\(\int (c+d x)^3 \sec ^2(a+b x) \sin (3 a+3 b x) \, dx\) [389]

Optimal result

Integrand size = 25, antiderivative size = 230 \[ \int (c+d x)^3 \sec ^2(a+b x) \sin (3 a+3 b x) \, dx=-\frac {6 i d (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b^2}+\frac {24 d^2 (c+d x) \cos (a+b x)}{b^3}-\frac {4 (c+d x)^3 \cos (a+b x)}{b}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}-\frac {6 d^3 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^4}+\frac {6 d^3 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^4}-\frac {(c+d x)^3 \sec (a+b x)}{b}-\frac {24 d^3 \sin (a+b x)}{b^4}+\frac {12 d (c+d x)^2 \sin (a+b x)}{b^2} \] Output:

-6*I*d*(d*x+c)^2*arctan(exp(I*(b*x+a)))/b^2+24*d^2*(d*x+c)*cos(b*x+a)/b^3- 
4*(d*x+c)^3*cos(b*x+a)/b+6*I*d^2*(d*x+c)*polylog(2,-I*exp(I*(b*x+a)))/b^3- 
6*I*d^2*(d*x+c)*polylog(2,I*exp(I*(b*x+a)))/b^3-6*d^3*polylog(3,-I*exp(I*( 
b*x+a)))/b^4+6*d^3*polylog(3,I*exp(I*(b*x+a)))/b^4-(d*x+c)^3*sec(b*x+a)/b- 
24*d^3*sin(b*x+a)/b^4+12*d*(d*x+c)^2*sin(b*x+a)/b^2
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(532\) vs. \(2(230)=460\).

Time = 1.57 (sec) , antiderivative size = 532, normalized size of antiderivative = 2.31 \[ \int (c+d x)^3 \sec ^2(a+b x) \sin (3 a+3 b x) \, dx=-\frac {\sec (a+b x) \left (3 b^3 c^3-12 b c d^2+9 b^3 c^2 d x-12 b d^3 x+9 b^3 c d^2 x^2+3 b^3 d^3 x^3+6 i b^2 c^2 d \arctan \left (e^{i (a+b x)}\right ) \cos (a+b x)+2 b^3 c^3 \cos (2 (a+b x))-12 b c d^2 \cos (2 (a+b x))+6 b^3 c^2 d x \cos (2 (a+b x))-12 b d^3 x \cos (2 (a+b x))+6 b^3 c d^2 x^2 \cos (2 (a+b x))+2 b^3 d^3 x^3 \cos (2 (a+b x))-6 b^2 c d^2 x \cos (a+b x) \log \left (1-i e^{i (a+b x)}\right )-3 b^2 d^3 x^2 \cos (a+b x) \log \left (1-i e^{i (a+b x)}\right )+6 b^2 c d^2 x \cos (a+b x) \log \left (1+i e^{i (a+b x)}\right )+3 b^2 d^3 x^2 \cos (a+b x) \log \left (1+i e^{i (a+b x)}\right )-6 i b d^2 (c+d x) \cos (a+b x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )+6 i b d^2 (c+d x) \cos (a+b x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )+6 d^3 \cos (a+b x) \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )-6 d^3 \cos (a+b x) \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )-6 b^2 c^2 d \sin (2 (a+b x))+12 d^3 \sin (2 (a+b x))-12 b^2 c d^2 x \sin (2 (a+b x))-6 b^2 d^3 x^2 \sin (2 (a+b x))\right )}{b^4} \] Input:

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

Output:

