\(\int \frac {\cot ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx\) [192]

Optimal result

Integrand size = 23, antiderivative size = 223 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {2 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac {71 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{32 \sqrt {2} a^{3/2} d}+\frac {7 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{32 a^2 d}-\frac {\cot (c+d x) \sqrt {a+a \sec (c+d x)}}{4 a^2 d \left (2+\frac {\tan ^2(c+d x)}{1+\sec (c+d x)}\right )^2}-\frac {13 \cot (c+d x) \sqrt {a+a \sec (c+d x)}}{16 a^2 d \left (2+\frac {\tan ^2(c+d x)}{1+\sec (c+d x)}\right )} \] Output:

-2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d+71/64*2^(1/ 
2)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d 
+7/32*cot(d*x+c)*(a+a*sec(d*x+c))^(1/2)/a^2/d-1/4*cot(d*x+c)*(a+a*sec(d*x+ 
c))^(1/2)/a^2/d/(2+tan(d*x+c)^2/(1+sec(d*x+c)))^2-13/16*cot(d*x+c)*(a+a*se 
c(d*x+c))^(1/2)/a^2/d/(2+tan(d*x+c)^2/(1+sec(d*x+c)))
 

Mathematica [A] (warning: unable to verify)

Time = 5.97 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {27 \cot (2 (c+d x))+12 \csc (c+d x)+13 \csc (2 (c+d x))+256 \arctan \left (\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {\frac {1}{1+\sec (c+d x)}}}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\frac {\sec (c+d x)}{(1+\sec (c+d x))^2}} \sqrt {1+\sec (c+d x)}-142 \arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sec ^{\frac {3}{2}}(c+d x) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {1+\sec (c+d x)}}{32 d (a (1+\sec (c+d x)))^{3/2}} \] Input:

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

Output:

-1/32*(27*Cot[2*(c + d*x)] + 12*Csc[c + d*x] + 13*Csc[2*(c + d*x)] + 256*A 
rcTan[Tan[(c + d*x)/2]/Sqrt[(1 + Sec[c + d*x])^(-1)]]*Cos[(c + d*x)/2]^4*S 
ec[c + d*x]^(3/2)*Sqrt[Sec[c + d*x]/(1 + Sec[c + d*x])^2]*Sqrt[1 + Sec[c + 
 d*x]] - 142*ArcSin[Tan[(c + d*x)/2]]*Cos[(c + d*x)/2]^4*Sqrt[Sec[(c + d*x 
)/2]^2]*Sec[c + d*x]^(3/2)*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[1 + Sec[c + 
d*x]])/(d*(a*(1 + Sec[c + d*x]))^(3/2))
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.02, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {3042, 4375, 374, 27, 441, 25, 27, 445, 25, 27, 397, 216}

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.

\(\displaystyle \int \frac {\cot ^2(c+d x)}{(a \sec (c+d x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\cot \left (c+d x+\frac {\pi }{2}\right )^2 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\)

\(\Big \downarrow \) 4375

\(\displaystyle -\frac {2 \int \frac {\cot ^2(c+d x) (\sec (c+d x) a+a)}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^3}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{a^2 d}\)

\(\Big \downarrow \) 374

\(\displaystyle -\frac {2 \left (\frac {\int \frac {a \cot ^2(c+d x) (\sec (c+d x) a+a) \left (3-\frac {5 a \tan ^2(c+d x)}{\sec (c+d x) a+a}\right )}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^2}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{8 a}+\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{8 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )^2}\right )}{a^2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 \left (\frac {1}{8} \int \frac {\cot ^2(c+d x) (\sec (c+d x) a+a) \left (3-\frac {5 a \tan ^2(c+d x)}{\sec (c+d x) a+a}\right )}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )^2}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )+\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{8 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )^2}\right )}{a^2 d}\)

\(\Big \downarrow \) 441

\(\displaystyle -\frac {2 \left (\frac {1}{8} \left (\frac {\int -\frac {a \cot ^2(c+d x) (\sec (c+d x) a+a) \left (\frac {39 a \tan ^2(c+d x)}{\sec (c+d x) a+a}+7\right )}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{4 a}+\frac {13 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )}\right )+\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{8 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )^2}\right )}{a^2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2 \left (\frac {1}{8} \left (\frac {13 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )}-\frac {\int \frac {a \cot ^2(c+d x) (\sec (c+d x) a+a) \left (\frac {39 a \tan ^2(c+d x)}{\sec (c+d x) a+a}+7\right )}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{4 a}\right )+\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{8 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )^2}\right )}{a^2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 \left (\frac {1}{8} \left (\frac {13 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )}-\frac {1}{4} \int \frac {\cot ^2(c+d x) (\sec (c+d x) a+a) \left (\frac {39 a \tan ^2(c+d x)}{\sec (c+d x) a+a}+7\right )}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )\right )+\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{8 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )^2}\right )}{a^2 d}\)

