3.1.28 \(\int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))} \, dx\) [28]

3.1.28.1 Optimal result
3.1.28.2 Mathematica [C] (verified)
3.1.28.3 Rubi [A] (warning: unable to verify)
3.1.28.4 Maple [B] (verified)
3.1.28.5 Fricas [A] (verification not implemented)
3.1.28.6 Sympy [F]
3.1.28.7 Maxima [F(-2)]
3.1.28.8 Giac [F]
3.1.28.9 Mupad [B] (verification not implemented)

3.1.28.1 Optimal result

Integrand size = 25, antiderivative size = 135 \[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))} \, dx=-\frac {\arctan \left (\frac {\sqrt {e \cot (c+d x)}}{\sqrt {e}}\right )}{a d e^{5/2}}+\frac {\text {arctanh}\left (\frac {\sqrt {e}+\sqrt {e} \cot (c+d x)}{\sqrt {2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d e^{5/2}}+\frac {2}{3 a d e (e \cot (c+d x))^{3/2}}-\frac {2}{a d e^2 \sqrt {e \cot (c+d x)}} \]

output
-arctan((e*cot(d*x+c))^(1/2)/e^(1/2))/a/d/e^(5/2)+2/3/a/d/e/(e*cot(d*x+c)) 
^(3/2)+1/2*arctanh(1/2*(e^(1/2)+cot(d*x+c)*e^(1/2))*2^(1/2)/(e*cot(d*x+c)) 
^(1/2))/a/d/e^(5/2)*2^(1/2)-2/a/d/e^2/(e*cot(d*x+c))^(1/2)
 
3.1.28.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.62 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.64 \[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))} \, dx=\frac {\operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},-\cot (c+d x)\right )+\operatorname {Hypergeometric2F1}\left (-\frac {3}{4},1,\frac {1}{4},-\cot ^2(c+d x)\right )-3 \cot (c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\cot ^2(c+d x)\right )}{3 a d e (e \cot (c+d x))^{3/2}} \]

input
Integrate[1/((e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])),x]
 
output
(Hypergeometric2F1[-3/2, 1, -1/2, -Cot[c + d*x]] + Hypergeometric2F1[-3/4, 
 1, 1/4, -Cot[c + d*x]^2] - 3*Cot[c + d*x]*Hypergeometric2F1[-1/4, 1, 3/4, 
 -Cot[c + d*x]^2])/(3*a*d*e*(e*Cot[c + d*x])^(3/2))
 
3.1.28.3 Rubi [A] (warning: unable to verify)

Time = 1.22 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.08, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.680, Rules used = {3042, 4052, 27, 3042, 4132, 27, 2030, 3042, 4056, 25, 3042, 4015, 221, 4117, 27, 73, 216}

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.

\(\displaystyle \int \frac {1}{(a \cot (c+d x)+a) (e \cot (c+d x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\left (a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right ) \left (-e \tan \left (c+d x+\frac {\pi }{2}\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3 a d e (e \cot (c+d x))^{3/2}}-\frac {\int \frac {a \cot ^2(c+d x) e^2+a e^2+a \cot (c+d x) e^2}{(e \cot (c+d x))^{3/2} (\cot (c+d x) a+a)}dx}{a e^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

\(\displaystyle \frac {2}{3 a d e (e \cot (c+d x))^{3/2}}-\frac {\frac {2 \int -\frac {a^2 e^4 \cot ^2(c+d x)}{2 \sqrt {e \cot (c+d x)} (\cot (c+d x) a+a)}dx}{a e^3}+\frac {2 e}{d \sqrt {e \cot (c+d x)}}}{a e^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3 a d e (e \cot (c+d x))^{3/2}}-\frac {\frac {2 e}{d \sqrt {e \cot (c+d x)}}-a e \int \frac {\cot ^2(c+d x)}{\sqrt {e \cot (c+d x)} (\cot (c+d x) a+a)}dx}{a e^3}\)

\(\Big \downarrow \) 2030

\(\displaystyle \frac {2}{3 a d e (e \cot (c+d x))^{3/2}}-\frac {\frac {2 e}{d \sqrt {e \cot (c+d x)}}-\frac {a \int \frac {(e \cot (c+d x))^{3/2}}{\cot (c+d x) a+a}dx}{e}}{a e^3}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4056

