\(\int \frac {\sec (c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\) [911]

Optimal result

Integrand size = 39, antiderivative size = 148 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {C \text {arctanh}(\sin (c+d x))}{b^2 d}+\frac {2 \left (a A b^2-b^3 B-a^3 C+2 a b^2 C\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{(a-b)^{3/2} b^2 (a+b)^{3/2} d}-\frac {\left (A b^2-a (b B-a C)\right ) \tan (c+d x)}{b \left (a^2-b^2\right ) d (a+b \sec (c+d x))} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 3.14 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.41 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {2 (b+a \cos (c+d x)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-C (b+a \cos (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+C (b+a \cos (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 \left (b^3 B+a^3 C-a b^2 (A+2 C)\right ) \arctan \left (\frac {(i \cos (c)+\sin (c)) \left (a \sin (c)+(-b+a \cos (c)) \tan \left (\frac {d x}{2}\right )\right )}{\sqrt {a^2-b^2} \sqrt {(\cos (c)-i \sin (c))^2}}\right ) (b+a \cos (c+d x)) (i \cos (c)+\sin (c))}{\left (a^2-b^2\right )^{3/2} \sqrt {(\cos (c)-i \sin (c))^2}}+\frac {b \left (A b^2+a (-b B+a C)\right ) (b \sin (c)-a \sin (d x))}{a (a-b) (a+b) \left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right )}\right )}{b^2 d (A+2 C+2 B \cos (c+d x)+A \cos (2 (c+d x))) (a+b \sec (c+d x))^2} \] Input:

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

Output:

(2*(b + a*Cos[c + d*x])*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*(-(C*(b + 
a*Cos[c + d*x])*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]) + C*(b + a*Cos[c 
 + d*x])*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*(b^3*B + a^3*C - a* 
b^2*(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])*(I*Cos[c] + Sin[c]))/((a^2 - b^2)^(3/2)*Sqrt[(Cos[c] - I*Sin[c])^2]) 
 + (b*(A*b^2 + a*(-(b*B) + a*C))*(b*Sin[c] - a*Sin[d*x]))/(a*(a - b)*(a + 
b)*(Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2]))))/(b^2*d*(A + 2*C + 2*B*Co 
s[c + d*x] + A*Cos[2*(c + d*x)])*(a + b*Sec[c + d*x])^2)
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.16, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.256, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4568

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4486

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4257

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

\(\Big \downarrow \) 4318

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3138

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

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]
 

Maple [A] (verified)

Time = 0.57 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.35

method	result	size
derivativedivides	\(\frac {-\frac {2 \left (-\frac {b \left (A \,b^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (a A \,b^{2}-B \,b^{3}-a^{3} C +2 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}}{d}\)	\(200\)
default	\(\frac {-\frac {2 \left (-\frac {b \left (A \,b^{2}-B a b +C \,a^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\left (a^{2}-b^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} b -a -b \right )}-\frac {\left (a A \,b^{2}-B \,b^{3}-a^{3} C +2 C a \,b^{2}\right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\left (a +b \right ) \left (a -b \right ) \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{b^{2}}+\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b^{2}}-\frac {C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b^{2}}}{d}\)	\(200\)
risch	\(-\frac {2 i \left (A \,b^{2}-B a b +C \,a^{2}\right ) \left (b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}{\left (a^{2}-b^{2}\right ) d b a \left ({\mathrm e}^{2 i \left (d x +c \right )} a +2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}+\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) A}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) B}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d \,b^{2}}-\frac {2 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i a^{2}-i b^{2}-\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right ) C a}{\sqrt {a^{2}-b^{2}}\, \left (a +b \right ) \left (a -b \right ) d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{b^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{b^{2} d}\)	\(795\)

Input:

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

Output:

1/d*(-2/b^2*(-b*(A*b^2-B*a*b+C*a^2)/(a^2-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)-(A*a*b^2-B*b^3-C*a^3+2*C*a*b^2) 
/(a+b)/(a-b)/((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^2*ln(tan(1/2*d*x+1/2*c)+1)-C/b^2*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. 336 vs. \(2 (137) = 274\).

Time = 3.78 (sec) , antiderivative size = 728, normalized size of antiderivative = 4.92 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(sec(d*x+c)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^2,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 [A] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.69 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (C a^{3} - A a b^{2} - 2 \, C a b^{2} + B b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b^{2}} - \frac {C \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b^{2}} + \frac {2 \, {\left (C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{2} b - b^{3}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \] Input:

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

Output:

(2*(C*a^3 - A*a*b^2 - 2*C*a*b^2 + B*b^3)*(pi*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^2*b^2 - b^4)*sqrt(-a^2 + b^2)) + C*log(abs(tan(1/2 
*d*x + 1/2*c) + 1))/b^2 - C*log(abs(tan(1/2*d*x + 1/2*c) - 1))/b^2 + 2*(C* 
a^2*tan(1/2*d*x + 1/2*c) - B*a*b*tan(1/2*d*x + 1/2*c) + A*b^2*tan(1/2*d*x 
+ 1/2*c))/((a^2*b - b^3)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c 
)^2 - a - b)))/d
 

