Integrand size = 36, antiderivative size = 259 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(5 i A-2 B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {(19 A+9 i B) \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {(41 A+15 i B) \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {7 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d} \]
(5*I*A-2*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d+1/8*(I*A+B )*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(5/2)/d*2^(1/2)+ 1/12*(41*A+15*I*B)*cot(d*x+c)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-7/4*(3*A+I*B) *cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^3/d+1/5*(A+I*B)*cot(d*x+c)/d/(a+I*a *tan(d*x+c))^(5/2)+1/30*(19*A+9*I*B)*cot(d*x+c)/a/d/(a+I*a*tan(d*x+c))^(3/ 2)
Time = 6.37 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.04 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {12 a^{15/2} (A+i B) \cot (c+d x)-\frac {1}{2} a^7 (i+\cot (c+d x)) \tan ^2(c+d x) \left (2 \sqrt {a} \left (-105 (3 A+i B)+5 i (85 A+27 i B) \cot (c+d x)+12 (6 A+i B) \cot ^2(c+d x)\right )+120 (5 A+2 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right ) (1-i \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}+15 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) (1-i \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}\right )}{60 a^{15/2} d (a+i a \tan (c+d x))^{5/2}} \]
(12*a^(15/2)*(A + I*B)*Cot[c + d*x] - (a^7*(I + Cot[c + d*x])*Tan[c + d*x] ^2*(2*Sqrt[a]*(-105*(3*A + I*B) + (5*I)*(85*A + (27*I)*B)*Cot[c + d*x] + 1 2*(6*A + I*B)*Cot[c + d*x]^2) + 120*(5*A + (2*I)*B)*ArcTanh[Sqrt[a + I*a*T an[c + d*x]]/Sqrt[a]]*(1 - I*Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]] + 15 *Sqrt[2]*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]*( 1 - I*Cot[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]]))/2)/(60*a^(15/2)*d*(a + I* a*Tan[c + d*x])^(5/2))
Time = 1.92 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.09, number of steps used = 21, number of rules used = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 4079, 27, 3042, 4079, 27, 3042, 4079, 27, 3042, 4081, 25, 3042, 4083, 3042, 3961, 219, 4082, 73, 221}
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.
| \(\displaystyle \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (c+d x)}{\tan (c+d x)^2 (a+i a \tan (c+d x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\int \frac {\cot ^2(c+d x) (2 a (6 A+i B)-7 a (i A-B) \tan (c+d x))}{2 (i \tan (c+d x) a+a)^{3/2}}dx}{5 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\cot ^2(c+d x) (2 a (6 A+i B)-7 a (i A-B) \tan (c+d x))}{(i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 a (6 A+i B)-7 a (i A-B) \tan (c+d x)}{\tan (c+d x)^2 (i \tan (c+d x) a+a)^{3/2}}dx}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {\int \frac {5 \cot ^2(c+d x) \left (2 a^2 (11 A+3 i B)-a^2 (19 i A-9 B) \tan (c+d x)\right )}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {5 \int \frac {\cot ^2(c+d x) \left (2 a^2 (11 A+3 i B)-a^2 (19 i A-9 B) \tan (c+d x)\right )}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {5 \int \frac {2 a^2 (11 A+3 i B)-a^2 (19 i A-9 B) \tan (c+d x)}{\tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4079 |
| \(\displaystyle \frac {\frac {5 \left (\frac {\int \frac {3}{2} \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (14 a^3 (3 A+i B)-a^3 (41 i A-15 B) \tan (c+d x)\right )dx}{a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \int \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (14 a^3 (3 A+i B)-a^3 (41 i A-15 B) \tan (c+d x)\right )dx}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (14 a^3 (3 A+i B)-a^3 (41 i A-15 B) \tan (c+d x)\right )}{\tan (c+d x)^2}dx}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4081 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (\frac {\int -\cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 (5 i A-2 B) a^4+7 (3 A+i B) \tan (c+d x) a^4\right )dx}{a}-\frac {14 a^3 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (4 (5 i A-2 B) a^4+7 (3 A+i B) \tan (c+d x) a^4\right )dx}{a}-\frac {14 a^3 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (4 (5 i A-2 B) a^4+7 (3 A+i B) \tan (c+d x) a^4\right )}{\tan (c+d x)}dx}{a}-\frac {14 a^3 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4083 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {a^4 (A-i B) \int \sqrt {i \tan (c+d x) a+a}dx+4 a^3 (-2 B+5 i A) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{a}-\frac {14 a^3 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {a^4 (A-i B) \int \sqrt {i \tan (c+d x) a+a}dx+4 a^3 (-2 B+5 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{a}-\frac {14 a^3 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 3961 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {4 a^3 (-2 B+5 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {2 i a^5 (A-i B) \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}}{a}-\frac {14 a^3 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {4 a^3 (-2 B+5 i A) \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {i \sqrt {2} a^{9/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^3 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {\frac {4 a^5 (-2 B+5 i A) \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {i \sqrt {2} a^{9/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^3 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\frac {5 \left (\frac {3 \left (-\frac {-\frac {8 i a^4 (-2 B+5 i A) \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {i \sqrt {2} a^{9/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^3 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {5 \left (\frac {a^2 (41 A+15 i B) \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {3 \left (-\frac {-\frac {8 a^{9/2} (-2 B+5 i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {i \sqrt {2} a^{9/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^3 (3 A+i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}\right )}{6 a^2}+\frac {a (19 A+9 i B) \cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}}{10 a^2}+\frac {(A+i B) \cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}\) |
((A + I*B)*Cot[c + d*x])/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((a*(19*A + (9*I)*B)*Cot[c + d*x])/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (5*((a^2*(41*A + (15*I)*B)*Cot[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (3*(-(((-8*a^( 9/2)*((5*I)*A - 2*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d - (I*S qrt[2]*a^(9/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[ a])])/d)/a) - (14*a^3*(3*A + I*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]) /d))/(2*a^2)))/(6*a^2))/(10*a^2)
3.2.10.3.1 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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^2, 0]
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]
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*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
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[b*(B/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[( A*b + a*B)/(b*c + a*d) Int[(a + b*Tan[e + f*x])^m, x], x] - Simp[(B*c - A *d)/(b*c + a*d) Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*T an[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Time = 0.30 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.75
| method | result | size |
| derivativedivides | \(\frac {2 i a^{2} \left (-\frac {7 i B +17 A}{8 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {3 i B +5 A}{12 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B +A}{10 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {9}{2}}}+\frac {\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {\left (2 i B +5 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{4}}\right )}{d}\) | \(195\) |
| default | \(\frac {2 i a^{2} \left (-\frac {7 i B +17 A}{8 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {3 i B +5 A}{12 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {i B +A}{10 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\left (i B -A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {9}{2}}}+\frac {\frac {i A \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {\left (2 i B +5 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{4}}\right )}{d}\) | \(195\) |
2*I/d*a^2*(-1/8/a^4*(17*A+7*I*B)/(a+I*a*tan(d*x+c))^(1/2)-1/12/a^3*(5*A+3* I*B)/(a+I*a*tan(d*x+c))^(3/2)-1/10/a^2*(A+I*B)/(a+I*a*tan(d*x+c))^(5/2)-1/ 16*(-A+I*B)/a^(9/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a ^(1/2))+1/a^4*(1/2*I*A*(a+I*a*tan(d*x+c))^(1/2)/a/tan(d*x+c)+1/2*(5*A+2*I* B)/a^(1/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 834 vs. \(2 (204) = 408\).
