\(\int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx\) [43]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [C] (verification not implemented)
   Maxima [B] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 24, antiderivative size = 183 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=-\frac {d e^{a+b x} \cos (c+d x)}{8 \left (b^2+d^2\right )}-\frac {3 d e^{a+b x} \cos (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}+\frac {5 d e^{a+b x} \cos (5 c+5 d x)}{16 \left (b^2+25 d^2\right )}+\frac {b e^{a+b x} \sin (c+d x)}{8 \left (b^2+d^2\right )}+\frac {b e^{a+b x} \sin (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {b e^{a+b x} \sin (5 c+5 d x)}{16 \left (b^2+25 d^2\right )} \]

[Out]

-1/8*d*exp(b*x+a)*cos(d*x+c)/(b^2+d^2)-3/16*d*exp(b*x+a)*cos(3*d*x+3*c)/(b^2+9*d^2)+5/16*d*exp(b*x+a)*cos(5*d*
x+5*c)/(b^2+25*d^2)+1/8*b*exp(b*x+a)*sin(d*x+c)/(b^2+d^2)+1/16*b*exp(b*x+a)*sin(3*d*x+3*c)/(b^2+9*d^2)-1/16*b*
exp(b*x+a)*sin(5*d*x+5*c)/(b^2+25*d^2)

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {4557, 4517} \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\frac {b e^{a+b x} \sin (c+d x)}{8 \left (b^2+d^2\right )}+\frac {b e^{a+b x} \sin (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {b e^{a+b x} \sin (5 c+5 d x)}{16 \left (b^2+25 d^2\right )}-\frac {d e^{a+b x} \cos (c+d x)}{8 \left (b^2+d^2\right )}-\frac {3 d e^{a+b x} \cos (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}+\frac {5 d e^{a+b x} \cos (5 c+5 d x)}{16 \left (b^2+25 d^2\right )} \]

[In]

Int[E^(a + b*x)*Cos[c + d*x]^2*Sin[c + d*x]^3,x]

[Out]

-1/8*(d*E^(a + b*x)*Cos[c + d*x])/(b^2 + d^2) - (3*d*E^(a + b*x)*Cos[3*c + 3*d*x])/(16*(b^2 + 9*d^2)) + (5*d*E
^(a + b*x)*Cos[5*c + 5*d*x])/(16*(b^2 + 25*d^2)) + (b*E^(a + b*x)*Sin[c + d*x])/(8*(b^2 + d^2)) + (b*E^(a + b*
x)*Sin[3*c + 3*d*x])/(16*(b^2 + 9*d^2)) - (b*E^(a + b*x)*Sin[5*c + 5*d*x])/(16*(b^2 + 25*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}{8} e^{a+b x} \sin (c+d x)+\frac {1}{16} e^{a+b x} \sin (3 c+3 d x)-\frac {1}{16} e^{a+b x} \sin (5 c+5 d x)\right ) \, dx \\ & = \frac {1}{16} \int e^{a+b x} \sin (3 c+3 d x) \, dx-\frac {1}{16} \int e^{a+b x} \sin (5 c+5 d x) \, dx+\frac {1}{8} \int e^{a+b x} \sin (c+d x) \, dx \\ & = -\frac {d e^{a+b x} \cos (c+d x)}{8 \left (b^2+d^2\right )}-\frac {3 d e^{a+b x} \cos (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}+\frac {5 d e^{a+b x} \cos (5 c+5 d x)}{16 \left (b^2+25 d^2\right )}+\frac {b e^{a+b x} \sin (c+d x)}{8 \left (b^2+d^2\right )}+\frac {b e^{a+b x} \sin (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {b e^{a+b x} \sin (5 c+5 d x)}{16 \left (b^2+25 d^2\right )} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.60 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\frac {1}{16} e^{a+b x} \left (\frac {2 (-d \cos (c+d x)+b \sin (c+d x))}{b^2+d^2}+\frac {-3 d \cos (3 (c+d x))+b \sin (3 (c+d x))}{b^2+9 d^2}+\frac {5 d \cos (5 (c+d x))-b \sin (5 (c+d x))}{b^2+25 d^2}\right ) \]

