\(\int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx\) [589]

Optimal result

Integrand size = 23, antiderivative size = 221 \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {(a-b) \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a-b) \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 b^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} \left (a^2+b^2\right ) d}-\frac {(a+b) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {\tan (c+d x)}}{1+\tan (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {2 b}{a^2 d \sqrt {\tan (c+d x)}} \] Output:

-1/2*(a-b)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)/d-1/2*(a- 
b)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)/(a^2+b^2)/d+2*b^(7/2)*arctan 
(b^(1/2)*tan(d*x+c)^(1/2)/a^(1/2))/a^(5/2)/(a^2+b^2)/d-1/2*(a+b)*arctanh(2 
^(1/2)*tan(d*x+c)^(1/2)/(1+tan(d*x+c)))*2^(1/2)/(a^2+b^2)/d-2/3/a/d/tan(d* 
x+c)^(3/2)+2*b/a^2/d/tan(d*x+c)^(1/2)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 1.23 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {\frac {6 \sqrt {2} a^2 (a-b) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )}{a^2+b^2}+\frac {24 b^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \left (a^2+b^2\right )}+\frac {3 \sqrt {2} a^2 (a+b) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )}{a^2+b^2}-\frac {8 a}{\tan ^{\frac {3}{2}}(c+d x)}+\frac {24 b}{\sqrt {\tan (c+d x)}}}{12 a^2 d} \] Input:

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

Output:

((6*Sqrt[2]*a^2*(a - b)*(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 
 + Sqrt[2]*Sqrt[Tan[c + d*x]]]))/(a^2 + b^2) + (24*b^(7/2)*ArcTan[(Sqrt[b] 
*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[a]*(a^2 + b^2)) + (3*Sqrt[2]*a^2*(a + 
 b)*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Log[1 + Sqrt[2]* 
Sqrt[Tan[c + d*x]] + Tan[c + d*x]]))/(a^2 + b^2) - (8*a)/Tan[c + d*x]^(3/2 
) + (24*b)/Sqrt[Tan[c + d*x]])/(12*a^2*d)
 

Rubi [A] (verified)

Time = 1.19 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.15, number of steps used = 23, number of rules used = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.957, Rules used = {3042, 4052, 27, 3042, 4132, 27, 3042, 4137, 3042, 4017, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4052

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4132

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4137

\(\displaystyle -\frac {\frac {\frac {\int \frac {a^3-a^2 b \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}-\frac {b^4 \int \frac {\tan ^2(c+d x)+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 b}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\frac {\frac {\int \frac {a^3-a^2 b \tan (c+d x)}{\sqrt {\tan (c+d x)}}dx}{a^2+b^2}-\frac {b^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 b}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b^4 \int \frac {\tan (c+d x)^2+1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}dx}{a^2+b^2}}{a}-\frac {2 b}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 4117

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {b^4 \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))}d\tan (c+d x)}{d \left (a^2+b^2\right )}}{a}-\frac {2 b}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 73

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {2 b^4 \int \frac {1}{a+b \tan (c+d x)}d\sqrt {\tan (c+d x)}}{d \left (a^2+b^2\right )}}{a}-\frac {2 b}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {\frac {\frac {2 a^2 \left (\frac {1}{2} (a-b) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2}}\right )+\frac {1}{2} (a+b) \left (\frac {\log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}-\frac {2 b^{7/2} \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} d \left (a^2+b^2\right )}}{a}-\frac {2 b}{a d \sqrt {\tan (c+d x)}}}{a}-\frac {2}{3 a d \tan ^{\frac {3}{2}}(c+d x)}\)

Input:

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

Output:

-((((-2*b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[a]*(a^ 
2 + b^2)*d) + (2*a^2*(((a - b)*(-(ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]]/S 
qrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]/Sqrt[2]))/2 + ((a + b)*(- 
1/2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/Sqrt[2] + Log[1 + S 
qrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]/(2*Sqrt[2])))/2))/((a^2 + b^2)*d 
))/a - (2*b)/(a*d*Sqrt[Tan[c + d*x]]))/a) - 2/(3*a*d*Tan[c + d*x]^(3/2))
 

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 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])))
 

rule 4137

Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*tan[(e_.) 
+ (f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Sim 
p[1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*Simp[a*(A - C) - (A*b - b*C)*T 
an[e + f*x], x], x], x] + Simp[(A*b^2 + 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, 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 [A] (verified)

Time = 0.13 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\frac {2 b^{4} \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{a^{2} \left (a^{2}+b^{2}\right ) \sqrt {a b}}-\frac {2}{3 a \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 b}{a^{2} \sqrt {\tan \left (d x +c \right )}}+\frac {-\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{a^{2}+b^{2}}}{d}\)	\(255\)
default	\(\frac {\frac {2 b^{4} \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right )}{a^{2} \left (a^{2}+b^{2}\right ) \sqrt {a b}}-\frac {2}{3 a \tan \left (d x +c \right )^{\frac {3}{2}}}+\frac {2 b}{a^{2} \sqrt {\tan \left (d x +c \right )}}+\frac {-\frac {a \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}+\frac {b \sqrt {2}\, \left (\ln \left (\frac {\tan \left (d x +c \right )-\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}{\tan \left (d x +c \right )+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+1}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )\right )}{4}}{a^{2}+b^{2}}}{d}\)	\(255\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 807 vs. \(2 (187) = 374\).

