\(\int \sec (c+d x) (a+b \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [655]

Optimal result

Integrand size = 31, antiderivative size = 234 \[ \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))}{8 d}-\frac {\left (3 a^4 C-4 b^4 (5 A+4 C)-4 a^2 b^2 (20 A+13 C)\right ) \tan (c+d x)}{30 b d}+\frac {a \left (100 A b^2-6 a^2 C+71 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{120 d}-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}-\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d} \] Output:

1/8*a*(4*a^2*(2*A+C)+3*b^2*(4*A+3*C))*arctanh(sin(d*x+c))/d-1/30*(3*a^4*C- 
4*b^4*(5*A+4*C)-4*a^2*b^2*(20*A+13*C))*tan(d*x+c)/b/d+1/120*a*(100*A*b^2-6 
*C*a^2+71*C*b^2)*sec(d*x+c)*tan(d*x+c)/d-1/60*(3*C*a^2-4*b^2*(5*A+4*C))*(a 
+b*sec(d*x+c))^2*tan(d*x+c)/b/d-1/20*a*C*(a+b*sec(d*x+c))^3*tan(d*x+c)/b/d 
+1/5*C*(a+b*sec(d*x+c))^4*tan(d*x+c)/b/d
 

Mathematica [A] (verified)

Time = 4.54 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.69 \[ \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {120 a^3 A \coth ^{-1}(\sin (c+d x))+15 a \left (12 A b^2+4 a^2 C+9 b^2 C\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 a \left (12 A b^2+4 a^2 C+9 b^2 C\right ) \sec (c+d x)+90 a b^2 C \sec ^3(c+d x)+8 b \left (15 \left (3 a^2+b^2\right ) (A+C)+5 \left (A b^2+3 a^2 C+2 b^2 C\right ) \tan ^2(c+d x)+3 b^2 C \tan ^4(c+d x)\right )\right )}{120 d} \] Input:

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

Output:

(120*a^3*A*ArcCoth[Sin[c + d*x]] + 15*a*(12*A*b^2 + 4*a^2*C + 9*b^2*C)*Arc 
Tanh[Sin[c + d*x]] + Tan[c + d*x]*(15*a*(12*A*b^2 + 4*a^2*C + 9*b^2*C)*Sec 
[c + d*x] + 90*a*b^2*C*Sec[c + d*x]^3 + 8*b*(15*(3*a^2 + b^2)*(A + C) + 5* 
(A*b^2 + 3*a^2*C + 2*b^2*C)*Tan[c + d*x]^2 + 3*b^2*C*Tan[c + d*x]^4)))/(12 
0*d)
 

Rubi [A] (verified)

Time = 1.39 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.05, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 4571, 3042, 4490, 3042, 4490, 3042, 4485, 3042, 4274, 3042, 4254, 24, 4257}

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.

\(\displaystyle \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx\)

\(\Big \downarrow \) 4571

\(\displaystyle \frac {\int \sec (c+d x) (a+b \sec (c+d x))^3 (b (5 A+4 C)-a C \sec (c+d x))dx}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^3 \left (b (5 A+4 C)-a C \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{4} \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (a b (20 A+13 C)-\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \sec (c+d x)\right )dx-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (a b (20 A+13 C)+\left (4 b^2 (5 A+4 C)-3 a^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 4490

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left ((60 A+33 C) a^2+8 b^2 (5 A+4 C)\right )+a \left (-6 C a^2+100 A b^2+71 b^2 C\right ) \sec (c+d x)\right )dx-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (b \left ((60 A+33 C) a^2+8 b^2 (5 A+4 C)\right )+a \left (-6 C a^2+100 A b^2+71 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 4485

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \sec (c+d x) \left (15 a b \left (4 (2 A+C) a^2+3 b^2 (4 A+3 C)\right )-4 \left (3 C a^4-4 b^2 (20 A+13 C) a^2-4 b^4 (5 A+4 C)\right ) \sec (c+d x)\right )dx+\frac {a b \left (-6 a^2 C+100 A b^2+71 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (15 a b \left (4 (2 A+C) a^2+3 b^2 (4 A+3 C)\right )-4 \left (3 C a^4-4 b^2 (20 A+13 C) a^2-4 b^4 (5 A+4 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a b \left (-6 a^2 C+100 A b^2+71 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 4274