-((Sec[a + b*x]*(3*b^3*c^3 - 12*b*c*d^2 + 9*b^3*c^2*d*x - 12*b*d^3*x + 9*b 
^3*c*d^2*x^2 + 3*b^3*d^3*x^3 + (6*I)*b^2*c^2*d*ArcTan[E^(I*(a + b*x))]*Cos 
[a + b*x] + 2*b^3*c^3*Cos[2*(a + b*x)] - 12*b*c*d^2*Cos[2*(a + b*x)] + 6*b 
^3*c^2*d*x*Cos[2*(a + b*x)] - 12*b*d^3*x*Cos[2*(a + b*x)] + 6*b^3*c*d^2*x^ 
2*Cos[2*(a + b*x)] + 2*b^3*d^3*x^3*Cos[2*(a + b*x)] - 6*b^2*c*d^2*x*Cos[a 
+ b*x]*Log[1 - I*E^(I*(a + b*x))] - 3*b^2*d^3*x^2*Cos[a + b*x]*Log[1 - I*E 
^(I*(a + b*x))] + 6*b^2*c*d^2*x*Cos[a + b*x]*Log[1 + I*E^(I*(a + b*x))] + 
3*b^2*d^3*x^2*Cos[a + b*x]*Log[1 + I*E^(I*(a + b*x))] - (6*I)*b*d^2*(c + d 
*x)*Cos[a + b*x]*PolyLog[2, (-I)*E^(I*(a + b*x))] + (6*I)*b*d^2*(c + d*x)* 
Cos[a + b*x]*PolyLog[2, I*E^(I*(a + b*x))] + 6*d^3*Cos[a + b*x]*PolyLog[3, 
 (-I)*E^(I*(a + b*x))] - 6*d^3*Cos[a + b*x]*PolyLog[3, I*E^(I*(a + b*x))] 
- 6*b^2*c^2*d*Sin[2*(a + b*x)] + 12*d^3*Sin[2*(a + b*x)] - 12*b^2*c*d^2*x* 
Sin[2*(a + b*x)] - 6*b^2*d^3*x^2*Sin[2*(a + b*x)]))/b^4)
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, 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.

\(\displaystyle \int (c+d x)^3 \sin (3 a+3 b x) \sec ^2(a+b x) \, dx\)

\(\Big \downarrow \) 4931

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {6 i d (c+d x)^2 \arctan \left (e^{i (a+b x)}\right )}{b^2}-\frac {6 d^3 \operatorname {PolyLog}\left (3,-i e^{i (a+b x)}\right )}{b^4}+\frac {6 d^3 \operatorname {PolyLog}\left (3,i e^{i (a+b x)}\right )}{b^4}-\frac {24 d^3 \sin (a+b x)}{b^4}+\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (a+b x)}\right )}{b^3}-\frac {6 i d^2 (c+d x) \operatorname {PolyLog}\left (2,i e^{i (a+b x)}\right )}{b^3}+\frac {24 d^2 (c+d x) \cos (a+b x)}{b^3}+\frac {12 d (c+d x)^2 \sin (a+b x)}{b^2}-\frac {4 (c+d x)^3 \cos (a+b x)}{b}-\frac {(c+d x)^3 \sec (a+b x)}{b}\)

Input:

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

Output:

((-6*I)*d*(c + d*x)^2*ArcTan[E^(I*(a + b*x))])/b^2 + (24*d^2*(c + d*x)*Cos 
[a + b*x])/b^3 - (4*(c + d*x)^3*Cos[a + b*x])/b + ((6*I)*d^2*(c + d*x)*Pol 
yLog[2, (-I)*E^(I*(a + b*x))])/b^3 - ((6*I)*d^2*(c + d*x)*PolyLog[2, I*E^( 
I*(a + b*x))])/b^3 - (6*d^3*PolyLog[3, (-I)*E^(I*(a + b*x))])/b^4 + (6*d^3 
*PolyLog[3, I*E^(I*(a + b*x))])/b^4 - ((c + d*x)^3*Sec[a + b*x])/b - (24*d 
^3*Sin[a + b*x])/b^4 + (12*d*(c + d*x)^2*Sin[a + b*x])/b^2
 

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. 676 vs. \(2 (213 ) = 426\).

Time = 8.58 (sec) , antiderivative size = 677, normalized size of antiderivative = 2.94

method	result	size
risch	\(-\frac {2 \left (d^{3} x^{3} b^{3}+3 b^{3} c \,d^{2} x^{2}+3 i b^{2} d^{3} x^{2}+3 b^{3} c^{2} d x +6 i b^{2} c \,d^{2} x +b^{3} c^{3}+3 i b^{2} c^{2} d -6 b \,d^{3} x -6 c \,d^{2} b -6 i d^{3}\right ) {\mathrm e}^{i \left (b x +a \right )}}{b^{4}}-\frac {2 \left (d^{3} x^{3} b^{3}+3 b^{3} c \,d^{2} x^{2}-3 i b^{2} d^{3} x^{2}+3 b^{3} c^{2} d x -6 i b^{2} c \,d^{2} x +b^{3} c^{3}-3 i b^{2} c^{2} d -6 b \,d^{3} x -6 c \,d^{2} b +6 i d^{3}\right ) {\mathrm e}^{-i \left (b x +a \right )}}{b^{4}}-\frac {2 \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right ) {\mathrm e}^{i \left (b x +a \right )}}{b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )}+\frac {3 d^{3} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b^{2}}-\frac {6 d^{3} \operatorname {polylog}\left (3, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 d^{2} c \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {6 c \,d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b^{2}}+\frac {12 i d^{2} c a \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 c \,d^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{3}}+\frac {3 d^{3} a^{2} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{4}}-\frac {6 i d^{3} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}-\frac {6 i d^{2} c \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {6 i d \,c^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {6 c \,d^{2} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{3}}-\frac {6 i d^{3} a^{2} \arctan \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}-\frac {3 d^{3} \ln \left (i {\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b^{2}}+\frac {6 i d^{2} c \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {3 a^{2} d^{3} \ln \left (1-i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 d^{3} \operatorname {polylog}\left (3, i {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{4}}+\frac {6 i d^{3} \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{3}}\)	\(677\)

