3.2.49 \(\int \frac {\sec ^2(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^4} \, dx\) [149]

3.2.49.1 Optimal result

Integrand size = 33, antiderivative size = 138 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {(23 A-54 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}+\frac {4 (2 A+9 C) \tan (c+d x)}{105 a^4 d (1+\sec (c+d x))}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 (3 A-4 C) \tan (c+d x)}{35 a d (a+a \sec (c+d x))^3} \]

output

1/105*(23*A-54*C)*tan(d*x+c)/a^4/d/(1+sec(d*x+c))^2+4/105*(2*A+9*C)*tan(d* 
x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A+C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x 
+c))^4-2/35*(3*A-4*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^3
 

3.2.49.2 Mathematica [A] (verified)

Time = 2.64 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.09 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (70 (4 A+3 C) \sin \left (\frac {d x}{2}\right )-175 A \sin \left (c+\frac {d x}{2}\right )+168 A \sin \left (c+\frac {3 d x}{2}\right )+126 C \sin \left (c+\frac {3 d x}{2}\right )-105 A \sin \left (2 c+\frac {3 d x}{2}\right )+91 A \sin \left (2 c+\frac {5 d x}{2}\right )+42 C \sin \left (2 c+\frac {5 d x}{2}\right )+13 A \sin \left (3 c+\frac {7 d x}{2}\right )+6 C \sin \left (3 c+\frac {7 d x}{2}\right )\right )}{6720 a^4 d} \]

input

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

output

(Sec[c/2]*Sec[(c + d*x)/2]^7*(70*(4*A + 3*C)*Sin[(d*x)/2] - 175*A*Sin[c + 
(d*x)/2] + 168*A*Sin[c + (3*d*x)/2] + 126*C*Sin[c + (3*d*x)/2] - 105*A*Sin 
[2*c + (3*d*x)/2] + 91*A*Sin[2*c + (5*d*x)/2] + 42*C*Sin[2*c + (5*d*x)/2] 
+ 13*A*Sin[3*c + (7*d*x)/2] + 6*C*Sin[3*c + (7*d*x)/2]))/(6720*a^4*d)
 

3.2.49.3 Rubi [A] (verified)

Time = 0.87 (sec) , antiderivative size = 149, normalized size of antiderivative = 1.08, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4573, 25, 3042, 4496, 25, 3042, 4488, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

\(\displaystyle -\frac {\int -\frac {\sec ^2(c+d x) (a (5 A-2 C)-a (A-6 C) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {\sec ^2(c+d x) (a (5 A-2 C)-a (A-6 C) \sec (c+d x))}{(\sec (c+d x) a+a)^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (a (5 A-2 C)-a (A-6 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^3}dx}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4496

\(\displaystyle \frac {-\frac {\int -\frac {\sec (c+d x) \left (6 a^2 (3 A-4 C)-5 a^2 (A-6 C) \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {2 a (3 A-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (6 a^2 (3 A-4 C)-5 a^2 (A-6 C) \sec (c+d x)\right )}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {2 a (3 A-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (6 a^2 (3 A-4 C)-5 a^2 (A-6 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {2 a (3 A-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4488

\(\displaystyle \frac {\frac {\frac {4}{3} a (2 A+9 C) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx+\frac {(23 A-54 C) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {2 a (3 A-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\frac {4}{3} a (2 A+9 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {(23 A-54 C) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {2 a (3 A-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {\frac {\frac {4 a (2 A+9 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)}+\frac {(23 A-54 C) \tan (c+d x)}{3 d (\sec (c+d x)+1)^2}}{5 a^2}-\frac {2 a (3 A-4 C) \tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}}{7 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{7 d (a \sec (c+d x)+a)^4}\)

input

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

output

-1/7*((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^4) + (( 
-2*a*(3*A - 4*C)*Tan[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) + (((23*A - 54 
*C)*Tan[c + d*x])/(3*d*(1 + Sec[c + d*x])^2) + (4*a*(2*A + 9*C)*Tan[c + d* 
x])/(3*d*(a + a*Sec[c + d*x])))/(5*a^2))/(7*a^2)
 

