3.6.92 \(\int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx\) [592]

3.6.92.1 Optimal result
3.6.92.2 Mathematica [A] (verified)
3.6.92.3 Rubi [A] (warning: unable to verify)
3.6.92.4 Maple [A] (verified)
3.6.92.5 Fricas [B] (verification not implemented)
3.6.92.6 Sympy [F]
3.6.92.7 Maxima [A] (verification not implemented)
3.6.92.8 Giac [F]
3.6.92.9 Mupad [F(-1)]

3.6.92.1 Optimal result

Integrand size = 33, antiderivative size = 278 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=\frac {(a (A-B)+b (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {(a (A-B)+b (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {2 \sqrt {a} (A b-a B) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {b} \left (a^2+b^2\right ) d}-\frac {(b (A-B)-a (A+B)) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(b (A-B)-a (A+B)) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d} \]

output
-1/2*(a*(A-B)+b*(A+B))*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2^( 
1/2)-1/2*(a*(A-B)+b*(A+B))*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d* 
2^(1/2)-1/4*(b*(A-B)-a*(A+B))*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(a 
^2+b^2)/d*2^(1/2)+1/4*(b*(A-B)-a*(A+B))*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c) 
^(1/2))/(a^2+b^2)/d*2^(1/2)+2*(A*b-B*a)*arctan(a^(1/2)*cot(d*x+c)^(1/2)/b^ 
(1/2))*a^(1/2)/(a^2+b^2)/d/b^(1/2)
 
3.6.92.2 Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.77 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=-\frac {\sqrt {\cot (c+d x)} \left (2 \sqrt {2} (a (A-B)+b (A+B)) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )+\frac {8 \sqrt {a} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {b}}-\sqrt {2} (b (-A+B)+a (A+B)) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right ) d} \]

input
Integrate[(A + B*Tan[c + d*x])/(Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])),x 
]
 
output
-1/4*(Sqrt[Cot[c + d*x]]*(2*Sqrt[2]*(a*(A - B) + b*(A + B))*(ArcTan[1 - Sq 
rt[2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) + (8*S 
qrt[a]*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/Sqrt[b] - 
 Sqrt[2]*(b*(-A + B) + a*(A + B))*(Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Ta 
n[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]))*Sqrt[Ta 
n[c + d*x]])/((a^2 + b^2)*d)
 
3.6.92.3 Rubi [A] (warning: unable to verify)

Time = 1.09 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.81, number of steps used = 19, number of rules used = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.545, Rules used = {3042, 4064, 3042, 4096, 3042, 4017, 25, 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.

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))}dx\)

\(\Big \downarrow \) 4064

\(\displaystyle \int \frac {A \cot (c+d x)+B}{\sqrt {\cot (c+d x)} (a \cot (c+d x)+b)}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4096

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4017

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1482

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

\(\Big \downarrow \) 1476

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1479

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 1103

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

\(\Big \downarrow \) 4117

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

\(\Big \downarrow \) 73

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

\(\Big \downarrow \) 218

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

input
Int[(A + B*Tan[c + d*x])/(Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])),x]
 
output
(-2*Sqrt[a]*(A*b - a*B)*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[b]])/(Sqrt[b]*( 
a^2 + b^2)*d) + (2*(-1/2*((a*(A - B) + b*(A + B))*(-(ArcTan[1 - Sqrt[2]*Sq 
rt[Cot[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2 
])) + ((b*(A - B) - a*(A + B))*(-1/2*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + 
Cot[c + d*x]]/Sqrt[2] + Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] 
/(2*Sqrt[2])))/2))/((a^2 + b^2)*d)
 

3.6.92.3.1 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 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 4064
Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*( 
x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp 
[g^(m + n)   Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d + c 
*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !Integer 
Q[p] && IntegerQ[m] && IntegerQ[n]
 

rule 4096
Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_ 
.)*(x_)])^(n_))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
1/(a^2 + b^2)   Int[(c + d*Tan[e + f*x])^n*Simp[a*A + b*B - (A*b - a*B)*Tan 
[e + f*x], x], x], x] + Simp[b*((A*b - a*B)/(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, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && N 
eQ[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]
 
