\(\int \frac {\sec ^2(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^3} \, dx\) [918]

Optimal result

Integrand size = 41, antiderivative size = 242 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{b^3 d}+\frac {\left (a^2 b^3 B+2 b^5 B-2 a^5 C+5 a^3 b^2 C-3 a b^4 (A+2 C)\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{5/2} b^3 (a+b)^{5/2} d}+\frac {a \left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^2}+\frac {\left (2 A b^4+a^3 b B-4 a b^3 B-3 a^4 C+a^2 b^2 (A+6 C)\right ) \tan (c+d x)}{2 b^2 \left (a^2-b^2\right )^2 d (a+b \sec (c+d x))} \] Output:

C*arctanh(sin(d*x+c))/b^3/d+(B*a^2*b^3+2*B*b^5-2*a^5*C+5*C*a^3*b^2-3*a*b^4 
*(A+2*C))*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/(a-b)^(5/2)/ 
b^3/(a+b)^(5/2)/d+1/2*a*(A*b^2-a*(B*b-C*a))*tan(d*x+c)/b^2/(a^2-b^2)/d/(a+ 
b*sec(d*x+c))^2+1/2*(2*A*b^4+B*a^3*b-4*B*a*b^3-3*a^4*C+a^2*b^2*(A+6*C))*ta 
n(d*x+c)/b^2/(a^2-b^2)^2/d/(a+b*sec(d*x+c))
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.83 (sec) , antiderivative size = 514, normalized size of antiderivative = 2.12 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\frac {(b+a \cos (c+d x)) \sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-4 C (b+a \cos (c+d x))^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 C (b+a \cos (c+d x))^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 \left (-a^2 b^3 B-2 b^5 B+2 a^5 C-5 a^3 b^2 C+3 a b^4 (A+2 C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x))^2 (i \cos (c)+\sin (c))}{\left (a^2-b^2\right )^{5/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {b \left (a \sec (c) \left (b \left (4 A b^4+a^3 b B-10 a b^3 B-7 a^4 C+a^2 b^2 (5 A+16 C)\right ) \sin (d x)+a \left (b \left (a^2 b B+2 b^3 B+a^3 C-a b^2 (3 A+4 C)\right ) \sin (2 c+d x)+\left (A b^4-3 a b^3 B-2 a^4 C+a^2 b^2 (2 A+5 C)\right ) \sin (c+2 d x)\right )\right )+\left (a^2+2 b^2\right ) \left (-A b^4+3 a b^3 B+2 a^4 C-a^2 b^2 (2 A+5 C)\right ) \tan (c)\right )}{a \left (a^2-b^2\right )^2}\right )}{2 b^3 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^3} \] Input:

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

Output:

((b + a*Cos[c + d*x])*Sec[c + d*x]*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2) 
*(-4*C*(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 4 
*C*(b + a*Cos[c + d*x])^2*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (4*(- 
(a^2*b^3*B) - 2*b^5*B + 2*a^5*C - 5*a^3*b^2*C + 3*a*b^4*(A + 2*C))*ArcTan[ 
((I*Cos[c] + Sin[c])*(a*Sin[c] + (-b + a*Cos[c])*Tan[(d*x)/2]))/(Sqrt[a^2 
- b^2]*Sqrt[(Cos[c] - I*Sin[c])^2])]*(b + a*Cos[c + d*x])^2*(I*Cos[c] + Si 
n[c]))/((a^2 - b^2)^(5/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) + (b*(a*Sec[c]*(b*( 
4*A*b^4 + a^3*b*B - 10*a*b^3*B - 7*a^4*C + a^2*b^2*(5*A + 16*C))*Sin[d*x] 
+ a*(b*(a^2*b*B + 2*b^3*B + a^3*C - a*b^2*(3*A + 4*C))*Sin[2*c + d*x] + (A 
*b^4 - 3*a*b^3*B - 2*a^4*C + a^2*b^2*(2*A + 5*C))*Sin[c + 2*d*x])) + (a^2 
+ 2*b^2)*(-(A*b^4) + 3*a*b^3*B + 2*a^4*C - a^2*b^2*(2*A + 5*C))*Tan[c]))/( 
a*(a^2 - b^2)^2)))/(2*b^3*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x 
)])*(a + b*Sec[c + d*x])^3)
 

