\(\int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx\) [383]

Optimal result

Integrand size = 23, antiderivative size = 242 \[ \int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx=\frac {3 d^3 x}{2 b^3}-\frac {(c+d x)^3}{b}-\frac {i (c+d x)^4}{4 d}+\frac {(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}+\frac {3 i d^3 \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 d^3 \cos (a+b x) \sin (a+b x)}{2 b^4}+\frac {3 d (c+d x)^2 \cos (a+b x) \sin (a+b x)}{b^2}-\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{b^3}+\frac {2 (c+d x)^3 \sin ^2(a+b x)}{b} \] Output:

3/2*d^3*x/b^3-(d*x+c)^3/b-1/4*I*(d*x+c)^4/d+(d*x+c)^3*ln(1+exp(2*I*(b*x+a) 
))/b-3/2*I*d*(d*x+c)^2*polylog(2,-exp(2*I*(b*x+a)))/b^2+3/2*d^2*(d*x+c)*po 
lylog(3,-exp(2*I*(b*x+a)))/b^3+3/4*I*d^3*polylog(4,-exp(2*I*(b*x+a)))/b^4- 
3/2*d^3*cos(b*x+a)*sin(b*x+a)/b^4+3*d*(d*x+c)^2*cos(b*x+a)*sin(b*x+a)/b^2- 
3*d^2*(d*x+c)*sin(b*x+a)^2/b^3+2*(d*x+c)^3*sin(b*x+a)^2/b
 

Mathematica [B] (warning: unable to verify)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1730\) vs. \(2(242)=484\).

Time = 6.33 (sec) , antiderivative size = 1730, normalized size of antiderivative = 7.15 \[ \int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx =\text {Too large to display} \] Input:

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

Output:

((I/4)*c*d^2*(2*b^2*x^2*(2*b*x - (3*I)*(1 + E^((2*I)*a))*Log[1 + E^((-2*I) 
*(a + b*x))]) + 6*b*(1 + E^((2*I)*a))*x*PolyLog[2, -E^((-2*I)*(a + b*x))] 
- (3*I)*(1 + E^((2*I)*a))*PolyLog[3, -E^((-2*I)*(a + b*x))])*Sec[a])/(b^3* 
E^(I*a)) + ((I/8)*d^3*E^(I*a)*((2*b^4*x^4)/E^((2*I)*a) - (4*I)*b^3*(1 + E^ 
((-2*I)*a))*x^3*Log[1 + E^((-2*I)*(a + b*x))] + 6*b^2*(1 + E^((-2*I)*a))*x 
^2*PolyLog[2, -E^((-2*I)*(a + b*x))] - (6*I)*b*(1 + E^((-2*I)*a))*x*PolyLo 
g[3, -E^((-2*I)*(a + b*x))] - 3*(1 + E^((-2*I)*a))*PolyLog[4, -E^((-2*I)*( 
a + b*x))])*Sec[a])/b^4 + (c^3*Sec[a]*(Cos[a]*Log[Cos[a]*Cos[b*x] - Sin[a] 
*Sin[b*x]] + b*x*Sin[a]))/(b*(Cos[a]^2 + Sin[a]^2)) + (3*c^2*d*Csc[a]*((b^ 
2*x^2)/E^(I*ArcTan[Cot[a]]) - (Cot[a]*(I*b*x*(-Pi - 2*ArcTan[Cot[a]]) - Pi 
*Log[1 + E^((-2*I)*b*x)] - 2*(b*x - ArcTan[Cot[a]])*Log[1 - E^((2*I)*(b*x 
- ArcTan[Cot[a]]))] + Pi*Log[Cos[b*x]] - 2*ArcTan[Cot[a]]*Log[Sin[b*x - Ar 
cTan[Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x - ArcTan[Cot[a]]))]))/Sqrt[1 + 
 Cot[a]^2])*Sec[a])/(2*b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin[a]^2)]) + Sec[a]* 
(Cos[2*a + 2*b*x]/(16*b^4) - ((I/16)*Sin[2*a + 2*b*x])/b^4)*(-8*b^3*c^3*Co 
s[a] + (12*I)*b^2*c^2*d*Cos[a] + 12*b*c*d^2*Cos[a] - (6*I)*d^3*Cos[a] - 24 
*b^3*c^2*d*x*Cos[a] + (24*I)*b^2*c*d^2*x*Cos[a] + 12*b*d^3*x*Cos[a] - 24*b 
^3*c*d^2*x^2*Cos[a] + (12*I)*b^2*d^3*x^2*Cos[a] - 8*b^3*d^3*x^3*Cos[a] - ( 
8*I)*b^4*c^3*x*Cos[a + 2*b*x] - (12*I)*b^4*c^2*d*x^2*Cos[a + 2*b*x] - (8*I 
)*b^4*c*d^2*x^3*Cos[a + 2*b*x] - (2*I)*b^4*d^3*x^4*Cos[a + 2*b*x] + (8*...
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4931, 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.

