Integrand size = 26, antiderivative size = 181 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {3 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {13 \cot (c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {7 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a^2 d} \]
3*I*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(3/2)/d+1/4*I*arctanh(1/2* (a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/d*2^(1/2)+13/6*cot(d*x+c )/a/d/(a+I*a*tan(d*x+c))^(1/2)-7/2*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a^2 /d+1/3*cot(d*x+c)/d/(a+I*a*tan(d*x+c))^(3/2)
Time = 2.21 (sec) , antiderivative size = 146, normalized size of antiderivative = 0.81 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\frac {36 i \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {3 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{a^{3/2}}-\frac {2 \cot (c+d x) \left (-21+29 i \cot (c+d x)+6 \cot ^2(c+d x)\right ) \sqrt {a+i a \tan (c+d x)}}{a^2 (i+\cot (c+d x))^2}}{12 d} \]
(((36*I)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/a^(3/2) + ((3*I)*Sqr t[2]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/a^(3/2) - (2*C ot[c + d*x]*(-21 + (29*I)*Cot[c + d*x] + 6*Cot[c + d*x]^2)*Sqrt[a + I*a*Ta n[c + d*x]])/(a^2*(I + Cot[c + d*x])^2))/(12*d)
Time = 1.28 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.08, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.654, Rules used = {3042, 4042, 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.
\(\displaystyle \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^2 (a+i a \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4042 |
\(\displaystyle \frac {\int \frac {\cot ^2(c+d x) (8 a-5 i a \tan (c+d x))}{2 \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\cot ^2(c+d x) (8 a-5 i a \tan (c+d x))}{\sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {8 a-5 i a \tan (c+d x)}{\tan (c+d x)^2 \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4079 |
\(\displaystyle \frac {\frac {\int \frac {3}{2} \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (14 a^2-13 i a^2 \tan (c+d x)\right )dx}{a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {3 \int \cot ^2(c+d x) \sqrt {i \tan (c+d x) a+a} \left (14 a^2-13 i a^2 \tan (c+d x)\right )dx}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 \int \frac {\sqrt {i \tan (c+d x) a+a} \left (14 a^2-13 i a^2 \tan (c+d x)\right )}{\tan (c+d x)^2}dx}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {\frac {3 \left (\frac {\int -\cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (7 \tan (c+d x) a^3+6 i a^3\right )dx}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\frac {3 \left (-\frac {\int \cot (c+d x) \sqrt {i \tan (c+d x) a+a} \left (7 \tan (c+d x) a^3+6 i a^3\right )dx}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 \left (-\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (7 \tan (c+d x) a^3+6 i a^3\right )}{\tan (c+d x)}dx}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4083 |
\(\displaystyle \frac {\frac {3 \left (-\frac {a^3 \int \sqrt {i \tan (c+d x) a+a}dx+6 i a^2 \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {3 \left (-\frac {a^3 \int \sqrt {i \tan (c+d x) a+a}dx+6 i a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {\frac {3 \left (-\frac {6 i a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {2 i a^4 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {3 \left (-\frac {6 i a^2 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx-\frac {i \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 4082 |
\(\displaystyle \frac {\frac {3 \left (-\frac {\frac {6 i a^4 \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {i \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\frac {3 \left (-\frac {\frac {12 a^3 \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}-\frac {i \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\frac {3 \left (-\frac {-\frac {12 i a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}-\frac {i \sqrt {2} a^{7/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}}{a}-\frac {14 a^2 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{d}\right )}{2 a^2}+\frac {13 a \cot (c+d x)}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}\) |
Cot[c + d*x]/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + ((13*a*Cot[c + d*x])/(d* Sqrt[a + I*a*Tan[c + d*x]]) + (3*(-((((-12*I)*a^(7/2)*ArcTanh[Sqrt[a + I*a *Tan[c + d*x]]/Sqrt[a]])/d - (I*Sqrt[2]*a^(7/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d)/a) - (14*a^2*Cot[c + d*x]*Sqrt[a + I*a*Tan [c + d*x]])/d))/(2*a^2))/(6*a^2)
3.2.24.3.1 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_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(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)) In t[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
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 = 1.16 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {2 i a^{3} \left (-\frac {5}{4 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {1}{6 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {9}{2}}}+\frac {\frac {i \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{4}}\right )}{d}\) | \(137\) |
default | \(\frac {2 i a^{3} \left (-\frac {5}{4 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {1}{6 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {9}{2}}}+\frac {\frac {i \sqrt {a +i a \tan \left (d x +c \right )}}{2 a \tan \left (d x +c \right )}+\frac {3 \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{2 \sqrt {a}}}{a^{4}}\right )}{d}\) | \(137\) |
2*I/d*a^3*(-5/4/a^4/(a+I*a*tan(d*x+c))^(1/2)-1/6/a^3/(a+I*a*tan(d*x+c))^(3 /2)+1/8/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+I*a*tan(d*x+c))^(1/2)/a/tan(d*x+c)+3/2/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. 601 vs. \(2 (138) = 276\).
