Integrand size = 30, antiderivative size = 211 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )}{4 a^2 f}-\frac {\left (2 c d-i \left (2 c^2+d^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right )}{8 a^2 (c+i d)^{3/2} f}+\frac {(2 i c-d) \sqrt {c+d \tan (e+f x)}}{8 a^2 (c+i d) f (1+i \tan (e+f x))}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2} \]
-1/8*(2*c*d-I*(2*c^2+d^2))*arctanh((c+d*tan(f*x+e))^(1/2)/(c+I*d)^(1/2))/a ^2/(c+I*d)^(3/2)/f-1/4*I*arctanh((c+d*tan(f*x+e))^(1/2)/(c-I*d)^(1/2))*(c- I*d)^(1/2)/a^2/f+1/8*(2*I*c-d)*(c+d*tan(f*x+e))^(1/2)/a^2/(c+I*d)/f/(1+I*t an(f*x+e))+1/4*I*(c+d*tan(f*x+e))^(1/2)/f/(a+I*a*tan(f*x+e))^2
Time = 2.59 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.95 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i \left (2 \sqrt {c-i d} \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d}}\right )+\frac {\left (2 c^2+2 i c d+d^2\right ) \text {arctanh}\left (\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {c+i d}}\right ) \sec ^2(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x)))+\sqrt {c+i d} (4 c+3 i d+(2 i c-d) \tan (e+f x)) \sqrt {c+d \tan (e+f x)}}{(c+i d)^{3/2} (-i+\tan (e+f x))^2}\right )}{8 a^2 f} \]
((-1/8*I)*(2*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]] + ((2*c^2 + (2*I)*c*d + d^2)*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I* d]]*Sec[e + f*x]^2*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)]) + Sqrt[c + I*d] *(4*c + (3*I)*d + ((2*I)*c - d)*Tan[e + f*x])*Sqrt[c + d*Tan[e + f*x]])/(( c + I*d)^(3/2)*(-I + Tan[e + f*x])^2)))/(a^2*f)
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.
| \(\displaystyle \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2}dx\) |
| \(\Big \downarrow \) 4040 |
| \(\displaystyle \frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}-\frac {\int -\frac {a (4 c-i d)+3 a d \tan (e+f x)}{(i \tan (e+f x) a+a) \sqrt {c+d \tan (e+f x)}}dx}{8 a^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {a (4 c-i d)+3 a d \tan (e+f x)}{(i \tan (e+f x) a+a) \sqrt {c+d \tan (e+f x)}}dx}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {a (4 c-i d)+3 a d \tan (e+f x)}{(i \tan (e+f x) a+a) \sqrt {c+d \tan (e+f x)}}dx}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}-\frac {\int -\frac {\left (4 i c^2-2 d c+3 i d^2\right ) a^2+(2 i c-d) d \tan (e+f x) a^2}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int -\frac {a^2 \left (2 c d-i \left (4 c^2+3 d^2\right )\right )-a^2 (2 i c-d) d \tan (e+f x)}{\sqrt {c+d \tan (e+f x)}}dx}{2 a^2 (-d+i c)}+\frac {(-d+2 i c) \sqrt {c+d \tan (e+f x)}}{f (c+i d) (1+i \tan (e+f x))}}{8 a^2}+\frac {i \sqrt {c+d \tan (e+f x)}}{4 f (a+i a \tan (e+f x))^2}\) |
3.12.5.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*Sqrt[(c_.) + (d_.)*tan[(e_. ) + (f_.)*(x_)]], x_Symbol] :> Simp[(-b)*(a + b*Tan[e + f*x])^m*(Sqrt[c + d *Tan[e + f*x]]/(2*a*f*m)), x] + Simp[1/(4*a^2*m) Int[(a + b*Tan[e + f*x]) ^(m + 1)*(Simp[2*a*c*m + b*d + a*d*(2*m + 1)*Tan[e + f*x], x]/Sqrt[c + d*Ta n[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] & & EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && IntegersQ[2*m]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*( b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d)) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Free Q[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n, 0]
Time = 0.69 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.27
| method | result | size |