Input:

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

Output:

-2*(d^3*x^3*b^3+3*b^3*c*d^2*x^2+3*b^3*c^2*d*x+b^3*c^3+3*I*b^2*d^3*x^2-6*b* 
d^3*x+6*I*b^2*c*d^2*x-6*c*d^2*b+3*I*b^2*c^2*d-6*I*d^3)/b^4*exp(I*(b*x+a))- 
2*(d^3*x^3*b^3+3*b^3*c*d^2*x^2+3*b^3*c^2*d*x+b^3*c^3-3*I*b^2*d^3*x^2-6*b*d 
^3*x-6*I*b^2*c*d^2*x-6*c*d^2*b-3*I*b^2*c^2*d+6*I*d^3)/b^4*exp(-I*(b*x+a))- 
2*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)*exp(I*(b*x+a))/b/(exp(2*I*(b*x+a))+1 
)+3/b^2*d^3*ln(1-I*exp(I*(b*x+a)))*x^2-6*d^3*polylog(3,-I*exp(I*(b*x+a)))/ 
b^4+6/b^2*c*d^2*ln(1-I*exp(I*(b*x+a)))*x-6/b^2*c*d^2*ln(I*exp(I*(b*x+a))+1 
)*x+12*I/b^3*d^2*c*a*arctan(exp(I*(b*x+a)))-6/b^3*c*d^2*ln(I*exp(I*(b*x+a) 
)+1)*a+3/b^4*a^2*d^3*ln(I*exp(I*(b*x+a))+1)-6*I/b^3*d^3*polylog(2,I*exp(I* 
(b*x+a)))*x-6*I/b^3*d^2*c*polylog(2,I*exp(I*(b*x+a)))-6*I/b^2*d*c^2*arctan 
(exp(I*(b*x+a)))+6/b^3*c*d^2*ln(1-I*exp(I*(b*x+a)))*a-6*I/b^4*d^3*a^2*arct 
an(exp(I*(b*x+a)))-3/b^2*d^3*ln(I*exp(I*(b*x+a))+1)*x^2+6*I/b^3*d^2*c*poly 
log(2,-I*exp(I*(b*x+a)))-3/b^4*a^2*d^3*ln(1-I*exp(I*(b*x+a)))+6*d^3*polylo 
g(3,I*exp(I*(b*x+a)))/b^4+6*I/b^3*d^3*polylog(2,-I*exp(I*(b*x+a)))*x
 

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. 896 vs. \(2 (204) = 408\).

