\(\int (a+b \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\) [665]

Optimal result

Integrand size = 25, antiderivative size = 227 \[ \int (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^4 A x+\frac {a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (6 a^4 C+2 b^4 (5 A+4 C)+a^2 b^2 (85 A+56 C)\right ) \tan (c+d x)}{15 d}+\frac {a b \left (40 A b^2+6 a^2 C+29 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{30 d}+\frac {\left (3 a^2 C+b^2 (5 A+4 C)\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{15 d}+\frac {a C (a+b \sec (c+d x))^3 \tan (c+d x)}{5 d}+\frac {C (a+b \sec (c+d x))^4 \tan (c+d x)}{5 d} \] Output:

a^4*A*x+1/2*a*b*(4*a^2*(2*A+C)+b^2*(4*A+3*C))*arctanh(sin(d*x+c))/d+1/15*( 
6*a^4*C+2*b^4*(5*A+4*C)+a^2*b^2*(85*A+56*C))*tan(d*x+c)/d+1/30*a*b*(40*A*b 
^2+6*C*a^2+29*C*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/15*(3*C*a^2+b^2*(5*A+4*C))* 
(a+b*sec(d*x+c))^2*tan(d*x+c)/d+1/5*a*C*(a+b*sec(d*x+c))^3*tan(d*x+c)/d+1/ 
5*C*(a+b*sec(d*x+c))^4*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 3.39 (sec) , antiderivative size = 223, normalized size of antiderivative = 0.98 \[ \int (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {30 a^4 A d x+120 a^3 A b \coth ^{-1}(\sin (c+d x))+45 a b^3 C \text {arctanh}(\sin (c+d x))+60 a b \left (A b^2+a^2 C\right ) \text {arctanh}(\sin (c+d x))+30 a^2 \left (6 A b^2+a^2 C\right ) \tan (c+d x)+45 a b^3 C \sec (c+d x) \tan (c+d x)+60 a b \left (A b^2+a^2 C\right ) \sec (c+d x) \tan (c+d x)+30 a b^3 C \sec ^3(c+d x) \tan (c+d x)+10 b^2 \left (A b^2+6 a^2 C\right ) \tan (c+d x) \left (3+\tan ^2(c+d x)\right )+2 b^4 C \tan (c+d x) \left (15+10 \tan ^2(c+d x)+3 \tan ^4(c+d x)\right )}{30 d} \] Input:

Integrate[(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]
 

Output:

(30*a^4*A*d*x + 120*a^3*A*b*ArcCoth[Sin[c + d*x]] + 45*a*b^3*C*ArcTanh[Sin 
[c + d*x]] + 60*a*b*(A*b^2 + a^2*C)*ArcTanh[Sin[c + d*x]] + 30*a^2*(6*A*b^ 
2 + a^2*C)*Tan[c + d*x] + 45*a*b^3*C*Sec[c + d*x]*Tan[c + d*x] + 60*a*b*(A 
*b^2 + a^2*C)*Sec[c + d*x]*Tan[c + d*x] + 30*a*b^3*C*Sec[c + d*x]^3*Tan[c 
+ d*x] + 10*b^2*(A*b^2 + 6*a^2*C)*Tan[c + d*x]*(3 + Tan[c + d*x]^2) + 2*b^ 
4*C*Tan[c + d*x]*(15 + 10*Tan[c + d*x]^2 + 3*Tan[c + d*x]^4))/(30*d)
 

Rubi [A] (verified)

Time = 1.01 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4545, 3042, 4544, 27, 3042, 4544, 3042, 4536, 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.