| derivativedivides | \(\frac {2 d^{3} \left (\frac {\frac {\frac {d \left (2 i c^{2}-i d^{2}-3 c d \right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2 i c^{2}-2 i d^{2}-4 c d}-\frac {i d \left (7 i c^{2} d -3 i d^{3}+2 c^{3}-8 c \,d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (i c^{2}-i d^{2}-2 c d \right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}-\frac {i \left (2 i c^{3}-i c \,d^{2}-4 c^{2} d -d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (i c^{2}-i d^{2}-2 c d \right ) \sqrt {-i d -c}}}{8 d^{3}}-\frac {i \sqrt {i d -c}\, \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 d^{3}}\right )}{f \,a^{2}}\) | \(269\) |
| default | \(\frac {2 d^{3} \left (\frac {\frac {\frac {d \left (2 i c^{2}-i d^{2}-3 c d \right ) \left (c +d \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{2 i c^{2}-2 i d^{2}-4 c d}-\frac {i d \left (7 i c^{2} d -3 i d^{3}+2 c^{3}-8 c \,d^{2}\right ) \sqrt {c +d \tan \left (f x +e \right )}}{2 \left (i c^{2}-i d^{2}-2 c d \right )}}{\left (-d \tan \left (f x +e \right )+i d \right )^{2}}-\frac {i \left (2 i c^{3}-i c \,d^{2}-4 c^{2} d -d^{3}\right ) \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {-i d -c}}\right )}{2 \left (i c^{2}-i d^{2}-2 c d \right ) \sqrt {-i d -c}}}{8 d^{3}}-\frac {i \sqrt {i d -c}\, \arctan \left (\frac {\sqrt {c +d \tan \left (f x +e \right )}}{\sqrt {i d -c}}\right )}{8 d^{3}}\right )}{f \,a^{2}}\) | \(269\) |
2/f/a^2*d^3*(1/8/d^3*((1/2*d*(2*I*c^2-I*d^2-3*c*d)/(I*c^2-I*d^2-2*c*d)*(c+ d*tan(f*x+e))^(3/2)-1/2*I*d*(7*I*c^2*d-3*I*d^3+2*c^3-8*c*d^2)/(I*c^2-I*d^2 -2*c*d)*(c+d*tan(f*x+e))^(1/2))/(-d*tan(f*x+e)+I*d)^2-1/2*I*(-d^3+2*I*c^3- I*c*d^2-4*c^2*d)/(I*c^2-I*d^2-2*c*d)/(-I*d-c)^(1/2)*arctan((c+d*tan(f*x+e) )^(1/2)/(-I*d-c)^(1/2)))-1/8*I*(I*d-c)^(1/2)/d^3*arctan((c+d*tan(f*x+e))^( 1/2)/(I*d-c)^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1086 vs. \(2 (162) = 324\).
Time = 0.37 (sec) , antiderivative size = 1086, normalized size of antiderivative = 5.15 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\text {Too large to display} \]
1/32*(2*(-I*a^2*c + a^2*d)*f*sqrt(-(c - I*d)/(a^4*f^2))*e^(4*I*f*x + 4*I*e )*log(-2*((I*a^2*f*e^(2*I*f*x + 2*I*e) + I*a^2*f)*sqrt(((c - I*d)*e^(2*I*f *x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c - I*d)/(a^4*f^2 )) - (c - I*d)*e^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + 2*(I*a^2*c - a^2*d)*f*sqrt(-(c - I*d)/(a^4*f^2))*e^(4*I*f*x + 4*I*e)*log(-2*((-I*a^2 *f*e^(2*I*f*x + 2*I*e) - I*a^2*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(c - I*d)/(a^4*f^2)) - (c - I*d)*e ^(2*I*f*x + 2*I*e) - c)*e^(-2*I*f*x - 2*I*e)) + (-I*a^2*c + a^2*d)*f*sqrt( -(-4*I*c^4 + 8*c^3*d + 4*c*d^3 - I*d^4)/((-I*a^4*c^3 + 3*a^4*c^2*d + 3*I*a ^4*c*d^2 - a^4*d^3)*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/8*(-2*I*c^3 + 4*c^2*d + I*c*d^2 + d^3 + ((a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*e^(2*I*f*x + 2*I*e ) + (a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I* e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(-4*I*c^4 + 8*c^3*d + 4*c*d ^3 - I*d^4)/((-I*a^4*c^3 + 3*a^4*c^2*d + 3*I*a^4*c*d^2 - a^4*d^3)*f^2)) + (-2*I*c^3 + 2*c^2*d - I*c*d^2)*e^(2*I*f*x + 2*I*e))*e^(-2*I*f*x - 2*I*e)/( (a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f)) + (I*a^2*c - a^2*d)*f*sqrt(-(-4*I*c^ 4 + 8*c^3*d + 4*c*d^3 - I*d^4)/((-I*a^4*c^3 + 3*a^4*c^2*d + 3*I*a^4*c*d^2 - a^4*d^3)*f^2))*e^(4*I*f*x + 4*I*e)*log(-1/8*(-2*I*c^3 + 4*c^2*d + I*c*d^ 2 + d^3 - ((a^2*c^2 + 2*I*a^2*c*d - a^2*d^2)*f*e^(2*I*f*x + 2*I*e) + (a^2* c^2 + 2*I*a^2*c*d - a^2*d^2)*f)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c...
