\(\int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx\) [645]

Optimal result

Integrand size = 25, antiderivative size = 193 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\frac {\arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a-b)^{3/2} d}-\frac {\text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(i a+b)^{3/2} d}-\frac {2}{a d \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {2 b \left (a^2+2 b^2\right ) \sqrt {\tan (c+d x)}}{a^2 \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 2.92 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.05 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=-\frac {\frac {\sqrt [4]{-1} \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{3/2}}+\frac {\frac {\sqrt [4]{-1} (a-i b) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}+\frac {2 \left (a \left (a^2+b^2\right )+b \left (a^2+2 b^2\right ) \tan (c+d x)\right )}{a^2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}}{a^2+b^2}}{d} \] Input:

Integrate[1/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2)),x]
 

Output:

-((((-1)^(1/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[ 
a + b*Tan[c + d*x]]])/(-a + I*b)^(3/2) + (((-1)^(1/4)*(a - I*b)*ArcTan[((- 
1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt 
[a + I*b] + (2*(a*(a^2 + b^2) + b*(a^2 + 2*b^2)*Tan[c + d*x]))/(a^2*Sqrt[T 
an[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]))/(a^2 + b^2))/d)
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.22, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {3042, 4052, 27, 3042, 4132, 27, 3042, 4099, 3042, 4098, 104, 216, 219}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4099

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4098

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

\(\Big \downarrow \) 104

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 219

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

Input:

Int[1/(Tan[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(3/2)),x]
 

Output:

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

Defintions of rubi rules used

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 104

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x 
_)), x_] :> With[{q = Denominator[m]}, Simp[q   Subst[Int[x^(q*(m + 1) - 1) 
/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] 
] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L 
tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*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 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 3042

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

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 4098

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

rule 4099

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
 (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si 
mp[(A + I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 - I*T 
an[e + f*x]), x], x] + Simp[(A - I*B)/2   Int[(a + b*Tan[e + f*x])^m*(c + d 
*Tan[e + f*x])^n*(1 + I*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A 
, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2 + B^2, 
0]
 

rule 4132

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
 (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) 
 + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*b^2 - a*(b*B - a*C))*(a + b*Tan[e + 
 f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 + 
b^2))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 + b^2))   Int[(a + b*Tan[e + 
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1) - b^2*d* 
(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d 
)*(A*b - a*B - b*C)*Tan[e + f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Ta 
n[e + f*x]^2, 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] && LtQ[m, -1] && 
!(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

Maple [B] (warning: unable to verify)

result has leaf size over 500,000. Avoiding possible recursion issues.

Time = 1.03 (sec) , antiderivative size = 799509, normalized size of antiderivative = 4142.53

\[\text {output too large to display}\]

Input:

int(1/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(3/2),x)
 

Output:

result too large to display
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 7597 vs. \(2 (161) = 322\).

Time = 1.23 (sec) , antiderivative size = 7597, normalized size of antiderivative = 39.36 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \] Input:

integrate(1/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

integrate(1/tan(d*x+c)**(3/2)/(a+b*tan(d*x+c))**(3/2),x)
 

Output:

Integral(1/((a + b*tan(c + d*x))**(3/2)*tan(c + d*x)**(3/2)), x)
 

Maxima [F]

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

integrate(1/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/((b*tan(d*x + c) + a)^(3/2)*tan(d*x + c)^(3/2)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/tan(d*x+c)^(3/2)/(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:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\int \frac {1}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:

int(1/(tan(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(3/2)),x)
 

Output:

int(1/(tan(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{3/2}} \, dx=\frac {-4 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, \tan \left (d x +c \right ) b -2 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}\, a -\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{2} b^{2}+2 \tan \left (d x +c \right ) a b +a^{2}}d x \right ) \tan \left (d x +c \right )^{2} a^{2} b d -\left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a +\tan \left (d x +c \right ) b}}{\tan \left (d x +c \right )^{2} b^{2}+2 \tan \left (d x +c \right ) a b +a^{2}}d x \right ) \tan \left (d x +c \right ) a^{3} d}{\tan \left (d x +c \right ) a^{2} d \left (a +\tan \left (d x +c \right ) b \right )} \] Input:

int(1/tan(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(3/2),x)
 

Output:

( - 4*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*tan(c + d*x)*b - 2*sqrt( 
tan(c + d*x))*sqrt(tan(c + d*x)*b + a)*a - int((sqrt(tan(c + d*x))*sqrt(ta 
n(c + d*x)*b + a))/(tan(c + d*x)**2*b**2 + 2*tan(c + d*x)*a*b + a**2),x)*t 
an(c + d*x)**2*a**2*b*d - int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*b + a) 
)/(tan(c + d*x)**2*b**2 + 2*tan(c + d*x)*a*b + a**2),x)*tan(c + d*x)*a**3* 
d)/(tan(c + d*x)*a**2*d*(tan(c + d*x)*b + a))