[In]

Integrate[E^(a + b*x)*Cos[c + d*x]^2*Sin[c + d*x]^3,x]

[Out]

(E^(a + b*x)*((2*(-(d*Cos[c + d*x]) + b*Sin[c + d*x]))/(b^2 + d^2) + (-3*d*Cos[3*(c + d*x)] + b*Sin[3*(c + d*x
)])/(b^2 + 9*d^2) + (5*d*Cos[5*(c + d*x)] - b*Sin[5*(c + d*x)])/(b^2 + 25*d^2)))/16

Maple [A] (verified)

Time = 1.21 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.91

method result size
default \(-\frac {d \,{\mathrm e}^{x b +a} \cos \left (d x +c \right )}{8 \left (b^{2}+d^{2}\right )}-\frac {3 d \,{\mathrm e}^{x b +a} \cos \left (3 d x +3 c \right )}{16 \left (b^{2}+9 d^{2}\right )}+\frac {5 d \,{\mathrm e}^{x b +a} \cos \left (5 d x +5 c \right )}{16 \left (b^{2}+25 d^{2}\right )}+\frac {b \,{\mathrm e}^{x b +a} \sin \left (d x +c \right )}{8 b^{2}+8 d^{2}}+\frac {b \,{\mathrm e}^{x b +a} \sin \left (3 d x +3 c \right )}{16 b^{2}+144 d^{2}}-\frac {b \,{\mathrm e}^{x b +a} \sin \left (5 d x +5 c \right )}{16 \left (b^{2}+25 d^{2}\right )}\) \(166\)
parallelrisch \(\frac {{\mathrm e}^{x b +a} \left (3 \left (-\frac {1}{2} b^{4} d -13 d^{3} b^{2}-\frac {25}{2} d^{5}\right ) \cos \left (3 d x +3 c \right )+5 \left (\frac {1}{2} b^{4} d +5 d^{3} b^{2}+\frac {9}{2} d^{5}\right ) \cos \left (5 d x +5 c \right )+\left (\frac {1}{2} b^{5}+13 d^{2} b^{3}+\frac {25}{2} d^{4} b \right ) \sin \left (3 d x +3 c \right )+\left (b^{2}+9 d^{2}\right ) \left (\frac {\left (-b^{3}-b \,d^{2}\right ) \sin \left (5 d x +5 c \right )}{2}+\left (b^{2}+25 d^{2}\right ) \left (b \sin \left (d x +c \right )-d \cos \left (d x +c \right )\right )\right )\right )}{8 b^{6}+280 b^{4} d^{2}+2072 b^{2} d^{4}+1800 d^{6}}\) \(191\)
risch \(\frac {i {\mathrm e}^{x b +a} \left (i \left (4 b^{5}+136 d^{2} b^{3}+900 d^{4} b \right ) \sin \left (d x +c \right )-4 i d \left (b^{4}+34 b^{2} d^{2}+225 d^{4}\right ) \cos \left (d x +c \right )+10 i d \left (b^{4}+10 b^{2} d^{2}+9 d^{4}\right ) \cos \left (5 d x +5 c \right )-i \left (2 b^{5}+20 d^{2} b^{3}+18 d^{4} b \right ) \sin \left (5 d x +5 c \right )-6 i d \left (b^{4}+26 b^{2} d^{2}+25 d^{4}\right ) \cos \left (3 d x +3 c \right )-i \left (-2 b^{5}-52 d^{2} b^{3}-50 d^{4} b \right ) \sin \left (3 d x +3 c \right )\right )}{32 \left (5 i d +b \right ) \left (3 i d +b \right ) \left (i d +b \right ) \left (i d -b \right ) \left (3 i d -b \right ) \left (5 i d -b \right )}\) \(245\)

[In]

int(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^3,x,method=_RETURNVERBOSE)

[Out]