\[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=- \frac {\int \frac {\sqrt {c + d \tan {\left (e + f x \right )}}}{\tan ^{2}{\left (e + f x \right )} - 2 i \tan {\left (e + f x \right )} - 1}\, dx}{a^{2}} \]
Exception generated. \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 476 vs. \(2 (162) = 324\).
Time = 0.54 (sec) , antiderivative size = 476, normalized size of antiderivative = 2.26 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\frac {1}{8} \, d^{3} {\left (\frac {16 \, {\left (2 \, c^{2} + 2 i \, c d + d^{2}\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} + i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{-8 \, {\left (-i \, a^{2} c d^{3} f + a^{2} d^{4} f\right )} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} {\left (\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}} + \frac {2 \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} c - 2 \, \sqrt {d \tan \left (f x + e\right ) + c} c^{2} + i \, {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} d - 5 i \, \sqrt {d \tan \left (f x + e\right ) + c} c d + 3 \, \sqrt {d \tan \left (f x + e\right ) + c} d^{2}}{{\left (a^{2} c d^{2} f + i \, a^{2} d^{3} f\right )} {\left (d \tan \left (f x + e\right ) - i \, d\right )}^{2}} + \frac {4 \, {\left (i \, c + d\right )} \arctan \left (\frac {2 \, {\left (\sqrt {d \tan \left (f x + e\right ) + c} c - \sqrt {c^{2} + d^{2}} \sqrt {d \tan \left (f x + e\right ) + c}\right )}}{c \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} - i \, \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d - \sqrt {c^{2} + d^{2}} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}}}\right )}{a^{2} \sqrt {-2 \, c + 2 \, \sqrt {c^{2} + d^{2}}} d^{3} f {\left (-\frac {i \, d}{c - \sqrt {c^{2} + d^{2}}} + 1\right )}}\right )} \]
1/8*d^3*(16*(2*c^2 + 2*I*c*d + d^2)*arctan(2*(sqrt(d*tan(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2*c + 2*sqrt(c^2 + d^2 )) + I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d^2)*sqrt(-2*c + 2*sq rt(c^2 + d^2))))/((8*I*a^2*c*d^3*f - 8*a^2*d^4*f)*sqrt(-2*c + 2*sqrt(c^2 + d^2))*(I*d/(c - sqrt(c^2 + d^2)) + 1)) + (2*(d*tan(f*x + e) + c)^(3/2)*c - 2*sqrt(d*tan(f*x + e) + c)*c^2 + I*(d*tan(f*x + e) + c)^(3/2)*d - 5*I*sq rt(d*tan(f*x + e) + c)*c*d + 3*sqrt(d*tan(f*x + e) + c)*d^2)/((a^2*c*d^2*f + I*a^2*d^3*f)*(d*tan(f*x + e) - I*d)^2) + 4*(I*c + d)*arctan(2*(sqrt(d*t an(f*x + e) + c)*c - sqrt(c^2 + d^2)*sqrt(d*tan(f*x + e) + c))/(c*sqrt(-2* c + 2*sqrt(c^2 + d^2)) - I*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d - sqrt(c^2 + d ^2)*sqrt(-2*c + 2*sqrt(c^2 + d^2))))/(a^2*sqrt(-2*c + 2*sqrt(c^2 + d^2))*d ^3*f*(-I*d/(c - sqrt(c^2 + d^2)) + 1)))
Time = 8.86 (sec) , antiderivative size = 14675, normalized size of antiderivative = 69.55 \[ \int \frac {\sqrt {c+d \tan (e+f x)}}{(a+i a \tan (e+f x))^2} \, dx=\text {Too large to display} \]
log(((-(3*d^9 - c*d^8*9i + 12*c^2*d^7 - c^3*d^6*16i + 8*c^4*d^5 - c^5*d^4* 8i - a^4*c^2*f^2*((((3*d^11 + 27*c^2*d^9 + 28*c^4*d^7 + 8*c^6*d^5)*1i)/(a^ 4*c^4*f^2 + a^4*d^4*f^2 + 2*a^4*c^2*d^2*f^2) - (17*c^3*d^8 - 3*c*d^10 + 24 *c^5*d^6 + 8*c^7*d^4)/(a^4*c^4*f^2 + a^4*d^4*f^2 + 2*a^4*c^2*d^2*f^2))^2 + 4*(256*d^6 + 256*c^2*d^4)*((((5*c^3*d^9)/16 - (c*d^11)/8 + (7*c^5*d^7)/16 + (c^7*d^5)/8)*1i)/(a^8*c^4*f^4 + a^8*d^4*f^4 + 2*a^8*c^2*d^2*f^4) - (d^1 2/64 - (11*c^2*d^10)/32 - (11*c^4*d^8)/64 + (c^6*d^6)/8 + (c^8*d^4)/16)/(a ^8*c^4*f^4 + a^8*d^4*f^4 + 2*a^8*c^2*d^2*f^4)))^(1/2)*1i + a^4*d^2*f^2*((( (3*d^11 + 27*c^2*d^9 + 28*c^4*d^7 + 8*c^6*d^5)*1i)/(a^4*c^4*f^2 + a^4*d^4* f^2 + 2*a^4*c^2*d^2*f^2) - (17*c^3*d^8 - 3*c*d^10 + 24*c^5*d^6 + 8*c^7*d^4 )/(a^4*c^4*f^2 + a^4*d^4*f^2 + 2*a^4*c^2*d^2*f^2))^2 + 4*(256*d^6 + 256*c^ 2*d^4)*((((5*c^3*d^9)/16 - (c*d^11)/8 + (7*c^5*d^7)/16 + (c^7*d^5)/8)*1i)/ (a^8*c^4*f^4 + a^8*d^4*f^4 + 2*a^8*c^2*d^2*f^4) - (d^12/64 - (11*c^2*d^10) /32 - (11*c^4*d^8)/64 + (c^6*d^6)/8 + (c^8*d^4)/16)/(a^8*c^4*f^4 + a^8*d^4 *f^4 + 2*a^8*c^2*d^2*f^4)))^(1/2)*1i + 2*a^4*c*d*f^2*((((3*d^11 + 27*c^2*d ^9 + 28*c^4*d^7 + 8*c^6*d^5)*1i)/(a^4*c^4*f^2 + a^4*d^4*f^2 + 2*a^4*c^2*d^ 2*f^2) - (17*c^3*d^8 - 3*c*d^10 + 24*c^5*d^6 + 8*c^7*d^4)/(a^4*c^4*f^2 + a ^4*d^4*f^2 + 2*a^4*c^2*d^2*f^2))^2 + 4*(256*d^6 + 256*c^2*d^4)*((((5*c^3*d ^9)/16 - (c*d^11)/8 + (7*c^5*d^7)/16 + (c^7*d^5)/8)*1i)/(a^8*c^4*f^4 + a^8 *d^4*f^4 + 2*a^8*c^2*d^2*f^4) - (d^12/64 - (11*c^2*d^10)/32 - (11*c^4*d...