\(\Big \downarrow \) 445

\(\displaystyle -\frac {2 \left (\frac {1}{8} \left (\frac {1}{4} \left (\frac {1}{2} \int -\frac {a \left (57-\frac {7 a \tan ^2(c+d x)}{\sec (c+d x) a+a}\right )}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )-\frac {7}{2} \cot (c+d x) \sqrt {a \sec (c+d x)+a}\right )+\frac {13 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )}\right )+\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{8 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )^2}\right )}{a^2 d}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {2 \left (\frac {1}{8} \left (\frac {1}{4} \left (-\frac {1}{2} \int \frac {a \left (57-\frac {7 a \tan ^2(c+d x)}{\sec (c+d x) a+a}\right )}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )-\frac {7}{2} \cot (c+d x) \sqrt {a \sec (c+d x)+a}\right )+\frac {13 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )}\right )+\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{8 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )^2}\right )}{a^2 d}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {2 \left (\frac {1}{8} \left (\frac {1}{4} \left (-\frac {1}{2} a \int \frac {57-\frac {7 a \tan ^2(c+d x)}{\sec (c+d x) a+a}}{\left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1\right ) \left (\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2\right )}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )-\frac {7}{2} \cot (c+d x) \sqrt {a \sec (c+d x)+a}\right )+\frac {13 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )}\right )+\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{8 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )^2}\right )}{a^2 d}\)

\(\Big \downarrow \) 397

\(\displaystyle -\frac {2 \left (\frac {1}{8} \left (\frac {1}{4} \left (-\frac {1}{2} a \left (64 \int \frac {1}{\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+1}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )-71 \int \frac {1}{\frac {a \tan ^2(c+d x)}{\sec (c+d x) a+a}+2}d\left (-\frac {\tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )\right )-\frac {7}{2} \cot (c+d x) \sqrt {a \sec (c+d x)+a}\right )+\frac {13 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )}\right )+\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{8 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )^2}\right )}{a^2 d}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {2 \left (\frac {1}{8} \left (\frac {1}{4} \left (-\frac {1}{2} a \left (\frac {71 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {2} \sqrt {a}}-\frac {64 \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a}}\right )-\frac {7}{2} \cot (c+d x) \sqrt {a \sec (c+d x)+a}\right )+\frac {13 \cot (c+d x) \sqrt {a \sec (c+d x)+a}}{4 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )}\right )+\frac {\cot (c+d x) \sqrt {a \sec (c+d x)+a}}{8 \left (\frac {a \tan ^2(c+d x)}{a \sec (c+d x)+a}+2\right )^2}\right )}{a^2 d}\)

Input:

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

Output:

(-2*((Cot[c + d*x]*Sqrt[a + a*Sec[c + d*x]])/(8*(2 + (a*Tan[c + d*x]^2)/(a 
 + a*Sec[c + d*x]))^2) + ((-1/2*(a*((-64*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqr 
t[a + a*Sec[c + d*x]]])/Sqrt[a] + (71*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[ 
2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[2]*Sqrt[a]))) - (7*Cot[c + d*x]*Sqrt[ 
a + a*Sec[c + d*x]])/2)/4 + (13*Cot[c + d*x]*Sqrt[a + a*Sec[c + d*x]])/(4* 
(2 + (a*Tan[c + d*x]^2)/(a + a*Sec[c + d*x]))))/8))/(a^2*d)
 

Defintions of rubi rules used

rule 25

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 374

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-b)*(e*x)^(m + 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q 
 + 1)/(a*e*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(e*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[b*c*(m + 1) + 2*(b*c - 
a*d)*(p + 1) + d*b*(m + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, 
c, d, e, m, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IntBinomialQ[a, b, 
 c, d, e, m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 441

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*(g*x)^(m + 1)*(a 
+ b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(a*g*2*(b*c - a*d)*(p + 1))), x] + Si 
mp[1/(a*2*(b*c - a*d)*(p + 1))   Int[(g*x)^m*(a + b*x^2)^(p + 1)*(c + d*x^2 
)^q*Simp[c*(b*e - a*f)*(m + 1) + e*2*(b*c - a*d)*(p + 1) + d*(b*e - a*f)*(m 
 + 2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m, q}, 
 x] && LtQ[p, -1]
 

rule 445

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_ 
.)*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[e*(g*x)^(m + 1)*(a + b*x^2)^(p 
+ 1)*((c + d*x^2)^(q + 1)/(a*c*g*(m + 1))), x] + Simp[1/(a*c*g^2*(m + 1)) 
 Int[(g*x)^(m + 2)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*f*c*(m + 1) - e*(b*c 
+ a*d)*(m + 2 + 1) - e*2*(b*c*p + a*d*q) - b*e*d*(m + 2*(p + q + 2) + 1)*x^ 
2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && LtQ[m, -1]
 

rule 3042

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

rule 4375

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n 
_.), x_Symbol] :> Simp[-2*(a^(m/2 + n + 1/2)/d)   Subst[Int[x^m*((2 + a*x^2 
)^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x] 
]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && I 
ntegerQ[n - 1/2]
 