Time = 0.13 (sec) , antiderivative size = 896, normalized size of antiderivative = 3.90 \[ \int (c+d x)^3 \sec ^2(a+b x) \sin (3 a+3 b x) \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^3*sec(b*x+a)^2*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*b^3*c^2*d*x + 2*b^3*c^3 + 6*d^3* 
cos(b*x + a)*polylog(3, I*cos(b*x + a) + sin(b*x + a)) - 6*d^3*cos(b*x + a 
)*polylog(3, I*cos(b*x + a) - sin(b*x + a)) + 6*d^3*cos(b*x + a)*polylog(3 
, -I*cos(b*x + a) + sin(b*x + a)) - 6*d^3*cos(b*x + a)*polylog(3, -I*cos(b 
*x + a) - sin(b*x + a)) + 8*(b^3*d^3*x^3 + 3*b^3*c*d^2*x^2 + b^3*c^3 - 6*b 
*c*d^2 + 3*(b^3*c^2*d - 2*b*d^3)*x)*cos(b*x + a)^2 + 6*(I*b*d^3*x + I*b*c* 
d^2)*cos(b*x + a)*dilog(I*cos(b*x + a) + sin(b*x + a)) + 6*(I*b*d^3*x + I* 
b*c*d^2)*cos(b*x + a)*dilog(I*cos(b*x + a) - sin(b*x + a)) + 6*(-I*b*d^3*x 
 - I*b*c*d^2)*cos(b*x + a)*dilog(-I*cos(b*x + a) + sin(b*x + a)) + 6*(-I*b 
*d^3*x - I*b*c*d^2)*cos(b*x + a)*dilog(-I*cos(b*x + a) - sin(b*x + a)) - 3 
*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*cos(b*x + a)*log(cos(b*x + a) + I*sin 
(b*x + a) + I) + 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*cos(b*x + a)*log(co 
s(b*x + a) - I*sin(b*x + a) + I) - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b* 
c*d^2 - a^2*d^3)*cos(b*x + a)*log(I*cos(b*x + a) + sin(b*x + a) + 1) + 3*( 
b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*cos(b*x + a)*log(I*co 
s(b*x + a) - sin(b*x + a) + 1) - 3*(b^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c* 
d^2 - a^2*d^3)*cos(b*x + a)*log(-I*cos(b*x + a) + sin(b*x + a) + 1) + 3*(b 
^2*d^3*x^2 + 2*b^2*c*d^2*x + 2*a*b*c*d^2 - a^2*d^3)*cos(b*x + a)*log(-I*co 
s(b*x + a) - sin(b*x + a) + 1) - 3*(b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*cos 
(b*x + a)*log(-cos(b*x + a) + I*sin(b*x + a) + I) + 3*(b^2*c^2*d - 2*a*...
 

Sympy [F(-1)]

Timed out. \[ \int (c+d x)^3 \sec ^2(a+b x) \sin (3 a+3 b x) \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

\[ \int (c+d x)^3 \sec ^2(a+b x) \sin (3 a+3 b x) \, dx=\int { {\left (d x + c\right )}^{3} \sec \left (b x + a\right )^{2} \sin \left (3 \, b x + 3 \, a\right ) \,d x } \] Input:

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

Output:

-2*((cos(3*b*x + 3*a) + cos(b*x + a))*cos(4*b*x + 4*a) + (3*cos(2*b*x + 2* 
a) + 1)*cos(3*b*x + 3*a) + 3*cos(2*b*x + 2*a)*cos(b*x + a) + (sin(3*b*x + 
3*a) + sin(b*x + a))*sin(4*b*x + 4*a) + 3*sin(3*b*x + 3*a)*sin(2*b*x + 2*a 
) + 3*sin(2*b*x + 2*a)*sin(b*x + a) + cos(b*x + a))*c^3/(b*cos(3*b*x + 3*a 
)^2 + 2*b*cos(3*b*x + 3*a)*cos(b*x + a) + b*cos(b*x + a)^2 + b*sin(3*b*x + 
 3*a)^2 + 2*b*sin(3*b*x + 3*a)*sin(b*x + a) + b*sin(b*x + a)^2) - 3/2*(4*( 
cos(a)^2 + sin(a)^2)*b*x*cos(b*x + a) + 12*(b*x*cos(2*b*x + 3*a)*cos(b*x + 
 2*a) + b*x*cos(b*x + 2*a)*cos(a) + b*x*sin(2*b*x + 3*a)*sin(b*x + 2*a) + 
b*x*sin(b*x + 2*a)*sin(a))*cos(3*b*x + 3*a)^2 + 4*(b*x*cos(b*x + a) - sin( 
b*x + a))*cos(2*b*x + 3*a)^2 + 12*(b*x*cos(2*b*x + 3*a)*cos(b*x + 2*a) + b 
*x*cos(b*x + 2*a)*cos(a) + b*x*sin(2*b*x + 3*a)*sin(b*x + 2*a) + b*x*sin(b 
*x + 2*a)*sin(a))*sin(3*b*x + 3*a)^2 + 4*(b*x*cos(b*x + a) - sin(b*x + a)) 
*sin(2*b*x + 3*a)^2 + 4*((b*x*cos(2*b*x + 3*a) + b*x*cos(a) + sin(2*b*x + 
3*a) + sin(a))*cos(3*b*x + 3*a)^2 + (b*x*cos(a) + sin(a))*cos(b*x + a)^2 + 
 (b*x*cos(2*b*x + 3*a) + b*x*cos(a) + sin(2*b*x + 3*a) + sin(a))*sin(3*b*x 
 + 3*a)^2 + (b*x*cos(a) + sin(a))*sin(b*x + a)^2 + 2*(b*x*cos(2*b*x + 3*a) 
*cos(b*x + a) + (b*x*cos(a) + sin(a))*cos(b*x + a) + cos(b*x + a)*sin(2*b* 
x + 3*a))*cos(3*b*x + 3*a) + (b*x*cos(b*x + a)^2 + b*x*sin(b*x + a)^2)*cos 
(2*b*x + 3*a) + 2*(b*x*cos(2*b*x + 3*a)*sin(b*x + a) + (b*x*cos(a) + sin(a 
))*sin(b*x + a) + sin(2*b*x + 3*a)*sin(b*x + a))*sin(3*b*x + 3*a) + (co...
 