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a b \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \int \sec (c+d x)dx-4 \left (3 a^4 C-4 a^2 b^2 (20 A+13 C)-4 b^4 (5 A+4 C)\right ) \int \sec ^2(c+d x)dx\right )+\frac {a b \left (-6 a^2 C+100 A b^2+71 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a b \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-4 \left (3 a^4 C-4 a^2 b^2 (20 A+13 C)-4 b^4 (5 A+4 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx\right )+\frac {a b \left (-6 a^2 C+100 A b^2+71 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a b \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {4 \left (3 a^4 C-4 a^2 b^2 (20 A+13 C)-4 b^4 (5 A+4 C)\right ) \int 1d(-\tan (c+d x))}{d}\right )+\frac {a b \left (-6 a^2 C+100 A b^2+71 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {1}{2} \left (15 a b \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {4 \left (3 a^4 C-4 a^2 b^2 (20 A+13 C)-4 b^4 (5 A+4 C)\right ) \tan (c+d x)}{d}\right )+\frac {a b \left (-6 a^2 C+100 A b^2+71 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{3} \left (\frac {a b \left (-6 a^2 C+100 A b^2+71 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {1}{2} \left (\frac {15 a b \left (4 a^2 (2 A+C)+3 b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))}{d}-\frac {4 \left (3 a^4 C-4 a^2 b^2 (20 A+13 C)-4 b^4 (5 A+4 C)\right ) \tan (c+d x)}{d}\right )\right )-\frac {\left (3 a^2 C-4 b^2 (5 A+4 C)\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}\right )-\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}}{5 b}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d}\)

Input:

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

Output:

(C*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*b*d) + (-1/4*(a*C*(a + b*Sec[c 
+ d*x])^3*Tan[c + d*x])/d + (-1/3*((3*a^2*C - 4*b^2*(5*A + 4*C))*(a + b*Se 
c[c + d*x])^2*Tan[c + d*x])/d + ((a*b*(100*A*b^2 - 6*a^2*C + 71*b^2*C)*Sec 
[c + d*x]*Tan[c + d*x])/(2*d) + ((15*a*b*(4*a^2*(2*A + C) + 3*b^2*(4*A + 3 
*C))*ArcTanh[Sin[c + d*x]])/d - (4*(3*a^4*C - 4*b^4*(5*A + 4*C) - 4*a^2*b^ 
2*(20*A + 13*C))*Tan[c + d*x])/d)/2)/3)/4)/(5*b)
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4485

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[ 
e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Simp[1/(n + 1)   Int[(d*Csc 
[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x 
], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[ 
n, -1]
 

rule 4490

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[1/(m + 1)   Int[Csc[e + f*x]* 
(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1 
))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a* 
B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
 

rule 4571

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[ 
(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x] 
*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   In 
t[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C* 
Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 

Maple [A] (verified)

