\(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [112]

Optimal result

Integrand size = 31, antiderivative size = 138 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {(A-B) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {(4 A-11 B) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3}+\frac {(8 A+13 B) \tan (c+d x)}{105 d \left (a^2+a^2 \sec (c+d x)\right )^2}+\frac {(8 A+13 B) \tan (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )} \] Output:

-1/7*(A-B)*tan(d*x+c)/d/(a+a*sec(d*x+c))^4+1/35*(4*A-11*B)*tan(d*x+c)/a/d/ 
(a+a*sec(d*x+c))^3+1/105*(8*A+13*B)*tan(d*x+c)/d/(a^2+a^2*sec(d*x+c))^2+1/ 
105*(8*A+13*B)*tan(d*x+c)/d/(a^4+a^4*sec(d*x+c))
 

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.18 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (140 (2 A+B) \sin \left (\frac {d x}{2}\right )-35 (5 A+4 B) \sin \left (c+\frac {d x}{2}\right )+168 A \sin \left (c+\frac {3 d x}{2}\right )+168 B \sin \left (c+\frac {3 d x}{2}\right )-105 A \sin \left (2 c+\frac {3 d x}{2}\right )+91 A \sin \left (2 c+\frac {5 d x}{2}\right )+56 B \sin \left (2 c+\frac {5 d x}{2}\right )+13 A \sin \left (3 c+\frac {7 d x}{2}\right )+8 B \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{420 a^4 d (1+\cos (c+d x))^4} \] Input:

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

Output:

(Cos[(c + d*x)/2]*Sec[c/2]*(140*(2*A + B)*Sin[(d*x)/2] - 35*(5*A + 4*B)*Si 
n[c + (d*x)/2] + 168*A*Sin[c + (3*d*x)/2] + 168*B*Sin[c + (3*d*x)/2] - 105 
*A*Sin[2*c + (3*d*x)/2] + 91*A*Sin[2*c + (5*d*x)/2] + 56*B*Sin[2*c + (5*d* 
x)/2] + 13*A*Sin[3*c + (7*d*x)/2] + 8*B*Sin[3*c + (7*d*x)/2]))/(420*a^4*d* 
(1 + Cos[c + d*x])^4)
 

Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.99, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, Rules used = {3042, 4496, 25, 3042, 4488, 3042, 4283, 3042, 4281}

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.

\(\displaystyle \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a \sec (c+d x)+a)^4} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4496

\(\displaystyle -\frac {\int -\frac {\sec (c+d x) (4 a (A-B)+7 a B \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\sec (c+d x) (4 a (A-B)+7 a B \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (4 a (A-B)+7 a B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4488

\(\displaystyle \frac {\frac {1}{5} (8 A+13 B) \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^2}dx+\frac {a (4 A-11 B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} (8 A+13 B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx+\frac {a (4 A-11 B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4283

\(\displaystyle \frac {\frac {1}{5} (8 A+13 B) \left (\frac {\int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )+\frac {a (4 A-11 B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{5} (8 A+13 B) \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )+\frac {a (4 A-11 B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {\frac {a (4 A-11 B) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {1}{5} (8 A+13 B) \left (\frac {\tan (c+d x)}{3 a d (a \sec (c+d x)+a)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{7 a^2}-\frac {(A-B) \tan (c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

Input:

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

Output:

-1/7*((A - B)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + ((a*(4*A - 11*B)* 
Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + ((8*A + 13*B)*(Tan[c + d*x]/( 
3*d*(a + a*Sec[c + d*x])^2) + Tan[c + d*x]/(3*a*d*(a + a*Sec[c + d*x]))))/ 
5)/(7*a^2)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

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

rule 4281

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} 
, x] && EqQ[a^2 - b^2, 0]
 

rule 4283

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ 
Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] 
 + Simp[(m + 1)/(a*(2*m + 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 
1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1) 
] && IntegerQ[2*m]
 