Giac [F]

\[ \int (c+d x)^3 \sec ^2(a+b x) \sin (3 a+3 b x) \, dx=\int { {\left (d x + c\right )}^{3} \sec \left (b x + a\right )^{2} \sin \left (3 \, b x + 3 \, a\right ) \,d x } \] Input:

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

Output:

integrate((d*x + c)^3*sec(b*x + a)^2*sin(3*b*x + 3*a), x)
 

Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 \sec ^2(a+b x) \sin (3 a+3 b x) \, dx=\text {Hanged} \] Input:

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

Output:

\text{Hanged}
 

Reduce [F]

\[ \int (c+d x)^3 \sec ^2(a+b x) \sin (3 a+3 b x) \, dx=\text {too large to display} \] Input:

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

Output:

(35937*cos(3*a + 3*b*x)*cos(a + b*x)*tan((3*a + 3*b*x)/2)**2*tan((a + b*x) 
/2)**2*b**3*c**3 + 107811*cos(3*a + 3*b*x)*cos(a + b*x)*tan((3*a + 3*b*x)/ 
2)**2*tan((a + b*x)/2)**2*b**3*c**2*d*x + 107811*cos(3*a + 3*b*x)*cos(a + 
b*x)*tan((3*a + 3*b*x)/2)**2*tan((a + b*x)/2)**2*b**3*c*d**2*x**2 + 35937* 
cos(3*a + 3*b*x)*cos(a + b*x)*tan((3*a + 3*b*x)/2)**2*tan((a + b*x)/2)**2* 
b**3*d**3*x**3 - 5346*cos(3*a + 3*b*x)*cos(a + b*x)*tan((3*a + 3*b*x)/2)** 
2*tan((a + b*x)/2)**2*b*c*d**2 - 5346*cos(3*a + 3*b*x)*cos(a + b*x)*tan((3 
*a + 3*b*x)/2)**2*tan((a + b*x)/2)**2*b*d**3*x - 35937*cos(3*a + 3*b*x)*co 
s(a + b*x)*tan((3*a + 3*b*x)/2)**2*b**3*c**3 - 107811*cos(3*a + 3*b*x)*cos 
(a + b*x)*tan((3*a + 3*b*x)/2)**2*b**3*c**2*d*x - 107811*cos(3*a + 3*b*x)* 
cos(a + b*x)*tan((3*a + 3*b*x)/2)**2*b**3*c*d**2*x**2 - 35937*cos(3*a + 3* 
b*x)*cos(a + b*x)*tan((3*a + 3*b*x)/2)**2*b**3*d**3*x**3 + 5346*cos(3*a + 
3*b*x)*cos(a + b*x)*tan((3*a + 3*b*x)/2)**2*b*c*d**2 + 5346*cos(3*a + 3*b* 
x)*cos(a + b*x)*tan((3*a + 3*b*x)/2)**2*b*d**3*x + 35937*cos(3*a + 3*b*x)* 
cos(a + b*x)*tan((a + b*x)/2)**2*b**3*c**3 + 107811*cos(3*a + 3*b*x)*cos(a 
 + b*x)*tan((a + b*x)/2)**2*b**3*c**2*d*x + 107811*cos(3*a + 3*b*x)*cos(a 
+ b*x)*tan((a + b*x)/2)**2*b**3*c*d**2*x**2 + 35937*cos(3*a + 3*b*x)*cos(a 
 + b*x)*tan((a + b*x)/2)**2*b**3*d**3*x**3 - 5346*cos(3*a + 3*b*x)*cos(a + 
 b*x)*tan((a + b*x)/2)**2*b*c*d**2 - 5346*cos(3*a + 3*b*x)*cos(a + b*x)*ta 
n((a + b*x)/2)**2*b*d**3*x - 35937*cos(3*a + 3*b*x)*cos(a + b*x)*b**3*c...