Time = 0.29 (sec) , antiderivative size = 834, normalized size of antiderivative = 3.22 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
-1/120*(15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c ))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a^3*d *e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2)) + (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I *c)/(I*A + B)) - 15*sqrt(1/2)*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d* x + 5*I*c))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))*log(4*(sqrt(2)*sqrt(1/2 )*(a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sq rt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2)) - (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I *d*x - I*c)/(I*A + B)) - 30*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(-(25*A^2 + 20*I*A*B - 4*B^2)/(a^5*d^2))*log(-16*(3*(5*I*A - 2*B)*a^2*e^(2*I*d*x + 2*I*c) + (5*I*A - 2*B)*a^2 + 2*sqrt(2)*(a^4*d*e^(3* I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))* sqrt(-(25*A^2 + 20*I*A*B - 4*B^2)/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/(-5*I*A + 2*B)) + 30*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt (-(25*A^2 + 20*I*A*B - 4*B^2)/(a^5*d^2))*log(-16*(3*(5*I*A - 2*B)*a^2*e^(2 *I*d*x + 2*I*c) + (5*I*A - 2*B)*a^2 - 2*sqrt(2)*(a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(25*A^2 + 20*I*A*B - 4*B^2)/(a^5*d^2)))*e^(-2*I*d*x - 2*I*c)/(-5*I*A + 2*B)) - sqr t(2)*((-403*I*A + 123*B)*e^(8*I*d*x + 8*I*c) + (-151*I*A + 21*B)*e^(6*I*d* x + 6*I*c) - 40*(-7*I*A + 3*B)*e^(4*I*d*x + 4*I*c) + (31*I*A - 21*B)*e^...
\[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \cot ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.31 (sec) , antiderivative size = 240, normalized size of antiderivative = 0.93 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \, a {\left (\frac {4 \, {\left (105 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} {\left (3 \, A + i \, B\right )} - 5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (41 \, A + 15 i \, B\right )} a - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (19 \, A + 9 i \, B\right )} a^{2} - 12 \, {\left (A + i \, B\right )} a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} - {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4}} + \frac {15 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {7}{2}}} + \frac {120 \, {\left (5 \, A + 2 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )}}{240 \, d} \]
-1/240*I*a*(4*(105*(I*a*tan(d*x + c) + a)^3*(3*A + I*B) - 5*(I*a*tan(d*x + c) + a)^2*(41*A + 15*I*B)*a - 2*(I*a*tan(d*x + c) + a)*(19*A + 9*I*B)*a^2 - 12*(A + I*B)*a^3)/((I*a*tan(d*x + c) + a)^(7/2)*a^3 - (I*a*tan(d*x + c) + a)^(5/2)*a^4) + 15*sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*t an(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(7/2) + 120*(5*A + 2*I*B)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*t an(d*x + c) + a) + sqrt(a)))/a^(7/2))/d
\[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Time = 10.09 (sec) , antiderivative size = 3002, normalized size of antiderivative = 11.59 \[ \int \frac {\cot ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]
2*atanh((12*a*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*((129*B^2)/(256*a^5*d^2) - (801*A^2)/(256*a^5*d^2) - ((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(1 6*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3* B*a^2*256479i)/(4*d^4))^(1/2)/(64*a^6) - (A*B*319i)/(128*a^5*d^2))^(1/2)*( (638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2 )/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2 ))/((A^3*d*31161i)/8 + (2159*B^3*d)/8 - (A*B^2*d*15867i)/8 - (38621*A^2*B* d)/8 + (A*d^3*((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307 555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i )/(4*d^4))^(1/2)*41i)/(2*a) - (15*B*d^3*((638401*A^4*a^2)/(16*d^4) + (1612 9*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4 *d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2))/(2*a)) + (799*A^2*a^2*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((129*B^2)/(256*a^5*d^2) - (801*A^2)/(256*a^5*d^ 2) - ((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B ^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4) )^(1/2)/(64*a^6) - (A*B*319i)/(128*a^5*d^2))^(1/2))/((A^3*d*31161i)/8 + (2 159*B^3*d)/8 - (A*B^2*d*15867i)/8 - (38621*A^2*B*d)/8 + (A*d^3*((638401*A^ 4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - (307555*A^2*B^2*a^2)/(8*d^4) - (A*B^3*a^2*40767i)/(4*d^4) + (A^3*B*a^2*256479i)/(4*d^4))^(1/2)*41i)/(2* a) - (15*B*d^3*((638401*A^4*a^2)/(16*d^4) + (16129*B^4*a^2)/(16*d^4) - ...