-1/8*d*exp(b*x+a)*cos(d*x+c)/(b^2+d^2)-3/16*d*exp(b*x+a)*cos(3*d*x+3*c)/(b^2+9*d^2)+5/16*d*exp(b*x+a)*cos(5*d*
x+5*c)/(b^2+25*d^2)+1/8*b*exp(b*x+a)*sin(d*x+c)/(b^2+d^2)+1/16*b*exp(b*x+a)*sin(3*d*x+3*c)/(b^2+9*d^2)-1/16*b*
exp(b*x+a)*sin(5*d*x+5*c)/(b^2+25*d^2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.10 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\frac {{\left (2 \, b^{3} d^{2} + 26 \, b d^{4} - {\left (b^{5} + 10 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cos \left (d x + c\right )^{4} + {\left (b^{5} + 14 \, b^{3} d^{2} + 13 \, b d^{4}\right )} \cos \left (d x + c\right )^{2}\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) + {\left (5 \, {\left (b^{4} d + 10 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cos \left (d x + c\right )^{5} - {\left (7 \, b^{4} d + 82 \, b^{2} d^{3} + 75 \, d^{5}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (b^{4} d + 13 \, b^{2} d^{3}\right )} \cos \left (d x + c\right )\right )} e^{\left (b x + a\right )}}{b^{6} + 35 \, b^{4} d^{2} + 259 \, b^{2} d^{4} + 225 \, d^{6}} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^3,x, algorithm="fricas")

[Out]

((2*b^3*d^2 + 26*b*d^4 - (b^5 + 10*b^3*d^2 + 9*b*d^4)*cos(d*x + c)^4 + (b^5 + 14*b^3*d^2 + 13*b*d^4)*cos(d*x +
 c)^2)*e^(b*x + a)*sin(d*x + c) + (5*(b^4*d + 10*b^2*d^3 + 9*d^5)*cos(d*x + c)^5 - (7*b^4*d + 82*b^2*d^3 + 75*
d^5)*cos(d*x + c)^3 + 2*(b^4*d + 13*b^2*d^3)*cos(d*x + c))*e^(b*x + a))/(b^6 + 35*b^4*d^2 + 259*b^2*d^4 + 225*
d^6)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 25.82 (sec) , antiderivative size = 2958, normalized size of antiderivative = 16.16 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\text {Too large to display} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)**2*sin(d*x+c)**3,x)

[Out]

