\(\int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx\) [543]

Optimal result

Integrand size = 21, antiderivative size = 150 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}+\frac {2 b^2}{a \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \] Output:

-2*arctanh((a+b*tan(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+arctanh((a+b*tan(d*x+ 
c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(3/2)/d+arctanh((a+b*tan(d*x+c))^(1/2)/(a 
+I*b)^(1/2))/(a+I*b)^(3/2)/d+2*b^2/a/(a^2+b^2)/d/(a+b*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.10 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\frac {-\frac {2 \left (a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {a (a+i b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}+\frac {a (a-i b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}+\frac {2 b^2}{\sqrt {a+b \tan (c+d x)}}}{a \left (a^2+b^2\right ) d} \] Input:

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

Output:

((-2*(a^2 + b^2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/Sqrt[a] + (a*( 
a + I*b)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/Sqrt[a - I*b] + 
(a*(a - I*b)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/Sqrt[a + I*b 
] + (2*b^2)/Sqrt[a + b*Tan[c + d*x]])/(a*(a^2 + b^2)*d)
 

Rubi [A] (warning: unable to verify)

Time = 1.25 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.16, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.762, Rules used = {3042, 4052, 27, 3042, 4136, 25, 3042, 4022, 3042, 4020, 25, 73, 221, 4117, 73, 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.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\tan (c+d x) (a+b \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

\(\displaystyle \frac {2 b^2}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\left (a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a (b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (-b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 b^2}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\left (a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {1}{2} a (b+i a) \int \frac {1-i \tan (c+d x)}{\sqrt {a+b \tan (c+d x)}}dx+\frac {1}{2} a (-b+i a) \int \frac {i \tan (c+d x)+1}{\sqrt {a+b \tan (c+d x)}}dx}{a \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4020

\(\displaystyle \frac {2 b^2}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\left (a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+\frac {i a (-b+i a) \int -\frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}+\frac {i a (b+i a) \int -\frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}}{a \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2 b^2}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\left (a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {i a (-b+i a) \int \frac {1}{(1-i \tan (c+d x)) \sqrt {a+b \tan (c+d x)}}d(i \tan (c+d x))}{2 d}-\frac {i a (b+i a) \int \frac {1}{(i \tan (c+d x)+1) \sqrt {a+b \tan (c+d x)}}d(-i \tan (c+d x))}{2 d}}{a \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 b^2}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\left (a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx-\frac {a (b+i a) \int \frac {1}{-\frac {i \tan ^2(c+d x)}{b}-\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {a (-b+i a) \int \frac {1}{\frac {i \tan ^2(c+d x)}{b}+\frac {i a}{b}+1}d\sqrt {a+b \tan (c+d x)}}{b d}}{a \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 b^2}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\left (a^2+b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+\frac {a (-b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {a (b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{a \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 4117

\(\displaystyle \frac {2 b^2}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {\left (a^2+b^2\right ) \int \frac {\cot (c+d x)}{\sqrt {a+b \tan (c+d x)}}d\tan (c+d x)}{d}+\frac {a (-b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {a (b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{a \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2 b^2}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {\frac {2 \left (a^2+b^2\right ) \int \frac {1}{\frac {a+b \tan (c+d x)}{b}-\frac {a}{b}}d\sqrt {a+b \tan (c+d x)}}{b d}+\frac {a (-b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {a (b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{a \left (a^2+b^2\right )}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 b^2}{a d \left (a^2+b^2\right ) \sqrt {a+b \tan (c+d x)}}+\frac {-\frac {2 \left (a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d}+\frac {a (-b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {a (b+i a) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}}{a \left (a^2+b^2\right )}\)

Input:

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

Output:

((a*(I*a - b)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/(Sqrt[a - I*b]*d) - (a*( 
I*a + b)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/(Sqrt[a + I*b]*d) - (2*(a^2 + 
 b^2)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(Sqrt[a]*d))/(a*(a^2 + b^ 
2)) + (2*b^2)/(a*(a^2 + b^2)*d*Sqrt[a + b*Tan[c + d*x]])
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, 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 4020

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[c*(d/f)   Subst[Int[(a + (b/d)*x)^m/(d^2 + 
c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[ 
b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]
 

rule 4022

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)]), x_Symbol] :> Simp[(c + I*d)/2   Int[(a + b*Tan[e + f*x])^m*( 
1 - I*Tan[e + f*x]), x], x] + Simp[(c - I*d)/2   Int[(a + b*Tan[e + f*x])^m 
*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c 
 - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] &&  !IntegerQ[m]
 

rule 4052

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c 
+ d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + Simp[1 
/((m + 1)*(a^2 + b^2)*(b*c - a*d))   Int[(a + b*Tan[e + f*x])^(m + 1)*(c + 
d*Tan[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - 
 a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x], x], x] / 
; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] 
 && NeQ[c^2 + d^2, 0] && IntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || Integ 
erQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4117