3.6.92.4 Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 244, normalized size of antiderivative = 0.88

method result size
derivativedivides \(\frac {-\frac {2 \left (\frac {\left (A a +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{8}+\frac {\left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{8}\right )}{a^{2}+b^{2}}+\frac {2 \left (A b -B a \right ) a \arctan \left (\frac {a \sqrt {\cot \left (d x +c \right )}}{\sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right ) \sqrt {a b}}}{d}\) \(244\)
default \(\frac {-\frac {2 \left (\frac {\left (A a +B b \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{8}+\frac {\left (A b -B a \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\cot \left (d x +c \right )-\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}{1+\cot \left (d x +c \right )+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}}\right )+2 \arctan \left (1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )+2 \arctan \left (-1+\sqrt {2}\, \sqrt {\cot \left (d x +c \right )}\right )\right )}{8}\right )}{a^{2}+b^{2}}+\frac {2 \left (A b -B a \right ) a \arctan \left (\frac {a \sqrt {\cot \left (d x +c \right )}}{\sqrt {a b}}\right )}{\left (a^{2}+b^{2}\right ) \sqrt {a b}}}{d}\) \(244\)

input
int((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c)),x,method=_RETURNVER 
BOSE)
 
output
1/d*(-2/(a^2+b^2)*(1/8*(A*a+B*b)*2^(1/2)*(ln((1+cot(d*x+c)+2^(1/2)*cot(d*x 
+c)^(1/2))/(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2)))+2*arctan(1+2^(1/2)*cot 
(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2)))+1/8*(A*b-B*a)*2^(1/2 
)*(ln((1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(1+cot(d*x+c)+2^(1/2)*cot(d* 
x+c)^(1/2)))+2*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*cot( 
d*x+c)^(1/2))))+2*(A*b-B*a)*a/(a^2+b^2)/(a*b)^(1/2)*arctan(a*cot(d*x+c)^(1 
/2)/(a*b)^(1/2)))
 
3.6.92.5 Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 2983 vs. \(2 (240) = 480\).

Time = 3.09 (sec) , antiderivative size = 5992, normalized size of antiderivative = 21.55 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=\text {Too large to display} \]

input
integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c)),x, algorithm= 
"fricas")
 
output
Too large to include
 
3.6.92.6 Sympy [F]

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

input
integrate((A+B*tan(d*x+c))/cot(d*x+c)**(1/2)/(a+b*tan(d*x+c)),x)
 
output
Integral((A + B*tan(c + d*x))/((a + b*tan(c + d*x))*sqrt(cot(c + d*x))), x 
)
 
3.6.92.7 Maxima [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.79 \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=-\frac {\frac {8 \, {\left (B a^{2} - A a b\right )} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{2} + b^{2}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{2} + b^{2}}}{4 \, d} \]

input
integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c)),x, algorithm= 
"maxima")
 
output
-1/4*(8*(B*a^2 - A*a*b)*arctan(a/(sqrt(a*b)*sqrt(tan(d*x + c))))/((a^2 + b 
^2)*sqrt(a*b)) + (2*sqrt(2)*((A - B)*a + (A + B)*b)*arctan(1/2*sqrt(2)*(sq 
rt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*((A - B)*a + (A + B)*b)*arctan( 
-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + sqrt(2)*((A + B)*a - (A - 
 B)*b)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - sqrt(2)*((A 
+ B)*a - (A - B)*b)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1)) 
/(a^2 + b^2))/d
 
3.6.92.8 Giac [F]

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

input
integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c)),x, algorithm= 
"giac")
 
output
integrate((B*tan(d*x + c) + A)/((b*tan(d*x + c) + a)*sqrt(cot(d*x + c))), 
x)
 
3.6.92.9 Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\sqrt {\cot (c+d x)} (a+b \tan (c+d x))} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )} \,d x \]

input
int((A + B*tan(c + d*x))/(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))),x)
 
output
int((A + B*tan(c + d*x))/(cot(c + d*x)^(1/2)*(a + b*tan(c + d*x))), x)