3.2.49.3.1 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 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)]
 

rule 4573

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) 
*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m 
+ 1))), x] + Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*C 
sc[e + f*x])^n*Simp[b*C*n + A*b*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - 
n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && EqQ[ 
a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

3.2.49.4 Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.57

method	result	size
parallelrisch	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\frac {7 \left (-A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{5}+7 \left (-\frac {A}{3}+C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+7 A +7 C \right )}{56 a^{4} d}\)	\(78\)
derivativedivides	\(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (-A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (-A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(88\)
default	\(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{7}+\frac {\left (-A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}+\frac {\left (-A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{8 d \,a^{4}}\)	\(88\)
risch	\(\frac {2 i \left (105 A \,{\mathrm e}^{5 i \left (d x +c \right )}+175 A \,{\mathrm e}^{4 i \left (d x +c \right )}+280 A \,{\mathrm e}^{3 i \left (d x +c \right )}+210 C \,{\mathrm e}^{3 i \left (d x +c \right )}+168 A \,{\mathrm e}^{2 i \left (d x +c \right )}+126 C \,{\mathrm e}^{2 i \left (d x +c \right )}+91 A \,{\mathrm e}^{i \left (d x +c \right )}+42 C \,{\mathrm e}^{i \left (d x +c \right )}+13 A +6 C \right )}{105 d \,a^{4} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{7}}\)	\(126\)
norman	\(\frac {-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{56 a d}+\frac {\left (5 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {\left (11 A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{70 a d}-\frac {\left (11 A -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{140 a d}-\frac {\left (19 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{40 a d}+\frac {\left (73 A -39 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{840 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} a^{3}}\)	\(193\)

input

int(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^4,x,method=_RETURNVER 
BOSE)
 

output

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

3.2.49.5 Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {{\left ({\left (13 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (13 \, A + 6 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (32 \, A + 39 \, C\right )} \cos \left (d x + c\right ) + 8 \, A + 36 \, C\right )} \sin \left (d x + c\right )}{105 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

input

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

output

1/105*((13*A + 6*C)*cos(d*x + c)^3 + 4*(13*A + 6*C)*cos(d*x + c)^2 + (32*A 
 + 39*C)*cos(d*x + c) + 8*A + 36*C)*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)
 

3.2.49.6 Sympy [F]

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

input

integrate(sec(d*x+c)**2*(A+C*sec(d*x+c)**2)/(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(C*sec(c + d*x)**4/(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
 

3.2.49.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {A {\left (\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}} + \frac {3 \, C {\left (\frac {35 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {35 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}\right )}}{a^{4}}}{840 \, d} \]

input

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

output

1/840*(A*(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 + 3*C*(35*sin(d*x + c)/(cos(d*x + c) + 1) + 35*s 
in(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^ 
5 + 5*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/a^4)/d
 

3.2.49.8 Giac [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.85 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 21 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 63 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 35 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{840 \, a^{4} d} \]

input

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

output

1/840*(15*A*tan(1/2*d*x + 1/2*c)^7 + 15*C*tan(1/2*d*x + 1/2*c)^7 - 21*A*ta 
n(1/2*d*x + 1/2*c)^5 + 63*C*tan(1/2*d*x + 1/2*c)^5 - 35*A*tan(1/2*d*x + 1/ 
2*c)^3 + 105*C*tan(1/2*d*x + 1/2*c)^3 + 105*A*tan(1/2*d*x + 1/2*c) + 105*C 
*tan(1/2*d*x + 1/2*c))/(a^4*d)
 

3.2.49.9 Mupad [B] (verification not implemented)

Time = 15.72 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.60 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx=\frac {\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A-3\,C\right )}{24\,a^4}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A-3\,C\right )}{40\,a^4}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{8\,a^4}}{d} \]

input

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

output

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