3.4.69 \(\int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx\) [369]

3.4.69.1 Optimal result

Integrand size = 34, antiderivative size = 154 \[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=-\frac {2 B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{(a-i b)^{3/2} d}+\frac {B \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{(a+i b)^{3/2} d}+\frac {2 b^2 B}{a \left (a^2+b^2\right ) d \sqrt {a+b \tan (c+d x)}} \]

output

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

3.4.69.2 Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.08 \[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\frac {B \left (-\frac {2 \left (a^2+b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {a (a+i b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{\sqrt {a-i b}}+\frac {a (a-i b) \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{\sqrt {a+i b}}+\frac {2 b^2}{\sqrt {a+b \tan (c+d x)}}\right )}{a \left (a^2+b^2\right ) d} \]

input

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

output

(B*((-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)
 

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

Time = 1.24 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.14, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2011, 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.

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

\(\Big \downarrow \) 2011

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

input

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

output

B*(((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]]))
 

3.4.69.3.1 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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + b*x])
 

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]
 

3.4.69.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1778\) vs. \(2(130)=260\).

Time = 0.33 (sec) , antiderivative size = 1779, normalized size of antiderivative = 11.55

method	result	size
default	\(\text {Expression too large to display}\)	\(1779\)

input

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

output

B*(-1/4/d/(a^2+b^2)^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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^2-1/4/ 
d*b^2/(a^2+b^2)^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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)+1/2/d/(a^2+b 
^2)^(5/2)*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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a^3+1/2/d*b^2/(a^2+b 
^2)^(5/2)*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))*(2*(a^2+b^2)^(1/2)+2*a)^(1/2)*a-2/d*b^4/(a^2+b^2)^ 
(5/2)/(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))+1/d/(a^2+b^2)^(3/ 
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))*a^2-1/d/(a^2+b^2)^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))*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+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))-1/d*b^2/(a^2+b^2)^ 
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))*a-2/d*b^2/(a^2+b^2)^( 
5/2)/(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))*a^2+1/4/d/(a^2+...
 

3.4.69.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2281 vs. \(2 (126) = 252\).

Time = 0.39 (sec) , antiderivative size = 4578, normalized size of antiderivative = 29.73 \[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input

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

output

[1/2*(4*sqrt(b*tan(d*x + c) + a)*B*a*b^2 + ((a^4*b + a^2*b^3)*d*tan(d*x + 
c) + (a^5 + a^3*b^2)*d)*sqrt((B^2*a^3 - 3*B^2*a*b^2 + (a^6 + 3*a^4*b^2 + 3 
*a^2*b^4 + b^6)*d^2*sqrt(-(9*B^4*a^4*b^2 - 6*B^4*a^2*b^4 + B^4*b^6)/((a^12 
 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)* 
d^4)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*log(-(3*B^3*a^2 - B^3*b^ 
2)*sqrt(b*tan(d*x + c) + a) + (2*(a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*d^3 
*sqrt(-(9*B^4*a^4*b^2 - 6*B^4*a^2*b^4 + B^4*b^6)/((a^12 + 6*a^10*b^2 + 15* 
a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4)) - (3*B^2*a^4 
- 4*B^2*a^2*b^2 + B^2*b^4)*d)*sqrt((B^2*a^3 - 3*B^2*a*b^2 + (a^6 + 3*a^4*b 
^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*B^4*a^4*b^2 - 6*B^4*a^2*b^4 + B^4*b^6)/ 
((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + 
b^12)*d^4)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))) - ((a^4*b + a^2*b 
^3)*d*tan(d*x + c) + (a^5 + a^3*b^2)*d)*sqrt((B^2*a^3 - 3*B^2*a*b^2 + (a^6 
 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2*sqrt(-(9*B^4*a^4*b^2 - 6*B^4*a^2*b^4 + 
 B^4*b^6)/((a^12 + 6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a 
^2*b^10 + b^12)*d^4)))/((a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)*d^2))*log(-(3* 
B^3*a^2 - B^3*b^2)*sqrt(b*tan(d*x + c) + a) - (2*(a^7 + 3*a^5*b^2 + 3*a^3* 
b^4 + a*b^6)*d^3*sqrt(-(9*B^4*a^4*b^2 - 6*B^4*a^2*b^4 + B^4*b^6)/((a^12 + 
6*a^10*b^2 + 15*a^8*b^4 + 20*a^6*b^6 + 15*a^4*b^8 + 6*a^2*b^10 + b^12)*d^4 
)) - (3*B^2*a^4 - 4*B^2*a^2*b^2 + B^2*b^4)*d)*sqrt((B^2*a^3 - 3*B^2*a*b...
 