\(\displaystyle \int (c+d x)^3 \sin (3 a+3 b x) \sec (a+b x) \, dx\)

\(\Big \downarrow \) 4931

\(\displaystyle \int \left (3 (c+d x)^3 \sin (a+b x) \cos (a+b x)-(c+d x)^3 \sin ^2(a+b x) \tan (a+b x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {3 i d^3 \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{4 b^4}-\frac {3 d^3 \sin (a+b x) \cos (a+b x)}{2 b^4}+\frac {3 d^2 (c+d x) \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{2 b^3}-\frac {3 d^2 (c+d x) \sin ^2(a+b x)}{b^3}-\frac {3 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{2 b^2}+\frac {3 d (c+d x)^2 \sin (a+b x) \cos (a+b x)}{b^2}+\frac {(c+d x)^3 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {2 (c+d x)^3 \sin ^2(a+b x)}{b}+\frac {3 d^3 x}{2 b^3}-\frac {(c+d x)^3}{b}-\frac {i (c+d x)^4}{4 d}\)

Input:

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

Output:

(3*d^3*x)/(2*b^3) - (c + d*x)^3/b - ((I/4)*(c + d*x)^4)/d + ((c + d*x)^3*L 
og[1 + E^((2*I)*(a + b*x))])/b - (((3*I)/2)*d*(c + d*x)^2*PolyLog[2, -E^(( 
2*I)*(a + b*x))])/b^2 + (3*d^2*(c + d*x)*PolyLog[3, -E^((2*I)*(a + b*x))]) 
/(2*b^3) + (((3*I)/4)*d^3*PolyLog[4, -E^((2*I)*(a + b*x))])/b^4 - (3*d^3*C 
os[a + b*x]*Sin[a + b*x])/(2*b^4) + (3*d*(c + d*x)^2*Cos[a + b*x]*Sin[a + 
b*x])/b^2 - (3*d^2*(c + d*x)*Sin[a + b*x]^2)/b^3 + (2*(c + d*x)^3*Sin[a + 
b*x]^2)/b
 

Defintions of rubi rules used

rule 2009

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

rule 4931

Int[((e_.) + (f_.)*(x_))^(m_.)*(F_)[(a_.) + (b_.)*(x_)]^(p_.)*(G_)[(c_.) + 
(d_.)*(x_)]^(q_.), x_Symbol] :> Int[ExpandTrigExpand[(e + f*x)^m*G[c + d*x] 
^q, F, c + d*x, p, b/d, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && Member 
Q[{Sin, Cos}, F] && MemberQ[{Sec, Csc}, G] && IGtQ[p, 0] && IGtQ[q, 0] && E 
qQ[b*c - a*d, 0] && IGtQ[b/d, 1]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 647 vs. \(2 (219 ) = 438\).

Time = 2.81 (sec) , antiderivative size = 648, normalized size of antiderivative = 2.68

method	result	size
risch	\(\frac {3 c \,d^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {3 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{2 b^{3}}-\frac {6 i d \,c^{2} a x}{b}+\frac {6 i c \,d^{2} a^{2} x}{b^{2}}-\frac {3 i c \,d^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {3 i d^{3} a^{4}}{2 b^{4}}-i d^{2} c \,x^{3}-\frac {3 i d \,c^{2} x^{2}}{2}+\frac {2 d^{3} a^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+i c^{3} x +\frac {i c^{4}}{4 d}-\frac {3 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{2 b^{2}}-\frac {3 i c^{2} d \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{2}}+\frac {3 c^{2} d \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x}{b}-\frac {\left (4 d^{3} x^{3} b^{3}+12 b^{3} c \,d^{2} x^{2}-6 i b^{2} d^{3} x^{2}+12 b^{3} c^{2} d x -12 i b^{2} c \,d^{2} x +4 b^{3} c^{3}-6 i b^{2} c^{2} d -6 b \,d^{3} x -6 c \,d^{2} b +3 i d^{3}\right ) {\mathrm e}^{-2 i \left (b x +a \right )}}{8 b^{4}}-\frac {\left (4 d^{3} x^{3} b^{3}+12 b^{3} c \,d^{2} x^{2}+6 i b^{2} d^{3} x^{2}+12 b^{3} c^{2} d x +12 i b^{2} c \,d^{2} x +4 b^{3} c^{3}+6 i b^{2} c^{2} d -6 b \,d^{3} x -6 c \,d^{2} b -3 i d^{3}\right ) {\mathrm e}^{2 i \left (b x +a \right )}}{8 b^{4}}+\frac {c^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}+\frac {d^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{3}}{b}+\frac {3 c \,d^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{3}}+\frac {3 i d^{3} \operatorname {polylog}\left (4, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{4 b^{4}}-\frac {6 c \,d^{2} a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 c^{2} d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {3 i d \,c^{2} a^{2}}{b^{2}}-\frac {2 i d^{3} a^{3} x}{b^{3}}+\frac {4 i c \,d^{2} a^{3}}{b^{3}}-\frac {i d^{3} x^{4}}{4}-\frac {2 c^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}\)	\(648\)

