Integrand size = 33, antiderivative size = 278 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {(b (A-B)-a (A+B)) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}+\frac {(b (A-B)-a (A+B)) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right ) d}-\frac {2 \sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{\sqrt {a} \left (a^2+b^2\right ) d}-\frac {(a (A-B)+b (A+B)) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d}+\frac {(a (A-B)+b (A+B)) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right ) d} \]
1/2*(b*(A-B)-a*(A+B))*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2^(1 /2)+1/2*(b*(A-B)-a*(A+B))*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)/d*2 ^(1/2)-1/4*(a*(A-B)+b*(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*(a*(A-B)+b*(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))*b^(1/2)/(a^2+b^2)/d/a^(1/2)
Time = 0.43 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\sqrt {\cot (c+d x)} \left (-2 \sqrt {2} (b (-A+B)+a (A+B)) \left (\arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )+\frac {8 \sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a}}-\sqrt {2} (a (A-B)+b (A+B)) \left (\log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )\right ) \sqrt {\tan (c+d x)}}{4 \left (a^2+b^2\right ) d} \]
(Sqrt[Cot[c + d*x]]*(-2*Sqrt[2]*(b*(-A + B) + a*(A + B))*(ArcTan[1 - Sqrt[ 2]*Sqrt[Tan[c + d*x]]] - ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) + (8*Sqrt [b]*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/Sqrt[a] - Sq rt[2]*(a*(A - B) + b*(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]]))*Sqrt[Tan[c + d*x]])/(4*(a^2 + b^2)*d)
Time = 1.06 (sec) , antiderivative size = 224, normalized size of antiderivative = 0.81, number of steps used = 20, number of rules used = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.576, Rules used = {3042, 4064, 3042, 4095, 25, 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.
| \(\displaystyle \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)}dx\) |
| \(\Big \downarrow \) 4064 |
| \(\displaystyle \int \frac {\sqrt {\cot (c+d x)} (A \cot (c+d x)+B)}{a \cot (c+d x)+b}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (B-A \tan \left (c+d x+\frac {\pi }{2}\right )\right )}{b-a \tan \left (c+d x+\frac {\pi }{2}\right )}dx\) |
| \(\Big \downarrow \) 4095 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\cot ^2(c+d x)+1}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))}dx}{a^2+b^2}+\frac {\int -\frac {A b-a B-(a A+b B) \cot (c+d x)}{\sqrt {\cot (c+d x)}}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\cot ^2(c+d x)+1}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))}dx}{a^2+b^2}-\frac {\int \frac {A b-a B-(a A+b B) \cot (c+d x)}{\sqrt {\cot (c+d x)}}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}-\frac {\int \frac {A b-a B-(-a A-b B) \tan \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )}}dx}{a^2+b^2}\) |
| \(\Big \downarrow \) 4017 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}-\frac {2 \int -\frac {A b-a B-(a A+b B) \cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}+\frac {2 \int \frac {A b-a B-(a A+b B) \cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 1482 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}-\frac {2 \left (-\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \int \frac {\cot (c+d x)+1}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 1476 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}-\frac {2 \left (-\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {1}{2} \int \frac {1}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}+\frac {1}{2} \int \frac {1}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}-\frac {2 \left (-\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\int \frac {1}{-\cot (c+d x)-1}d\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}-\frac {\int \frac {1}{-\cot (c+d x)-1}d\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}-\frac {2 \left (-\frac {1}{2} (a (A-B)+b (A+B)) \int \frac {1-\cot (c+d x)}{\cot ^2(c+d x)+1}d\sqrt {\cot (c+d x)}-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 1479 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}-\frac {2 \left (-\frac {1}{2} (a (A-B)+b (A+B)) \left (-\frac {\int -\frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}-\frac {\int -\frac {\sqrt {2} \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}-\frac {2 \left (-\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}+\frac {\int \frac {\sqrt {2} \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}-\frac {2 \left (-\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\int \frac {\sqrt {2}-2 \sqrt {\cot (c+d x)}}{\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}}{2 \sqrt {2}}+\frac {1}{2} \int \frac {\sqrt {2} \sqrt {\cot (c+d x)}+1}{\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1}d\sqrt {\cot (c+d x)}\right )-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {\tan \left (c+d x+\frac {\pi }{2}\right )^2+1}{\sqrt {-\tan \left (c+d x+\frac {\pi }{2}\right )} \left (b-a \tan \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a^2+b^2}-\frac {2 \left (-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 4117 |
| \(\displaystyle \frac {b (A b-a B) \int \frac {1}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))}d(-\cot (c+d x))}{d \left (a^2+b^2\right )}-\frac {2 \left (-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle -\frac {2 b (A b-a B) \int \frac {1}{a \cot ^2(c+d x)+b}d\sqrt {\cot (c+d x)}}{d \left (a^2+b^2\right )}-\frac {2 \left (-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {2 \sqrt {b} (A b-a B) \arctan \left (\frac {\sqrt {a} \cot (c+d x)}{\sqrt {b}}\right )}{\sqrt {a} d \left (a^2+b^2\right )}-\frac {2 \left (-\frac {1}{2} (b (A-B)-a (A+B)) \left (\frac {\arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2}}-\frac {\arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2}}\right )-\frac {1}{2} (a (A-B)+b (A+B)) \left (\frac {\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}-\frac {\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2}}\right )\right )}{d \left (a^2+b^2\right )}\) |