Piecewise((x*exp(a)*sin(c)**3*cos(c)**2, Eq(b, 0) & Eq(d, 0)), (-x*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**5/32 + 5
*I*x*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**4*cos(c + d*x)/32 + 5*x*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**3*cos(c + d
*x)**2/16 - 5*I*x*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**3/16 - 5*x*exp(a)*exp(-5*I*d*x)*sin(c + d
*x)*cos(c + d*x)**4/32 + I*x*exp(a)*exp(-5*I*d*x)*cos(c + d*x)**5/32 + 31*I*exp(a)*exp(-5*I*d*x)*sin(c + d*x)*
*5/(960*d) + 25*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**4*cos(c + d*x)/(192*d) - I*exp(a)*exp(-5*I*d*x)*sin(c + d*x
)**3*cos(c + d*x)**2/(6*d) - exp(a)*exp(-5*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) + 41*I*exp(a)*exp(-5*I
*d*x)*sin(c + d*x)*cos(c + d*x)**4/(192*d) + 47*exp(a)*exp(-5*I*d*x)*cos(c + d*x)**5/(960*d), Eq(b, -5*I*d)),
(-x*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**5/32 + 3*I*x*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**4*cos(c + d*x)/32 + x*e
xp(a)*exp(-3*I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/16 + I*x*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**2*cos(c + d*x)
**3/16 + 3*x*exp(a)*exp(-3*I*d*x)*sin(c + d*x)*cos(c + d*x)**4/32 - I*x*exp(a)*exp(-3*I*d*x)*cos(c + d*x)**5/3
2 - 7*I*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**5/(192*d) - 9*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**4*cos(c + d*x)/(64
*d) + I*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/(6*d) - exp(a)*exp(-3*I*d*x)*sin(c + d*x)**2*cos(
c + d*x)**3/(3*d) + 25*I*exp(a)*exp(-3*I*d*x)*sin(c + d*x)*cos(c + d*x)**4/(64*d) + 23*exp(a)*exp(-3*I*d*x)*co
s(c + d*x)**5/(192*d), Eq(b, -3*I*d)), (x*exp(a)*exp(-I*d*x)*sin(c + d*x)**5/16 - I*x*exp(a)*exp(-I*d*x)*sin(c
 + d*x)**4*cos(c + d*x)/16 + x*exp(a)*exp(-I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/8 - I*x*exp(a)*exp(-I*d*x)*s
in(c + d*x)**2*cos(c + d*x)**3/8 + x*exp(a)*exp(-I*d*x)*sin(c + d*x)*cos(c + d*x)**4/16 - I*x*exp(a)*exp(-I*d*
x)*cos(c + d*x)**5/16 - 5*I*exp(a)*exp(-I*d*x)*sin(c + d*x)**5/(96*d) + exp(a)*exp(-I*d*x)*sin(c + d*x)**4*cos
(c + d*x)/(96*d) - I*exp(a)*exp(-I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/(6*d) - exp(a)*exp(-I*d*x)*sin(c + d*x
)**2*cos(c + d*x)**3/(3*d) - 7*I*exp(a)*exp(-I*d*x)*sin(c + d*x)*cos(c + d*x)**4/(96*d) - 13*exp(a)*exp(-I*d*x
)*cos(c + d*x)**5/(96*d), Eq(b, -I*d)), (x*exp(a)*exp(I*d*x)*sin(c + d*x)**5/16 + I*x*exp(a)*exp(I*d*x)*sin(c
+ d*x)**4*cos(c + d*x)/16 + x*exp(a)*exp(I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/8 + I*x*exp(a)*exp(I*d*x)*sin(
c + d*x)**2*cos(c + d*x)**3/8 + x*exp(a)*exp(I*d*x)*sin(c + d*x)*cos(c + d*x)**4/16 + I*x*exp(a)*exp(I*d*x)*co
s(c + d*x)**5/16 + 5*I*exp(a)*exp(I*d*x)*sin(c + d*x)**5/(96*d) + exp(a)*exp(I*d*x)*sin(c + d*x)**4*cos(c + d*
x)/(96*d) + I*exp(a)*exp(I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/(6*d) - exp(a)*exp(I*d*x)*sin(c + d*x)**2*cos(
c + d*x)**3/(3*d) + 7*I*exp(a)*exp(I*d*x)*sin(c + d*x)*cos(c + d*x)**4/(96*d) - 13*exp(a)*exp(I*d*x)*cos(c + d