\(\displaystyle \int (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4545

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

\(\displaystyle \frac {1}{5} \left (\frac {1}{4} \int 4 (a+b \sec (c+d x))^2 \left (5 A a^2+b (10 A+7 C) \sec (c+d x) a+\left (3 C a^2+b^2 (5 A+4 C)\right ) \sec ^2(c+d x)\right )dx+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\int (a+b \sec (c+d x))^2 \left (5 A a^2+b (10 A+7 C) \sec (c+d x) a+\left (3 C a^2+b^2 (5 A+4 C)\right ) \sec ^2(c+d x)\right )dx+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2 \left (5 A a^2+b (10 A+7 C) \csc \left (c+d x+\frac {\pi }{2}\right ) a+\left (3 C a^2+b^2 (5 A+4 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )dx+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 4544

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int (a+b \sec (c+d x)) \left (15 A a^3+\left (6 C a^2+40 A b^2+29 b^2 C\right ) \sec ^2(c+d x) a+b \left (9 (5 A+3 C) a^2+2 b^2 (5 A+4 C)\right ) \sec (c+d x)\right )dx+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right ) \left (15 A a^3+\left (6 C a^2+40 A b^2+29 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2 a+b \left (9 (5 A+3 C) a^2+2 b^2 (5 A+4 C)\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 4536

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {1}{2} \int \left (30 A a^4+15 b \left (4 (2 A+C) a^2+b^2 (4 A+3 C)\right ) \sec (c+d x) a+2 \left (6 C a^4+b^2 (85 A+56 C) a^2+2 b^4 (5 A+4 C)\right ) \sec ^2(c+d x)\right )dx+\frac {a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} \left (\frac {\left (3 a^2 C+5 A b^2+4 b^2 C\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{3 d}+\frac {1}{3} \left (\frac {a b \left (6 a^2 C+40 A b^2+29 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{2 d}+\frac {1}{2} \left (30 a^4 A x+\frac {15 a b \left (4 a^2 (2 A+C)+b^2 (4 A+3 C)\right ) \text {arctanh}(\sin (c+d x))}{d}+\frac {2 \left (6 a^4 C+a^2 b^2 (85 A+56 C)+2 b^4 (5 A+4 C)\right ) \tan (c+d x)}{d}\right )\right )+\frac {a C \tan (c+d x) (a+b \sec (c+d x))^3}{d}\right )+\frac {C \tan (c+d x) (a+b \sec (c+d x))^4}{5 d}\)

Input:

Int[(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]
 

Output:

(C*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*d) + (((5*A*b^2 + 3*a^2*C + 4*b 
^2*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(3*d) + (a*C*(a + b*Sec[c + d*x 
])^3*Tan[c + d*x])/d + ((a*b*(40*A*b^2 + 6*a^2*C + 29*b^2*C)*Sec[c + d*x]* 
Tan[c + d*x])/(2*d) + (30*a^4*A*x + (15*a*b*(4*a^2*(2*A + C) + b^2*(4*A + 
3*C))*ArcTanh[Sin[c + d*x]])/d + (2*(6*a^4*C + 2*b^4*(5*A + 4*C) + a^2*b^2 
*(85*A + 56*C))*Tan[c + d*x])/d)/2)/3)/5
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

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

rule 3042

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

rule 4536

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

rule 4544

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + 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/(f*(m + 1))), x] + Simp[1/(m + 1)   Int[( 
a + b*Csc[e + f*x])^(m - 1)*Simp[a*A*(m + 1) + ((A*b + a*B)*(m + 1) + b*C*m 
)*Csc[e + f*x] + (b*B*(m + 1) + a*C*m)*Csc[e + f*x]^2, x], x], x] /; FreeQ[ 
{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 0]
 

rule 4545

Int[((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/(f*(m + 1))), x] + Simp[1/(m + 1)   Int[(a + b*Csc[e + f*x])^(m - 1)*Simp 
[a*A*(m + 1) + (A*b*(m + 1) + b*C*m)*Csc[e + f*x] + a*C*m*Csc[e + f*x]^2, x 
], x], x] /; FreeQ[{a, b, e, f, A, C}, x] && NeQ[a^2 - b^2, 0] && IGtQ[2*m, 
 0]
 

Maple [A] (verified)

