3*e^a*sin(4
*c))*e^(b*x)*sin(2*d*x + 6*c) - 2*(b^3*cos(4*c)*e^a + 16*b*d^2*cos(4*c)*e^a + 2*b^2*d*e^a*sin(4*c) + 32*d^3*e^
a*sin(4*c))*e^(b*x)*sin(2*d*x - 2*c))/(b^4*cos(4*c)^2 + b^4*sin(4*c)^2 + 64*(cos(4*c)^2 + sin(4*c)^2)*d^4 + 20
*(b^2*cos(4*c)^2 + b^2*sin(4*c)^2)*d^2)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.86 \[ \int e^{a+b x} \cos ^3(c+d x) \sin (c+d x) \, dx=-\frac {1}{8} \, {\left (\frac {4 \, d \cos \left (4 \, d x + 4 \, c\right )}{b^{2} + 16 \, d^{2}} - \frac {b \sin \left (4 \, d x + 4 \, c\right )}{b^{2} + 16 \, d^{2}}\right )} e^{\left (b x + a\right )} - \frac {1}{4} \, {\left (\frac {2 \, d \cos \left (2 \, d x + 2 \, c\right )}{b^{2} + 4 \, d^{2}} - \frac {b \sin \left (2 \, d x + 2 \, c\right )}{b^{2} + 4 \, d^{2}}\right )} e^{\left (b x + a\right )} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)^3*sin(d*x+c),x, algorithm="giac")

[Out]

-1/8*(4*d*cos(4*d*x + 4*c)/(b^2 + 16*d^2) - b*sin(4*d*x + 4*c)/(b^2 + 16*d^2))*e^(b*x + a) - 1/4*(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 = 1.09 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.39 \[ \int e^{a+b x} \cos ^3(c+d x) \sin (c+d x) \, dx=-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (2\,d\,x\right )-\sin \left (2\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (2\,c\right )-\sin \left (2\,c\right )\,1{}\mathrm {i}\right )}{8\,\left (2\,d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (4\,d\,x\right )-\sin \left (4\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (4\,c\right )-\sin \left (4\,c\right )\,1{}\mathrm {i}\right )}{16\,\left (4\,d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (2\,d\,x\right )+\sin \left (2\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (2\,c\right )+\sin \left (2\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,\left (b+d\,2{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (4\,d\,x\right )+\sin \left (4\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (4\,c\right )+\sin \left (4\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,\left (b+d\,4{}\mathrm {i}\right )} \]

[In]

int(cos(c + d*x)^3*exp(a + b*x)*sin(c + d*x),x)

[Out]

- (exp(a + b*x)*(cos(2*d*x) - sin(2*d*x)*1i)*(cos(2*c) - sin(2*c)*1i))/(8*(b*1i + 2*d)) - (exp(a + b*x)*(cos(4
*d*x) - sin(4*d*x)*1i)*(cos(4*c) - sin(4*c)*1i))/(16*(b*1i + 4*d)) - (exp(a + b*x)*(cos(2*d*x) + sin(2*d*x)*1i
)*(cos(2*c) + sin(2*c)*1i)*1i)/(8*(b + d*2i)) - (exp(a + b*x)*(cos(4*d*x) + sin(4*d*x)*1i)*(cos(4*c) + sin(4*c
)*1i)*1i)/(16*(b + d*4i))