\(\int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx\) [55]

Optimal result

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

-1/8*d*exp(b*x+a)*cos(d*x+c)/(b^2+d^2)-3*d*exp(b*x+a)*cos(3*d*x+3*c)/(16*b 
^2+144*d^2)+5*d*exp(b*x+a)*cos(5*d*x+5*c)/(16*b^2+400*d^2)+b*exp(b*x+a)*si 
n(d*x+c)/(8*b^2+8*d^2)+b*exp(b*x+a)*sin(3*d*x+3*c)/(16*b^2+144*d^2)-b*exp( 
b*x+a)*sin(5*d*x+5*c)/(16*b^2+400*d^2)
 

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.60 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\frac {1}{16} e^{a+b x} \left (\frac {2 (-d \cos (c+d x)+b \sin (c+d x))}{b^2+d^2}+\frac {-3 d \cos (3 (c+d x))+b \sin (3 (c+d x))}{b^2+9 d^2}+\frac {5 d \cos (5 (c+d x))-b \sin (5 (c+d x))}{b^2+25 d^2}\right ) \] Input:

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

Output:

(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
 

Rubi [A] (verified)

Time = 0.36 (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.

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

\(\Big \downarrow \) 4972

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

-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))
 

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 = 3.09 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.91

method	result	size
default	\(-\frac {3 d \,{\mathrm e}^{b x +a} \cos \left (3 d x +3 c \right )}{16 \left (b^{2}+9 d^{2}\right )}+\frac {b \,{\mathrm e}^{b x +a} \sin \left (3 d x +3 c \right )}{16 b^{2}+144 d^{2}}-\frac {d \,{\mathrm e}^{b x +a} \cos \left (d x +c \right )}{8 \left (b^{2}+d^{2}\right )}+\frac {b \,{\mathrm e}^{b x +a} \sin \left (d x +c \right )}{8 b^{2}+8 d^{2}}+\frac {5 d \,{\mathrm e}^{b x +a} \cos \left (5 d x +5 c \right )}{16 \left (b^{2}+25 d^{2}\right )}-\frac {b \,{\mathrm e}^{b x +a} \sin \left (5 d x +5 c \right )}{16 \left (b^{2}+25 d^{2}\right )}\)	\(166\)
parallelrisch	\(\frac {{\mathrm e}^{b x +a} \left (3 \left (-\frac {1}{2} b^{4} d -13 d^{3} b^{2}-\frac {25}{2} d^{5}\right ) \cos \left (3 d x +3 c \right )+5 \left (\frac {1}{2} b^{4} d +5 d^{3} b^{2}+\frac {9}{2} d^{5}\right ) \cos \left (5 d x +5 c \right )+\left (\frac {1}{2} b^{5}+13 b^{3} d^{2}+\frac {25}{2} d^{4} b \right ) \sin \left (3 d x +3 c \right )+\left (b^{2}+9 d^{2}\right ) \left (\frac {\left (-b^{3}-b \,d^{2}\right ) \sin \left (5 d x +5 c \right )}{2}+\left (b^{2}+25 d^{2}\right ) \left (b \sin \left (d x +c \right )-d \cos \left (d x +c \right )\right )\right )\right )}{8 b^{6}+280 b^{4} d^{2}+2072 b^{2} d^{4}+1800 d^{6}}\)	\(191\)
risch	\(\frac {i {\mathrm e}^{b x +a} \left (-4 i d \left (b^{4}+34 b^{2} d^{2}+225 d^{4}\right ) \cos \left (d x +c \right )+i \left (4 b^{5}+136 b^{3} d^{2}+900 d^{4} b \right ) \sin \left (d x +c \right )+10 i d \left (b^{4}+10 b^{2} d^{2}+9 d^{4}\right ) \cos \left (5 d x +5 c \right )-i \left (2 b^{5}+20 b^{3} d^{2}+18 d^{4} b \right ) \sin \left (5 d x +5 c \right )-6 i d \left (b^{4}+26 b^{2} d^{2}+25 d^{4}\right ) \cos \left (3 d x +3 c \right )-i \left (-2 b^{5}-52 b^{3} d^{2}-50 d^{4} b \right ) \sin \left (3 d x +3 c \right )\right )}{32 \left (5 i d +b \right ) \left (3 i d +b \right ) \left (i d +b \right ) \left (i d -b \right ) \left (3 i d -b \right ) \left (5 i d -b \right )}\)	\(245\)
orering	\(\text {Expression too large to display}\)	\(1398\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.10 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\frac {{\left (2 \, b^{3} d^{2} + 26 \, b d^{4} - {\left (b^{5} + 10 \, b^{3} d^{2} + 9 \, b d^{4}\right )} \cos \left (d x + c\right )^{4} + {\left (b^{5} + 14 \, b^{3} d^{2} + 13 \, b d^{4}\right )} \cos \left (d x + c\right )^{2}\right )} e^{\left (b x + a\right )} \sin \left (d x + c\right ) + {\left (5 \, {\left (b^{4} d + 10 \, b^{2} d^{3} + 9 \, d^{5}\right )} \cos \left (d x + c\right )^{5} - {\left (7 \, b^{4} d + 82 \, b^{2} d^{3} + 75 \, d^{5}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (b^{4} d + 13 \, b^{2} d^{3}\right )} \cos \left (d x + c\right )\right )} e^{\left (b x + a\right )}}{b^{6} + 35 \, b^{4} d^{2} + 259 \, b^{2} d^{4} + 225 \, d^{6}} \] Input:

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

Output:

((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 = 28.18 (sec) , antiderivative size = 2751, normalized size of antiderivative = 15.03 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

