\(\int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\) [630]

Optimal result

Integrand size = 35, antiderivative size = 383 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\frac {(i a-b)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\left (6 a^2 A b-16 A b^3-a^3 B-24 a b^2 B\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{8 b^{3/2} d}+\frac {(i a+b)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {\left (6 a A b-a^2 B-8 b^2 B\right ) \sqrt {a+b \tan (c+d x)}}{8 b d \sqrt {\cot (c+d x)}}+\frac {(6 A b-a B) (a+b \tan (c+d x))^{3/2}}{12 b d \sqrt {\cot (c+d x)}}+\frac {B (a+b \tan (c+d x))^{5/2}}{3 b d \sqrt {\cot (c+d x)}} \] Output:

(I*a-b)^(3/2)*(A+I*B)*arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c 
))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d+1/8*(6*A*a^2*b-16*A*b^3-B*a^ 
3-24*B*a*b^2)*arctanh(b^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot 
(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/b^(3/2)/d+(I*a+b)^(3/2)*(A-I*B)*arctanh((I* 
a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d 
*x+c)^(1/2)/d+1/8*(6*A*a*b-B*a^2-8*B*b^2)*(a+b*tan(d*x+c))^(1/2)/b/d/cot(d 
*x+c)^(1/2)+1/12*(6*A*b-B*a)*(a+b*tan(d*x+c))^(3/2)/b/d/cot(d*x+c)^(1/2)+1 
/3*B*(a+b*tan(d*x+c))^(5/2)/b/d/cot(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 4.06 (sec) , antiderivative size = 367, normalized size of antiderivative = 0.96 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (24 \sqrt [4]{-1} (-a+i b)^{3/2} b (i A+B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )+24 (-1)^{3/4} (a+i b)^{3/2} b (A+i B) \arctan \left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-3 \left (-6 a A b+a^2 B+8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}+2 (6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}+8 B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}-\frac {3 \sqrt {a} \left (-6 a^2 A b+16 A b^3+a^3 B+24 a b^2 B\right ) \text {arcsinh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+\frac {b \tan (c+d x)}{a}}}{\sqrt {b} \sqrt {a+b \tan (c+d x)}}\right )}{24 b d} \] Input:

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

Output:

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(24*(-1)^(1/4)*(-a + I*b)^(3/2)*b*( 
I*A + B)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b* 
Tan[c + d*x]]] + 24*(-1)^(3/4)*(a + I*b)^(3/2)*b*(A + I*B)*ArcTan[((-1)^(1 
/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - 3*(-6*a* 
A*b + a^2*B + 8*b^2*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]] + 2*(6* 
A*b - a*B)*Sqrt[Tan[c + d*x]]*(a + b*Tan[c + d*x])^(3/2) + 8*B*Sqrt[Tan[c 
+ d*x]]*(a + b*Tan[c + d*x])^(5/2) - (3*Sqrt[a]*(-6*a^2*A*b + 16*A*b^3 + a 
^3*B + 24*a*b^2*B)*ArcSinh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]]*Sqrt[1 + 
(b*Tan[c + d*x])/a])/(Sqrt[b]*Sqrt[a + b*Tan[c + d*x]])))/(24*b*d)
 

Rubi [A] (verified)

Time = 3.57 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.91, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 4729, 3042, 4090, 27, 3042, 4130, 27, 3042, 4130, 27, 3042, 4138, 2035, 2257, 2009}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4729

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {1}{2} \int \frac {3 \sqrt {a+b \tan (c+d x)} \left (-\left (\left (-B a^2+6 A b a-8 b^2 B\right ) \tan ^2(c+d x)\right )+8 b (A b+a B) \tan (c+d x)+a (2 A b+a B)\right )}{2 \sqrt {\tan (c+d x)}}dx-\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-\left (\left (-B a^2+6 A b a-8 b^2 B\right ) \tan ^2(c+d x)\right )+8 b (A b+a B) \tan (c+d x)+a (2 A b+a B)\right )}{\sqrt {\tan (c+d x)}}dx-\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \int \frac {\sqrt {a+b \tan (c+d x)} \left (-\left (\left (-B a^2+6 A b a-8 b^2 B\right ) \tan (c+d x)^2\right )+8 b (A b+a B) \tan (c+d x)+a (2 A b+a B)\right )}{\sqrt {\tan (c+d x)}}dx-\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\right )\)

\(\Big \downarrow \) 4130

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\int \frac {-\left (\left (-B a^3+6 A b a^2-24 b^2 B a-16 A b^3\right ) \tan ^2(c+d x)\right )+16 b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (B a^2+10 A b a-8 b^2 B\right )}{2 \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\right )\)

\(\Big \downarrow \) 27

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\frac {1}{2} \int \frac {-\left (\left (-B a^3+6 A b a^2-24 b^2 B a-16 A b^3\right ) \tan ^2(c+d x)\right )+16 b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (B a^2+10 A b a-8 b^2 B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\frac {1}{2} \int \frac {-\left (\left (-B a^3+6 A b a^2-24 b^2 B a-16 A b^3\right ) \tan (c+d x)^2\right )+16 b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (B a^2+10 A b a-8 b^2 B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}dx-\frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\right )\)

