\(\int (c+d x)^4 \sec (a+b x) \sin (3 a+3 b x) \, dx\) [382]

Optimal result

Integrand size = 23, antiderivative size = 293 \[ \int (c+d x)^4 \sec (a+b x) \sin (3 a+3 b x) \, dx=\frac {3 d^2 (c+d x)^2}{b^3}-\frac {(c+d x)^4}{b}-\frac {i (c+d x)^5}{5 d}+\frac {(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}-\frac {2 i d (c+d x)^3 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac {3 d^2 (c+d x)^2 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{b^3}+\frac {3 i d^3 (c+d x) \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{b^4}-\frac {3 d^4 \operatorname {PolyLog}\left (5,-e^{2 i (a+b x)}\right )}{2 b^5}-\frac {6 d^3 (c+d x) \cos (a+b x) \sin (a+b x)}{b^4}+\frac {4 d (c+d x)^3 \cos (a+b x) \sin (a+b x)}{b^2}+\frac {3 d^4 \sin ^2(a+b x)}{b^5}-\frac {6 d^2 (c+d x)^2 \sin ^2(a+b x)}{b^3}+\frac {2 (c+d x)^4 \sin ^2(a+b x)}{b} \] Output:

3*d^2*(d*x+c)^2/b^3-(d*x+c)^4/b-1/5*I*(d*x+c)^5/d+(d*x+c)^4*ln(1+exp(2*I*( 
b*x+a)))/b-2*I*d*(d*x+c)^3*polylog(2,-exp(2*I*(b*x+a)))/b^2+3*d^2*(d*x+c)^ 
2*polylog(3,-exp(2*I*(b*x+a)))/b^3+3*I*d^3*(d*x+c)*polylog(4,-exp(2*I*(b*x 
+a)))/b^4-3/2*d^4*polylog(5,-exp(2*I*(b*x+a)))/b^5-6*d^3*(d*x+c)*cos(b*x+a 
)*sin(b*x+a)/b^4+4*d*(d*x+c)^3*cos(b*x+a)*sin(b*x+a)/b^2+3*d^4*sin(b*x+a)^ 
2/b^5-6*d^2*(d*x+c)^2*sin(b*x+a)^2/b^3+2*(d*x+c)^4*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. \(2529\) vs. \(2(293)=586\).

Time = 6.38 (sec) , antiderivative size = 2529, normalized size of antiderivative = 8.63 \[ \int (c+d x)^4 \sec (a+b x) \sin (3 a+3 b x) \, dx=\text {Result too large to show} \] Input:

Integrate[(c + d*x)^4*Sec[a + b*x]*Sin[3*a + 3*b*x],x]
 

Output:

((I/2)*c^2*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/2)*c*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*Po 
lyLog[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 + ((I/20)*d^4*E^(I*a)*((4*b^5*x^5)/E^((2*I)*a) 
 - (10*I)*b^4*(1 + E^((-2*I)*a))*x^4*Log[1 + E^((-2*I)*(a + b*x))] + 20*b^ 
3*(1 + E^((-2*I)*a))*x^3*PolyLog[2, -E^((-2*I)*(a + b*x))] - (30*I)*b^2*(1 
 + E^((-2*I)*a))*x^2*PolyLog[3, -E^((-2*I)*(a + b*x))] - 30*b*(1 + E^((-2* 
I)*a))*x*PolyLog[4, -E^((-2*I)*(a + b*x))] + (15*I)*(1 + E^((-2*I)*a))*Pol 
yLog[5, -E^((-2*I)*(a + b*x))])*Sec[a])/b^5 + (c^4*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)) + 
(2*c^3*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]])*Lo 
g[1 - E^((2*I)*(b*x - ArcTan[Cot[a]]))] + Pi*Log[Cos[b*x]] - 2*ArcTan[Cot[ 
a]]*Log[Sin[b*x - ArcTan[Cot[a]]]] + I*PolyLog[2, E^((2*I)*(b*x - ArcTan[C 
ot[a]]))]))/Sqrt[1 + Cot[a]^2])*Sec[a])/(b^2*Sqrt[Csc[a]^2*(Cos[a]^2 + Sin 
[a]^2)]) + Sec[a]*(Cos[2*a + 2*b*x]/(40*b^5) - ((I/40)*Sin[2*a + 2*b*x]...
 

Rubi [A] (verified)

Time = 0.73 (sec) , antiderivative size = 293, 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.