Rubi [A] (verified)

Time = 1.53 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.16, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.317, Rules used = {3042, 4578, 25, 3042, 4568, 3042, 4486, 3042, 4257, 4318, 3042, 3138, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4578

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4568

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4486

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

\(\Big \downarrow \) 4318

\(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {-\frac {\frac {\left (-2 a^5 C+5 a^3 b^2 C+a^2 b^3 B-3 a b^4 (A+2 C)+2 b^5 B\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b}+\frac {2 C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (-3 a^4 C+a^3 b B+a^2 b^2 (A+6 C)-4 a b^3 B+2 A b^4\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {a \tan (c+d x) \left (A b^2-a (b B-a C)\right )}{2 b^2 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^2}-\frac {-\frac {\frac {\left (-2 a^5 C+5 a^3 b^2 C+a^2 b^3 B-3 a b^4 (A+2 C)+2 b^5 B\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b}+\frac {2 C \left (a^2-b^2\right )^2 \text {arctanh}(\sin (c+d x))}{d}}{b \left (a^2-b^2\right )}-\frac {\tan (c+d x) \left (-3 a^4 C+a^3 b B+a^2 b^2 (A+6 C)-4 a b^3 B+2 A b^4\right )}{d \left (a^2-b^2\right ) (a+b \sec (c+d x))}}{2 b^2 \left (a^2-b^2\right )}\)

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(a*(A*b^2 - a*(b*B - a*C))*Tan[c + d*x])/(2*b^2*(a^2 - b^2)*d*(a + b*Sec[c 
 + d*x])^2) - (-(((2*(a^2 - b^2)^2*C*ArcTanh[Sin[c + d*x]])/d + (2*(a^2*b^ 
3*B + 2*b^5*B - 2*a^5*C + 5*a^3*b^2*C - 3*a*b^4*(A + 2*C))*ArcTanh[(Sqrt[a 
 - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(Sqrt[a - b]*Sqrt[a + b]*d))/(b*(a^2 
 - b^2))) - ((2*A*b^4 + a^3*b*B - 4*a*b^3*B - 3*a^4*C + a^2*b^2*(A + 6*C)) 
*Tan[c + d*x])/((a^2 - b^2)*d*(a + b*Sec[c + d*x])))/(2*b^2*(a^2 - b^2))
 

Defintions of rubi rules used

rule 25

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

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
 

rule 3042

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

rule 3138

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ 
e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d)   Subst[Int[1/(a + b + 
(a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] 
 && NeQ[a^2 - b^2, 0]
 

rule 4257

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

rule 4318

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo 
l] :> Simp[1/b   Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, 
f}, x] && NeQ[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 4568

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

rule 4578

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

Maple [A] (verified)

Time = 0.89 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.50

method	result	size
derivativedivides	\(\frac {\frac {\frac {2 \left (-\frac {\left (2 A \,a^{2} b^{2}+a A \,b^{3}+2 A \,b^{4}-B \,a^{2} b^{2}-4 B a \,b^{3}-2 a^{4} C +a^{3} b C +6 C \,a^{2} b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b \left (2 A \,a^{2} b^{2}-a A \,b^{3}+2 A \,b^{4}+B \,a^{2} b^{2}-4 B a \,b^{3}-2 a^{4} C -a^{3} b C +6 C \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (3 a A \,b^{4}-B \,a^{2} b^{3}-2 B \,b^{5}+2 a^{5} C -5 C \,a^{3} b^{2}+6 C a \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{3}}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}}{d}\)	\(364\)
default	\(\frac {\frac {\frac {2 \left (-\frac {\left (2 A \,a^{2} b^{2}+a A \,b^{3}+2 A \,b^{4}-B \,a^{2} b^{2}-4 B a \,b^{3}-2 a^{4} C +a^{3} b C +6 C \,a^{2} b^{2}\right ) b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{2 \left (a -b \right ) \left (a^{2}+2 a b +b^{2}\right )}+\frac {b \left (2 A \,a^{2} b^{2}-a A \,b^{3}+2 A \,b^{4}+B \,a^{2} b^{2}-4 B a \,b^{3}-2 a^{4} C -a^{3} b C +6 C \,a^{2} b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 \left (a +b \right ) \left (a -b \right )^{2}}\right )}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )^{2}}-\frac {\left (3 a A \,b^{4}-B \,a^{2} b^{3}-2 B \,b^{5}+2 a^{5} C -5 C \,a^{3} b^{2}+6 C a \,b^{4}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}}{b^{3}}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{3}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{3}}}{d}\)	\(364\)
risch	\(\text {Expression too large to display}\)	\(1489\)

