\(\int e^{a+b x} \cos ^3(c+d x) \sin ^2(c+d x) \, dx\) [57]

Optimal result

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

b*exp(b*x+a)*cos(d*x+c)/(8*b^2+8*d^2)-b*exp(b*x+a)*cos(3*d*x+3*c)/(16*b^2+ 
144*d^2)-b*exp(b*x+a)*cos(5*d*x+5*c)/(16*b^2+400*d^2)+d*exp(b*x+a)*sin(d*x 
+c)/(8*b^2+8*d^2)-3*d*exp(b*x+a)*sin(3*d*x+3*c)/(16*b^2+144*d^2)-5*d*exp(b 
*x+a)*sin(5*d*x+5*c)/(16*b^2+400*d^2)
 

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.60 \[ \int e^{a+b x} \cos ^3(c+d x) \sin ^2(c+d x) \, dx=\frac {1}{16} e^{a+b x} \left (\frac {2 (b \cos (c+d x)+d \sin (c+d x))}{b^2+d^2}-\frac {b \cos (3 (c+d x))+3 d \sin (3 (c+d x))}{b^2+9 d^2}-\frac {b \cos (5 (c+d x))+5 d \sin (5 (c+d x))}{b^2+25 d^2}\right ) \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {4972, 2009}

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.

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

\(\Big \downarrow \) 4972

\(\displaystyle \int \left (\frac {1}{8} e^{a+b x} \cos (c+d x)-\frac {1}{16} e^{a+b x} \cos (3 c+3 d x)-\frac {1}{16} e^{a+b x} \cos (5 c+5 d x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d e^{a+b x} \sin (c+d x)}{8 \left (b^2+d^2\right )}-\frac {3 d e^{a+b x} \sin (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {5 d e^{a+b x} \sin (5 c+5 d x)}{16 \left (b^2+25 d^2\right )}+\frac {b e^{a+b x} \cos (c+d x)}{8 \left (b^2+d^2\right )}-\frac {b e^{a+b x} \cos (3 c+3 d x)}{16 \left (b^2+9 d^2\right )}-\frac {b e^{a+b x} \cos (5 c+5 d x)}{16 \left (b^2+25 d^2\right )}\)

Input:

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

Output:

(b*E^(a + b*x)*Cos[c + d*x])/(8*(b^2 + d^2)) - (b*E^(a + b*x)*Cos[3*c + 3* 
d*x])/(16*(b^2 + 9*d^2)) - (b*E^(a + b*x)*Cos[5*c + 5*d*x])/(16*(b^2 + 25* 
d^2)) + (d*E^(a + b*x)*Sin[c + d*x])/(8*(b^2 + d^2)) - (3*d*E^(a + b*x)*Si 
n[3*c + 3*d*x])/(16*(b^2 + 9*d^2)) - (5*d*E^(a + b*x)*Sin[5*c + 5*d*x])/(1 
6*(b^2 + 25*d^2))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 4972

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]
 

Maple [A] (verified)

Time = 3.07 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {b \,{\mathrm e}^{b x +a} \cos \left (5 d x +5 c \right )}{16 \left (b^{2}+25 d^{2}\right )}-\frac {5 d \,{\mathrm e}^{b x +a} \sin \left (5 d x +5 c \right )}{16 \left (b^{2}+25 d^{2}\right )}-\frac {b \,{\mathrm e}^{b x +a} \cos \left (3 d x +3 c \right )}{16 \left (b^{2}+9 d^{2}\right )}-\frac {3 d \,{\mathrm e}^{b x +a} \sin \left (3 d x +3 c \right )}{16 \left (b^{2}+9 d^{2}\right )}+\frac {b \,{\mathrm e}^{b x +a} \cos \left (d x +c \right )}{8 b^{2}+8 d^{2}}+\frac {d \,{\mathrm e}^{b x +a} \sin \left (d x +c \right )}{8 b^{2}+8 d^{2}}\)	\(166\)
parallelrisch	\(\frac {\left (\left (-\frac {1}{2} b^{5}-13 b^{3} d^{2}-\frac {25}{2} d^{4} b \right ) \cos \left (3 d x +3 c \right )+\left (-\frac {1}{2} b^{5}-5 b^{3} d^{2}-\frac {9}{2} d^{4} b \right ) \cos \left (5 d x +5 c \right )+3 \left (-\frac {1}{2} b^{4} d -13 d^{3} b^{2}-\frac {25}{2} d^{5}\right ) \sin \left (3 d x +3 c \right )+\left (b^{2}+9 d^{2}\right ) \left (\frac {5 \left (-b^{2} d -d^{3}\right ) \sin \left (5 d x +5 c \right )}{2}+\left (b^{2}+25 d^{2}\right ) \left (d \sin \left (d x +c \right )+b \cos \left (d x +c \right )\right )\right )\right ) {\mathrm e}^{b x +a}}{8 b^{6}+280 b^{4} d^{2}+2072 b^{2} d^{4}+1800 d^{6}}\)	\(189\)
risch	\(-\frac {{\mathrm e}^{b x +a} \left (4 d \left (b^{4}+34 b^{2} d^{2}+225 d^{4}\right ) \sin \left (d x +c \right )+\left (4 b^{5}+136 b^{3} d^{2}+900 d^{4} b \right ) \cos \left (d x +c \right )+\left (-2 b^{5}-20 b^{3} d^{2}-18 d^{4} b \right ) \cos \left (5 d x +5 c \right )-10 d \left (b^{4}+10 b^{2} d^{2}+9 d^{4}\right ) \sin \left (5 d x +5 c \right )+\left (-2 b^{5}-52 b^{3} d^{2}-50 d^{4} b \right ) \cos \left (3 d x +3 c \right )-6 d \left (b^{4}+26 b^{2} d^{2}+25 d^{4}\right ) \sin \left (3 d x +3 c \right )\right )}{32 \left (5 i d +b \right ) \left (3 i d +b \right ) \left (i d +b \right ) \left (i d -b \right ) \left (3 i d -b \right ) \left (5 i d -b \right )}\)	\(235\)
orering	\(\text {Expression too large to display}\)	\(1398\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.09 \[ \int e^{a+b x} \cos ^3(c+d x) \sin ^2(c+d x) \, dx=\frac {{\left (6 \, b^{2} d^{3} + 30 \, d^{5} - 5 \, {\left (b^{4} d + 10 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (b^{4} d + 6 \, b^{2} d^{3} + 5 \, d^{5}\right )} \cos \left (d x + c\right )^{2}\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) - {\left ({\left (b^{5} + 10 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cos \left (d x + c\right )^{5} - {\left (b^{5} + 6 \, b^{3} d^{2} + 5 \, b d^{4}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (b^{3} d^{2} + 5 \, b d^{4}\right )} \cos \left (d x + c\right )\right )} e^{\left (b x + a\right )}}{b^{6} + 35 \, b^{4} d^{2} + 259 \, b^{2} d^{4} + 225 \, d^{6}} \] Input:

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

Output:

((6*b^2*d^3 + 30*d^5 - 5*(b^4*d + 10*b^2*d^3 + 9*d^5)*cos(d*x + c)^4 + 3*( 
b^4*d + 6*b^2*d^3 + 5*d^5)*cos(d*x + c)^2)*e^(b*x + a)*sin(d*x + c) - ((b^ 
5 + 10*b^3*d^2 + 9*b*d^4)*cos(d*x + c)^5 - (b^5 + 6*b^3*d^2 + 5*b*d^4)*cos 
(d*x + c)^3 - 6*(b^3*d^2 + 5*b*d^4)*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 = 28.39 (sec) , antiderivative size = 2958, normalized size of antiderivative = 16.16 \[ \int e^{a+b x} \cos ^3(c+d x) \sin ^2(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

Piecewise((x*exp(a)*sin(c)**2*cos(c)**3, Eq(b, 0) & Eq(d, 0)), (-I*x*exp(a 
)*exp(-5*I*d*x)*sin(c + d*x)**5/32 - 5*x*exp(a)*exp(-5*I*d*x)*sin(c + d*x) 
**4*cos(c + d*x)/32 + 5*I*x*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**3*cos(c + d 
*x)**2/16 + 5*x*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**3/16 - 
5*I*x*exp(a)*exp(-5*I*d*x)*sin(c + d*x)*cos(c + d*x)**4/32 - x*exp(a)*exp( 
-5*I*d*x)*cos(c + d*x)**5/32 - 47*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**5/(96 
0*d) + 41*I*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**4*cos(c + d*x)/(192*d) + ex 
p(a)*exp(-5*I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/(3*d) - I*exp(a)*exp(-5 
*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**3/(6*d) - 25*exp(a)*exp(-5*I*d*x)*si 
n(c + d*x)*cos(c + d*x)**4/(192*d) + 31*I*exp(a)*exp(-5*I*d*x)*cos(c + d*x 
)**5/(960*d), Eq(b, -5*I*d)), (I*x*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**5/32 
 + 3*x*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**4*cos(c + d*x)/32 - I*x*exp(a)*e 
xp(-3*I*d*x)*sin(c + d*x)**3*cos(c + d*x)**2/16 + x*exp(a)*exp(-3*I*d*x)*s 
in(c + d*x)**2*cos(c + d*x)**3/16 - 3*I*x*exp(a)*exp(-3*I*d*x)*sin(c + d*x 
)*cos(c + d*x)**4/32 - x*exp(a)*exp(-3*I*d*x)*cos(c + d*x)**5/32 - 23*exp( 
a)*exp(-3*I*d*x)*sin(c + d*x)**5/(192*d) + 25*I*exp(a)*exp(-3*I*d*x)*sin(c 
 + d*x)**4*cos(c + d*x)/(64*d) + exp(a)*exp(-3*I*d*x)*sin(c + d*x)**3*cos( 
c + d*x)**2/(3*d) + I*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**3 
/(6*d) + 9*exp(a)*exp(-3*I*d*x)*sin(c + d*x)*cos(c + d*x)**4/(64*d) - 7*I* 
exp(a)*exp(-3*I*d*x)*cos(c + d*x)**5/(192*d), Eq(b, -3*I*d)), (I*x*exp(...
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1144 vs. \(2 (165) = 330\).