Time = 10.71 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.92

method	result	size
parts	\(-\frac {\left (A \,b^{3}+3 a^{2} b C \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 a A \,b^{2}+a^{3} C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{3}}{d}-\frac {C \,b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {3 A \,a^{2} b \tan \left (d x +c \right )}{d}+\frac {3 C a \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(215\)
derivativedivides	\(\frac {a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \,a^{2} b \tan \left (d x +c \right )-3 a^{2} b C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a A \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C a \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-A \,b^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \,b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(246\)
default	\(\frac {a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \,a^{2} b \tan \left (d x +c \right )-3 a^{2} b C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 a A \,b^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 C a \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-A \,b^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \,b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(246\)
parallelrisch	\(\frac {-a \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (a^{2} \left (A +\frac {C}{2}\right )+\frac {3 \left (A +\frac {3 C}{4}\right ) b^{2}}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+a \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (a^{2} \left (A +\frac {C}{2}\right )+\frac {3 \left (A +\frac {3 C}{4}\right ) b^{2}}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+9 \left (\left (A +\frac {10 C}{9}\right ) a^{2}+\frac {10 b^{2} \left (A +\frac {4 C}{5}\right )}{27}\right ) b \sin \left (3 d x +3 c \right )+3 \left (a^{2} \left (A +\frac {2 C}{3}\right )+\frac {2 b^{2} \left (A +\frac {4 C}{5}\right )}{9}\right ) b \sin \left (5 d x +5 c \right )+6 \left (\frac {C \,a^{2}}{3}+b^{2} \left (A +\frac {7 C}{4}\right )\right ) a \sin \left (2 d x +2 c \right )+3 a \left (\frac {C \,a^{2}}{3}+\left (A +\frac {3 C}{4}\right ) b^{2}\right ) \sin \left (4 d x +4 c \right )+6 \left (\left (A +\frac {4 C}{3}\right ) a^{2}+\frac {4 b^{2} \left (A +2 C \right )}{9}\right ) b \sin \left (d x +c \right )}{d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(313\)
norman	\(\frac {-\frac {\left (24 A \,a^{2} b -12 a A \,b^{2}+8 A \,b^{3}-4 a^{3} C +24 a^{2} b C -15 C a \,b^{2}+8 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (24 A \,a^{2} b +12 a A \,b^{2}+8 A \,b^{3}+4 a^{3} C +24 a^{2} b C +15 C a \,b^{2}+8 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (144 A \,a^{2} b -36 a A \,b^{2}+32 A \,b^{3}-12 a^{3} C +96 a^{2} b C -9 C a \,b^{2}+16 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (144 A \,a^{2} b +36 a A \,b^{2}+32 A \,b^{3}+12 a^{3} C +96 a^{2} b C +9 C a \,b^{2}+16 C \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}-\frac {4 b \left (135 a^{2} A +25 A \,b^{2}+75 C \,a^{2}+29 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a \left (8 a^{2} A +12 A \,b^{2}+4 C \,a^{2}+9 C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (8 a^{2} A +12 A \,b^{2}+4 C \,a^{2}+9 C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(397\)
risch	\(-\frac {i \left (-240 a^{2} b C -64 C \,b^{3}-80 A \,b^{3}-360 A \,a^{2} b -135 b^{2} C a \,{\mathrm e}^{i \left (d x +c \right )}-180 A a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+180 A a \,b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+135 C a \,b^{2} {\mathrm e}^{9 i \left (d x +c \right )}-360 A \,a^{2} b \,{\mathrm e}^{8 i \left (d x +c \right )}+360 A a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}+630 C a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-1440 A \,a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-720 C \,a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-2160 A \,a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-1680 C \,a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-360 A a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-630 C a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-1440 A \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-1200 C \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-560 A \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}-400 A \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-320 C \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}-640 C \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+60 C \,a^{3} {\mathrm e}^{9 i \left (d x +c \right )}-60 C \,a^{3} {\mathrm e}^{i \left (d x +c \right )}+120 C \,a^{3} {\mathrm e}^{7 i \left (d x +c \right )}-120 C \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-240 A \,b^{3} {\mathrm e}^{6 i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{2}}{2 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}-\frac {9 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{2}}{8 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{2}}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}+\frac {9 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{2}}{8 d}\)	\(602\)

Input:

int(sec(d*x+c)*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBO 
SE)
 

Output:

-(A*b^3+3*C*a^2*b)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(3*A*a*b^2+C*a^3)/ 
d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+1/d*A*ln(sec(d 
*x+c)+tan(d*x+c))*a^3-C*b^3/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*t 
an(d*x+c)+3*A*a^2*b/d*tan(d*x+c)+3*C*a*b^2/d*(-(-1/4*sec(d*x+c)^3-3/8*sec( 
d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.96 \[ \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, {\left (2 \, A + C\right )} a^{3} + 3 \, {\left (4 \, A + 3 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, {\left (2 \, A + C\right )} a^{3} + 3 \, {\left (4 \, A + 3 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (90 \, C a b^{2} \cos \left (d x + c\right ) + 8 \, {\left (15 \, {\left (3 \, A + 2 \, C\right )} a^{2} b + 2 \, {\left (5 \, A + 4 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, C b^{3} + 15 \, {\left (4 \, C a^{3} + 3 \, {\left (4 \, A + 3 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, C a^{2} b + {\left (5 \, A + 4 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \] Input:

integrate(sec(d*x+c)*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="f 
ricas")
 

Output:

1/240*(15*(4*(2*A + C)*a^3 + 3*(4*A + 3*C)*a*b^2)*cos(d*x + c)^5*log(sin(d 
*x + c) + 1) - 15*(4*(2*A + C)*a^3 + 3*(4*A + 3*C)*a*b^2)*cos(d*x + c)^5*l 
og(-sin(d*x + c) + 1) + 2*(90*C*a*b^2*cos(d*x + c) + 8*(15*(3*A + 2*C)*a^2 
*b + 2*(5*A + 4*C)*b^3)*cos(d*x + c)^4 + 24*C*b^3 + 15*(4*C*a^3 + 3*(4*A + 
 3*C)*a*b^2)*cos(d*x + c)^3 + 8*(15*C*a^2*b + (5*A + 4*C)*b^3)*cos(d*x + c 
)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)
 

Sympy [F]

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

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

Output:

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**3*sec(c + d*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.24 \[ \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{3} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{3} - 45 \, C a b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, A a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 720 \, A a^{2} b \tan \left (d x + c\right )}{240 \, d} \] Input:

integrate(sec(d*x+c)*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="m 
axima")
 

Output:

1/240*(240*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2*b + 80*(tan(d*x + c)^3 
+ 3*tan(d*x + c))*A*b^3 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*ta 
n(d*x + c))*C*b^3 - 45*C*a*b^2*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin 
(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin( 
d*x + c) - 1)) - 60*C*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d 
*x + c) + 1) + log(sin(d*x + c) - 1)) - 180*A*a*b^2*(2*sin(d*x + c)/(sin(d 
*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 240*A*a^ 
3*log(sec(d*x + c) + tan(d*x + c)) + 720*A*a^2*b*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 656 vs. \(2 (222) = 444\).

Time = 0.28 (sec) , antiderivative size = 656, normalized size of antiderivative = 2.80 \[ \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

integrate(sec(d*x+c)*(a+b*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="g 
iac")
 

Output:

1/120*(15*(8*A*a^3 + 4*C*a^3 + 12*A*a*b^2 + 9*C*a*b^2)*log(abs(tan(1/2*d*x 
 + 1/2*c) + 1)) - 15*(8*A*a^3 + 4*C*a^3 + 12*A*a*b^2 + 9*C*a*b^2)*log(abs( 
tan(1/2*d*x + 1/2*c) - 1)) + 2*(60*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 360*A*a^ 
2*b*tan(1/2*d*x + 1/2*c)^9 - 360*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 180*A*a* 
b^2*tan(1/2*d*x + 1/2*c)^9 + 225*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 120*A*b^ 
3*tan(1/2*d*x + 1/2*c)^9 - 120*C*b^3*tan(1/2*d*x + 1/2*c)^9 - 120*C*a^3*ta 
n(1/2*d*x + 1/2*c)^7 + 1440*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 960*C*a^2*b*t 
an(1/2*d*x + 1/2*c)^7 - 360*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 90*C*a*b^2*ta 
n(1/2*d*x + 1/2*c)^7 + 320*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 160*C*b^3*tan(1/ 
2*d*x + 1/2*c)^7 - 2160*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 1200*C*a^2*b*tan( 
1/2*d*x + 1/2*c)^5 - 400*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 464*C*b^3*tan(1/2* 
d*x + 1/2*c)^5 + 120*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 1440*A*a^2*b*tan(1/2*d 
*x + 1/2*c)^3 + 960*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 360*A*a*b^2*tan(1/2*d 
*x + 1/2*c)^3 + 90*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 320*A*b^3*tan(1/2*d*x 
+ 1/2*c)^3 + 160*C*b^3*tan(1/2*d*x + 1/2*c)^3 - 60*C*a^3*tan(1/2*d*x + 1/2 
*c) - 360*A*a^2*b*tan(1/2*d*x + 1/2*c) - 360*C*a^2*b*tan(1/2*d*x + 1/2*c) 
- 180*A*a*b^2*tan(1/2*d*x + 1/2*c) - 225*C*a*b^2*tan(1/2*d*x + 1/2*c) - 12 
0*A*b^3*tan(1/2*d*x + 1/2*c) - 120*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d* 
x + 1/2*c)^2 - 1)^5)/d
 

Mupad [B] (verification not implemented)

