3.2.40 \(\int \frac {\sec ^2(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\) [140]

3.2.40.1 Optimal result

Integrand size = 33, antiderivative size = 123 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(3 A-7 C) \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {(6 A-29 C) \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]

output

C*arctanh(sin(d*x+c))/a^3/d-1/5*(A+C)*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d 
*x+c))^3-1/15*(3*A-7*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2+1/15*(6*A-29*C)* 
tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))
 

3.2.40.2 Mathematica [A] (verified)

Time = 3.76 (sec) , antiderivative size = 236, normalized size of antiderivative = 1.92 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (240 C \cos ^5\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \left (-5 (3 A-29 C) \sin \left (\frac {d x}{2}\right )+15 (A-5 C) \sin \left (c+\frac {d x}{2}\right )-15 A \sin \left (c+\frac {3 d x}{2}\right )+95 C \sin \left (c+\frac {3 d x}{2}\right )-15 C \sin \left (2 c+\frac {3 d x}{2}\right )-3 A \sin \left (2 c+\frac {5 d x}{2}\right )+22 C \sin \left (2 c+\frac {5 d x}{2}\right )\right )\right )}{15 a^3 d (A+2 C+A \cos (2 (c+d x))) (1+\sec (c+d x))^3} \]

input

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

output

-1/15*(Cos[(c + d*x)/2]*Sec[c + d*x]*(A + C*Sec[c + d*x]^2)*(240*C*Cos[(c 
+ d*x)/2]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/ 
2] + Sin[(c + d*x)/2]]) + Sec[c/2]*(-5*(3*A - 29*C)*Sin[(d*x)/2] + 15*(A - 
 5*C)*Sin[c + (d*x)/2] - 15*A*Sin[c + (3*d*x)/2] + 95*C*Sin[c + (3*d*x)/2] 
 - 15*C*Sin[2*c + (3*d*x)/2] - 3*A*Sin[2*c + (5*d*x)/2] + 22*C*Sin[2*c + ( 
5*d*x)/2])))/(a^3*d*(A + 2*C + A*Cos[2*(c + d*x)])*(1 + Sec[c + d*x])^3)
 

3.2.40.3 Rubi [A] (verified)

Time = 0.89 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.07, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4573, 25, 3042, 4496, 25, 3042, 4486, 3042, 4257, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4496

\(\displaystyle \frac {-\frac {\int -\frac {\sec (c+d x) \left (2 (3 A-7 C) a^2+15 C \sec (c+d x) a^2\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (3 A-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (2 (3 A-7 C) a^2+15 C \sec (c+d x) a^2\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {a (3 A-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 (3 A-7 C) a^2+15 C \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {a (3 A-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4486

\(\displaystyle \frac {\frac {a^2 (6 A-29 C) \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx+15 a C \int \sec (c+d x)dx}{3 a^2}-\frac {a (3 A-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {a^2 (6 A-29 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+15 a C \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{3 a^2}-\frac {a (3 A-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {a^2 (6 A-29 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx+\frac {15 a C \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {a (3 A-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

\(\Big \downarrow \) 4281

\(\displaystyle \frac {\frac {\frac {a^2 (6 A-29 C) \tan (c+d x)}{d (a \sec (c+d x)+a)}+\frac {15 a C \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {a (3 A-7 C) \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\)

input

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

output

-1/5*((A + C)*Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) + (- 
1/3*(a*(3*A - 7*C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^2) + ((15*a*C*Arc 
Tanh[Sin[c + d*x]])/d + (a^2*(6*A - 29*C)*Tan[c + d*x])/(d*(a + a*Sec[c + 
d*x])))/(3*a^2))/(5*a^2)
 

3.2.40.3.1 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 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, 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 4486

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

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)]
 