Input:

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

Output:

-6/b^3*c*d^2*a^2*ln(exp(I*(b*x+a)))-6*I/b*d*c^2*a*x+6*I/b^2*c*d^2*a^2*x-3/ 
2*I/b^2*d^3*polylog(2,-exp(2*I*(b*x+a)))*x^2-3/2*I/b^2*c^2*d*polylog(2,-ex 
p(2*I*(b*x+a)))-1/8*(4*d^3*x^3*b^3-6*I*b^2*d^3*x^2+12*b^3*c*d^2*x^2-12*I*b 
^2*c*d^2*x+12*b^3*c^2*d*x-6*I*b^2*c^2*d+4*b^3*c^3-6*b*d^3*x+3*I*d^3-6*c*d^ 
2*b)/b^4*exp(-2*I*(b*x+a))-3*I/b^2*d^2*c*polylog(2,-exp(2*I*(b*x+a)))*x+1/ 
b*d^3*ln(exp(2*I*(b*x+a))+1)*x^3+3/2/b^3*c*d^2*polylog(3,-exp(2*I*(b*x+a)) 
)+3/2/b^3*d^3*polylog(3,-exp(2*I*(b*x+a)))*x+3/b*c^2*d*ln(exp(2*I*(b*x+a)) 
+1)*x+3/b*c*d^2*ln(exp(2*I*(b*x+a))+1)*x^2+I*c^3*x+1/4*I/d*c^4-3/2*I/b^4*d 
^3*a^4-I*d^2*c*x^3-3/2*I*d*c^2*x^2+2/b^4*d^3*a^3*ln(exp(I*(b*x+a)))+6/b^2* 
c^2*d*a*ln(exp(I*(b*x+a)))-3*I/b^2*d*c^2*a^2-2*I/b^3*d^3*a^3*x+4*I/b^3*c*d 
^2*a^3-1/8*(4*d^3*x^3*b^3+6*I*b^2*d^3*x^2+12*b^3*c*d^2*x^2+12*I*b^2*c*d^2* 
x+12*b^3*c^2*d*x+6*I*b^2*c^2*d+4*b^3*c^3-6*b*d^3*x-3*I*d^3-6*c*d^2*b)/b^4* 
exp(2*I*(b*x+a))+1/b*c^3*ln(exp(2*I*(b*x+a))+1)-1/4*I*d^3*x^4-2/b*c^3*ln(e 
xp(I*(b*x+a)))+3/4*I*d^3*polylog(4,-exp(2*I*(b*x+a)))/b^4
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1126 vs. \(2 (215) = 430\).