Time = 15.33 (sec) , antiderivative size = 445, normalized size of antiderivative = 1.90 \[ \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a\,\mathrm {atanh}\left (\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (8\,A\,a^2+12\,A\,b^2+4\,C\,a^2+9\,C\,b^2\right )}{2\,\left (4\,A\,a^3+2\,C\,a^3+6\,A\,a\,b^2+\frac {9\,C\,a\,b^2}{2}\right )}\right )\,\left (8\,A\,a^2+12\,A\,b^2+4\,C\,a^2+9\,C\,b^2\right )}{4\,d}-\frac {\left (2\,A\,b^3-C\,a^3+2\,C\,b^3-3\,A\,a\,b^2+6\,A\,a^2\,b-\frac {15\,C\,a\,b^2}{4}+6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,C\,a^3-\frac {16\,A\,b^3}{3}-\frac {8\,C\,b^3}{3}+6\,A\,a\,b^2-24\,A\,a^2\,b+\frac {3\,C\,a\,b^2}{2}-16\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,b^3}{3}+\frac {116\,C\,b^3}{15}+36\,A\,a^2\,b+20\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {16\,A\,b^3}{3}-2\,C\,a^3-\frac {8\,C\,b^3}{3}-6\,A\,a\,b^2-24\,A\,a^2\,b-\frac {3\,C\,a\,b^2}{2}-16\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,b^3+C\,a^3+2\,C\,b^3+3\,A\,a\,b^2+6\,A\,a^2\,b+\frac {15\,C\,a\,b^2}{4}+6\,C\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:

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

Output:

(a*atanh((a*tan(c/2 + (d*x)/2)*(8*A*a^2 + 12*A*b^2 + 4*C*a^2 + 9*C*b^2))/( 
2*(4*A*a^3 + 2*C*a^3 + 6*A*a*b^2 + (9*C*a*b^2)/2)))*(8*A*a^2 + 12*A*b^2 + 
4*C*a^2 + 9*C*b^2))/(4*d) - (tan(c/2 + (d*x)/2)^9*(2*A*b^3 - C*a^3 + 2*C*b 
^3 - 3*A*a*b^2 + 6*A*a^2*b - (15*C*a*b^2)/4 + 6*C*a^2*b) - tan(c/2 + (d*x) 
/2)^3*((16*A*b^3)/3 + 2*C*a^3 + (8*C*b^3)/3 + 6*A*a*b^2 + 24*A*a^2*b + (3* 
C*a*b^2)/2 + 16*C*a^2*b) - tan(c/2 + (d*x)/2)^7*((16*A*b^3)/3 - 2*C*a^3 + 
(8*C*b^3)/3 - 6*A*a*b^2 + 24*A*a^2*b - (3*C*a*b^2)/2 + 16*C*a^2*b) + tan(c 
/2 + (d*x)/2)^5*((20*A*b^3)/3 + (116*C*b^3)/15 + 36*A*a^2*b + 20*C*a^2*b) 
+ tan(c/2 + (d*x)/2)*(2*A*b^3 + C*a^3 + 2*C*b^3 + 3*A*a*b^2 + 6*A*a^2*b + 
(15*C*a*b^2)/4 + 6*C*a^2*b))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d* 
x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d* 
x)/2)^10 - 1))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1035, normalized size of antiderivative = 4.42 \[ \int \sec (c+d x) (a+b \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

( - 120*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**4 - 60*c 
os(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3*c - 180*cos(c + 
 d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b**2 - 135*cos(c + d* 
x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**2*c + 240*cos(c + d*x)*l 
og(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4 + 120*cos(c + d*x)*log(tan(( 
c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*c + 360*cos(c + d*x)*log(tan((c + d* 
x)/2) - 1)*sin(c + d*x)**2*a**2*b**2 + 270*cos(c + d*x)*log(tan((c + d*x)/ 
2) - 1)*sin(c + d*x)**2*a*b**2*c - 120*cos(c + d*x)*log(tan((c + d*x)/2) - 
 1)*a**4 - 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3*c - 180*cos(c + 
d*x)*log(tan((c + d*x)/2) - 1)*a**2*b**2 - 135*cos(c + d*x)*log(tan((c + d 
*x)/2) - 1)*a*b**2*c + 120*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + 
d*x)**4*a**4 + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a 
**3*c + 180*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a**2*b* 
*2 + 135*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a*b**2*c - 
 240*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4 - 120*cos 
(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*c - 360*cos(c + d 
*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**2 - 270*cos(c + d*x) 
*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**2*c + 120*cos(c + d*x)*log 
(tan((c + d*x)/2) + 1)*a**4 + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a* 
*3*c + 180*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b**2 + 135*cos(c...