\(\int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\) [825]

Optimal result

Integrand size = 23, antiderivative size = 197 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=-\frac {(a+b) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a+b) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{b^{3/2} \left (a^2+b^2\right ) d}+\frac {(a-b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\cot (c+d x)}}{1+\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2}{b d \sqrt {\cot (c+d x)}} \] Output:

1/2*(a+b)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)/d+1/2*(a+b 
)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)/d+2*a^(5/2)*arctan( 
a^(1/2)*cot(d*x+c)^(1/2)/b^(1/2))/b^(3/2)/(a^2+b^2)/d+1/2*(a-b)*arctanh(2^ 
(1/2)*cot(d*x+c)^(1/2)/(1+cot(d*x+c)))*2^(1/2)/(a^2+b^2)/d+2/b/d/cot(d*x+c 
)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [C] (verified)

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

Time = 0.29 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {8 a^2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {a \cot (c+d x)}{b}\right )+b \left (8 b \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},1,\frac {3}{4},-\cot ^2(c+d x)\right )+\sqrt {2} a \sqrt {\cot (c+d x)} \left (-2 \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )+2 \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )-\log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )+\log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )\right )\right )}{4 b \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)}} \] Input:

Integrate[1/(Cot[c + d*x]^(5/2)*(a + b*Tan[c + d*x])),x]
 

Output:

(8*a^2*Hypergeometric2F1[-1/2, 1, 1/2, -((a*Cot[c + d*x])/b)] + b*(8*b*Hyp 
ergeometric2F1[-1/4, 1, 3/4, -Cot[c + d*x]^2] + Sqrt[2]*a*Sqrt[Cot[c + d*x 
]]*(-2*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] + 2*ArcTan[1 + Sqrt[2]*Sqrt[ 
Cot[c + d*x]]] - Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] + Log[ 
1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])))/(4*b*(a^2 + b^2)*d*Sqrt[ 
Cot[c + d*x]])
 

Rubi [A] (warning: unable to verify)

Time = 1.18 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.13, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.957, Rules used = {3042, 4156, 3042, 4052, 27, 3042, 4136, 3042, 4017, 25, 27, 1482, 1476, 1082, 217, 1479, 25, 27, 1103, 4117, 73, 218}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4156

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4136

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {2}{b d \sqrt {\cot (c+d x)}}-\frac {\frac {2 a^{5/2} \arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {b}}\right )}{\sqrt {b} d \left (a^2+b^2\right )}-\frac {2 b \left (\frac {1}{2} (a+b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a-b) \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}}{b}\)

Input:

Int[1/(Cot[c + d*x]^(5/2)*(a + b*Tan[c + d*x])),x]
 

Output:

2/(b*d*Sqrt[Cot[c + d*x]]) - ((2*a^(5/2)*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqr 
t[b]])/(Sqrt[b]*(a^2 + b^2)*d) - (2*b*(((a + b)*(-(ArcTan[1 - Sqrt[2]*Sqrt 
[Cot[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2]) 
)/2 + ((a - b)*(-1/2*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/Sq 
rt[2] + Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]]/(2*Sqrt[2])))/2 
))/((a^2 + b^2)*d))/b
 

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 217

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

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{a, b, c}, x]
 

rule 1103

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, 
e}, x] && EqQ[2*c*d - b*e, 0]
 

rule 1476

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
2*(d/e), 2]}, Simp[e/(2*c)   Int[1/Simp[d/e + q*x + x^2, x], x], x] + Simp[ 
e/(2*c)   Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e}, x] 
 && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
 

rule 1479

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
-2*(d/e), 2]}, Simp[e/(2*c*q)   Int[(q - 2*x)/Simp[d/e + q*x - x^2, x], x], 
 x] + Simp[e/(2*c*q)   Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /; F 
reeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
 

rule 1482

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[ 
a*c, 2]}, Simp[(d*q + a*e)/(2*a*c)   Int[(q + c*x^2)/(a + c*x^4), x], x] + 
Simp[(d*q - a*e)/(2*a*c)   Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ[{a 
, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(- 
a)*c]
 