\(\displaystyle \int (c+d x)^4 \sin (3 a+3 b x) \sec (a+b x) \, dx\)

\(\Big \downarrow \) 4931

\(\displaystyle \int \left (3 (c+d x)^4 \sin (a+b x) \cos (a+b x)-(c+d x)^4 \sin ^2(a+b x) \tan (a+b x)\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {3 d^4 \operatorname {PolyLog}\left (5,-e^{2 i (a+b x)}\right )}{2 b^5}+\frac {3 d^4 \sin ^2(a+b x)}{b^5}+\frac {3 i d^3 (c+d x) \operatorname {PolyLog}\left (4,-e^{2 i (a+b x)}\right )}{b^4}-\frac {6 d^3 (c+d x) \sin (a+b x) \cos (a+b x)}{b^4}+\frac {3 d^2 (c+d x)^2 \operatorname {PolyLog}\left (3,-e^{2 i (a+b x)}\right )}{b^3}-\frac {6 d^2 (c+d x)^2 \sin ^2(a+b x)}{b^3}-\frac {2 i d (c+d x)^3 \operatorname {PolyLog}\left (2,-e^{2 i (a+b x)}\right )}{b^2}+\frac {4 d (c+d x)^3 \sin (a+b x) \cos (a+b x)}{b^2}+\frac {(c+d x)^4 \log \left (1+e^{2 i (a+b x)}\right )}{b}+\frac {2 (c+d x)^4 \sin ^2(a+b x)}{b}+\frac {3 d^2 (c+d x)^2}{b^3}-\frac {(c+d x)^4}{b}-\frac {i (c+d x)^5}{5 d}\)

Input:

Int[(c + d*x)^4*Sec[a + b*x]*Sin[3*a + 3*b*x],x]
 

Output:

(3*d^2*(c + d*x)^2)/b^3 - (c + d*x)^4/b - ((I/5)*(c + d*x)^5)/d + ((c + d* 
x)^4*Log[1 + E^((2*I)*(a + b*x))])/b - ((2*I)*d*(c + d*x)^3*PolyLog[2, -E^ 
((2*I)*(a + b*x))])/b^2 + (3*d^2*(c + d*x)^2*PolyLog[3, -E^((2*I)*(a + b*x 
))])/b^3 + ((3*I)*d^3*(c + d*x)*PolyLog[4, -E^((2*I)*(a + b*x))])/b^4 - (3 
*d^4*PolyLog[5, -E^((2*I)*(a + b*x))])/(2*b^5) - (6*d^3*(c + d*x)*Cos[a + 
b*x]*Sin[a + b*x])/b^4 + (4*d*(c + d*x)^3*Cos[a + b*x]*Sin[a + b*x])/b^2 + 
 (3*d^4*Sin[a + b*x]^2)/b^5 - (6*d^2*(c + d*x)^2*Sin[a + b*x]^2)/b^3 + (2* 
(c + d*x)^4*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. 964 vs. \(2 (276 ) = 552\).

Time = 6.35 (sec) , antiderivative size = 965, normalized size of antiderivative = 3.29