*x)**5/(96*d), Eq(b, I*d)), (-x*exp(a)*exp(3*I*d*x)*sin(c + d*x)**5/32 - 3*I*x*exp(a)*exp(3*I*d*x)*sin(c + d*x
)**4*cos(c + d*x)/32 + x*exp(a)*exp(3*I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/16 - I*x*exp(a)*exp(3*I*d*x)*sin(
c + d*x)**2*cos(c + d*x)**3/16 + 3*x*exp(a)*exp(3*I*d*x)*sin(c + d*x)*cos(c + d*x)**4/32 + I*x*exp(a)*exp(3*I*
d*x)*cos(c + d*x)**5/32 + 7*I*exp(a)*exp(3*I*d*x)*sin(c + d*x)**5/(192*d) - 9*exp(a)*exp(3*I*d*x)*sin(c + d*x)
**4*cos(c + d*x)/(64*d) - I*exp(a)*exp(3*I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/(6*d) - exp(a)*exp(3*I*d*x)*si
n(c + d*x)**2*cos(c + d*x)**3/(3*d) - 25*I*exp(a)*exp(3*I*d*x)*sin(c + d*x)*cos(c + d*x)**4/(64*d) + 23*exp(a)
*exp(3*I*d*x)*cos(c + d*x)**5/(192*d), Eq(b, 3*I*d)), (-x*exp(a)*exp(5*I*d*x)*sin(c + d*x)**5/32 - 5*I*x*exp(a
)*exp(5*I*d*x)*sin(c + d*x)**4*cos(c + d*x)/32 + 5*x*exp(a)*exp(5*I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/16 +
5*I*x*exp(a)*exp(5*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**3/16 - 5*x*exp(a)*exp(5*I*d*x)*sin(c + d*x)*cos(c + d*
x)**4/32 - I*x*exp(a)*exp(5*I*d*x)*cos(c + d*x)**5/32 - 31*I*exp(a)*exp(5*I*d*x)*sin(c + d*x)**5/(960*d) + 25*
exp(a)*exp(5*I*d*x)*sin(c + d*x)**4*cos(c + d*x)/(192*d) + I*exp(a)*exp(5*I*d*x)*sin(c + d*x)**3*cos(c + d*x)*
*2/(6*d) - exp(a)*exp(5*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**3/(3*d) - 41*I*exp(a)*exp(5*I*d*x)*sin(c + d*x)*c
os(c + d*x)**4/(192*d) + 47*exp(a)*exp(5*I*d*x)*cos(c + d*x)**5/(960*d), Eq(b, 5*I*d)), (b**5*exp(a)*exp(b*x)*
sin(c + d*x)**3*cos(c + d*x)**2/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6) + 2*b**4*d*exp(a)*exp(b*x)*si
n(c + d*x)**4*cos(c + d*x)/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6) - 3*b**4*d*exp(a)*exp(b*x)*sin(c +
 d*x)**2*cos(c + d*x)**3/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6) + 2*b**3*d**2*exp(a)*exp(b*x)*sin(c
+ d*x)**5/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6) + 18*b**3*d**2*exp(a)*exp(b*x)*sin(c + d*x)**3*cos(
c + d*x)**2/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6) + 6*b**3*d**2*exp(a)*exp(b*x)*sin(c + d*x)*cos(c
+ d*x)**4/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6) + 26*b**2*d**3*exp(a)*exp(b*x)*sin(c + d*x)**4*cos(
c + d*x)/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6) - 30*b**2*d**3*exp(a)*exp(b*x)*sin(c + d*x)**2*cos(c
 + d*x)**3/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6) - 6*b**2*d**3*exp(a)*exp(b*x)*cos(c + d*x)**5/(b**
6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6) + 26*b*d**4*exp(a)*exp(b*x)*sin(c + d*x)**5/(b**6 + 35*b**4*d**2
+ 259*b**2*d**4 + 225*d**6) + 65*b*d**4*exp(a)*exp(b*x)*sin(c + d*x)**3*cos(c + d*x)**2/(b**6 + 35*b**4*d**2 +
 259*b**2*d**4 + 225*d**6) + 30*b*d**4*exp(a)*exp(b*x)*sin(c + d*x)*cos(c + d*x)**4/(b**6 + 35*b**4*d**2 + 259
*b**2*d**4 + 225*d**6) - 75*d**5*exp(a)*exp(b*x)*sin(c + d*x)**2*cos(c + d*x)**3/(b**6 + 35*b**4*d**2 + 259*b*
*2*d**4 + 225*d**6) - 30*d**5*exp(a)*exp(b*x)*cos(c + d*x)**5/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6)
, True))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1148 vs. \(2 (165) = 330\).

