\(\int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx\) [349]

Optimal result

Integrand size = 33, antiderivative size = 224 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {\left (8 a^2 A-3 A b^2+4 a b B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{5/2} d}-\frac {(A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b} d}-\frac {(A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b} d}+\frac {(3 A b-4 a B) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{4 a^2 d}-\frac {A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 a d} \] Output:

1/4*(8*A*a^2-3*A*b^2+4*B*a*b)*arctanh((a+b*tan(d*x+c))^(1/2)/a^(1/2))/a^(5 
/2)/d-(A-I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a-I*b)^(1/2))/(a-I*b)^(1/2)/ 
d-(A+I*B)*arctanh((a+b*tan(d*x+c))^(1/2)/(a+I*b)^(1/2))/(a+I*b)^(1/2)/d+1/ 
4*(3*A*b-4*B*a)*cot(d*x+c)*(a+b*tan(d*x+c))^(1/2)/a^2/d-1/2*A*cot(d*x+c)^2 
*(a+b*tan(d*x+c))^(1/2)/a/d
 

Mathematica [A] (verified)

Time = 6.20 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.61 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\frac {2 b^3 \left (\frac {A \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} b^3}-\frac {3 A \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{8 a^{5/2} b}+\frac {B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{2 a^{3/2} b^2}-\frac {\left (A \sqrt {-b^2}+b B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-\sqrt {-b^2}}}\right )}{2 b^3 \sqrt {-b^2} \sqrt {a-\sqrt {-b^2}}}-\frac {b \left (A \sqrt {-b^2}-b B\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+\sqrt {-b^2}}}\right )}{2 \left (-b^2\right )^{5/2} \sqrt {a+\sqrt {-b^2}}}+\frac {3 A \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{8 a^2 b^2}-\frac {B \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{2 a b^3}-\frac {A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{4 a b^3}\right )}{d} \] Input:

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

Output:

(2*b^3*((A*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(Sqrt[a]*b^3) - (3*A 
*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/(8*a^(5/2)*b) + (B*ArcTanh[Sqr 
t[a + b*Tan[c + d*x]]/Sqrt[a]])/(2*a^(3/2)*b^2) - ((A*Sqrt[-b^2] + b*B)*Ar 
cTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - Sqrt[-b^2]]])/(2*b^3*Sqrt[-b^2]*Sq 
rt[a - Sqrt[-b^2]]) - (b*(A*Sqrt[-b^2] - b*B)*ArcTanh[Sqrt[a + b*Tan[c + d 
*x]]/Sqrt[a + Sqrt[-b^2]]])/(2*(-b^2)^(5/2)*Sqrt[a + Sqrt[-b^2]]) + (3*A*C 
ot[c + d*x]*Sqrt[a + b*Tan[c + d*x]])/(8*a^2*b^2) - (B*Cot[c + d*x]*Sqrt[a 
 + b*Tan[c + d*x]])/(2*a*b^3) - (A*Cot[c + d*x]^2*Sqrt[a + b*Tan[c + d*x]] 
)/(4*a*b^3)))/d
 

Rubi [A] (warning: unable to verify)

Time = 1.72 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.03, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 4092, 27, 3042, 4132, 27, 3042, 4136, 27, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4092

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

\(\displaystyle -\frac {A \cot ^2(c+d x) \sqrt {a+b \tan (c+d x)}}{2 a d}-\frac {-\frac {(3 A b-4 a B) \cot (c+d x) \sqrt {a+b \tan (c+d x)}}{a d}+\frac {\left (8 a^2 A+4 a b B-3 A b^2\right ) \int \frac {\tan (c+d x)^2+1}{\tan (c+d x) \sqrt {a+b \tan (c+d x)}}dx+8 \left (\frac {a^2 (B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a-i b}}\right )}{d \sqrt {a-i b}}-\frac {a^2 (-B+i A) \arctan \left (\frac {\tan (c+d x)}{\sqrt {a+i b}}\right )}{d \sqrt {a+i b}}\right )}{2 a}}{4 a}\)

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 4092

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*(A*b - a*B)*(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[b* 
B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(m + n + 2 
)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n 
+ 2)*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] && LtQ[m, -1] 
&& (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(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 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])))
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(4174\) vs. \(2(190)=380\).

Time = 0.16 (sec) , antiderivative size = 4175, normalized size of antiderivative = 18.64

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(4175\)
default	\(\text {Expression too large to display}\)	\(4175\)

Input:

int(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

2*A*arctanh((a+b*tan(d*x+c))^(1/2)/a^(1/2))/a^(1/2)/d+3/4/d/tan(d*x+c)^2/a 
^2*(a+b*tan(d*x+c))^(3/2)*A-5/4/d/tan(d*x+c)^2/a*(a+b*tan(d*x+c))^(1/2)*A+ 
1/d/b/tan(d*x+c)^2*(a+b*tan(d*x+c))^(1/2)*B-3/4/d*b^2/a^(5/2)*arctanh((a+b 
*tan(d*x+c))^(1/2)/a^(1/2))*A+1/d*b/a^(3/2)*arctanh((a+b*tan(d*x+c))^(1/2) 
/a^(1/2))*B-1/d/b/tan(d*x+c)^2/a*(a+b*tan(d*x+c))^(3/2)*B+1/4/d/b/(a^2+b^2 
)^(3/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^( 
1/2)+(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3+1/4/d*b/(a^2+b^2 
)^(3/2)*ln(b*tan(d*x+c)+a+(a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^( 
1/2)+(a^2+b^2)^(1/2))*B*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a+1/d/b/(a^2+b^2)^(1 
/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan((2*(a+b*tan(d*x+c))^(1/2)+(2*(a^2 
+b^2)^(1/2)+2*a)^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*a^2+1/d/b/(a^2+b^ 
2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^(1/ 
2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*a^4+1/4/d/b^ 
2/(a^2+b^2)*ln((a+b*tan(d*x+c))^(1/2)*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)-b*tan( 
d*x+c)-a-(a^2+b^2)^(1/2))*A*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3+1/d*b^2/(a^2 
+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2*a)^ 
(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*A*a+3/d*b/( 
a^2+b^2)^(3/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2)^(1/2)+2* 
a)^(1/2)-2*(a+b*tan(d*x+c))^(1/2))/(2*(a^2+b^2)^(1/2)-2*a)^(1/2))*B*a^2+1/ 
d/b^2*(a^2+b^2)^(1/2)/(2*(a^2+b^2)^(1/2)-2*a)^(1/2)*arctan(((2*(a^2+b^2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1833 vs. \(2 (184) = 368\).

Time = 7.17 (sec) , antiderivative size = 3685, normalized size of antiderivative = 16.45 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x, algorith 
m="fricas")
 

Output:

Too large to include
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x, algorith 
m="maxima")
 

Output:

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

Giac [F(-1)]

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

integrate(cot(d*x+c)^3*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^(1/2),x, algorith 
m="giac")
 

Output:

Timed out
 

Mupad [B] (verification not implemented)

Time = 6.07 (sec) , antiderivative size = 13182, normalized size of antiderivative = 58.85 \[ \int \frac {\cot ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+b \tan (c+d x)}} \, dx=\text {Too large to display} \] Input:

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

Output:

- (((5*A*b^2 - 4*B*a*b)*(a + b*tan(c + d*x))^(1/2))/(4*a) - ((3*A*b^2 - 4* 
B*a*b)*(a + b*tan(c + d*x))^(3/2))/(4*a^2))/(d*(a + b*tan(c + d*x))^2 + a^ 
2*d - 2*a*d*(a + b*tan(c + d*x))) - atan(((((((640*A*a^4*b^10*d^4 - 384*A* 
a^2*b^12*d^4 + 768*A*a^6*b^8*d^4 + 512*B*a^3*b^11*d^4 + 256*B*a^5*b^9*d^4) 
/(2*a^4*d^5) + ((512*a^4*b^10*d^4 + 768*a^6*b^8*d^4)*(a + b*tan(c + d*x))^ 
(1/2)*(-(((8*A^2*a*d^2 - 8*B^2*a*d^2 + 16*A*B*b*d^2)^2/4 - (16*a^2*d^4 + 1 
6*b^2*d^4)*(A^4 + 2*A^2*B^2 + B^4))^(1/2) - 4*A^2*a*d^2 + 4*B^2*a*d^2 - 8* 
A*B*b*d^2)/(16*(a^2*d^4 + b^2*d^4)))^(1/2))/(a^4*d^4))*(-(((8*A^2*a*d^2 - 
8*B^2*a*d^2 + 16*A*B*b*d^2)^2/4 - (16*a^2*d^4 + 16*b^2*d^4)*(A^4 + 2*A^2*B 
^2 + B^4))^(1/2) - 4*A^2*a*d^2 + 4*B^2*a*d^2 - 8*A*B*b*d^2)/(16*(a^2*d^4 + 
 b^2*d^4)))^(1/2) - ((a + b*tan(c + d*x))^(1/2)*(576*A^2*a^5*b^8*d^2 - 192 
*A^2*a^3*b^10*d^2 + 64*B^2*a^3*b^10*d^2 - 320*B^2*a^5*b^8*d^2 + 36*A^2*a*b 
^12*d^2 - 96*A*B*a^2*b^11*d^2 + 768*A*B*a^4*b^9*d^2))/(a^4*d^4))*(-(((8*A^ 
2*a*d^2 - 8*B^2*a*d^2 + 16*A*B*b*d^2)^2/4 - (16*a^2*d^4 + 16*b^2*d^4)*(A^4 
 + 2*A^2*B^2 + B^4))^(1/2) - 4*A^2*a*d^2 + 4*B^2*a*d^2 - 8*A*B*b*d^2)/(16* 
(a^2*d^4 + b^2*d^4)))^(1/2) + (64*B^3*a^2*b^11*d^2 - 192*A^3*a^5*b^8*d^2 + 
 256*B^3*a^4*b^9*d^2 + 36*A^2*B*b^13*d^2 + 36*A^3*a*b^12*d^2 - 96*A*B^2*a* 
b^12*d^2 - 384*A*B^2*a^3*b^10*d^2 + 576*A*B^2*a^5*b^8*d^2 + 96*A^2*B*a^2*b 
^11*d^2 - 768*A^2*B*a^4*b^9*d^2)/(2*a^4*d^5))*(-(((8*A^2*a*d^2 - 8*B^2*a*d 
^2 + 16*A*B*b*d^2)^2/4 - (16*a^2*d^4 + 16*b^2*d^4)*(A^4 + 2*A^2*B^2 + B...
 

Reduce [F]

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

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

Output:

int(sqrt(tan(c + d*x)*b + a)*cot(c + d*x)**3,x)