(2*Sqrt[b]*(A*b - a*B)*ArcTan[(Sqrt[a]*Cot[c + d*x])/Sqrt[b]])/(Sqrt[a]*(a ^2 + b^2)*d) - (2*(-1/2*((b*(A - B) - a*(A + B))*(-(ArcTan[1 - Sqrt[2]*Sqr t[Cot[c + d*x]]]/Sqrt[2]) + ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]/Sqrt[2] )) - ((a*(A - B) + b*(A + B))*(-1/2*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + C ot[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)
3.6.91.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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])
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]
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]
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]
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]
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]
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]
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]
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]
Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[1 /(a^2 + b^2) Int[Simp[A*(a*c + b*d) + B*(b*c - a*d) - (A*(b*c - a*d) - B* (a*c + b*d))*Tan[e + f*x], x]/Sqrt[c + d*Tan[e + f*x]], x], x] - Simp[(b*c - a*d)*((B*a - A*b)/(a^2 + b^2)) Int[(1 + Tan[e + f*x]^2)/((a + b*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(530\) vs. \(2(240)=480\).
Time = 0.40 (sec) , antiderivative size = 531, normalized size of antiderivative = 1.91
| method | result | size |
| derivativedivides | \(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \sqrt {\tan \left (d x +c \right )}\, \left (A \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a -2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a -2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b -A \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +B \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +B \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +8 A \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right ) b^{2}-8 B \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right ) a b \right )}{4 d \left (a^{2}+b^{2}\right ) \sqrt {a b}}\) | \(531\) |
| default | \(\frac {\sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \sqrt {\tan \left (d x +c \right )}\, \left (A \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a -2 A \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a -2 A \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b -A \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +B \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 B \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +2 B \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, b +B \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right ) \sqrt {2}\, \sqrt {a b}\, a +8 A \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right ) b^{2}-8 B \arctan \left (\frac {b \sqrt {\tan \left (d x +c \right )}}{\sqrt {a b}}\right ) a b \right )}{4 d \left (a^{2}+b^{2}\right ) \sqrt {a b}}\) | \(531\) |
1/4/d*(1/tan(d*x+c))^(1/2)*tan(d*x+c)^(1/2)*(A*ln(-(1+2^(1/2)*tan(d*x+c)^( 1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*2^(1/2)*(a*b)^(1 /2)*a+2*A*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a-2*A*arc tan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*b+2*A*arctan(-1+2^(1/2 )*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a-2*A*arctan(-1+2^(1/2)*tan(d*x+c) ^(1/2))*2^(1/2)*(a*b)^(1/2)*b-A*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1 )/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*2^(1/2)*(a*b)^(1/2)*b+B*ln(-(1+ 2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)- 1))*2^(1/2)*(a*b)^(1/2)*b+2*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*( a*b)^(1/2)*a+2*B*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*b+ 2*B*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*a+2*B*arctan(- 1+2^(1/2)*tan(d*x+c)^(1/2))*2^(1/2)*(a*b)^(1/2)*b+B*ln(-(2^(1/2)*tan(d*x+c )^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*2^(1/2)*(a* b)^(1/2)*a+8*A*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*b^2-8*B*arctan(b*tan (d*x+c)^(1/2)/(a*b)^(1/2))*a*b)/(a^2+b^2)/(a*b)^(1/2)
Leaf count of result is larger than twice the leaf count of optimal. 2995 vs. \(2 (240) = 480\).
Time = 4.33 (sec) , antiderivative size = 6020, normalized size of antiderivative = 21.65 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\text {Too large to display} \]
\[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\cot {\left (c + d x \right )}}}{a + b \tan {\left (c + d x \right )}}\, dx \]
Time = 0.32 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\frac {8 \, {\left (B a b - A b^{2}\right )} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{2} + b^{2}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A + B\right )} a - {\left (A - B\right )} b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left ({\left (A - B\right )} a + {\left (A + B\right )} b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{2} + b^{2}}}{4 \, d} \]
1/4*(8*(B*a*b - A*b^2)*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)*(sqr t(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
\[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {\cot \left (d x + c\right )}}{b \tan \left (d x + c\right ) + a} \,d x } \]
Timed out. \[ \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{a+b\,\mathrm {tan}\left (c+d\,x\right )} \,d x \]