method	result	size
risch	\(-\frac {8 i d^{3} c \,a^{3} x}{b^{3}}-\frac {6 i d^{3} c \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {2 c^{4} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {\left (2 d^{4} x^{4} b^{4}+8 b^{4} c \,d^{3} x^{3}-4 i b^{3} d^{4} x^{3}+12 b^{4} c^{2} d^{2} x^{2}-12 i b^{3} c \,d^{3} x^{2}+8 b^{4} c^{3} d x -12 i b^{3} c^{2} d^{2} x +2 b^{4} c^{4}-4 i b^{3} c^{3} d -6 b^{2} d^{4} x^{2}-12 b^{2} c \,d^{3} x +6 i b \,d^{4} x -6 b^{2} c^{2} d^{2}+6 i b c \,d^{3}+3 d^{4}\right ) {\mathrm e}^{-2 i \left (b x +a \right )}}{4 b^{5}}-\frac {\left (2 d^{4} x^{4} b^{4}+8 b^{4} c \,d^{3} x^{3}+4 i b^{3} d^{4} x^{3}+12 b^{4} c^{2} d^{2} x^{2}+12 i b^{3} c \,d^{3} x^{2}+8 b^{4} c^{3} d x +12 i b^{3} c^{2} d^{2} x +2 b^{4} c^{4}+4 i b^{3} c^{3} d -6 b^{2} d^{4} x^{2}-12 b^{2} c \,d^{3} x -6 i b \,d^{4} x -6 b^{2} c^{2} d^{2}-6 i b c \,d^{3}+3 d^{4}\right ) {\mathrm e}^{2 i \left (b x +a \right )}}{4 b^{5}}-\frac {i d^{4} x^{5}}{5}+i c^{4} x +\frac {i c^{5}}{5 d}+\frac {c^{4} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}{b}-\frac {3 d^{4} \operatorname {polylog}\left (5, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{2 b^{5}}-\frac {8 i d \,c^{3} a x}{b}-\frac {6 i d^{2} c^{2} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {12 i d^{2} c^{2} a^{2} x}{b^{2}}-i d^{3} c \,x^{4}-2 i d^{2} c^{2} x^{3}-2 i d \,c^{3} x^{2}-\frac {2 d^{4} a^{4} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{5}}+\frac {3 d^{4} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{2}}{b^{3}}+\frac {3 d^{2} c^{2} \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{3}}+\frac {d^{4} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{4}}{b}+\frac {8 i d^{4} a^{5}}{5 b^{5}}+\frac {3 i d^{4} \operatorname {polylog}\left (4, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{4}}+\frac {8 i d^{2} c^{2} a^{3}}{b^{3}}-\frac {2 i d^{4} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x^{3}}{b^{2}}-\frac {6 i d^{3} c \,a^{4}}{b^{4}}+\frac {3 i d^{3} c \operatorname {polylog}\left (4, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{4}}+\frac {2 i d^{4} a^{4} x}{b^{4}}-\frac {4 i d \,c^{3} a^{2}}{b^{2}}-\frac {2 i d \,c^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{2 i \left (b x +a \right )}\right )}{b^{2}}+\frac {8 a^{3} d^{3} c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {12 a^{2} d^{2} c^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {6 d^{3} c \operatorname {polylog}\left (3, -{\mathrm e}^{2 i \left (b x +a \right )}\right ) x}{b^{3}}+\frac {4 d^{3} c \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{3}}{b}+\frac {4 d \,c^{3} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x}{b}+\frac {6 d^{2} c^{2} \ln \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {8 a d \,c^{3} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}\)	\(965\)

Input:

int((d*x+c)^4*sec(b*x+a)*sin(3*b*x+3*a),x,method=_RETURNVERBOSE)
 

Output:

-2/b*c^4*ln(exp(I*(b*x+a)))+12*I/b^2*d^2*c^2*a^2*x-8*I/b*d*c^3*a*x-1/4*(2* 
d^4*x^4*b^4-4*I*b^3*d^4*x^3+8*b^4*c*d^3*x^3-12*I*b^3*c*d^3*x^2+12*b^4*c^2* 
d^2*x^2-12*I*b^3*c^2*d^2*x+8*b^4*c^3*d*x-4*I*b^3*c^3*d+2*b^4*c^4-6*b^2*d^4 
*x^2+6*I*b*d^4*x-12*b^2*c*d^3*x+6*I*b*c*d^3-6*b^2*c^2*d^2+3*d^4)/b^5*exp(- 
2*I*(b*x+a))-6*I/b^2*d^2*c^2*polylog(2,-exp(2*I*(b*x+a)))*x-6*I/b^2*d^3*c* 
polylog(2,-exp(2*I*(b*x+a)))*x^2-8*I/b^3*d^3*c*a^3*x-I*d^3*c*x^4-2*I*d^2*c 
^2*x^3-2*I*d*c^3*x^2+3*I/b^4*d^4*polylog(4,-exp(2*I*(b*x+a)))*x+8*I/b^3*d^ 
2*c^2*a^3-2*I/b^2*d^4*polylog(2,-exp(2*I*(b*x+a)))*x^3-6*I/b^4*d^3*c*a^4+3 
*I/b^4*d^3*c*polylog(4,-exp(2*I*(b*x+a)))+2*I/b^4*d^4*a^4*x-4*I/b^2*d*c^3* 
a^2-2*I/b^2*d*c^3*polylog(2,-exp(2*I*(b*x+a)))-2/b^5*d^4*a^4*ln(exp(I*(b*x 
+a)))+3/b^3*d^4*polylog(3,-exp(2*I*(b*x+a)))*x^2+3/b^3*d^2*c^2*polylog(3,- 
exp(2*I*(b*x+a)))+1/b*d^4*ln(exp(2*I*(b*x+a))+1)*x^4+8/5*I/b^5*d^4*a^5+8/b 
^4*a^3*d^3*c*ln(exp(I*(b*x+a)))-12/b^3*a^2*d^2*c^2*ln(exp(I*(b*x+a)))+6/b^ 
3*d^3*c*polylog(3,-exp(2*I*(b*x+a)))*x+4/b*d^3*c*ln(exp(2*I*(b*x+a))+1)*x^ 
3+4/b*d*c^3*ln(exp(2*I*(b*x+a))+1)*x+6/b*d^2*c^2*ln(exp(2*I*(b*x+a))+1)*x^ 
2+8/b^2*a*d*c^3*ln(exp(I*(b*x+a)))-1/4*(2*d^4*x^4*b^4+4*I*b^3*d^4*x^3+8*b^ 
4*c*d^3*x^3+12*I*b^3*c*d^3*x^2+12*b^4*c^2*d^2*x^2+12*I*b^3*c^2*d^2*x+8*b^4 
*c^3*d*x+4*I*b^3*c^3*d+2*b^4*c^4-6*b^2*d^4*x^2-6*I*b*d^4*x-12*b^2*c*d^3*x- 
6*I*b*c*d^3-6*b^2*c^2*d^2+3*d^4)/b^5*exp(2*I*(b*x+a))-1/5*I*d^4*x^5+I*c^4* 
x+1/5*I/d*c^5+1/b*c^4*ln(exp(2*I*(b*x+a))+1)-3/2*d^4*polylog(5,-exp(2*I...
 