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)])^(n_.)*((A_) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> 
 Simp[A/f   Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; 
FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
 

rule 4136

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) 
+ (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^ 
n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Simp[ 
(A*b^2 - a*b*B + a^2*C)/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*((1 + Tan[ 
e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 
 C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] & 
&  !GtQ[n, 0] &&  !LeQ[n, -1]
 

Maple [F(-1)]

Timed out.

hanged

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2039 vs. \(2 (122) = 244\).

Time = 0.18 (sec) , antiderivative size = 4097, normalized size of antiderivative = 27.31 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

Integral(cot(c + d*x)/(a + b*tan(c + d*x))**(3/2), x)
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro 
unding error%%%{%%%{1,[4,13,4]%%%}+%%%{6,[4,11,4]%%%}+%%%{15,[4,9,4]%%%}+% 
%%{20,[4,
 

Mupad [B] (verification not implemented)

Time = 1.78 (sec) , antiderivative size = 6384, normalized size of antiderivative = 42.56 \[ \int \frac {\cot (c+d x)}{(a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

log(((((8*a^3*d^2 - 24*a*b^2*d^2)^2/64 - b^6*d^4 - a^6*d^4 - 3*a^2*b^4*d^4 
 - 3*a^4*b^2*d^4)^(1/2) + a^3*d^2 - 3*a*b^2*d^2)/(4*(a^6*d^4 + b^6*d^4 + 3 
*a^2*b^4*d^4 + 3*a^4*b^2*d^4)))^(1/2)*((((((8*a^3*d^2 - 24*a*b^2*d^2)^2/64 
 - b^6*d^4 - a^6*d^4 - 3*a^2*b^4*d^4 - 3*a^4*b^2*d^4)^(1/2) + a^3*d^2 - 3* 
a*b^2*d^2)/(4*(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4)))^(1/2)* 
(512*a^8*b^28*d^8 - (a + b*tan(c + d*x))^(1/2)*((((8*a^3*d^2 - 24*a*b^2*d^ 
2)^2/64 - b^6*d^4 - a^6*d^4 - 3*a^2*b^4*d^4 - 3*a^4*b^2*d^4)^(1/2) + a^3*d 
^2 - 3*a*b^2*d^2)/(4*(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))) 
^(1/2)*(512*a^9*b^28*d^9 + 5376*a^11*b^26*d^9 + 25344*a^13*b^24*d^9 + 7065 
6*a^15*b^22*d^9 + 129024*a^17*b^20*d^9 + 161280*a^19*b^18*d^9 + 139776*a^2 
1*b^16*d^9 + 82944*a^23*b^14*d^9 + 32256*a^25*b^12*d^9 + 7424*a^27*b^10*d^ 
9 + 768*a^29*b^8*d^9) + 5248*a^10*b^26*d^8 + 23936*a^12*b^24*d^8 + 64000*a 
^14*b^22*d^8 + 111104*a^16*b^20*d^8 + 130816*a^18*b^18*d^8 + 105728*a^20*b 
^16*d^8 + 57856*a^22*b^14*d^8 + 20480*a^24*b^12*d^8 + 4224*a^26*b^10*d^8 + 
 384*a^28*b^8*d^8) + (a + b*tan(c + d*x))^(1/2)*(256*a^8*b^26*d^7 + 1472*a 
^10*b^24*d^7 + 3712*a^12*b^22*d^7 + 6272*a^14*b^20*d^7 + 9856*a^16*b^18*d^ 
7 + 14336*a^18*b^16*d^7 + 15232*a^20*b^14*d^7 + 10112*a^22*b^12*d^7 + 3712 
*a^24*b^10*d^7 + 576*a^26*b^8*d^7))*((((8*a^3*d^2 - 24*a*b^2*d^2)^2/64 - b 
^6*d^4 - a^6*d^4 - 3*a^2*b^4*d^4 - 3*a^4*b^2*d^4)^(1/2) + a^3*d^2 - 3*a*b^ 
2*d^2)/(4*(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4)))^(1/2) -...
 

Reduce [F]

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

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

Output:

int((sqrt(tan(c + d*x)*b + a)*cot(c + d*x))/(tan(c + d*x)**2*b**2 + 2*tan( 
c + d*x)*a*b + a**2),x)