Time = 0.27 (sec) , antiderivative size = 1148, normalized size of antiderivative = 6.27 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\text {Too large to display} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^3,x, algorithm="maxima")

[Out]

1/32*((5*b^4*d*cos(5*c)*e^a + 50*b^2*d^3*cos(5*c)*e^a + 45*d^5*cos(5*c)*e^a - b^5*e^a*sin(5*c) - 10*b^3*d^2*e^
a*sin(5*c) - 9*b*d^4*e^a*sin(5*c))*cos(5*d*x)*e^(b*x) + (5*b^4*d*cos(5*c)*e^a + 50*b^2*d^3*cos(5*c)*e^a + 45*d
^5*cos(5*c)*e^a + b^5*e^a*sin(5*c) + 10*b^3*d^2*e^a*sin(5*c) + 9*b*d^4*e^a*sin(5*c))*cos(5*d*x + 10*c)*e^(b*x)
 - (3*b^4*d*cos(5*c)*e^a + 78*b^2*d^3*cos(5*c)*e^a + 75*d^5*cos(5*c)*e^a + b^5*e^a*sin(5*c) + 26*b^3*d^2*e^a*s
in(5*c) + 25*b*d^4*e^a*sin(5*c))*cos(3*d*x + 8*c)*e^(b*x) - (3*b^4*d*cos(5*c)*e^a + 78*b^2*d^3*cos(5*c)*e^a +
75*d^5*cos(5*c)*e^a - b^5*e^a*sin(5*c) - 26*b^3*d^2*e^a*sin(5*c) - 25*b*d^4*e^a*sin(5*c))*cos(3*d*x - 2*c)*e^(
b*x) - 2*(b^4*d*cos(5*c)*e^a + 34*b^2*d^3*cos(5*c)*e^a + 225*d^5*cos(5*c)*e^a + b^5*e^a*sin(5*c) + 34*b^3*d^2*
e^a*sin(5*c) + 225*b*d^4*e^a*sin(5*c))*cos(d*x + 6*c)*e^(b*x) - 2*(b^4*d*cos(5*c)*e^a + 34*b^2*d^3*cos(5*c)*e^
a + 225*d^5*cos(5*c)*e^a - b^5*e^a*sin(5*c) - 34*b^3*d^2*e^a*sin(5*c) - 225*b*d^4*e^a*sin(5*c))*cos(d*x - 4*c)
*e^(b*x) - (b^5*cos(5*c)*e^a + 10*b^3*d^2*cos(5*c)*e^a + 9*b*d^4*cos(5*c)*e^a + 5*b^4*d*e^a*sin(5*c) + 50*b^2*
d^3*e^a*sin(5*c) + 45*d^5*e^a*sin(5*c))*e^(b*x)*sin(5*d*x) - (b^5*cos(5*c)*e^a + 10*b^3*d^2*cos(5*c)*e^a + 9*b
*d^4*cos(5*c)*e^a - 5*b^4*d*e^a*sin(5*c) - 50*b^2*d^3*e^a*sin(5*c) - 45*d^5*e^a*sin(5*c))*e^(b*x)*sin(5*d*x +
10*c) + (b^5*cos(5*c)*e^a + 26*b^3*d^2*cos(5*c)*e^a + 25*b*d^4*cos(5*c)*e^a - 3*b^4*d*e^a*sin(5*c) - 78*b^2*d^
3*e^a*sin(5*c) - 75*d^5*e^a*sin(5*c))*e^(b*x)*sin(3*d*x + 8*c) + (b^5*cos(5*c)*e^a + 26*b^3*d^2*cos(5*c)*e^a +
 25*b*d^4*cos(5*c)*e^a + 3*b^4*d*e^a*sin(5*c) + 78*b^2*d^3*e^a*sin(5*c) + 75*d^5*e^a*sin(5*c))*e^(b*x)*sin(3*d
*x - 2*c) + 2*(b^5*cos(5*c)*e^a + 34*b^3*d^2*cos(5*c)*e^a + 225*b*d^4*cos(5*c)*e^a - b^4*d*e^a*sin(5*c) - 34*b
^2*d^3*e^a*sin(5*c) - 225*d^5*e^a*sin(5*c))*e^(b*x)*sin(d*x + 6*c) + 2*(b^5*cos(5*c)*e^a + 34*b^3*d^2*cos(5*c)
*e^a + 225*b*d^4*cos(5*c)*e^a + b^4*d*e^a*sin(5*c) + 34*b^2*d^3*e^a*sin(5*c) + 225*d^5*e^a*sin(5*c))*e^(b*x)*s
in(d*x - 4*c))/(b^6*cos(5*c)^2 + b^6*sin(5*c)^2 + 225*(cos(5*c)^2 + sin(5*c)^2)*d^6 + 259*(b^2*cos(5*c)^2 + b^
2*sin(5*c)^2)*d^4 + 35*(b^4*cos(5*c)^2 + b^4*sin(5*c)^2)*d^2)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.85 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\frac {1}{16} \, {\left (\frac {5 \, d \cos \left (5 \, d x + 5 \, c\right )}{b^{2} + 25 \, d^{2}} - \frac {b \sin \left (5 \, d x + 5 \, c\right )}{b^{2} + 25 \, d^{2}}\right )} e^{\left (b x + a\right )} - \frac {1}{16} \, {\left (\frac {3 \, d \cos \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}} - \frac {b \sin \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}}\right )} e^{\left (b x + a\right )} - \frac {1}{8} \, {\left (\frac {d \cos \left (d x + c\right )}{b^{2} + d^{2}} - \frac {b \sin \left (d x + c\right )}{b^{2} + d^{2}}\right )} e^{\left (b x + a\right )} \]