Input:

int(sec(d*x+c)^2*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^3,x,meth 
od=_RETURNVERBOSE)
 

Output:

1/d*(2/b^3*((-1/2*(2*A*a^2*b^2+A*a*b^3+2*A*b^4-B*a^2*b^2-4*B*a*b^3-2*C*a^4 
+C*a^3*b+6*C*a^2*b^2)*b/(a-b)/(a^2+2*a*b+b^2)*tan(1/2*d*x+1/2*c)^3+1/2*b*( 
2*A*a^2*b^2-A*a*b^3+2*A*b^4+B*a^2*b^2-4*B*a*b^3-2*C*a^4-C*a^3*b+6*C*a^2*b^ 
2)/(a+b)/(a-b)^2*tan(1/2*d*x+1/2*c))/(tan(1/2*d*x+1/2*c)^2*a-tan(1/2*d*x+1 
/2*c)^2*b-a-b)^2-1/2*(3*A*a*b^4-B*a^2*b^3-2*B*b^5+2*C*a^5-5*C*a^3*b^2+6*C* 
a*b^4)/(a^4-2*a^2*b^2+b^4)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1 
/2*c)/((a+b)*(a-b))^(1/2)))-C/b^3*ln(tan(1/2*d*x+1/2*c)-1)+C/b^3*ln(tan(1/ 
2*d*x+1/2*c)+1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 721 vs. \(2 (232) = 464\).

Time = 16.04 (sec) , antiderivative size = 1501, normalized size of antiderivative = 6.20 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/4*((2*C*a^5*b^2 - 5*C*a^3*b^4 - B*a^2*b^5 + 3*(A + 2*C)*a*b^6 - 2*B*b^ 
7 + (2*C*a^7 - 5*C*a^5*b^2 - B*a^4*b^3 + 3*(A + 2*C)*a^3*b^4 - 2*B*a^2*b^5 
)*cos(d*x + c)^2 + 2*(2*C*a^6*b - 5*C*a^4*b^3 - B*a^3*b^4 + 3*(A + 2*C)*a^ 
2*b^5 - 2*B*a*b^6)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - 
 (a^2 - 2*b^2)*cos(d*x + c)^2 + 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin 
(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)) 
- 2*(C*a^6*b^2 - 3*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8 + (C*a^8 - 3*C*a^6*b^2 
+ 3*C*a^4*b^4 - C*a^2*b^6)*cos(d*x + c)^2 + 2*(C*a^7*b - 3*C*a^5*b^3 + 3*C 
*a^3*b^5 - C*a*b^7)*cos(d*x + c))*log(sin(d*x + c) + 1) + 2*(C*a^6*b^2 - 3 
*C*a^4*b^4 + 3*C*a^2*b^6 - C*b^8 + (C*a^8 - 3*C*a^6*b^2 + 3*C*a^4*b^4 - C* 
a^2*b^6)*cos(d*x + c)^2 + 2*(C*a^7*b - 3*C*a^5*b^3 + 3*C*a^3*b^5 - C*a*b^7 
)*cos(d*x + c))*log(-sin(d*x + c) + 1) + 2*(3*C*a^6*b^2 - B*a^5*b^3 - (A + 
 9*C)*a^4*b^4 + 5*B*a^3*b^5 - (A - 6*C)*a^2*b^6 - 4*B*a*b^7 + 2*A*b^8 + (2 
*C*a^7*b - (2*A + 7*C)*a^5*b^3 + 3*B*a^4*b^4 + (A + 5*C)*a^3*b^5 - 3*B*a^2 
*b^6 + A*a*b^7)*cos(d*x + c))*sin(d*x + c))/((a^8*b^3 - 3*a^6*b^5 + 3*a^4* 
b^7 - a^2*b^9)*d*cos(d*x + c)^2 + 2*(a^7*b^4 - 3*a^5*b^6 + 3*a^3*b^8 - a*b 
^10)*d*cos(d*x + c) + (a^6*b^5 - 3*a^4*b^7 + 3*a^2*b^9 - b^11)*d), -1/2*(( 
2*C*a^5*b^2 - 5*C*a^3*b^4 - B*a^2*b^5 + 3*(A + 2*C)*a*b^6 - 2*B*b^7 + (2*C 
*a^7 - 5*C*a^5*b^2 - B*a^4*b^3 + 3*(A + 2*C)*a^3*b^4 - 2*B*a^2*b^5)*cos(d* 
x + c)^2 + 2*(2*C*a^6*b - 5*C*a^4*b^3 - B*a^3*b^4 + 3*(A + 2*C)*a^2*b^5...
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 632 vs. \(2 (232) = 464\).