Time = 0.20 (sec) , antiderivative size = 1644, normalized size of antiderivative = 7.44 \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Too large to display} \] Input:

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

Output:

[1/6*(6*b^3*sqrt(-b/a)*log((2*a*sqrt(-b/a)*sqrt(tan(d*x + c)) + b*tan(d*x 
+ c) - a)/(b*tan(d*x + c) + a))*tan(d*x + c)^2 + 6*sqrt(1/2)*(a^4 + a^2*b^ 
2)*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))*tan(d*x + c)^2 + 6*sqrt(1/2 
)*(a^4 + a^2*b^2)*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))*tan(d*x + c)^ 
2 - 3*sqrt(1/2)*(a^4 + a^2*b^2)*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)*tan(d*x + c)^2 + 3*sqrt(1/2)*(a^4 + a^2*b^2)*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)*tan(d*x + c)^2 - 4*(a^3 + a*b^2 - 3*(a^2*b + b^3)* 
tan(d*x + c))*sqrt(tan(d*x + c)))/((a^4 + a^2*b^2)*d*tan(d*x + c)^2), 1...
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.91 \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\frac {\frac {24 \, b^{4} \arctan \left (\frac {b \sqrt {\tan \left (d x + c\right )}}{\sqrt {a b}}\right )}{{\left (a^{4} + a^{2} b^{2}\right )} \sqrt {a b}} - \frac {3 \, {\left (2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} {\left (a + b\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} {\left (a + b\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )}}{a^{2} + b^{2}} + \frac {8 \, {\left (3 \, b \tan \left (d x + c\right ) - a\right )}}{a^{2} \tan \left (d x + c\right )^{\frac {3}{2}}}}{12 \, d} \] Input:

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

Output:

1/12*(24*b^4*arctan(b*sqrt(tan(d*x + c))/sqrt(a*b))/((a^4 + a^2*b^2)*sqrt( 
a*b)) - 3*(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)) + tan(d*x + c) + 
1) - sqrt(2)*(a + b)*log(-sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1))/ 
(a^2 + b^2) + 8*(3*b*tan(d*x + c) - a)/(a^2*tan(d*x + c)^(3/2)))/d
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
 

Mupad [B] (verification not implemented)

Time = 3.29 (sec) , antiderivative size = 4806, normalized size of antiderivative = 21.75 \[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\text {Too large to display} \] Input:

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

Output:

(atan((((-a^5*b^7)^(1/2)*(tan(c + d*x)^(1/2)*(64*a^14*b^9*d^5 + 32*a^18*b^ 
5*d^5) - ((-a^5*b^7)^(1/2)*(32*a^19*b^5*d^6 - 384*a^15*b^9*d^6 + 32*a^21*b 
^3*d^6 + ((tan(c + d*x)^(1/2)*(512*a^15*b^10*d^7 + 448*a^19*b^6*d^7 - 128* 
a^21*b^4*d^7 - 64*a^23*b^2*d^7) + ((-a^5*b^7)^(1/2)*(512*a^16*b^10*d^8 + 5 
12*a^18*b^8*d^8 - 384*a^20*b^6*d^8 - 256*a^22*b^4*d^8 + 128*a^24*b^2*d^8 - 
 (tan(c + d*x)^(1/2)*(-a^5*b^7)^(1/2)*(512*a^18*b^9*d^9 + 512*a^20*b^7*d^9 
 - 512*a^22*b^5*d^9 - 512*a^24*b^3*d^9))/(a^5*d*(a^2 + b^2))))/(a^5*d*(a^2 
 + b^2)))*(-a^5*b^7)^(1/2))/(a^5*d*(a^2 + b^2))))/(a^5*d*(a^2 + b^2)))*1i) 
/(a^5*d*(a^2 + b^2)) + ((-a^5*b^7)^(1/2)*(tan(c + d*x)^(1/2)*(64*a^14*b^9* 
d^5 + 32*a^18*b^5*d^5) - ((-a^5*b^7)^(1/2)*(384*a^15*b^9*d^6 - 32*a^19*b^5 
*d^6 - 32*a^21*b^3*d^6 + ((tan(c + d*x)^(1/2)*(512*a^15*b^10*d^7 + 448*a^1 
9*b^6*d^7 - 128*a^21*b^4*d^7 - 64*a^23*b^2*d^7) - ((-a^5*b^7)^(1/2)*(512*a 
^16*b^10*d^8 + 512*a^18*b^8*d^8 - 384*a^20*b^6*d^8 - 256*a^22*b^4*d^8 + 12 
8*a^24*b^2*d^8 + (tan(c + d*x)^(1/2)*(-a^5*b^7)^(1/2)*(512*a^18*b^9*d^9 + 
512*a^20*b^7*d^9 - 512*a^22*b^5*d^9 - 512*a^24*b^3*d^9))/(a^5*d*(a^2 + b^2 
))))/(a^5*d*(a^2 + b^2)))*(-a^5*b^7)^(1/2))/(a^5*d*(a^2 + b^2))))/(a^5*d*( 
a^2 + b^2)))*1i)/(a^5*d*(a^2 + b^2)))/(64*a^14*b^8*d^4 - ((-a^5*b^7)^(1/2) 
*(tan(c + d*x)^(1/2)*(64*a^14*b^9*d^5 + 32*a^18*b^5*d^5) - ((-a^5*b^7)^(1/ 
2)*(32*a^19*b^5*d^6 - 384*a^15*b^9*d^6 + 32*a^21*b^3*d^6 + ((tan(c + d*x)^ 
(1/2)*(512*a^15*b^10*d^7 + 448*a^19*b^6*d^7 - 128*a^21*b^4*d^7 - 64*a^2...
 

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {5}{2}}(c+d x) (a+b \tan (c+d x))} \, dx=\int \frac {\sqrt {\tan \left (d x +c \right )}}{\tan \left (d x +c \right )^{4} b +\tan \left (d x +c \right )^{3} a}d x \] Input:

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

Output:

int(sqrt(tan(c + d*x))/(tan(c + d*x)**4*b + tan(c + d*x)**3*a),x)