\(\int \cot (c+d x) (a+b \tan (c+d x))^{5/2} \, dx\) [523]

Optimal result

Integrand size = 21, antiderivative size = 138 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=-\frac {2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {(a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )}{d}+\frac {(a+i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )}{d}+\frac {2 b^2 \sqrt {a+b \tan (c+d x)}}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.59 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\frac {-2 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a}}\right )+(a-i b)^{5/2} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a-i b}}\right )+a^2 \sqrt {a+i b} \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 i a \sqrt {a+i b} b \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )-\sqrt {a+i b} b^2 \text {arctanh}\left (\frac {\sqrt {a+b \tan (c+d x)}}{\sqrt {a+i b}}\right )+2 b^2 \sqrt {a+b \tan (c+d x)}}{d} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 1.23 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.04, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {3042, 4049, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4049

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4022

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4020

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

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 4049

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 - 2)*((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 - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n 
- 1) - b^2*(b*c*(m - 2) + a*d*(1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[ 
e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*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] && GtQ[m, 2] && (GeQ[n, -1] || I 
ntegerQ[m]) &&  !(IGtQ[n, 2] && ( !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]
 

Maple [B] (warning: unable to verify)

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

Time = 94.65 (sec) , antiderivative size = 1538880, normalized size of antiderivative = 11151.30

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

Input:

int(cot(d*x+c)*(a+b*tan(d*x+c))^(5/2),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. 1118 vs. \(2 (110) = 220\).

Time = 0.20 (sec) , antiderivative size = 2252, normalized size of antiderivative = 16.32 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/2*(2*a^(5/2)*log((b*tan(d*x + c) - 2*sqrt(b*tan(d*x + c) + a)*sqrt(a) + 
 2*a)/tan(d*x + c)) + 4*sqrt(b*tan(d*x + c) + a)*b^2 - d*sqrt((a^5 - 10*a^ 
3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a 
^2*b^8 + b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt( 
b*tan(d*x + c) + a) + (2*a*d^3*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b 
^6 - 20*a^2*b^8 + b^10)/d^4) - (5*a^6 - 15*a^4*b^2 + 11*a^2*b^4 - b^6)*d)* 
sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 1 
10*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4))/d^2)) + d*sqrt((a^5 - 10*a^3*b^2 + 5 
*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + 
b^10)/d^4))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x 
 + c) + a) - (2*a*d^3*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a 
^2*b^8 + b^10)/d^4) - (5*a^6 - 15*a^4*b^2 + 11*a^2*b^4 - b^6)*d)*sqrt((a^5 
 - 10*a^3*b^2 + 5*a*b^4 + d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^ 
6 - 20*a^2*b^8 + b^10)/d^4))/d^2)) + d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - 
d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + b^10)/d^4 
))/d^2)*log((5*a^8 - 14*a^4*b^4 - 8*a^2*b^6 + b^8)*sqrt(b*tan(d*x + c) + a 
) + (2*a*d^3*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^2*b^8 + 
b^10)/d^4) + (5*a^6 - 15*a^4*b^2 + 11*a^2*b^4 - b^6)*d)*sqrt((a^5 - 10*a^3 
*b^2 + 5*a*b^4 - d^2*sqrt(-(25*a^8*b^2 - 100*a^6*b^4 + 110*a^4*b^6 - 20*a^ 
2*b^8 + b^10)/d^4))/d^2)) - d*sqrt((a^5 - 10*a^3*b^2 + 5*a*b^4 - d^2*sq...
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

Exception generated. \[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(cot(d*x+c)*(a+b*tan(d*x+c))^(5/2),x, algorithm="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,[0,13,4]%%%}+%%%{6,[0,11,4]%%%}+%%%{15,[0,9,4]%%%}+% 
%%{20,[0,
 

Mupad [B] (verification not implemented)

Time = 8.35 (sec) , antiderivative size = 3333, normalized size of antiderivative = 24.15 \[ \int \cot (c+d x) (a+b \tan (c+d x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

log(((((-b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + a^5*d^2 + 5*a*b^4*d 
^2 - 10*a^3*b^2*d^2)/d^4)^(1/2)*(((((-b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2 
)^(1/2) + a^5*d^2 + 5*a*b^4*d^2 - 10*a^3*b^2*d^2)/d^4)^(1/2)*(((((-b^2*d^4 
*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + a^5*d^2 + 5*a*b^4*d^2 - 10*a^3*b^2* 
d^2)/d^4)^(1/2)*(((((-b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + a^5*d^ 
2 + 5*a*b^4*d^2 - 10*a^3*b^2*d^2)/d^4)^(1/2)*((128*a*b^8*(3*a^4 + b^4 + 4* 
a^2*b^2))/d - 128*b^8*(((-b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + a^ 
5*d^2 + 5*a*b^4*d^2 - 10*a^3*b^2*d^2)/d^4)^(1/2)*(3*a^2 + 2*b^2)*(a + b*ta 
n(c + d*x))^(1/2)))/2 + (64*a*b^8*(a + b*tan(c + d*x))^(1/2)*(9*a^6 + 19*b 
^6 - 5*a^2*b^4 - 51*a^4*b^2))/d^2))/2 - (96*a^2*b^8*(a^8 + b^8 + 16*a^2*b^ 
6 - 10*a^4*b^4 - 24*a^6*b^2))/d^3))/2 - (32*b^8*(a + b*tan(c + d*x))^(1/2) 
*(3*a^12 + b^12 + 6*a^2*b^10 + 15*a^4*b^8 + 18*a^6*b^6 + 45*a^8*b^4 - 24*a 
^10*b^2))/d^4))/2 + (32*a^3*b^10*(a^2 + b^2)^3*(6*a^4 + b^4 + 3*a^2*b^2))/ 
d^5)*((20*a^2*b^8*d^4 - b^10*d^4 - 110*a^4*b^6*d^4 + 100*a^6*b^4*d^4 - 25* 
a^8*b^2*d^4)^(1/2)/(4*d^4) + a^5/(4*d^2) + (5*a*b^4)/(4*d^2) - (5*a^3*b^2) 
/(2*d^2))^(1/2) - log(((((-b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + a 
^5*d^2 + 5*a*b^4*d^2 - 10*a^3*b^2*d^2)/d^4)^(1/2)*(((((-b^2*d^4*(5*a^4 + b 
^4 - 10*a^2*b^2)^2)^(1/2) + a^5*d^2 + 5*a*b^4*d^2 - 10*a^3*b^2*d^2)/d^4)^( 
1/2)*(((((-b^2*d^4*(5*a^4 + b^4 - 10*a^2*b^2)^2)^(1/2) + a^5*d^2 + 5*a*b^4 
*d^2 - 10*a^3*b^2*d^2)/d^4)^(1/2)*(((((-b^2*d^4*(5*a^4 + b^4 - 10*a^2*b...
 

Reduce [F]

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

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

Output:

int(sqrt(tan(c + d*x)*b + a)*cot(c + d*x)*tan(c + d*x)**2,x)*b**2 + 2*int( 
sqrt(tan(c + d*x)*b + a)*cot(c + d*x)*tan(c + d*x),x)*a*b + int(sqrt(tan(c 
 + d*x)*b + a)*cot(c + d*x),x)*a**2