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4015

\(\displaystyle \frac {2}{3 a d e (e \cot (c+d x))^{3/2}}-\frac {\frac {2 e}{d \sqrt {e \cot (c+d x)}}-\frac {a \left (\frac {e^4 \int \frac {1}{2 a^2 e^4-\left (a e^2+a \cot (c+d x) e^2\right )^2 \tan (c+d x)}d\frac {a e^2+a \cot (c+d x) e^2}{\sqrt {e \cot (c+d x)}}}{d}+\frac {1}{2} e^2 \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-e \tan \left (c+d x+\frac {\pi }{2}\right )} \left (a-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx\right )}{e}}{a e^3}\)

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4117

\(\displaystyle \frac {2}{3 a d e (e \cot (c+d x))^{3/2}}-\frac {\frac {2 e}{d \sqrt {e \cot (c+d x)}}-\frac {a \left (\frac {e^2 \int \frac {1}{a \sqrt {e \cot (c+d x)} (\cot (c+d x)+1)}d(-\cot (c+d x))}{2 d}+\frac {e^{3/2} \text {arctanh}\left (\frac {a e^2 \cot (c+d x)+a e^2}{\sqrt {2} a e^{3/2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d}\right )}{e}}{a e^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2}{3 a d e (e \cot (c+d x))^{3/2}}-\frac {\frac {2 e}{d \sqrt {e \cot (c+d x)}}-\frac {a \left (\frac {e^2 \int \frac {1}{\sqrt {e \cot (c+d x)} (\cot (c+d x)+1)}d(-\cot (c+d x))}{2 a d}+\frac {e^{3/2} \text {arctanh}\left (\frac {a e^2 \cot (c+d x)+a e^2}{\sqrt {2} a e^{3/2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d}\right )}{e}}{a e^3}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {2}{3 a d e (e \cot (c+d x))^{3/2}}-\frac {\frac {2 e}{d \sqrt {e \cot (c+d x)}}-\frac {a \left (\frac {e^{3/2} \text {arctanh}\left (\frac {a e^2 \cot (c+d x)+a e^2}{\sqrt {2} a e^{3/2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d}-\frac {e \int \frac {1}{\frac {\cot ^2(c+d x)}{e}+1}d\sqrt {e \cot (c+d x)}}{a d}\right )}{e}}{a e^3}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2}{3 a d e (e \cot (c+d x))^{3/2}}-\frac {\frac {2 e}{d \sqrt {e \cot (c+d x)}}-\frac {a \left (\frac {e^{3/2} \arctan \left (\frac {\cot (c+d x)}{\sqrt {e}}\right )}{a d}+\frac {e^{3/2} \text {arctanh}\left (\frac {a e^2 \cot (c+d x)+a e^2}{\sqrt {2} a e^{3/2} \sqrt {e \cot (c+d x)}}\right )}{\sqrt {2} a d}\right )}{e}}{a e^3}\)

input
Int[1/((e*Cot[c + d*x])^(5/2)*(a + a*Cot[c + d*x])),x]
 
output
2/(3*a*d*e*(e*Cot[c + d*x])^(3/2)) - (-((a*((e^(3/2)*ArcTan[Cot[c + d*x]/S 
qrt[e]])/(a*d) + (e^(3/2)*ArcTanh[(a*e^2 + a*e^2*Cot[c + d*x])/(Sqrt[2]*a* 
e^(3/2)*Sqrt[e*Cot[c + d*x]])])/(Sqrt[2]*a*d)))/e) + (2*e)/(d*Sqrt[e*Cot[c 
 + d*x]]))/(a*e^3)
 

3.1.28.3.1 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 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 2030
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4015
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[-2*(d^2/f)   Subst[Int[1/(2*c*d + b*x^2), x], x, (c 
- d*Tan[e + f*x])/Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && 
 EqQ[c^2 - d^2, 0]
 

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 4056
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) 
+ (f_.)*(x_)]), x_Symbol] :> Simp[1/(c^2 + d^2)   Int[Simp[a^2*c - b^2*c + 
2*a*b*d + (2*a*b*c - a^2*d + b^2*d)*Tan[e + f*x], x]/Sqrt[a + b*Tan[e + f*x 
]], x], x] + Simp[(b*c - a*d)^2/(c^2 + d^2)   Int[(1 + Tan[e + f*x]^2)/(Sqr 
t[a + b*Tan[e + f*x]]*(c + d*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 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])))
 
3.1.28.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(332\) vs. \(2(113)=226\).