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. 1652 vs. \(2 (272) = 544\).

Time = 0.18 (sec) , antiderivative size = 1652, normalized size of antiderivative = 5.64 \[ \int (c+d x)^4 \sec (a+b x) \sin (3 a+3 b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^4*sec(b*x+a)*sin(3*b*x+3*a),x, algorithm="fricas")
 

Output:

1/2*(2*b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 - 24*d^4*polylog(5, I*cos(b*x + a) + 
sin(b*x + a)) - 24*d^4*polylog(5, I*cos(b*x + a) - sin(b*x + a)) - 24*d^4* 
polylog(5, -I*cos(b*x + a) + sin(b*x + a)) - 24*d^4*polylog(5, -I*cos(b*x 
+ a) - sin(b*x + a)) + 6*(2*b^4*c^2*d^2 - b^2*d^4)*x^2 - 2*(2*b^4*d^4*x^4 
+ 8*b^4*c*d^3*x^3 + 2*b^4*c^4 - 6*b^2*c^2*d^2 + 3*d^4 + 6*(2*b^4*c^2*d^2 - 
 b^2*d^4)*x^2 + 4*(2*b^4*c^3*d - 3*b^2*c*d^3)*x)*cos(b*x + a)^2 + 4*(2*b^3 
*d^4*x^3 + 6*b^3*c*d^3*x^2 + 2*b^3*c^3*d - 3*b*c*d^3 + 3*(2*b^3*c^2*d^2 - 
b*d^4)*x)*cos(b*x + a)*sin(b*x + a) + 4*(2*b^4*c^3*d - 3*b^2*c*d^3)*x - 4* 
(-I*b^3*d^4*x^3 - 3*I*b^3*c*d^3*x^2 - 3*I*b^3*c^2*d^2*x - I*b^3*c^3*d)*dil 
og(I*cos(b*x + a) + sin(b*x + a)) - 4*(I*b^3*d^4*x^3 + 3*I*b^3*c*d^3*x^2 + 
 3*I*b^3*c^2*d^2*x + I*b^3*c^3*d)*dilog(I*cos(b*x + a) - sin(b*x + a)) - 4 
*(I*b^3*d^4*x^3 + 3*I*b^3*c*d^3*x^2 + 3*I*b^3*c^2*d^2*x + I*b^3*c^3*d)*dil 
og(-I*cos(b*x + a) + sin(b*x + a)) - 4*(-I*b^3*d^4*x^3 - 3*I*b^3*c*d^3*x^2 
 - 3*I*b^3*c^2*d^2*x - I*b^3*c^3*d)*dilog(-I*cos(b*x + a) - sin(b*x + a)) 
+ (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)* 
log(cos(b*x + a) + I*sin(b*x + a) + I) + (b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2* 
b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*log(cos(b*x + a) - I*sin(b*x + a) + 
 I) + (b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2*d^2*x^2 + 4*b^4*c^3*d*x + 
 4*a*b^3*c^3*d - 6*a^2*b^2*c^2*d^2 + 4*a^3*b*c*d^3 - a^4*d^4)*log(I*cos(b* 
x + a) + sin(b*x + a) + 1) + (b^4*d^4*x^4 + 4*b^4*c*d^3*x^3 + 6*b^4*c^2...
 