\(\Big \downarrow \) 4138

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\frac {\int \frac {-\left (\left (-B a^3+6 A b a^2-24 b^2 B a-16 A b^3\right ) \tan ^2(c+d x)\right )+16 b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (B a^2+10 A b a-8 b^2 B\right )}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\tan (c+d x)}{2 d}-\frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\right )\)

\(\Big \downarrow \) 2035

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\frac {\int \frac {-\left (\left (-B a^3+6 A b a^2-24 b^2 B a-16 A b^3\right ) \tan ^2(c+d x)\right )+16 b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)+a \left (B a^2+10 A b a-8 b^2 B\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}d\sqrt {\tan (c+d x)}}{d}-\frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\right )\)

\(\Big \downarrow \) 2257

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {\frac {3}{4} \left (\frac {\int \left (\frac {B a^3-6 A b a^2+24 b^2 B a+16 A b^3}{\sqrt {a+b \tan (c+d x)}}+\frac {16 \left (b \left (A a^2-2 b B a-A b^2\right )+b \left (B a^2+2 A b a-b^2 B\right ) \tan (c+d x)\right )}{\sqrt {a+b \tan (c+d x)} \left (\tan ^2(c+d x)+1\right )}\right )d\sqrt {\tan (c+d x)}}{d}-\frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}\right )-\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}}{6 b}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {B \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{5/2}}{3 b d}-\frac {-\frac {(6 A b-a B) \sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}}{2 d}+\frac {3}{4} \left (-\frac {\left (a^2 (-B)+6 a A b-8 b^2 B\right ) \sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}}{d}+\frac {-\frac {\left (a^3 (-B)+6 a^2 A b-24 a b^2 B-16 A b^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {b}}-8 b (-b+i a)^{3/2} (A+i B) \arctan \left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )-8 b (b+i a)^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d}\right )}{6 b}\right )\)

Input:

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

Output:

Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((B*Sqrt[Tan[c + d*x]]*(a + b*Tan[c 
+ d*x])^(5/2))/(3*b*d) - (-1/2*((6*A*b - a*B)*Sqrt[Tan[c + d*x]]*(a + b*Ta 
n[c + d*x])^(3/2))/d + (3*((-8*(I*a - b)^(3/2)*b*(A + I*B)*ArcTan[(Sqrt[I* 
a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]] - ((6*a^2*A*b - 16*A* 
b^3 - a^3*B - 24*a*b^2*B)*ArcTanh[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b* 
Tan[c + d*x]]])/Sqrt[b] - 8*b*(I*a + b)^(3/2)*(A - I*B)*ArcTanh[(Sqrt[I*a 
+ b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/d - ((6*a*A*b - a^2*B 
- 8*b^2*B)*Sqrt[Tan[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])/d))/4)/(6*b))
 

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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2035

Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k   Subst 
[Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti 
onQ[m] && AlgebraicFunctionQ[Fx, x]
 

rule 2257

Int[(Px_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol 
] :> Int[ExpandIntegrand[Px*(d + e*x^2)^q*(a + c*x^4)^p, x], x] /; FreeQ[{a 
, c, d, e, q}, x] && PolyQ[Px, x] && IntegerQ[p]
 

rule 3042

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

rule 4090

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[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(m + 
 n))), x] + Simp[1/(d*(m + n))   Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Ta 
n[e + f*x])^n*Simp[a^2*A*d*(m + n) - b*B*(b*c*(m - 1) + a*d*(n + 1)) + d*(m 
 + n)*(2*a*A*b + B*(a^2 - b^2))*Tan[e + f*x] - (b*B*(b*c - a*d)*(m - 1) - b 
*(A*b + a*B)*d*(m + n))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, 
 f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2 
, 0] && GtQ[m, 1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(IGtQ[n, 1] 
&& ( !IntegerQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
 

rule 4130

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[C*(a + b*Tan[e + f*x])^m*((c + d*Tan[ 
e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Simp[1/(d*(m + n + 1))   Int[(a 
+ b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C 
*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f*x] - (C* 
m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[ 
c, 0] && NeQ[a, 0])))
 

rule 4138

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] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, S 
imp[ff/f   Subst[Int[(a + b*ff*x)^m*(c + d*ff*x)^n*((A + B*ff*x + C*ff^2*x^ 
2)/(1 + ff^2*x^2)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f 
, A, B, C, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + 
 d^2, 0]
 

rule 4729

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a 
+ b*x])^m*(c*Tan[a + b*x])^m   Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], 
x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ[u, 
x]
 

Maple [F(-1)]

Timed out.

hanged

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 12332 vs. \(2 (312) = 624\).

Time = 4.94 (sec) , antiderivative size = 24697, normalized size of antiderivative = 64.48 \[ \int \frac {(a+b \tan (c+d x))^{3/2} (A+B \tan (c+d x))}{\cot ^{\frac {3}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

integrate((a+b*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c))/cot(d*x+c)^(3/2),x, algo 
rithm="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,[0,9,3]%%%}+%%%{4,[0,7,3]%%%}+%%%{6,[0,5,3]%%%}+%%%{ 
4,[0,3,3]
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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