Time = 0.04 (sec) , antiderivative size = 333, normalized size of antiderivative = 2.47

method result size
derivativedivides \(-\frac {2 e^{2} \left (\frac {-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}+\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}}{2 e^{4}}-\frac {1}{3 e^{3} \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{e^{4} \sqrt {e \cot \left (d x +c \right )}}+\frac {\arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2 e^{\frac {9}{2}}}\right )}{d a}\) \(333\)
default \(-\frac {2 e^{2} \left (\frac {-\frac {\left (e^{2}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 e}+\frac {\sqrt {2}\, \left (\ln \left (\frac {e \cot \left (d x +c \right )-\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}{e \cot \left (d x +c \right )+\left (e^{2}\right )^{\frac {1}{4}} \sqrt {e \cot \left (d x +c \right )}\, \sqrt {2}+\sqrt {e^{2}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )-2 \arctan \left (-\frac {\sqrt {2}\, \sqrt {e \cot \left (d x +c \right )}}{\left (e^{2}\right )^{\frac {1}{4}}}+1\right )\right )}{8 \left (e^{2}\right )^{\frac {1}{4}}}}{2 e^{4}}-\frac {1}{3 e^{3} \left (e \cot \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{e^{4} \sqrt {e \cot \left (d x +c \right )}}+\frac {\arctan \left (\frac {\sqrt {e \cot \left (d x +c \right )}}{\sqrt {e}}\right )}{2 e^{\frac {9}{2}}}\right )}{d a}\) \(333\)

input
int(1/(e*cot(d*x+c))^(5/2)/(a+a*cot(d*x+c)),x,method=_RETURNVERBOSE)
 
output
-2/d/a*e^2*(1/2/e^4*(-1/8/e*(e^2)^(1/4)*2^(1/2)*(ln((e*cot(d*x+c)+(e^2)^(1 
/4)*(e*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)-(e^2)^(1/4)*(e 
*cot(d*x+c))^(1/2)*2^(1/2)+(e^2)^(1/2)))+2*arctan(2^(1/2)/(e^2)^(1/4)*(e*c 
ot(d*x+c))^(1/2)+1)-2*arctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)) 
+1/8/(e^2)^(1/4)*2^(1/2)*(ln((e*cot(d*x+c)-(e^2)^(1/4)*(e*cot(d*x+c))^(1/2 
)*2^(1/2)+(e^2)^(1/2))/(e*cot(d*x+c)+(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)*2^(1 
/2)+(e^2)^(1/2)))+2*arctan(2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)-2*a 
rctan(-2^(1/2)/(e^2)^(1/4)*(e*cot(d*x+c))^(1/2)+1)))-1/3/e^3/(e*cot(d*x+c) 
)^(3/2)+1/e^4/(e*cot(d*x+c))^(1/2)+1/2/e^(9/2)*arctan((e*cot(d*x+c))^(1/2) 
/e^(1/2)))
 
3.1.28.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 500, normalized size of antiderivative = 3.70 \[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))} \, dx=\left [-\frac {3 \, \sqrt {2} \sqrt {-e} {\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1\right )}}{2 \, {\left (e \cos \left (2 \, d x + 2 \, c\right ) + e\right )}}\right ) + 3 \, \sqrt {-e} {\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (\frac {e \cos \left (2 \, d x + 2 \, c\right ) - e \sin \left (2 \, d x + 2 \, c\right ) + 2 \, \sqrt {-e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} \sin \left (2 \, d x + 2 \, c\right ) + e}{\cos \left (2 \, d x + 2 \, c\right ) + \sin \left (2 \, d x + 2 \, c\right ) + 1}\right ) + 4 \, \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + 3 \, \sin \left (2 \, d x + 2 \, c\right ) - 1\right )}}{6 \, {\left (a d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + a d e^{3}\right )}}, \frac {3 \, \sqrt {2} \sqrt {e} {\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) - \sin \left (2 \, d x + 2 \, c\right ) - 1\right )} + 2 \, e \sin \left (2 \, d x + 2 \, c\right ) + e\right ) - 12 \, \sqrt {e} {\left (\cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \arctan \left (\frac {\sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}}}{\sqrt {e}}\right ) - 8 \, \sqrt {\frac {e \cos \left (2 \, d x + 2 \, c\right ) + e}{\sin \left (2 \, d x + 2 \, c\right )}} {\left (\cos \left (2 \, d x + 2 \, c\right ) + 3 \, \sin \left (2 \, d x + 2 \, c\right ) - 1\right )}}{12 \, {\left (a d e^{3} \cos \left (2 \, d x + 2 \, c\right ) + a d e^{3}\right )}}\right ] \]