Time = 0.26 (sec) , antiderivative size = 601, normalized size of antiderivative = 3.32 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {3 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + i \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - i \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 9 \, {\left (-i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + i \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 9 \, {\left (i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - i \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {1}{a^{3} d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (a^{3} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{3} d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{3} d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-28 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 13 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}}{12 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \]
-1/12*(3*sqrt(1/2)*(-I*a^2*d*e^(5*I*d*x + 5*I*c) + I*a^2*d*e^(3*I*d*x + 3* I*c))*sqrt(1/(a^3*d^2))*log(4*(sqrt(2)*sqrt(1/2)*(a^2*d*e^(2*I*d*x + 2*I*c ) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a*e^(I*d* x + I*c))*e^(-I*d*x - I*c)) + 3*sqrt(1/2)*(I*a^2*d*e^(5*I*d*x + 5*I*c) - I *a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(-4*(sqrt(2)*sqrt(1/2)*(a ^2*d*e^(2*I*d*x + 2*I*c) + a^2*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1 /(a^3*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 9*(-I*a^2*d*e^(5*I*d* x + 5*I*c) + I*a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/(a^3*d^2))*log(16*(3*a^2* e^(2*I*d*x + 2*I*c) + 2*sqrt(2)*(a^3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d* x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a^3*d^2)) + a^2)*e^(-2 *I*d*x - 2*I*c)) + 9*(I*a^2*d*e^(5*I*d*x + 5*I*c) - I*a^2*d*e^(3*I*d*x + 3 *I*c))*sqrt(1/(a^3*d^2))*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) - 2*sqrt(2)*(a^ 3*d*e^(3*I*d*x + 3*I*c) + a^3*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I* c) + 1))*sqrt(1/(a^3*d^2)) + a^2)*e^(-2*I*d*x - 2*I*c)) - sqrt(2)*sqrt(a/( e^(2*I*d*x + 2*I*c) + 1))*(-28*I*e^(6*I*d*x + 6*I*c) - 13*I*e^(4*I*d*x + 4 *I*c) + 16*I*e^(2*I*d*x + 2*I*c) + I))/(a^2*d*e^(5*I*d*x + 5*I*c) - a^2*d* e^(3*I*d*x + 3*I*c))
\[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\cot ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.02 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {i \, a {\left (\frac {4 \, {\left (21 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 13 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a - 2 \, a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3}} + \frac {3 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} + \frac {36 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}}\right )}}{24 \, d} \]
-1/24*I*a*(4*(21*(I*a*tan(d*x + c) + a)^2 - 13*(I*a*tan(d*x + c) + a)*a - 2*a^2)/((I*a*tan(d*x + c) + a)^(5/2)*a^2 - (I*a*tan(d*x + c) + a)^(3/2)*a^ 3) + 3*sqrt(2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2 )*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(5/2) + 36*log((sqrt(I*a*tan(d* x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/a^(5/2))/d
\[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int { \frac {\cot \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
Time = 5.15 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.98 \[ \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx=-\frac {\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,13{}\mathrm {i}}{6\,d}+\frac {a\,1{}\mathrm {i}}{3\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,7{}\mathrm {i}}{2\,a\,d}}{a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\mathrm {atan}\left (\frac {\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2}\right )\,\sqrt {-a^3}\,3{}\mathrm {i}}{a^3\,d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^2}\right )\,\sqrt {-a^3}\,1{}\mathrm {i}}{4\,a^3\,d} \]
- (((a + a*tan(c + d*x)*1i)*13i)/(6*d) + (a*1i)/(3*d) - ((a + a*tan(c + d* x)*1i)^2*7i)/(2*a*d))/(a*(a + a*tan(c + d*x)*1i)^(3/2) - (a + a*tan(c + d* x)*1i)^(5/2)) - (atan(((-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/a^2)*(- a^3)^(1/2)*3i)/(a^3*d) - (2^(1/2)*atan((2^(1/2)*(-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^2))*(-a^3)^(1/2)*1i)/(4*a^3*d)