rule 3042

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

rule 4017

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_ 
)]], x_Symbol] :> Simp[2/f   Subst[Int[(b*c + d*x^2)/(b^2 + x^4), x], x, Sq 
rt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2, 0] & 
& NeQ[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 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]
 

rule 4156

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)]^(n_.))^(p_.), x_Symbol] :> Simp[d^(n*p)   Int[(d*Cot[e + f*x])^(m - n*p 
)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && 
  !IntegerQ[m] && IntegersQ[n, p]
 

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 328, normalized size of antiderivative = 1.66

method	result	size
derivativedivides	\(\frac {-\sqrt {a b}\, \ln \left (-\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, b^{2}-2 \sqrt {a b}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a b -2 \sqrt {a b}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b^{2}-2 \sqrt {a b}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a b -2 \sqrt {a b}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b^{2}-\sqrt {a b}\, \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right ) \sqrt {2}\, a b +8 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a b}\, a^{2}+8 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a b}\, b^{2}-8 a^{3} \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{4 d \left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {5}{2}} \tan \left (d x +c \right )^{\frac {5}{2}} b \left (a^{2}+b^{2}\right ) \sqrt {a b}}\)	\(328\)
default	\(\frac {-\sqrt {a b}\, \ln \left (-\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, b^{2}-2 \sqrt {a b}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a b -2 \sqrt {a b}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b^{2}-2 \sqrt {a b}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, a b -2 \sqrt {a b}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, b^{2}-\sqrt {a b}\, \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right ) \sqrt {2}\, a b +8 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a b}\, a^{2}+8 \sqrt {\tan \left (d x +c \right )}\, \sqrt {a b}\, b^{2}-8 a^{3} \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{4 d \left (\frac {1}{\tan \left (d x +c \right )}\right )^{\frac {5}{2}} \tan \left (d x +c \right )^{\frac {5}{2}} b \left (a^{2}+b^{2}\right ) \sqrt {a b}}\)	\(328\)

Input:

int(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

1/4/d*(-(a*b)^(1/2)*ln(-(tan(d*x+c)+2^(1/2)*tan(d*x+c)^(1/2)+1)/(2^(1/2)*t 
an(d*x+c)^(1/2)-tan(d*x+c)-1))*2^(1/2)*b^2-2*(a*b)^(1/2)*arctan(1+2^(1/2)* 
tan(d*x+c)^(1/2))*2^(1/2)*a*b-2*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1 
/2))*2^(1/2)*b^2-2*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2) 
*a*b-2*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*b^2-(a*b)^( 
1/2)*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(tan(d*x+c)+2^(1/2)*tan(d 
*x+c)^(1/2)+1))*2^(1/2)*a*b+8*tan(d*x+c)^(1/2)*(a*b)^(1/2)*a^2+8*tan(d*x+c 
)^(1/2)*(a*b)^(1/2)*b^2-8*a^3*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2)))/(1/t 
an(d*x+c))^(5/2)/tan(d*x+c)^(5/2)/b/(a^2+b^2)/(a*b)^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 749 vs. \(2 (168) = 336\).

Time = 0.14 (sec) , antiderivative size = 1524, normalized size of antiderivative = 7.74 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Too large to display} \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="fricas")
 

Output:

[-1/2*(2*sqrt(1/2)*(a^2*b + b^3)*d*sqrt((a^2 + 2*a*b + b^2)/((a^4 + 2*a^2* 
b^2 + b^4)*d^2))*arctan(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt((a^2 + 2*a*b + b 
^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt((a^2 - 2*a*b + b^2)/((a^4 + 2*a^2* 
b^2 + b^4)*d^2)) + 2*sqrt(1/2)*(a^3 - a^2*b + a*b^2 - b^3)*d*sqrt((a^2 + 2 
*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt(tan(d*x + c)))/(a^2 - b^2) 
) + 2*sqrt(1/2)*(a^2*b + b^3)*d*sqrt((a^2 + 2*a*b + b^2)/((a^4 + 2*a^2*b^2 
 + b^4)*d^2))*arctan(-((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt((a^2 + 2*a*b + b^2 
)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt((a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^ 
2 + b^4)*d^2)) - 2*sqrt(1/2)*(a^3 - a^2*b + a*b^2 - b^3)*d*sqrt((a^2 + 2*a 
*b + b^2)/((a^4 + 2*a^2*b^2 + b^4)*d^2))*sqrt(tan(d*x + c)))/(a^2 - b^2)) 
+ sqrt(1/2)*(a^2*b + b^3)*d*sqrt((a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b 
^4)*d^2))*log(2*sqrt(1/2)*(a^2 + b^2)*d*sqrt((a^2 - 2*a*b + b^2)/((a^4 + 2 
*a^2*b^2 + b^4)*d^2))*sqrt(tan(d*x + c)) - (a - b)*tan(d*x + c) - a + b) - 
 sqrt(1/2)*(a^2*b + b^3)*d*sqrt((a^2 - 2*a*b + b^2)/((a^4 + 2*a^2*b^2 + b^ 
4)*d^2))*log(-2*sqrt(1/2)*(a^2 + b^2)*d*sqrt((a^2 - 2*a*b + b^2)/((a^4 + 2 
*a^2*b^2 + b^4)*d^2))*sqrt(tan(d*x + c)) - (a - b)*tan(d*x + c) - a + b) - 
 2*a^2*sqrt(-a/b)*log(-(2*b*sqrt(-a/b)*sqrt(tan(d*x + c)) - b*tan(d*x + c) 
 + a)/(b*tan(d*x + c) + a)) - 4*(a^2 + b^2)*sqrt(tan(d*x + c)))/((a^2*b + 
b^3)*d), -1/2*(2*sqrt(1/2)*(a^2*b + b^3)*d*sqrt((a^2 + 2*a*b + b^2)/((a^4 
+ 2*a^2*b^2 + b^4)*d^2))*arctan(((a^4 + 2*a^2*b^2 + b^4)*d^2*sqrt((a^2 ...
 

Sympy [F]

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

integrate(1/cot(d*x+c)**(5/2)/(a+b*tan(d*x+c)),x)
 

Output:

Integral(1/((a + b*tan(c + d*x))*cot(c + d*x)**(5/2)), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.96 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {\frac {8 \, a^{3} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{2} b + b^{3}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (a + b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (a - b\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left (a - b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{2} + b^{2}} + \frac {8 \, \sqrt {\tan \left (d x + c\right )}}{b}}{4 \, d} \] Input:

integrate(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="maxima")
 

Output:

1/4*(8*a^3*arctan(a/(sqrt(a*b)*sqrt(tan(d*x + c))))/((a^2*b + b^3)*sqrt(a* 
b)) + (2*sqrt(2)*(a + b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c) 
))) + 2*sqrt(2)*(a + b)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c) 
))) + sqrt(2)*(a - b)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) 
 - sqrt(2)*(a - b)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))/ 
(a^2 + b^2) + 8*sqrt(tan(d*x + c))/b)/d
 

Giac [F]

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

integrate(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x, algorithm="giac")
 

Output:

integrate(1/((b*tan(d*x + c) + a)*cot(d*x + c)^(5/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )} \,d x \] Input:

int(1/(cot(c + d*x)^(5/2)*(a + b*tan(c + d*x))),x)
 

Output:

int(1/(cot(c + d*x)^(5/2)*(a + b*tan(c + d*x))), x)
 

Reduce [F]

\[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\int \frac {\sqrt {\cot \left (d x +c \right )}}{\cot \left (d x +c \right )^{3} \tan \left (d x +c \right ) b +\cot \left (d x +c \right )^{3} a}d x \] Input:

int(1/cot(d*x+c)^(5/2)/(a+b*tan(d*x+c)),x)
 

Output:

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