Time = 0.15 (sec) , antiderivative size = 1126, normalized size of antiderivative = 4.65 \[ \int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 - 6*I*d^3*polylog(4, I*cos(b*x + a) + 
 sin(b*x + a)) + 6*I*d^3*polylog(4, I*cos(b*x + a) - sin(b*x + a)) + 6*I*d 
^3*polylog(4, -I*cos(b*x + a) + sin(b*x + a)) - 6*I*d^3*polylog(4, -I*cos( 
b*x + a) - sin(b*x + a)) - 2*(2*b^3*d^3*x^3 + 6*b^3*c*d^2*x^2 + 2*b^3*c^3 
- 3*b*c*d^2 + 3*(2*b^3*c^2*d - b*d^3)*x)*cos(b*x + a)^2 + 3*(2*b^2*d^3*x^2 
 + 4*b^2*c*d^2*x + 2*b^2*c^2*d - d^3)*cos(b*x + a)*sin(b*x + a) + 3*(2*b^3 
*c^2*d - b*d^3)*x - 3*(-I*b^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2*d)*dil 
og(I*cos(b*x + a) + sin(b*x + a)) - 3*(I*b^2*d^3*x^2 + 2*I*b^2*c*d^2*x + I 
*b^2*c^2*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) - 3*(I*b^2*d^3*x^2 + 2*I* 
b^2*c*d^2*x + I*b^2*c^2*d)*dilog(-I*cos(b*x + a) + sin(b*x + a)) - 3*(-I*b 
^2*d^3*x^2 - 2*I*b^2*c*d^2*x - I*b^2*c^2*d)*dilog(-I*cos(b*x + a) - sin(b* 
x + a)) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*log(cos(b*x 
+ a) + I*sin(b*x + a) + I) + (b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^ 
3*d^3)*log(cos(b*x + a) - I*sin(b*x + a) + I) + (b^3*d^3*x^3 + 3*b^3*c*d^2 
*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(I*cos( 
b*x + a) + sin(b*x + a) + 1) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2* 
d*x + 3*a*b^2*c^2*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(I*cos(b*x + a) - sin(b* 
x + a) + 1) + (b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2 
*d - 3*a^2*b*c*d^2 + a^3*d^3)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + (b 
^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + 3*b^3*c^2*d*x + 3*a*b^2*c^2*d - 3*a^2*b*...
 

Sympy [F(-1)]

Timed out. \[ \int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 444 vs. \(2 (215) = 430\).

Time = 0.25 (sec) , antiderivative size = 444, normalized size of antiderivative = 1.83 \[ \int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx=-\frac {c^{3} {\left (2 \, \cos \left (2 \, b x + 2 \, a\right ) - \log \left (\cos \left (2 \, b x\right )^{2} + 2 \, \cos \left (2 \, b x\right ) \cos \left (2 \, a\right ) + \cos \left (2 \, a\right )^{2} + \sin \left (2 \, b x\right )^{2} - 2 \, \sin \left (2 \, b x\right ) \sin \left (2 \, a\right ) + \sin \left (2 \, a\right )^{2}\right )\right )}}{2 \, b} + \frac {-3 i \, b^{4} d^{3} x^{4} - 12 i \, b^{4} c d^{2} x^{3} - 18 i \, b^{4} c^{2} d x^{2} + 12 i \, d^{3} {\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )}) - 4 \, {\left (-4 i \, b^{3} d^{3} x^{3} - 9 i \, b^{3} c d^{2} x^{2} - 9 i \, b^{3} c^{2} d x\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 6 \, {\left (2 \, b^{3} d^{3} x^{3} + 6 \, b^{3} c d^{2} x^{2} - 3 \, b c d^{2} + 3 \, {\left (2 \, b^{3} c^{2} d - b d^{3}\right )} x\right )} \cos \left (2 \, b x + 2 \, a\right ) - 6 \, {\left (4 i \, b^{2} d^{3} x^{2} + 6 i \, b^{2} c d^{2} x + 3 i \, b^{2} c^{2} d\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 2 \, {\left (4 \, b^{3} d^{3} x^{3} + 9 \, b^{3} c d^{2} x^{2} + 9 \, b^{3} c^{2} d x\right )} \log \left (\cos \left (2 \, b x + 2 \, a\right )^{2} + \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) + 6 \, {\left (4 \, b d^{3} x + 3 \, b c d^{2}\right )} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 9 \, {\left (2 \, b^{2} d^{3} x^{2} + 4 \, b^{2} c d^{2} x + 2 \, b^{2} c^{2} d - d^{3}\right )} \sin \left (2 \, b x + 2 \, a\right )}{12 \, b^{4}} \] Input:

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

Output:

-1/2*c^3*(2*cos(2*b*x + 2*a) - log(cos(2*b*x)^2 + 2*cos(2*b*x)*cos(2*a) + 
cos(2*a)^2 + sin(2*b*x)^2 - 2*sin(2*b*x)*sin(2*a) + sin(2*a)^2))/b + 1/12* 
(-3*I*b^4*d^3*x^4 - 12*I*b^4*c*d^2*x^3 - 18*I*b^4*c^2*d*x^2 + 12*I*d^3*pol 
ylog(4, -e^(2*I*b*x + 2*I*a)) - 4*(-4*I*b^3*d^3*x^3 - 9*I*b^3*c*d^2*x^2 - 
9*I*b^3*c^2*d*x)*arctan2(sin(2*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 6*(2*b^ 
3*d^3*x^3 + 6*b^3*c*d^2*x^2 - 3*b*c*d^2 + 3*(2*b^3*c^2*d - b*d^3)*x)*cos(2 
*b*x + 2*a) - 6*(4*I*b^2*d^3*x^2 + 6*I*b^2*c*d^2*x + 3*I*b^2*c^2*d)*dilog( 
-e^(2*I*b*x + 2*I*a)) + 2*(4*b^3*d^3*x^3 + 9*b^3*c*d^2*x^2 + 9*b^3*c^2*d*x 
)*log(cos(2*b*x + 2*a)^2 + sin(2*b*x + 2*a)^2 + 2*cos(2*b*x + 2*a) + 1) + 
6*(4*b*d^3*x + 3*b*c*d^2)*polylog(3, -e^(2*I*b*x + 2*I*a)) + 9*(2*b^2*d^3* 
x^2 + 4*b^2*c*d^2*x + 2*b^2*c^2*d - d^3)*sin(2*b*x + 2*a))/b^4
 