Sympy [F(-1)]

Timed out. \[ \int (c+d x)^4 \sec (a+b x) \sin (3 a+3 b x) \, dx=\text {Timed out} \] Input:

integrate((d*x+c)**4*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. 610 vs. \(2 (272) = 544\).

Time = 0.31 (sec) , antiderivative size = 610, normalized size of antiderivative = 2.08 \[ \int (c+d x)^4 \sec (a+b x) \sin (3 a+3 b x) \, dx=-\frac {c^{4} {\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 {-6 i \, b^{5} d^{4} x^{5} - 30 i \, b^{5} c d^{3} x^{4} - 60 i \, b^{5} c^{2} d^{2} x^{3} - 60 i \, b^{5} c^{3} d x^{2} - 90 \, d^{4} {\rm Li}_{5}(-e^{\left (2 i \, b x + 2 i \, a\right )}) - 20 \, {\left (-3 i \, b^{4} d^{4} x^{4} - 8 i \, b^{4} c d^{3} x^{3} - 9 i \, b^{4} c^{2} d^{2} x^{2} - 6 i \, b^{4} c^{3} d x\right )} \arctan \left (\sin \left (2 \, b x + 2 \, a\right ), \cos \left (2 \, b x + 2 \, a\right ) + 1\right ) - 15 \, {\left (2 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} - 6 \, b^{2} c^{2} d^{2} + 3 \, d^{4} + 6 \, {\left (2 \, b^{4} c^{2} d^{2} - b^{2} d^{4}\right )} x^{2} + 4 \, {\left (2 \, b^{4} c^{3} d - 3 \, b^{2} c d^{3}\right )} x\right )} \cos \left (2 \, b x + 2 \, a\right ) - 60 \, {\left (2 i \, b^{3} d^{4} x^{3} + 4 i \, b^{3} c d^{3} x^{2} + 3 i \, b^{3} c^{2} d^{2} x + i \, b^{3} c^{3} d\right )} {\rm Li}_2\left (-e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + 10 \, {\left (3 \, b^{4} d^{4} x^{4} + 8 \, b^{4} c d^{3} x^{3} + 9 \, b^{4} c^{2} d^{2} x^{2} + 6 \, b^{4} c^{3} 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 ) - 60 \, {\left (-3 i \, b d^{4} x - 2 i \, b c d^{3}\right )} {\rm Li}_{4}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 30 \, {\left (6 \, b^{2} d^{4} x^{2} + 8 \, b^{2} c d^{3} x + 3 \, b^{2} c^{2} d^{2}\right )} {\rm Li}_{3}(-e^{\left (2 i \, b x + 2 i \, a\right )}) + 30 \, {\left (2 \, b^{3} d^{4} x^{3} + 6 \, b^{3} c d^{3} x^{2} + 2 \, b^{3} c^{3} d - 3 \, b c d^{3} + 3 \, {\left (2 \, b^{3} c^{2} d^{2} - b d^{4}\right )} x\right )} \sin \left (2 \, b x + 2 \, a\right )}{30 \, b^{5}} \] Input:

integrate((d*x+c)^4*sec(b*x+a)*sin(3*b*x+3*a),x, algorithm="maxima")
 

Output:

-1/2*c^4*(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 + 1/30* 
(-6*I*b^5*d^4*x^5 - 30*I*b^5*c*d^3*x^4 - 60*I*b^5*c^2*d^2*x^3 - 60*I*b^5*c 
^3*d*x^2 - 90*d^4*polylog(5, -e^(2*I*b*x + 2*I*a)) - 20*(-3*I*b^4*d^4*x^4 
- 8*I*b^4*c*d^3*x^3 - 9*I*b^4*c^2*d^2*x^2 - 6*I*b^4*c^3*d*x)*arctan2(sin(2 
*b*x + 2*a), cos(2*b*x + 2*a) + 1) - 15*(2*b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 - 
 6*b^2*c^2*d^2 + 3*d^4 + 6*(2*b^4*c^2*d^2 - b^2*d^4)*x^2 + 4*(2*b^4*c^3*d 
- 3*b^2*c*d^3)*x)*cos(2*b*x + 2*a) - 60*(2*I*b^3*d^4*x^3 + 4*I*b^3*c*d^3*x 
^2 + 3*I*b^3*c^2*d^2*x + I*b^3*c^3*d)*dilog(-e^(2*I*b*x + 2*I*a)) + 10*(3* 
b^4*d^4*x^4 + 8*b^4*c*d^3*x^3 + 9*b^4*c^2*d^2*x^2 + 6*b^4*c^3*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) - 60*(-3*I* 
b*d^4*x - 2*I*b*c*d^3)*polylog(4, -e^(2*I*b*x + 2*I*a)) + 30*(6*b^2*d^4*x^ 
2 + 8*b^2*c*d^3*x + 3*b^2*c^2*d^2)*polylog(3, -e^(2*I*b*x + 2*I*a)) + 30*( 
2*b^3*d^4*x^3 + 6*b^3*c*d^3*x^2 + 2*b^3*c^3*d - 3*b*c*d^3 + 3*(2*b^3*c^2*d 
^2 - b*d^4)*x)*sin(2*b*x + 2*a))/b^5
 

Giac [F]

\[ \int (c+d x)^4 \sec (a+b x) \sin (3 a+3 b x) \, dx=\int { {\left (d x + c\right )}^{4} \sec \left (b x + a\right ) \sin \left (3 \, b x + 3 \, a\right ) \,d x } \] Input:

integrate((d*x+c)^4*sec(b*x+a)*sin(3*b*x+3*a),x, algorithm="giac")
                                                                                    
                                                                                    
 

Output:

integrate((d*x + c)^4*sec(b*x + a)*sin(3*b*x + 3*a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^4 \sec (a+b x) \sin (3 a+3 b x) \, dx=\int \frac {\sin \left (3\,a+3\,b\,x\right )\,{\left (c+d\,x\right )}^4}{\cos \left (a+b\,x\right )} \,d x \] Input:

int((sin(3*a + 3*b*x)*(c + d*x)^4)/cos(a + b*x),x)
 

Output:

int((sin(3*a + 3*b*x)*(c + d*x)^4)/cos(a + b*x), x)
 

Reduce [F]

\[ \int (c+d x)^4 \sec (a+b x) \sin (3 a+3 b x) \, dx =\text {Too large to display} \] Input:

int((d*x+c)^4*sec(b*x+a)*sin(3*b*x+3*a),x)
 

Output:

(27*cos(3*a + 3*b*x)*b**4*c**4 + 108*cos(3*a + 3*b*x)*b**4*c**3*d*x + 162* 
cos(3*a + 3*b*x)*b**4*c**2*d**2*x**2 + 108*cos(3*a + 3*b*x)*b**4*c*d**3*x* 
*3 + 27*cos(3*a + 3*b*x)*b**4*d**4*x**4 - 36*cos(3*a + 3*b*x)*b**2*c**2*d* 
*2 - 72*cos(3*a + 3*b*x)*b**2*c*d**3*x - 36*cos(3*a + 3*b*x)*b**2*d**4*x** 
2 + 8*cos(3*a + 3*b*x)*d**4 - 324*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**5*c**4 - 324*int((tan((3*a + 3*b*x)/2)*x**4)/(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**5*d**4 - 1296*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)*b**5*c*d**3 - 1944*int((tan((3*a + 3*b*x)/2)*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**5*c**2*d**2 - 1296*int((tan((3*a + 3*b*x)/2)*x)/(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**5*c**3*d - 36*sin(3*a + 3*b*x)*b**3*c**3*d - 108 
*sin(3*a + 3*b*x)*b**3*c**2*d**2*x - 108*sin(3*a + 3*b*x)*b**3*c*d**3*x**2 
 - 36*sin(3*a + 3*b*x)*b**3*d**4*x**3 + 24*sin(3*a + 3*b*x)*b*c*d**3 + 24* 
sin(3*a + 3*b*x)*b*d**4*x + 108*a*b**3*c**3*d - 72*a*b*c*d**3 - 27*b**4*c* 
*4 + 36*b**2*c**2*d**2 - 8*d**4)/(81*b**5)