Time = 0.39 (sec) , antiderivative size = 632, normalized size of antiderivative = 2.61 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

-((2*C*a^5 - 5*C*a^3*b^2 - B*a^2*b^3 + 3*A*a*b^4 + 6*C*a*b^4 - 2*B*b^5)*(p 
i*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 
 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^4*b^3 - 2*a^2*b^5 
 + b^7)*sqrt(-a^2 + b^2)) - C*log(abs(tan(1/2*d*x + 1/2*c) + 1))/b^3 + C*l 
og(abs(tan(1/2*d*x + 1/2*c) - 1))/b^3 - (2*C*a^5*tan(1/2*d*x + 1/2*c)^3 - 
3*C*a^4*b*tan(1/2*d*x + 1/2*c)^3 - 2*A*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + B* 
a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - 5*C*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 + A*a^ 
2*b^3*tan(1/2*d*x + 1/2*c)^3 + 3*B*a^2*b^3*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^ 
2*b^3*tan(1/2*d*x + 1/2*c)^3 - A*a*b^4*tan(1/2*d*x + 1/2*c)^3 - 4*B*a*b^4* 
tan(1/2*d*x + 1/2*c)^3 + 2*A*b^5*tan(1/2*d*x + 1/2*c)^3 - 2*C*a^5*tan(1/2* 
d*x + 1/2*c) - 3*C*a^4*b*tan(1/2*d*x + 1/2*c) + 2*A*a^3*b^2*tan(1/2*d*x + 
1/2*c) + B*a^3*b^2*tan(1/2*d*x + 1/2*c) + 5*C*a^3*b^2*tan(1/2*d*x + 1/2*c) 
 + A*a^2*b^3*tan(1/2*d*x + 1/2*c) - 3*B*a^2*b^3*tan(1/2*d*x + 1/2*c) + 6*C 
*a^2*b^3*tan(1/2*d*x + 1/2*c) + A*a*b^4*tan(1/2*d*x + 1/2*c) - 4*B*a*b^4*t 
an(1/2*d*x + 1/2*c) + 2*A*b^5*tan(1/2*d*x + 1/2*c))/((a^4*b^2 - 2*a^2*b^4 
+ b^6)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b)^2))/d
 

Mupad [B] (verification not implemented)

Time = 24.12 (sec) , antiderivative size = 8128, normalized size of antiderivative = 33.59 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx=\text {Too large to display} \] Input:

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

Output:

- ((tan(c/2 + (d*x)/2)^3*(2*A*b^4 - 2*C*a^4 + 2*A*a^2*b^2 - B*a^2*b^2 + 6* 
C*a^2*b^2 + A*a*b^3 - 4*B*a*b^3 + C*a^3*b))/((a*b^2 - b^3)*(a + b)^2) - (t 
an(c/2 + (d*x)/2)*(2*A*b^4 - 2*C*a^4 + 2*A*a^2*b^2 + B*a^2*b^2 + 6*C*a^2*b 
^2 - A*a*b^3 - 4*B*a*b^3 - C*a^3*b))/((a + b)*(b^4 - 2*a*b^3 + a^2*b^2)))/ 
(d*(2*a*b - tan(c/2 + (d*x)/2)^2*(2*a^2 - 2*b^2) + tan(c/2 + (d*x)/2)^4*(a 
^2 - 2*a*b + b^2) + a^2 + b^2)) - (C*atan(((C*((8*tan(c/2 + (d*x)/2)*(4*B^ 
2*b^10 + 8*C^2*a^10 + 4*C^2*b^10 - 8*C^2*a*b^9 - 8*C^2*a^9*b + 9*A^2*a^2*b 
^8 + 4*B^2*a^2*b^8 + B^2*a^4*b^6 + 24*C^2*a^2*b^8 + 32*C^2*a^3*b^7 - 52*C^ 
2*a^4*b^6 - 48*C^2*a^5*b^5 + 57*C^2*a^6*b^4 + 32*C^2*a^7*b^3 - 32*C^2*a^8* 
b^2 - 12*A*B*a*b^9 - 24*B*C*a*b^9 - 6*A*B*a^3*b^7 + 36*A*C*a^2*b^8 - 30*A* 
C*a^4*b^6 + 12*A*C*a^6*b^4 + 8*B*C*a^3*b^7 + 2*B*C*a^5*b^5 - 4*B*C*a^7*b^3 
))/(a*b^10 + b^11 - 3*a^2*b^9 - 3*a^3*b^8 + 3*a^4*b^7 + 3*a^5*b^6 - a^6*b^ 
5 - a^7*b^4) + (C*((8*(4*B*b^15 + 4*C*b^15 + 6*A*a^2*b^13 + 12*A*a^3*b^12 
- 12*A*a^4*b^11 - 6*A*a^5*b^10 + 6*A*a^6*b^9 - 6*B*a^2*b^13 + 6*B*a^3*b^12 
 + 2*B*a^6*b^9 - 2*B*a^7*b^8 - 8*C*a^2*b^13 + 34*C*a^3*b^12 + 6*C*a^4*b^11 
 - 36*C*a^5*b^10 - 4*C*a^6*b^9 + 18*C*a^7*b^8 + 2*C*a^8*b^7 - 4*C*a^9*b^6 
- 6*A*a*b^14 - 4*B*a*b^14 - 12*C*a*b^14))/(a*b^12 + b^13 - 3*a^2*b^11 - 3* 
a^3*b^10 + 3*a^4*b^9 + 3*a^5*b^8 - a^6*b^7 - a^7*b^6) + (8*C*tan(c/2 + (d* 
x)/2)*(8*a*b^15 - 8*a^2*b^14 - 32*a^3*b^13 + 32*a^4*b^12 + 48*a^5*b^11 - 4 
8*a^6*b^10 - 32*a^7*b^9 + 32*a^8*b^8 + 8*a^9*b^7 - 8*a^10*b^6))/(b^3*(a...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 1905, normalized size of antiderivative = 7.87 \[ \int \frac {\sec ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 8*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ 
sqrt( - a**2 + b**2))*cos(c + d*x)*a**6*b*c + 20*sqrt( - a**2 + b**2)*atan 
((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d 
*x)*a**4*b**3*c - 8*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c 
 + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a**3*b**5 - 24*sqrt( - a* 
*2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b 
**2))*cos(c + d*x)*a**2*b**5*c + 8*sqrt( - a**2 + b**2)*atan((tan((c + d*x 
)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x)*a*b**7 + 4 
*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( 
 - a**2 + b**2))*sin(c + d*x)**2*a**7*c - 10*sqrt( - a**2 + b**2)*atan((ta 
n((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)* 
*2*a**5*b**2*c + 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c 
+ d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*a**4*b**4 + 12*sqrt( - 
a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + 
 b**2))*sin(c + d*x)**2*a**3*b**4*c - 4*sqrt( - a**2 + b**2)*atan((tan((c 
+ d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*sin(c + d*x)**2*a* 
*2*b**6 - 4*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/ 
2)*b)/sqrt( - a**2 + b**2))*a**7*c + 6*sqrt( - a**2 + b**2)*atan((tan((c + 
 d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**5*b**2*c - 4*sqr 
t( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( ...