rule 4488

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

rule 4496

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

Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.58

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {7 \left (-A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}+\frac {7 \left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}+7 A +7 B \right )}{56 a^{4} d}\)	\(80\)
derivativedivides	\(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (-A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{8 d \,a^{4}}\)	\(88\)
default	\(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (-A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (-A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{8 d \,a^{4}}\)	\(88\)
risch	\(\frac {2 i \left (105 A \,{\mathrm e}^{5 i \left (d x +c \right )}+175 A \,{\mathrm e}^{4 i \left (d x +c \right )}+140 B \,{\mathrm e}^{4 i \left (d x +c \right )}+280 A \,{\mathrm e}^{3 i \left (d x +c \right )}+140 B \,{\mathrm e}^{3 i \left (d x +c \right )}+168 A \,{\mathrm e}^{2 i \left (d x +c \right )}+168 B \,{\mathrm e}^{2 i \left (d x +c \right )}+91 \,{\mathrm e}^{i \left (d x +c \right )} A +56 B \,{\mathrm e}^{i \left (d x +c \right )}+13 A +8 B \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(138\)
norman	\(\frac {\frac {\left (A -B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{56 a d}+\frac {\left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {\left (7 A +5 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{24 a d}+\frac {\left (11 A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{60 a d}+\frac {\left (11 A +31 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{420 a d}-\frac {\left (17 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{280 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{2} a^{3}}\)	\(167\)

Input:

int(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x,method=_RETURNVERBO 
SE)
 

Output:

1/56*tan(1/2*d*x+1/2*c)*((A-B)*tan(1/2*d*x+1/2*c)^6+7/5*(-A-B)*tan(1/2*d*x 
+1/2*c)^4+7/3*(-A+B)*tan(1/2*d*x+1/2*c)^2+7*A+7*B)/a^4/d
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {{\left ({\left (13 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (13 \, A + 8 \, B\right )} \cos \left (d x + c\right )^{2} + 4 \, {\left (8 \, A + 13 \, B\right )} \cos \left (d x + c\right ) + 8 \, A + 13 \, B\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \] Input:

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="f 
ricas")
 

Output:

1/105*((13*A + 8*B)*cos(d*x + c)^3 + 4*(13*A + 8*B)*cos(d*x + c)^2 + 4*(8* 
A + 13*B)*cos(d*x + c) + 8*A + 13*B)*sin(d*x + c)/(a^4*d*cos(d*x + c)^4 + 
4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a 
^4*d)
 

Sympy [F]

\[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\int \frac {A \sec ^{2}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \sec ^{3}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \] Input:

integrate(sec(d*x+c)**2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))**4,x)
 

Output:

(Integral(A*sec(c + d*x)**2/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c 
 + d*x)**2 + 4*sec(c + d*x) + 1), x) + Integral(B*sec(c + d*x)**3/(sec(c + 
 d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x) 
)/a**4
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.26 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {B {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \] Input:

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="m 
axima")
 

Output:

1/840*(B*(105*sin(d*x + c)/(cos(d*x + c) + 1) + 35*sin(d*x + c)^3/(cos(d*x 
 + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/ 
(cos(d*x + c) + 1)^7)/a^4 + A*(105*sin(d*x + c)/(cos(d*x + c) + 1) - 35*si 
n(d*x + c)^3/(cos(d*x + c) + 1)^3 - 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 
 + 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 21 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 35 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \] Input:

integrate(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x, algorithm="g 
iac")
 

Output:

1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 - 15*B*tan(1/2*d*x + 1/2*c)^7 - 21*A*ta 
n(1/2*d*x + 1/2*c)^5 - 21*B*tan(1/2*d*x + 1/2*c)^5 - 35*A*tan(1/2*d*x + 1/ 
2*c)^3 + 35*B*tan(1/2*d*x + 1/2*c)^3 + 105*A*tan(1/2*d*x + 1/2*c) + 105*B* 
tan(1/2*d*x + 1/2*c))/(a^4*d)
 

Mupad [B] (verification not implemented)

Time = 11.22 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.61 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=-\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+B\right )}{40\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-B\right )}{24\,a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B\right )}{56\,a^4}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+B\right )}{8\,a^4}}{d} \] Input:

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

Output:

-((tan(c/2 + (d*x)/2)^5*(A + B))/(40*a^4) + (tan(c/2 + (d*x)/2)^3*(A - B)) 
/(24*a^4) - (tan(c/2 + (d*x)/2)^7*(A - B))/(56*a^4) - (tan(c/2 + (d*x)/2)* 
(A + B))/(8*a^4))/d
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.78 \[ \int \frac {\sec ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} a -15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} b -21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} a -21 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} b -35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +35 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b +105 a +105 b \right )}{840 a^{4} d} \] Input:

int(sec(d*x+c)^2*(A+B*sec(d*x+c))/(a+a*sec(d*x+c))^4,x)
 

Output:

(tan((c + d*x)/2)*(15*tan((c + d*x)/2)**6*a - 15*tan((c + d*x)/2)**6*b - 2 
1*tan((c + d*x)/2)**4*a - 21*tan((c + d*x)/2)**4*b - 35*tan((c + d*x)/2)** 
2*a + 35*tan((c + d*x)/2)**2*b + 105*a + 105*b))/(840*a**4*d)