[In]

integrate(exp(b*x+a)*cos(d*x+c)^2*sin(d*x+c)^3,x, algorithm="giac")

[Out]

1/16*(5*d*cos(5*d*x + 5*c)/(b^2 + 25*d^2) - b*sin(5*d*x + 5*c)/(b^2 + 25*d^2))*e^(b*x + a) - 1/16*(3*d*cos(3*d
*x + 3*c)/(b^2 + 9*d^2) - b*sin(3*d*x + 3*c)/(b^2 + 9*d^2))*e^(b*x + a) - 1/8*(d*cos(d*x + c)/(b^2 + d^2) - b*
sin(d*x + c)/(b^2 + d^2))*e^(b*x + a)

Mupad [B] (verification not implemented)

Time = 28.58 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.39 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (d\,x\right )-\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (c\right )-\sin \left (c\right )\,1{}\mathrm {i}\right )}{16\,\left (d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (d\,x\right )+\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (c\right )+\sin \left (c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,\left (b+d\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (3\,d\,x\right )-\sin \left (3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,c\right )-\sin \left (3\,c\right )\,1{}\mathrm {i}\right )}{32\,\left (3\,d+b\,1{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (5\,d\,x\right )-\sin \left (5\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,c\right )-\sin \left (5\,c\right )\,1{}\mathrm {i}\right )}{32\,\left (5\,d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (3\,d\,x\right )+\sin \left (3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,c\right )+\sin \left (3\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,\left (b+d\,3{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (5\,d\,x\right )+\sin \left (5\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,c\right )+\sin \left (5\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,\left (b+d\,5{}\mathrm {i}\right )} \]

[In]

int(cos(c + d*x)^2*exp(a + b*x)*sin(c + d*x)^3,x)

[Out]

(exp(a + b*x)*(cos(5*d*x) - sin(5*d*x)*1i)*(cos(5*c) - sin(5*c)*1i))/(32*(b*1i + 5*d)) - (exp(a + b*x)*(cos(d*
x) + sin(d*x)*1i)*(cos(c) + sin(c)*1i)*1i)/(16*(b + d*1i)) - (exp(a + b*x)*(cos(3*d*x) - sin(3*d*x)*1i)*(cos(3
*c) - sin(3*c)*1i))/(32*(b*1i + 3*d)) - (exp(a + b*x)*(cos(d*x) - sin(d*x)*1i)*(cos(c) - sin(c)*1i))/(16*(b*1i
 + d)) - (exp(a + b*x)*(cos(3*d*x) + sin(3*d*x)*1i)*(cos(3*c) + sin(3*c)*1i)*1i)/(32*(b + d*3i)) + (exp(a + b*
x)*(cos(5*d*x) + sin(5*d*x)*1i)*(cos(5*c) + sin(5*c)*1i)*1i)/(32*(b + d*5i))