3.2.40.4 Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71

method	result	size
parallelrisch	\(\frac {-60 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+60 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {20 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{3}-5 A +35 C \right )}{60 a^{3} d}\)	\(87\)
derivativedivides	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\)	\(105\)
default	\(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} A}{5}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5} C}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} C}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A -7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\)	\(105\)
risch	\(-\frac {2 i \left (15 C \,{\mathrm e}^{4 i \left (d x +c \right )}-15 A \,{\mathrm e}^{3 i \left (d x +c \right )}+75 C \,{\mathrm e}^{3 i \left (d x +c \right )}-15 A \,{\mathrm e}^{2 i \left (d x +c \right )}+145 C \,{\mathrm e}^{2 i \left (d x +c \right )}-15 A \,{\mathrm e}^{i \left (d x +c \right )}+95 C \,{\mathrm e}^{i \left (d x +c \right )}-3 A +22 C \right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{a^{3} d}\)	\(158\)
norman	\(\frac {\frac {\left (A -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 a d}-\frac {\left (A -7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}-\frac {\left (7 A -43 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{10 a d}+\frac {\left (9 A -11 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}+\frac {\left (-59 C +9 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3} a^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\)	\(209\)

input

int(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x,method=_RETURNVER 
BOSE)
 

output

1/60*(-60*C*ln(tan(1/2*d*x+1/2*c)-1)+60*C*ln(tan(1/2*d*x+1/2*c)+1)-3*tan(1 
/2*d*x+1/2*c)*((A+C)*tan(1/2*d*x+1/2*c)^4+20/3*C*tan(1/2*d*x+1/2*c)^2-5*A+ 
35*C))/a^3/d
 

3.2.40.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (C \cos \left (d x + c\right )^{3} + 3 \, C \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + C\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (3 \, A - 22 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A - 17 \, C\right )} \cos \left (d x + c\right ) + 3 \, A - 32 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

input

integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm= 
"fricas")
 

output

1/30*(15*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos(d*x + c) + C)*lo 
g(sin(d*x + c) + 1) - 15*(C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + 3*C*cos( 
d*x + c) + C)*log(-sin(d*x + c) + 1) + 2*((3*A - 22*C)*cos(d*x + c)^2 + 3* 
(3*A - 17*C)*cos(d*x + c) + 3*A - 32*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^ 
3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
 

3.2.40.6 Sympy [F]

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

input

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

output

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

3.2.40.7 Maxima [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.36 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=-\frac {C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {3 \, A {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, d} \]

input

integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm= 
"maxima")
 

output

-1/60*(C*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d* 
x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 60*log(sin(d* 
x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1 
) - 1)/a^3) - 3*A*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos 
(d*x + c) + 1)^5)/a^3)/d
 

3.2.40.8 Giac [A] (verification not implemented)

Time = 0.34 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.07 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

input

integrate(sec(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm= 
"giac")
 

output

1/60*(60*C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*C*log(abs(tan(1/2*d 
*x + 1/2*c) - 1))/a^3 - (3*A*a^12*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^12*tan(1/ 
2*d*x + 1/2*c)^5 + 20*C*a^12*tan(1/2*d*x + 1/2*c)^3 - 15*A*a^12*tan(1/2*d* 
x + 1/2*c) + 105*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
 

3.2.40.9 Mupad [B] (verification not implemented)

Time = 15.32 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.89 \[ \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx=\frac {2\,C\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{a^3\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A+C}{4\,a^3}-\frac {A-3\,C}{2\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{12\,a^3}-\frac {A-3\,C}{12\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]

input

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

output

(2*C*atanh(tan(c/2 + (d*x)/2)))/(a^3*d) - (tan(c/2 + (d*x)/2)*((A + C)/(4* 
a^3) - (A - 3*C)/(2*a^3)))/d - (tan(c/2 + (d*x)/2)^3*((A + C)/(12*a^3) - ( 
A - 3*C)/(12*a^3)))/d - (tan(c/2 + (d*x)/2)^5*(A + C))/(20*a^3*d)