Maple [A] (warning: unable to verify)

Time = 1.04 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.63

method	result	size
default	\(\frac {\sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \left (\left (-192 \cos \left (d x +c \right )^{3}-576 \cos \left (d x +c \right )^{2}-576 \cos \left (d x +c \right )-192\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\cot \left (d x +c \right )^{2}-2 \csc \left (d x +c \right ) \cot \left (d x +c \right )+\csc \left (d x +c \right )^{2}-1}}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (213 \cos \left (d x +c \right )^{3}+639 \cos \left (d x +c \right )^{2}+639 \cos \left (d x +c \right )+213\right ) \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\left (10 \cos \left (d x +c \right )^{3}+122 \cos \left (d x +c \right )^{2}+182 \cos \left (d x +c \right )+70\right ) \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cot \left (d x +c \right )+\left (-142 \cos \left (d x +c \right )^{3}-10 \cos \left (d x +c \right )^{2}+110 \cos \left (d x +c \right )+42\right ) \cot \left (d x +c \right )\right )}{192 d \,a^{2} \left (1+\cos \left (d x +c \right )\right ) \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right )}\)	\(364\)

Input:

int(cot(d*x+c)^2/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/192/d/a^2*(a*(1+sec(d*x+c)))^(1/2)/(1+cos(d*x+c))/(cos(d*x+c)^2+2*cos(d* 
x+c)+1)*((-192*cos(d*x+c)^3-576*cos(d*x+c)^2-576*cos(d*x+c)-192)*2^(1/2)*a 
rctanh(2^(1/2)/(cot(d*x+c)^2-2*csc(d*x+c)*cot(d*x+c)+csc(d*x+c)^2-1)^(1/2) 
*(csc(d*x+c)-cot(d*x+c)))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+(213*cos(d* 
x+c)^3+639*cos(d*x+c)^2+639*cos(d*x+c)+213)*ln((-2*cos(d*x+c)/(1+cos(d*x+c 
)))^(1/2)-cot(d*x+c)+csc(d*x+c))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+(10* 
cos(d*x+c)^3+122*cos(d*x+c)^2+182*cos(d*x+c)+70)*2^(1/2)*(-cos(d*x+c)/(1+c 
os(d*x+c)))^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cot(d*x+c)+(-142*co 
s(d*x+c)^3-10*cos(d*x+c)^2+110*cos(d*x+c)+42)*cot(d*x+c))
                                                                                    
                                                                                    
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 603, normalized size of antiderivative = 2.70 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/128*(71*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(-a)*log((2* 
sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin( 
d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2* 
cos(d*x + c) + 1))*sin(d*x + c) + 64*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1) 
*sqrt(-a)*log(-(8*a*cos(d*x + c)^3 - 4*(2*cos(d*x + c)^2 - cos(d*x + c))*s 
qrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 7*a*cos(d*x 
 + c) + a)/(cos(d*x + c) + 1))*sin(d*x + c) + 4*(27*cos(d*x + c)^3 + 12*co 
s(d*x + c)^2 - 7*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)))/(( 
a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)*sin(d*x + c)), -1/64* 
(71*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*arctan(sqrt(2)*s 
qrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)) 
)*sin(d*x + c) + 64*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*arctan(2 
*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) 
/(2*a*cos(d*x + c)^2 + a*cos(d*x + c) - a))*sin(d*x + c) + 2*(27*cos(d*x + 
 c)^3 + 12*cos(d*x + c)^2 - 7*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos( 
d*x + c)))/((a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)*sin(d*x 
+ c))]
 

Sympy [F]

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

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

Output:

Integral(cot(c + d*x)**2/(a*(sec(c + d*x) + 1))**(3/2), x)
 

Maxima [F]

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

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

Output:

integrate(cot(d*x + c)^2/(a*sec(d*x + c) + a)^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.64 \[ \int \frac {\cot ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {17 \, \sqrt {2}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {16 \, \sqrt {2}}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - a\right )} \sqrt {-a} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{64 \, d} \] Input:

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

Output:

-1/64*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*tan(1/2*d*x + 1/2*c) 
^2/(a^2*sgn(cos(d*x + c))) - 17*sqrt(2)/(a^2*sgn(cos(d*x + c))))*tan(1/2*d 
*x + 1/2*c) + 16*sqrt(2)/(((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/ 
2*d*x + 1/2*c)^2 + a))^2 - a)*sqrt(-a)*sgn(cos(d*x + c))))/d
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {cot}\left (c+d\,x\right )}^2}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {\cot ^2(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \cot \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{2}+2 \sec \left (d x +c \right )+1}d x \right )}{a^{2}} \] Input:

int(cot(d*x+c)^2/(a+a*sec(d*x+c))^(3/2),x)
 

Output:

(sqrt(a)*int((sqrt(sec(c + d*x) + 1)*cot(c + d*x)**2)/(sec(c + d*x)**2 + 2 
*sec(c + d*x) + 1),x))/a**2