input
integrate(1/(e*cot(d*x+c))^(5/2)/(a+a*cot(d*x+c)),x, algorithm="fricas")
 
output
[-1/6*(3*sqrt(2)*sqrt(-e)*(cos(2*d*x + 2*c) + 1)*arctan(1/2*sqrt(2)*sqrt(- 
e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*(cos(2*d*x + 2*c) + sin 
(2*d*x + 2*c) + 1)/(e*cos(2*d*x + 2*c) + e)) + 3*sqrt(-e)*(cos(2*d*x + 2*c 
) + 1)*log((e*cos(2*d*x + 2*c) - e*sin(2*d*x + 2*c) + 2*sqrt(-e)*sqrt((e*c 
os(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*sin(2*d*x + 2*c) + e)/(cos(2*d*x + 
2*c) + sin(2*d*x + 2*c) + 1)) + 4*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x 
+ 2*c))*(cos(2*d*x + 2*c) + 3*sin(2*d*x + 2*c) - 1))/(a*d*e^3*cos(2*d*x + 
2*c) + a*d*e^3), 1/12*(3*sqrt(2)*sqrt(e)*(cos(2*d*x + 2*c) + 1)*log(-sqrt( 
2)*sqrt(e)*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*(cos(2*d*x + 2* 
c) - sin(2*d*x + 2*c) - 1) + 2*e*sin(2*d*x + 2*c) + e) - 12*sqrt(e)*(cos(2 
*d*x + 2*c) + 1)*arctan(sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))/sq 
rt(e)) - 8*sqrt((e*cos(2*d*x + 2*c) + e)/sin(2*d*x + 2*c))*(cos(2*d*x + 2* 
c) + 3*sin(2*d*x + 2*c) - 1))/(a*d*e^3*cos(2*d*x + 2*c) + a*d*e^3)]
 
3.1.28.6 Sympy [F]

\[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))} \, dx=\frac {\int \frac {1}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}} \cot {\left (c + d x \right )} + \left (e \cot {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx}{a} \]

input
integrate(1/(e*cot(d*x+c))**(5/2)/(a+a*cot(d*x+c)),x)
 
output
Integral(1/((e*cot(c + d*x))**(5/2)*cot(c + d*x) + (e*cot(c + d*x))**(5/2) 
), x)/a
 
3.1.28.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))} \, dx=\text {Exception raised: ValueError} \]

input
integrate(1/(e*cot(d*x+c))^(5/2)/(a+a*cot(d*x+c)),x, algorithm="maxima")
 
output
Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 
3.1.28.8 Giac [F]

\[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))} \, dx=\int { \frac {1}{{\left (a \cot \left (d x + c\right ) + a\right )} \left (e \cot \left (d x + c\right )\right )^{\frac {5}{2}}} \,d x } \]

input
integrate(1/(e*cot(d*x+c))^(5/2)/(a+a*cot(d*x+c)),x, algorithm="giac")
 
output
integrate(1/((a*cot(d*x + c) + a)*(e*cot(d*x + c))^(5/2)), x)
 
3.1.28.9 Mupad [B] (verification not implemented)

Time = 13.31 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.98 \[ \int \frac {1}{(e \cot (c+d x))^{5/2} (a+a \cot (c+d x))} \, dx=\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {12\,\sqrt {2}\,a^3\,d^3\,e^{21/2}\,\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{12\,a^3\,d^3\,e^{11}+12\,a^3\,d^3\,e^{11}\,\mathrm {cot}\left (c+d\,x\right )}\right )}{2\,a\,d\,e^{5/2}}-\frac {\mathrm {atan}\left (\frac {\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}}{\sqrt {e}}\right )}{a\,d\,e^{5/2}}-\frac {\frac {2\,\mathrm {cot}\left (c+d\,x\right )}{e}-\frac {2}{3\,e}}{a\,d\,{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}} \]

input
int(1/((e*cot(c + d*x))^(5/2)*(a + a*cot(c + d*x))),x)
 
output
(2^(1/2)*atanh((12*2^(1/2)*a^3*d^3*e^(21/2)*(e*cot(c + d*x))^(1/2))/(12*a^ 
3*d^3*e^11 + 12*a^3*d^3*e^11*cot(c + d*x))))/(2*a*d*e^(5/2)) - atan((e*cot 
(c + d*x))^(1/2)/e^(1/2))/(a*d*e^(5/2)) - ((2*cot(c + d*x))/e - 2/(3*e))/( 
a*d*(e*cot(c + d*x))^(3/2))