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 + I*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**5/(64* 
d) + 3*exp(a)*exp(-5*I*d*x)*sin(c + d*x)**4*cos(c + d*x)/(64*d) - exp(a)*e 
xp(-5*I*d*x)*sin(c + d*x)**2*cos(c + d*x)**3/(6*d) + 25*I*exp(a)*exp(-5*I* 
d*x)*sin(c + d*x)*cos(c + d*x)**4/(192*d) + 31*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*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 + 3 
*I*exp(a)*exp(-3*I*d*x)*sin(c + d*x)**5/(64*d) + 7*exp(a)*exp(-3*I*d*x)*si 
n(c + d*x)**4*cos(c + d*x)/(64*d) - exp(a)*exp(-3*I*d*x)*sin(c + d*x)**2*c 
os(c + d*x)**3/(6*d) + 9*I*exp(a)*exp(-3*I*d*x)*sin(c + d*x)*cos(c + d*x)* 
*4/(64*d) + 7*exp(a)*exp(-3*I*d*x)*cos(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 ...
 

Maxima [B] (verification not implemented)

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

Time = 0.14 (sec) , antiderivative size = 1148, normalized size of antiderivative = 6.27 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\text {Too large to display} \] Input:

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

Output:

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*s 
in(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*co 
s(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 + 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) - 2 
6*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*si 
n(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)*s...
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.85 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\frac {1}{16} \, {\left (\frac {5 \, d \cos \left (5 \, d x + 5 \, c\right )}{b^{2} + 25 \, d^{2}} - \frac {b \sin \left (5 \, d x + 5 \, c\right )}{b^{2} + 25 \, d^{2}}\right )} e^{\left (b x + a\right )} - \frac {1}{16} \, {\left (\frac {3 \, d \cos \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}} - \frac {b \sin \left (3 \, d x + 3 \, c\right )}{b^{2} + 9 \, d^{2}}\right )} e^{\left (b x + a\right )} - \frac {1}{8} \, {\left (\frac {d \cos \left (d x + c\right )}{b^{2} + d^{2}} - \frac {b \sin \left (d x + c\right )}{b^{2} + d^{2}}\right )} e^{\left (b x + a\right )} \] Input:

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

Output:

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*si 
n(d*x + c)/(b^2 + d^2))*e^(b*x + a)
 

Mupad [B] (verification not implemented)

Time = 21.44 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.39 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (d\,x\right )-\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (c\right )-\sin \left (c\right )\,1{}\mathrm {i}\right )}{16\,\left (d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (d\,x\right )+\sin \left (d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (c\right )+\sin \left (c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{16\,\left (b+d\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (3\,d\,x\right )-\sin \left (3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,c\right )-\sin \left (3\,c\right )\,1{}\mathrm {i}\right )}{32\,\left (3\,d+b\,1{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (5\,d\,x\right )-\sin \left (5\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,c\right )-\sin \left (5\,c\right )\,1{}\mathrm {i}\right )}{32\,\left (5\,d+b\,1{}\mathrm {i}\right )}-\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (3\,d\,x\right )+\sin \left (3\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (3\,c\right )+\sin \left (3\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,\left (b+d\,3{}\mathrm {i}\right )}+\frac {{\mathrm {e}}^{a+b\,x}\,\left (\cos \left (5\,d\,x\right )+\sin \left (5\,d\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (5\,c\right )+\sin \left (5\,c\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{32\,\left (b+d\,5{}\mathrm {i}\right )} \] Input:

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

Output:

(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)*1 
i)*(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))
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.61 \[ \int e^{a+b x} \cos ^2(c+d x) \sin ^3(c+d x) \, dx=\frac {e^{b x +a} \left (5 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{4} d +50 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{2} d^{3}+45 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} d^{5}-3 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{4} d -18 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2} d^{3}-15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d^{5}-6 \cos \left (d x +c \right ) b^{2} d^{3}-30 \cos \left (d x +c \right ) d^{5}-\sin \left (d x +c \right )^{5} b^{5}-10 \sin \left (d x +c \right )^{5} b^{3} d^{2}-9 \sin \left (d x +c \right )^{5} b \,d^{4}+\sin \left (d x +c \right )^{3} b^{5}+6 \sin \left (d x +c \right )^{3} b^{3} d^{2}+5 \sin \left (d x +c \right )^{3} b \,d^{4}+6 \sin \left (d x +c \right ) b^{3} d^{2}+30 \sin \left (d x +c \right ) b \,d^{4}\right )}{b^{6}+35 b^{4} d^{2}+259 b^{2} d^{4}+225 d^{6}} \] Input:

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

Output:

(e**(a + b*x)*(5*cos(c + d*x)*sin(c + d*x)**4*b**4*d + 50*cos(c + d*x)*sin 
(c + d*x)**4*b**2*d**3 + 45*cos(c + d*x)*sin(c + d*x)**4*d**5 - 3*cos(c + 
d*x)*sin(c + d*x)**2*b**4*d - 18*cos(c + d*x)*sin(c + d*x)**2*b**2*d**3 - 
15*cos(c + d*x)*sin(c + d*x)**2*d**5 - 6*cos(c + d*x)*b**2*d**3 - 30*cos(c 
 + d*x)*d**5 - sin(c + d*x)**5*b**5 - 10*sin(c + d*x)**5*b**3*d**2 - 9*sin 
(c + d*x)**5*b*d**4 + sin(c + d*x)**3*b**5 + 6*sin(c + d*x)**3*b**3*d**2 + 
 5*sin(c + d*x)**3*b*d**4 + 6*sin(c + d*x)*b**3*d**2 + 30*sin(c + d*x)*b*d 
**4))/(b**6 + 35*b**4*d**2 + 259*b**2*d**4 + 225*d**6)