Time = 10.84 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.04

method	result	size
parts	\(a^{4} A x -\frac {\left (A \,b^{4}+6 C \,a^{2} b^{2}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (4 a A \,b^{3}+4 a^{3} b C \right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (6 A \,a^{2} b^{2}+a^{4} C \right ) \tan \left (d x +c \right )}{d}-\frac {C \,b^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {4 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 C a \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(236\)
derivativedivides	\(\frac {a^{4} A \left (d x +c \right )+a^{4} C \tan \left (d x +c \right )+4 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{3} b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 A \tan \left (d x +c \right ) a^{2} b^{2}-6 C \,a^{2} b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 C a \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-A \,b^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \,b^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(275\)
default	\(\frac {a^{4} A \left (d x +c \right )+a^{4} C \tan \left (d x +c \right )+4 A \,a^{3} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{3} b C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+6 A \tan \left (d x +c \right ) a^{2} b^{2}-6 C \,a^{2} b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+4 a A \,b^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 C a \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-A \,b^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \,b^{4} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(275\)
parallelrisch	\(\frac {-12 \left (\frac {\left (A +\frac {3 C}{4}\right ) b^{2}}{2}+a^{2} \left (A +\frac {C}{2}\right )\right ) \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 \left (\frac {\left (A +\frac {3 C}{4}\right ) b^{2}}{2}+a^{2} \left (A +\frac {C}{2}\right )\right ) \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) a b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+15 a^{4} A x d \cos \left (3 d x +3 c \right )+3 a^{4} A x d \cos \left (5 d x +5 c \right )+\left (9 a^{4} C +54 b^{2} \left (A +\frac {10 C}{9}\right ) a^{2}+10 b^{4} \left (A +\frac {4 C}{5}\right )\right ) \sin \left (3 d x +3 c \right )+\left (3 a^{4} C +18 b^{2} \left (A +\frac {2 C}{3}\right ) a^{2}+2 b^{4} \left (A +\frac {4 C}{5}\right )\right ) \sin \left (5 d x +5 c \right )+24 a b \left (C \,a^{2}+b^{2} \left (A +\frac {7 C}{4}\right )\right ) \sin \left (2 d x +2 c \right )+12 \left (C \,a^{2}+\left (A +\frac {3 C}{4}\right ) b^{2}\right ) a b \sin \left (4 d x +4 c \right )+30 a^{4} A x d \cos \left (d x +c \right )+36 \left (\frac {a^{4} C}{6}+\left (A +\frac {4 C}{3}\right ) b^{2} a^{2}+\frac {2 b^{4} \left (A +2 C \right )}{9}\right ) \sin \left (d x +c \right )}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(389\)
norman	\(\frac {a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}-a^{4} A x -\frac {4 \left (270 A \,a^{2} b^{2}+25 A \,b^{4}+45 a^{4} C +150 C \,a^{2} b^{2}+29 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {\left (12 A \,a^{2} b^{2}-4 a A \,b^{3}+2 A \,b^{4}+2 a^{4} C -4 a^{3} b C +12 C \,a^{2} b^{2}-5 C a \,b^{3}+2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{d}-\frac {\left (12 A \,a^{2} b^{2}+4 a A \,b^{3}+2 A \,b^{4}+2 a^{4} C +4 a^{3} b C +12 C \,a^{2} b^{2}+5 C a \,b^{3}+2 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 \left (72 A \,a^{2} b^{2}-12 a A \,b^{3}+8 A \,b^{4}+12 a^{4} C -12 a^{3} b C +48 C \,a^{2} b^{2}-3 C a \,b^{3}+4 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}+\frac {2 \left (72 A \,a^{2} b^{2}+12 a A \,b^{3}+8 A \,b^{4}+12 a^{4} C +12 a^{3} b C +48 C \,a^{2} b^{2}+3 C a \,b^{3}+4 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+5 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-10 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+10 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-5 a^{4} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a b \left (8 a^{2} A +4 A \,b^{2}+4 C \,a^{2}+3 C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a b \left (8 a^{2} A +4 A \,b^{2}+4 C \,a^{2}+3 C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(550\)
risch	\(a^{4} A x -\frac {i \left (-16 C \,b^{4}-20 A \,b^{4}-30 a^{4} C +120 A a \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+120 C \,a^{3} b \,{\mathrm e}^{7 i \left (d x +c \right )}+210 C a \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-120 C \,a^{2} b^{2}-180 A \,a^{2} b^{2}-1080 A \,a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-840 C \,a^{2} b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-120 A a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-120 C \,a^{3} b \,{\mathrm e}^{3 i \left (d x +c \right )}-210 C a \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+60 C \,a^{3} b \,{\mathrm e}^{9 i \left (d x +c \right )}+45 C a \,b^{3} {\mathrm e}^{9 i \left (d x +c \right )}-180 A \,a^{2} b^{2} {\mathrm e}^{8 i \left (d x +c \right )}+60 A a \,b^{3} {\mathrm e}^{9 i \left (d x +c \right )}-720 A \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-600 C \,a^{2} b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-720 A \,a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-360 C \,a^{2} b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-60 A a \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-60 C \,a^{3} b \,{\mathrm e}^{i \left (d x +c \right )}-45 b^{3} C a \,{\mathrm e}^{i \left (d x +c \right )}-140 A \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-180 C \,a^{4} {\mathrm e}^{4 i \left (d x +c \right )}-160 C \,b^{4} {\mathrm e}^{4 i \left (d x +c \right )}-120 C \,a^{4} {\mathrm e}^{2 i \left (d x +c \right )}-80 C \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-120 C \,a^{4} {\mathrm e}^{6 i \left (d x +c \right )}-100 A \,b^{4} {\mathrm e}^{2 i \left (d x +c \right )}-60 A \,b^{4} {\mathrm e}^{6 i \left (d x +c \right )}-30 C \,a^{4} {\mathrm e}^{8 i \left (d x +c \right )}\right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {2 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}+\frac {3 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}-\frac {4 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {2 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {2 a^{3} b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}-\frac {3 a \,b^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}\)	\(701\)