3.4.69.6 Sympy [F]

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

input

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

output

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

3.4.69.7 Maxima [F(-1)]

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

input

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

output

Timed out
 

3.4.69.8 Giac [F(-1)]

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

input

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

output

Timed out
 

3.4.69.9 Mupad [B] (verification not implemented)

Time = 14.09 (sec) , antiderivative size = 7172, normalized size of antiderivative = 46.57 \[ \int \frac {\cot (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

input

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

output

(log(((((96*B^4*a^2*b^4*d^4 - 16*B^4*b^6*d^4 - 144*B^4*a^4*b^2*d^4)^(1/2) 
+ 4*B^2*a^3*d^2 - 12*B^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3 
*a^4*b^2*d^4))^(1/2)*(((((((96*B^4*a^2*b^4*d^4 - 16*B^4*b^6*d^4 - 144*B^4* 
a^4*b^2*d^4)^(1/2) + 4*B^2*a^3*d^2 - 12*B^2*a*b^2*d^2)/(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*B*a^8*b^28*d^8 - ((((96*B^4*a 
^2*b^4*d^4 - 16*B^4*b^6*d^4 - 144*B^4*a^4*b^2*d^4)^(1/2) + 4*B^2*a^3*d^2 - 
 12*B^2*a*b^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1 
/2)*(a + b*tan(c + d*x))^(1/2)*(512*a^9*b^28*d^9 + 5376*a^11*b^26*d^9 + 25 
344*a^13*b^24*d^9 + 70656*a^15*b^22*d^9 + 129024*a^17*b^20*d^9 + 161280*a^ 
19*b^18*d^9 + 139776*a^21*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))/4 + 5248*B*a^10*b^26*d^8 + 
23936*B*a^12*b^24*d^8 + 64000*B*a^14*b^22*d^8 + 111104*B*a^16*b^20*d^8 + 1 
30816*B*a^18*b^18*d^8 + 105728*B*a^20*b^16*d^8 + 57856*B*a^22*b^14*d^8 + 2 
0480*B*a^24*b^12*d^8 + 4224*B*a^26*b^10*d^8 + 384*B*a^28*b^8*d^8))/4 + (a 
+ b*tan(c + d*x))^(1/2)*(256*B^2*a^8*b^26*d^7 + 1472*B^2*a^10*b^24*d^7 + 3 
712*B^2*a^12*b^22*d^7 + 6272*B^2*a^14*b^20*d^7 + 9856*B^2*a^16*b^18*d^7 + 
14336*B^2*a^18*b^16*d^7 + 15232*B^2*a^20*b^14*d^7 + 10112*B^2*a^22*b^12*d^ 
7 + 3712*B^2*a^24*b^10*d^7 + 576*B^2*a^26*b^8*d^7))*(((96*B^4*a^2*b^4*d^4 
- 16*B^4*b^6*d^4 - 144*B^4*a^4*b^2*d^4)^(1/2) + 4*B^2*a^3*d^2 - 12*B^2*a*b 
^2*d^2)/(a^6*d^4 + b^6*d^4 + 3*a^2*b^4*d^4 + 3*a^4*b^2*d^4))^(1/2))/4 -...