3.4.35 \(\int \cot (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [335]

3.4.35.1 Optimal result

Integrand size = 31, antiderivative size = 182 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=-\frac {2 a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {(a-i b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 b (A b+2 a B) \sqrt {a+b \tan (c+d x)}}{d}+\frac {2 b B (a+b \tan (c+d x))^{3/2}}{3 d} \]

output

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

3.4.35.2 Mathematica [A] (verified)

Time = 1.26 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.97 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {2 \left (-3 a^{5/2} A \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )+\frac {3}{2} (a-i b)^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+\frac {3}{2} (a+i b)^{5/2} (A+i B) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+3 b (A b+2 a B) \sqrt {a+b \tan (c+d x)}+b B (a+b \tan (c+d x))^{3/2}\right )}{3 d} \]

input

Integrate[Cot[c + d*x]*(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]
 

output

(2*(-3*a^(5/2)*A*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]] + (3*(a - I*b)^ 
(5/2)*(A - I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a - I*b]])/2 + (3*(a 
 + I*b)^(5/2)*(A + I*B)*ArcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a + I*b]])/2 
 + 3*b*(A*b + 2*a*B)*Sqrt[a + b*Tan[c + d*x]] + b*B*(a + b*Tan[c + d*x])^( 
3/2)))/(3*d)
 

3.4.35.3 Rubi [A] (warning: unable to verify)

Time = 1.64 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.93, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.581, Rules used = {3042, 4090, 27, 3042, 4130, 27, 3042, 4136, 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.

\(\displaystyle \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)}dx\)

\(\Big \downarrow \) 4090

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4130

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

Int[Cot[c + d*x]*(a + b*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]
 

output

((a - I*b)^(5/2)*(I*A + B)*ArcTan[Tan[c + d*x]/Sqrt[a - I*b]])/d - ((a + I 
*b)^(5/2)*(I*A - B)*ArcTan[Tan[c + d*x]/Sqrt[a + I*b]])/d - (2*a^(5/2)*A*A 
rcTanh[Sqrt[a + b*Tan[c + d*x]]/Sqrt[a]])/d + (2*b*(A*b + 2*a*B)*Sqrt[a + 
b*Tan[c + d*x]])/d + (2*b*B*(a + b*Tan[c + d*x])^(3/2))/(3*d)
 

3.4.35.3.1 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 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 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 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 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]
 

3.4.35.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2372\) vs. \(2(152)=304\).

Time = 0.25 (sec) , antiderivative size = 2373, normalized size of antiderivative = 13.04

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2373\)
default	\(\text {Expression too large to display}\)	\(2373\)

input

int(cot(d*x+c)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x,method=_RETURNVER 
BOSE)
 

output

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

3.4.35.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 4872 vs. \(2 (146) = 292\).

Time = 10.91 (sec) , antiderivative size = 9757, normalized size of antiderivative = 53.61 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

integrate(cot(d*x+c)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm= 
"fricas")
 

output

Too large to include
 

3.4.35.6 Sympy [F]

\[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int \left (A + B \tan {\left (c + d x \right )}\right ) \left (a + b \tan {\left (c + d x \right )}\right )^{\frac {5}{2}} \cot {\left (c + d x \right )}\, dx \]

input

integrate(cot(d*x+c)*(a+b*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)
 

output

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**(5/2)*cot(c + d*x), x)
 

3.4.35.7 Maxima [F]

\[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right ) \,d x } \]

input

integrate(cot(d*x+c)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm= 
"maxima")
 

output

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^(5/2)*cot(d*x + c), x)
 

3.4.35.8 Giac [F(-1)]

Timed out. \[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

input

integrate(cot(d*x+c)*(a+b*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm= 
"giac")
 

output

Timed out
 

3.4.35.9 Mupad [B] (verification not implemented)

Time = 14.54 (sec) , antiderivative size = 29441, normalized size of antiderivative = 161.76 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

input

int(cot(c + d*x)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x))^(5/2),x)
 

output

((2*A*b^2 - 2*B*a*b)/d + (6*B*a*b)/d)*(a + b*tan(c + d*x))^(1/2) - atan((( 
((32*(3*A^3*a^2*b^16*d^2 + 48*A^3*a^4*b^14*d^2 - 30*A^3*a^6*b^12*d^2 - 72* 
A^3*a^8*b^10*d^2 + 3*A^3*a^10*b^8*d^2 + 6*B^3*a^5*b^13*d^2 + 8*B^3*a^7*b^1 
1*d^2 + 3*B^3*a^9*b^9*d^2 - B^3*a*b^17*d^2 - A^2*B*a*b^17*d^2 + 3*A*B^2*a^ 
2*b^16*d^2 - 32*A*B^2*a^4*b^14*d^2 + 46*A*B^2*a^6*b^12*d^2 + 72*A*B^2*a^8* 
b^10*d^2 - 9*A*B^2*a^10*b^8*d^2 - 16*A^2*B*a^3*b^15*d^2 + 150*A^2*B*a^5*b^ 
13*d^2 + 96*A^2*B*a^7*b^11*d^2 - 69*A^2*B*a^9*b^9*d^2))/d^5 - (((32*(4*A*a 
*b^12*d^4 + 16*A*a^3*b^10*d^4 + 12*A*a^5*b^8*d^4 + 8*B*a^2*b^11*d^4 + 8*B* 
a^4*b^9*d^4))/d^5 - (32*(16*b^10*d^4 + 24*a^2*b^8*d^4)*(a + b*tan(c + d*x) 
)^(1/2)*((((8*B^2*a^5*d^2 - 8*A^2*a^5*d^2 + 80*A^2*a^3*b^2*d^2 - 80*B^2*a^ 
3*b^2*d^2 + 16*A*B*b^5*d^2 - 40*A^2*a*b^4*d^2 + 40*B^2*a*b^4*d^2 - 160*A*B 
*a^2*b^3*d^2 + 80*A*B*a^4*b*d^2)^2/64 - d^4*(A^4*a^10 + A^4*b^10 + B^4*a^1 
0 + B^4*b^10 + 2*A^2*B^2*a^10 + 2*A^2*B^2*b^10 + 5*A^4*a^2*b^8 + 10*A^4*a^ 
4*b^6 + 10*A^4*a^6*b^4 + 5*A^4*a^8*b^2 + 5*B^4*a^2*b^8 + 10*B^4*a^4*b^6 + 
10*B^4*a^6*b^4 + 5*B^4*a^8*b^2 + 10*A^2*B^2*a^2*b^8 + 20*A^2*B^2*a^4*b^6 + 
 20*A^2*B^2*a^6*b^4 + 10*A^2*B^2*a^8*b^2))^(1/2) + A^2*a^5*d^2 - B^2*a^5*d 
^2 - 10*A^2*a^3*b^2*d^2 + 10*B^2*a^3*b^2*d^2 - 2*A*B*b^5*d^2 + 5*A^2*a*b^4 
*d^2 - 5*B^2*a*b^4*d^2 + 20*A*B*a^2*b^3*d^2 - 10*A*B*a^4*b*d^2)/(4*d^4))^( 
1/2))/d^4)*((((8*B^2*a^5*d^2 - 8*A^2*a^5*d^2 + 80*A^2*a^3*b^2*d^2 - 80*B^2 
*a^3*b^2*d^2 + 16*A*B*b^5*d^2 - 40*A^2*a*b^4*d^2 + 40*B^2*a*b^4*d^2 - 1...