Time = 0.14 (sec) , antiderivative size = 1144, normalized size of antiderivative = 6.25 \[ \int e^{a+b x} \cos ^3(c+d x) \sin ^2(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

-1/32*((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))*co 
s(5*d*x)*e^(b*x) + (b^5*cos(5*c)*e^a + 10*b^3*d^2*cos(5*c)*e^a + 9*b*d^4*c 
os(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))*cos(5*d*x + 10*c)*e^(b*x) + (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*si 
n(5*c) - 75*d^5*e^a*sin(5*c))*cos(3*d*x + 8*c)*e^(b*x) + (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))*cos(3*d*x - 2*c)*e^(b*x) 
- 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))*cos( 
d*x + 6*c)*e^(b*x) - 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))*cos(d*x - 4*c)*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*si 
n(5*c) - 9*b*d^4*e^a*sin(5*c))*e^(b*x)*sin(5*d*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))*e^(b*x)*sin(5*d*x + 10*c) + (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))*e^(b*x)*...
 

Giac [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 152, normalized size of antiderivative = 0.83 \[ \int e^{a+b x} \cos ^3(c+d x) \sin ^2(c+d x) \, dx=-\frac {1}{16} \, {\left (\frac {b \cos \left (5 \, d x + 5 \, c\right )}{b^{2} + 25 \, d^{2}} + \frac {5 \, d \sin \left (5 \, d x + 5 \, c\right )}{b^{2} + 25 \, d^{2}}\right )} e^{\left (b x + a\right )} - \frac {1}{16} \, {\left (\frac {b \cos \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}} + \frac {3 \, d \sin \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}}\right )} e^{\left (b x + a\right )} + \frac {1}{8} \, {\left (\frac {b \cos \left (d x + c\right )}{b^{2} + d^{2}} + \frac {d \sin \left (d x + c\right )}{b^{2} + d^{2}}\right )} e^{\left (b x + a\right )} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 22.05 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.39 \[ \int e^{a+b x} \cos ^3(c+d x) \sin ^2(c+d x) \, dx=\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (d\,x\right )-\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (c\right )-\sin \left (c\right )\,1{}\mathrm {i}\right )}{16\,\left (b-d\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (3\,d\,x\right )+\sin \left (3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,c\right )+\sin \left (3\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,\left (-3\,d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (5\,d\,x\right )+\sin \left (5\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,c\right )+\sin \left (5\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,\left (-5\,d+b\,1{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (d\,x\right )+\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (c\right )+\sin \left (c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,\left (-d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (3\,d\,x\right )-\sin \left (3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,c\right )-\sin \left (3\,c\right )\,1{}\mathrm {i}\right )}{32\,\left (b-d\,3{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (5\,d\,x\right )-\sin \left (5\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,c\right )-\sin \left (5\,c\right )\,1{}\mathrm {i}\right )}{32\,\left (b-d\,5{}\mathrm {i}\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.61 \[ \int e^{a+b x} \cos ^3(c+d x) \sin ^2(c+d x) \, dx=\frac {e^{b x +a} \left (-\cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{5}-10 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{3} d^{2}-9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b \,d^{4}+\cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{5}+14 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{3} d^{2}+13 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b \,d^{4}+2 \cos \left (d x +c \right ) b^{3} d^{2}+26 \cos \left (d x +c \right ) b \,d^{4}-5 \sin \left (d x +c \right )^{5} b^{4} d -50 \sin \left (d x +c \right )^{5} b^{2} d^{3}-45 \sin \left (d x +c \right )^{5} d^{5}+7 \sin \left (d x +c \right )^{3} b^{4} d +82 \sin \left (d x +c \right )^{3} b^{2} d^{3}+75 \sin \left (d x +c \right )^{3} d^{5}-2 \sin \left (d x +c \right ) b^{4} d -26 \sin \left (d x +c \right ) b^{2} d^{3}\right )}{b^{6}+35 b^{4} d^{2}+259 b^{2} d^{4}+225 d^{6}} \] Input:

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

Output:

(e**(a + b*x)*( - cos(c + d*x)*sin(c + d*x)**4*b**5 - 10*cos(c + d*x)*sin( 
c + d*x)**4*b**3*d**2 - 9*cos(c + d*x)*sin(c + d*x)**4*b*d**4 + cos(c + d* 
x)*sin(c + d*x)**2*b**5 + 14*cos(c + d*x)*sin(c + d*x)**2*b**3*d**2 + 13*c 
os(c + d*x)*sin(c + d*x)**2*b*d**4 + 2*cos(c + d*x)*b**3*d**2 + 26*cos(c + 
 d*x)*b*d**4 - 5*sin(c + d*x)**5*b**4*d - 50*sin(c + d*x)**5*b**2*d**3 - 4 
5*sin(c + d*x)**5*d**5 + 7*sin(c + d*x)**3*b**4*d + 82*sin(c + d*x)**3*b** 
2*d**3 + 75*sin(c + d*x)**3*d**5 - 2*sin(c + d*x)*b**4*d - 26*sin(c + d*x) 
*b**2*d**3))/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6)