Input:

int((a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)
 

Output:

a^4*A*x-(A*b^4+6*C*a^2*b^2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(4*A*a*b^ 
3+4*C*a^3*b)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+( 
6*A*a^2*b^2+C*a^4)/d*tan(d*x+c)-C*b^4/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d 
*x+c)^2)*tan(d*x+c)+4*A*a^3*b/d*ln(sec(d*x+c)+tan(d*x+c))+4*C*a*b^3/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 = 0.10 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.11 \[ \int (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {60 \, A a^{4} d x \cos \left (d x + c\right )^{5} + 15 \, {\left (4 \, {\left (2 \, A + C\right )} a^{3} b + {\left (4 \, A + 3 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, {\left (2 \, A + C\right )} a^{3} b + {\left (4 \, A + 3 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (30 \, C a b^{3} \cos \left (d x + c\right ) + 6 \, C b^{4} + 2 \, {\left (15 \, C a^{4} + 30 \, {\left (3 \, A + 2 \, C\right )} a^{2} b^{2} + 2 \, {\left (5 \, A + 4 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, C a^{3} b + {\left (4 \, A + 3 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (30 \, C a^{2} b^{2} + {\left (5 \, A + 4 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{60 \, d \cos \left (d x + c\right )^{5}} \] Input:

integrate((a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")
 

Output:

1/60*(60*A*a^4*d*x*cos(d*x + c)^5 + 15*(4*(2*A + C)*a^3*b + (4*A + 3*C)*a* 
b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*(2*A + C)*a^3*b + (4*A + 
 3*C)*a*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(30*C*a*b^3*cos(d*x 
 + c) + 6*C*b^4 + 2*(15*C*a^4 + 30*(3*A + 2*C)*a^2*b^2 + 2*(5*A + 4*C)*b^4 
)*cos(d*x + c)^4 + 15*(4*C*a^3*b + (4*A + 3*C)*a*b^3)*cos(d*x + c)^3 + 2*( 
30*C*a^2*b^2 + (5*A + 4*C)*b^4)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + 
 c)^5)
 

Sympy [F]

\[ \int (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\int \left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{4}\, dx \] Input:

integrate((a+b*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)
 

Output:

Integral((A + C*sec(c + d*x)**2)*(a + b*sec(c + d*x))**4, x)
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 318, normalized size of antiderivative = 1.40 \[ \int (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {60 \, {\left (d x + c\right )} A a^{4} + 120 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} b^{2} + 20 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{4} + 4 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C b^{4} - 15 \, C a b^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a^{3} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{3} b \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 60 \, C a^{4} \tan \left (d x + c\right ) + 360 \, A a^{2} b^{2} \tan \left (d x + c\right )}{60 \, d} \] Input:

integrate((a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")
 