Mupad [B] (verification not implemented)

Time = 21.33 (sec) , antiderivative size = 4536, normalized size of antiderivative = 30.65 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx=\text {Too large to display} \] Input:

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

Output:

- (C*atan(((C*((32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + C^2*b^6 - 2*C 
^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2*a^3*b^3 - 5*C 
^2*a^4*b^2 - 2*A*B*a*b^5 - 4*B*C*a*b^5 + 4*A*C*a^2*b^4 - 2*A*C*a^4*b^2 + 2 
*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) + (C*((32*(A*a^4*b^5 - C* 
b^9 - A*a^2*b^7 - A*a^3*b^6 - B*b^9 + B*a^2*b^7 - B*a^3*b^6 + C*a^2*b^7 - 
3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + B*a*b^8 + 2*C*a*b^8))/(a*b^5 + b^6 - a 
^2*b^4 - a^3*b^3) + (32*C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^2*b^8 - 4*a^3* 
b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 - a^2*b^3 - a^ 
3*b^2))))/b^2)*1i)/b^2 + (C*((32*tan(c/2 + (d*x)/2)*(B^2*b^6 + 2*C^2*a^6 + 
 C^2*b^6 - 2*C^2*a*b^5 - 2*C^2*a^5*b + A^2*a^2*b^4 + 3*C^2*a^2*b^4 + 4*C^2 
*a^3*b^3 - 5*C^2*a^4*b^2 - 2*A*B*a*b^5 - 4*B*C*a*b^5 + 4*A*C*a^2*b^4 - 2*A 
*C*a^4*b^2 + 2*B*C*a^3*b^3))/(a*b^4 + b^5 - a^2*b^3 - a^3*b^2) - (C*((32*( 
A*a^4*b^5 - C*b^9 - A*a^2*b^7 - A*a^3*b^6 - B*b^9 + B*a^2*b^7 - B*a^3*b^6 
+ C*a^2*b^7 - 3*C*a^3*b^6 + C*a^5*b^4 + A*a*b^8 + B*a*b^8 + 2*C*a*b^8))/(a 
*b^5 + b^6 - a^2*b^4 - a^3*b^3) - (32*C*tan(c/2 + (d*x)/2)*(2*a*b^9 - 2*a^ 
2*b^8 - 4*a^3*b^7 + 4*a^4*b^6 + 2*a^5*b^5 - 2*a^6*b^4))/(b^2*(a*b^4 + b^5 
- a^2*b^3 - a^3*b^2))))/b^2)*1i)/b^2)/((64*(C^3*a^5 - B*C^2*b^5 + B^2*C*b^ 
5 + 2*C^3*a*b^4 - C^3*a^4*b + 2*C^3*a^2*b^3 - 3*C^3*a^3*b^2 + A*C^2*a*b^4 
- A*C^2*a^4*b - 3*B*C^2*a*b^4 + 3*A*C^2*a^2*b^3 - A*C^2*a^3*b^2 + A^2*C*a^ 
2*b^3 + B*C^2*a^2*b^3 + B*C^2*a^3*b^2 - 2*A*B*C*a*b^4))/(a*b^5 + b^6 - ...
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 817, normalized size of antiderivative = 5.52 \[ \int \frac {\sec (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 2*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*c + 2*sqrt( - a**2 + b**2)*atan((t 
an((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*cos(c + d*x) 
*a**3*b**2 + 4*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**2*c - 2*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**4 - 2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - 
tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a**3*b*c + 2*sqrt( - a**2 + b**2 
)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*a** 
2*b**3 + 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*b**3*c - 2*sqrt( - a**2 + b**2)*atan((tan((c 
+ d*x)/2)*a - tan((c + d*x)/2)*b)/sqrt( - a**2 + b**2))*b**5 - cos(c + d*x 
)*log(tan((c + d*x)/2) - 1)*a**5*c + 2*cos(c + d*x)*log(tan((c + d*x)/2) - 
 1)*a**3*b**2*c - cos(c + d*x)*log(tan((c + d*x)/2) - 1)*a*b**4*c + cos(c 
+ d*x)*log(tan((c + d*x)/2) + 1)*a**5*c - 2*cos(c + d*x)*log(tan((c + d*x) 
/2) + 1)*a**3*b**2*c + cos(c + d*x)*log(tan((c + d*x)/2) + 1)*a*b**4*c - l 
og(tan((c + d*x)/2) - 1)*a**4*b*c + 2*log(tan((c + d*x)/2) - 1)*a**2*b**3* 
c - log(tan((c + d*x)/2) - 1)*b**5*c + log(tan((c + d*x)/2) + 1)*a**4*b*c 
- 2*log(tan((c + d*x)/2) + 1)*a**2*b**3*c + log(tan((c + d*x)/2) + 1)*b**5 
*c - sin(c + d*x)*a**4*b*c + sin(c + d*x)*a**2*b**3*c)/(b**2*d*(cos(c +...