Giac [F]

\[ \int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx=\int { {\left (d x + c\right )}^{3} \sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right ) \,d x } \] Input:

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

Output:

integrate((d*x + c)^3*sec(b*x + a)*sin(3*b*x + 3*a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx=\int \frac {\sin \left (3\,a+3\,b\,x\right )\,{\left (c+d\,x\right )}^3}{\cos \left (a+b\,x\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (c+d x)^3 \sec (a+b x) \sin (3 a+3 b x) \, dx =\text {Too large to display} \] Input:

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

Output:

(9*cos(3*a + 3*b*x)*tan((3*a + 3*b*x)/2)**2*b**3*c**3 + 27*cos(3*a + 3*b*x 
)*tan((3*a + 3*b*x)/2)**2*b**3*c**2*d*x + 27*cos(3*a + 3*b*x)*tan((3*a + 3 
*b*x)/2)**2*b**3*c*d**2*x**2 + 9*cos(3*a + 3*b*x)*tan((3*a + 3*b*x)/2)**2* 
b**3*d**3*x**3 - 6*cos(3*a + 3*b*x)*tan((3*a + 3*b*x)/2)**2*b*c*d**2 - 6*c 
os(3*a + 3*b*x)*tan((3*a + 3*b*x)/2)**2*b*d**3*x + 9*cos(3*a + 3*b*x)*b**3 
*c**3 + 27*cos(3*a + 3*b*x)*b**3*c**2*d*x + 27*cos(3*a + 3*b*x)*b**3*c*d** 
2*x**2 + 9*cos(3*a + 3*b*x)*b**3*d**3*x**3 - 6*cos(3*a + 3*b*x)*b*c*d**2 - 
 6*cos(3*a + 3*b*x)*b*d**3*x - 108*int(tan((3*a + 3*b*x)/2)/(tan((3*a + 3* 
b*x)/2)**2*tan((a + b*x)/2)**2 - tan((3*a + 3*b*x)/2)**2 + tan((a + b*x)/2 
)**2 - 1),x)*tan((3*a + 3*b*x)/2)**2*b**4*c**3 - 108*int(tan((3*a + 3*b*x) 
/2)/(tan((3*a + 3*b*x)/2)**2*tan((a + b*x)/2)**2 - tan((3*a + 3*b*x)/2)**2 
 + tan((a + b*x)/2)**2 - 1),x)*b**4*c**3 - 324*int((tan((3*a + 3*b*x)/2)*t 
an((a + b*x)/2)**2*x**2)/(tan((3*a + 3*b*x)/2)**2*tan((a + b*x)/2)**2 - ta 
n((3*a + 3*b*x)/2)**2 + tan((a + b*x)/2)**2 - 1),x)*tan((3*a + 3*b*x)/2)** 
2*b**4*c*d**2 - 324*int((tan((3*a + 3*b*x)/2)*tan((a + b*x)/2)**2*x**2)/(t 
an((3*a + 3*b*x)/2)**2*tan((a + b*x)/2)**2 - tan((3*a + 3*b*x)/2)**2 + tan 
((a + b*x)/2)**2 - 1),x)*b**4*c*d**2 - 108*int((tan((3*a + 3*b*x)/2)*x**3) 
/(tan((3*a + 3*b*x)/2)**2*tan((a + b*x)/2)**2 - tan((3*a + 3*b*x)/2)**2 + 
tan((a + b*x)/2)**2 - 1),x)*tan((3*a + 3*b*x)/2)**2*b**4*d**3 - 108*int((t 
an((3*a + 3*b*x)/2)*x**3)/(tan((3*a + 3*b*x)/2)**2*tan((a + b*x)/2)**2 ...