Output:

1/60*(60*(d*x + c)*A*a^4 + 120*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2*b^2 
 + 20*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^4 + 4*(3*tan(d*x + c)^5 + 10*t 
an(d*x + c)^3 + 15*tan(d*x + c))*C*b^4 - 15*C*a*b^3*(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*b*(2*sin(d*x + c)/(sin(d*x 
+ c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*A*a*b^3* 
(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*b*log(sec(d*x + c) + tan(d*x + c)) + 60*C*a^4*tan( 
d*x + c) + 360*A*a^2*b^2*tan(d*x + c))/d
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 778 vs. \(2 (215) = 430\).

Time = 0.34 (sec) , antiderivative size = 778, normalized size of antiderivative = 3.43 \[ \int (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

integrate((a+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")
 

Output:

1/30*(30*(d*x + c)*A*a^4 + 15*(8*A*a^3*b + 4*C*a^3*b + 4*A*a*b^3 + 3*C*a*b 
^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(8*A*a^3*b + 4*C*a^3*b + 4*A*a 
*b^3 + 3*C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(30*C*a^4*tan(1/2 
*d*x + 1/2*c)^9 - 60*C*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 180*A*a^2*b^2*tan(1/ 
2*d*x + 1/2*c)^9 + 180*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*a*b^3*tan(1 
/2*d*x + 1/2*c)^9 - 75*C*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 30*A*b^4*tan(1/2*d 
*x + 1/2*c)^9 + 30*C*b^4*tan(1/2*d*x + 1/2*c)^9 - 120*C*a^4*tan(1/2*d*x + 
1/2*c)^7 + 120*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 720*A*a^2*b^2*tan(1/2*d*x 
+ 1/2*c)^7 - 480*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 120*A*a*b^3*tan(1/2*d* 
x + 1/2*c)^7 + 30*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 80*A*b^4*tan(1/2*d*x + 
1/2*c)^7 - 40*C*b^4*tan(1/2*d*x + 1/2*c)^7 + 180*C*a^4*tan(1/2*d*x + 1/2*c 
)^5 + 1080*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 + 600*C*a^2*b^2*tan(1/2*d*x + 
1/2*c)^5 + 100*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 116*C*b^4*tan(1/2*d*x + 1/2* 
c)^5 - 120*C*a^4*tan(1/2*d*x + 1/2*c)^3 - 120*C*a^3*b*tan(1/2*d*x + 1/2*c) 
^3 - 720*A*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 480*C*a^2*b^2*tan(1/2*d*x + 1/ 
2*c)^3 - 120*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 - 30*C*a*b^3*tan(1/2*d*x + 1/2 
*c)^3 - 80*A*b^4*tan(1/2*d*x + 1/2*c)^3 - 40*C*b^4*tan(1/2*d*x + 1/2*c)^3 
+ 30*C*a^4*tan(1/2*d*x + 1/2*c) + 60*C*a^3*b*tan(1/2*d*x + 1/2*c) + 180*A* 
a^2*b^2*tan(1/2*d*x + 1/2*c) + 180*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 60*A*a 
*b^3*tan(1/2*d*x + 1/2*c) + 75*C*a*b^3*tan(1/2*d*x + 1/2*c) + 30*A*b^4*...
 

Mupad [B] (verification not implemented)

Time = 14.54 (sec) , antiderivative size = 1738, normalized size of antiderivative = 7.66 \[ \int (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \] Input:

int((A + C/cos(c + d*x)^2)*(a + b/cos(c + d*x))^4,x)
 

Output:

((5*A*b^4*sin(3*c + 3*d*x))/24 + (A*b^4*sin(5*c + 5*d*x))/24 + (3*C*a^4*si 
n(3*c + 3*d*x))/16 + (C*a^4*sin(5*c + 5*d*x))/16 + (C*b^4*sin(3*c + 3*d*x) 
)/6 + (C*b^4*sin(5*c + 5*d*x))/30 + (A*b^4*sin(c + d*x))/6 + (C*a^4*sin(c 
+ d*x))/8 + (C*b^4*sin(c + d*x))/3 + (5*A*a^4*cos(c + d*x)*atan((4*A^2*a^6 
*sin(c/2 + (d*x)/2) + 16*A^2*b^6*sin(c/2 + (d*x)/2) + 9*C^2*b^6*sin(c/2 + 
(d*x)/2) + 64*A^2*a^2*b^4*sin(c/2 + (d*x)/2) + 64*A^2*a^4*b^2*sin(c/2 + (d 
*x)/2) + 24*C^2*a^2*b^4*sin(c/2 + (d*x)/2) + 16*C^2*a^4*b^2*sin(c/2 + (d*x 
)/2) + 24*A*C*b^6*sin(c/2 + (d*x)/2) + 80*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 
 64*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(4*A^2*a^6 + 16*A^ 
2*b^6 + 9*C^2*b^6 + 64*A^2*a^2*b^4 + 64*A^2*a^4*b^2 + 24*C^2*a^2*b^4 + 16* 
C^2*a^4*b^2 + 24*A*C*b^6 + 80*A*C*a^2*b^4 + 64*A*C*a^4*b^2))))/4 + (A*a*b^ 
3*sin(2*c + 2*d*x))/2 + (A*a*b^3*sin(4*c + 4*d*x))/4 + (3*A*a^2*b^2*sin(c 
+ d*x))/4 + (7*C*a*b^3*sin(2*c + 2*d*x))/8 + (C*a^3*b*sin(2*c + 2*d*x))/2 
+ (3*C*a*b^3*sin(4*c + 4*d*x))/16 + (C*a^3*b*sin(4*c + 4*d*x))/4 + C*a^2*b 
^2*sin(c + d*x) + (5*A*a^4*atan((4*A^2*a^6*sin(c/2 + (d*x)/2) + 16*A^2*b^6 
*sin(c/2 + (d*x)/2) + 9*C^2*b^6*sin(c/2 + (d*x)/2) + 64*A^2*a^2*b^4*sin(c/ 
2 + (d*x)/2) + 64*A^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*C^2*a^2*b^4*sin(c/2 
+ (d*x)/2) + 16*C^2*a^4*b^2*sin(c/2 + (d*x)/2) + 24*A*C*b^6*sin(c/2 + (d*x 
)/2) + 80*A*C*a^2*b^4*sin(c/2 + (d*x)/2) + 64*A*C*a^4*b^2*sin(c/2 + (d*x)/ 
2))/(cos(c/2 + (d*x)/2)*(4*A^2*a^6 + 16*A^2*b^6 + 9*C^2*b^6 + 64*A^2*a^...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1156, normalized size of antiderivative = 5.09 \[ \int (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

int((a+b*sec(d*x+c))^4*(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*b - 60 
*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**3*b*c - 60*cos( 
c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a**2*b**3 - 45*cos(c + 
d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a*b**3*c + 240*cos(c + d*x) 
*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a**4*b + 120*cos(c + d*x)*log(t 
an((c + d*x)/2) - 1)*sin(c + d*x)**2*a**3*b*c + 120*cos(c + d*x)*log(tan(( 
c + d*x)/2) - 1)*sin(c + d*x)**2*a**2*b**3 + 90*cos(c + d*x)*log(tan((c + 
d*x)/2) - 1)*sin(c + d*x)**2*a*b**3*c - 120*cos(c + d*x)*log(tan((c + d*x) 
/2) - 1)*a**4*b - 60*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**3*b*c - 60* 
cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a**2*b**3 - 45*cos(c + d*x)*log(tan 
((c + d*x)/2) - 1)*a*b**3*c + 120*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*s 
in(c + d*x)**4*a**4*b + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + 
d*x)**4*a**3*b*c + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)* 
*4*a**2*b**3 + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*a 
*b**3*c - 240*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**4* 
b - 120*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**3*b*c - 
120*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a**2*b**3 - 90* 
cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a*b**3*c + 120*cos( 
c + d*x)*log(tan((c + d*x)/2) + 1)*a**4*b + 60*cos(c + d*x)*log(tan((c + d 
*x)/2) + 1)*a**3*b*c + 60*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a**2*b...