Integrand size = 28, antiderivative size = 217 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {(-1)^{3/4} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}+\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{\sqrt {a} d}-\frac {1}{d \cot ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{a d \sqrt {\cot (c+d x)}} \]
-(-1)^(3/4)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^ (1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/d/a^(1/2)+(1/2+1/2*I)*arctanh((1+ I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan (d*x+c)^(1/2)/d/a^(1/2)-1/d/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(1/2)-2*I* (a+I*a*tan(d*x+c))^(1/2)/a/d/cot(d*x+c)^(1/2)
Time = 1.13 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.85 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sqrt {\cot (c+d x)} \left (2 \sqrt [4]{-1} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)}+2 a \tan (c+d x) (-2 i+\tan (c+d x))+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}\right )}{2 a d \sqrt {a+i a \tan (c+d x)}} \]
(Sqrt[Cot[c + d*x]]*(2*(-1)^(1/4)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]] *Sqrt[1 + I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]] + 2*a*Tan[c + d*x]*(-2*I + Ta n[c + d*x]) + Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I* a*Tan[c + d*x]]]*Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]))/(2*a* d*Sqrt[a + I*a*Tan[c + d*x]])
Time = 1.25 (sec) , antiderivative size = 205, normalized size of antiderivative = 0.94, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.536, Rules used = {3042, 4729, 3042, 4041, 27, 3042, 4080, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}
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.
| \(\displaystyle \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\cot (c+d x)^{5/2} \sqrt {a+i a \tan (c+d x)}}dx\) |
| \(\Big \downarrow \) 4729 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan ^{\frac {5}{2}}(c+d x)}{\sqrt {i \tan (c+d x) a+a}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \int \frac {\tan (c+d x)^{5/2}}{\sqrt {i \tan (c+d x) a+a}}dx\) |
| \(\Big \downarrow \) 4041 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (-\frac {\int -\frac {1}{2} \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (3 a-4 i a \tan (c+d x))dx}{a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (3 a-4 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\int \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a} (3 a-4 i a \tan (c+d x))dx}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 4080 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (\tan (c+d x) a^2+2 i a^2\right )}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {4 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\int \frac {\sqrt {i \tan (c+d x) a+a} \left (\tan (c+d x) a^2+2 i a^2\right )}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {4 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 4084 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a^2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+i a \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {4 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a^2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+i a \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {4 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 4027 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\frac {2 a^4 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}+i a \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx}{a}-\frac {4 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {i a \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {(1+i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {4 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 4082 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\frac {i a^3 \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {(1+i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {4 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 65 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\frac {2 i a^3 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}+\frac {(1+i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {4 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {\frac {\frac {(1+i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 (-1)^{3/4} a^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}}{a}-\frac {4 i a \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}}{2 a^2}-\frac {\tan ^{\frac {3}{2}}(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}\right )\) |
Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(-(Tan[c + d*x]^(3/2)/(d*Sqrt[a + I* a*Tan[c + d*x]])) + (((-2*(-1)^(3/4)*a^(5/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sq rt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d + ((1 + I)*a^(5/2)*ArcTan h[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/a - ((4*I)*a*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d)/(2*a^2))
3.8.73.3.1 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[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[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[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-(b*c - a*d))*(a + b*Tan[e + f*x])^m* ((c + d*Tan[e + f*x])^(n - 1)/(2*a*f*m)), x] + Simp[1/(2*a^2*m) Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[c*(a*c*m + b*d*(n - 1)) - d*(b*c*m + a*d*(n - 1)) - d*(b*d*(m - n + 1) - a*c*(m + n - 1))*Ta n[e + f*x], 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] && GtQ[n, 1] && (In tegerQ[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[B*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(f*(m + n))), x] + Simp[ 1/(a*(m + n)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 1)*Sim p[a*A*c*(m + n) - B*(b*c*m + a*d*n) + (a*A*d*(m + n) - B*(b*d*m - a*c*n))*T an[e + f*x], 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] && 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[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_)])^(n_), x_Symbol] :> Sim p[(A*b + a*B)/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] - Simp[B/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ e + f*x]), 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] && NeQ[A*b + a*B, 0]
Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Simp[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m Int[ActivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownTangentIntegrandQ[u, x]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 791 vs. \(2 (172 ) = 344\).
Time = 36.68 (sec) , antiderivative size = 792, normalized size of antiderivative = 3.65
(-1/4-1/4*I)/d*(-2*I*(cot(d*x+c)-csc(d*x+c))^(1/2)*sin(d*x+c)-I*ln((cot(d* x+c)-csc(d*x+c))^(1/2)-I)*cos(d*x+c)^2-I*ln((cot(d*x+c)-csc(d*x+c))^(1/2)- 1)*cos(d*x+c)-I*ln((cot(d*x+c)-csc(d*x+c))^(1/2)-1)*cos(d*x+c)^2-4*I*(cot( d*x+c)-csc(d*x+c))^(1/2)*cos(d*x+c)^2-2*I*(cot(d*x+c)-csc(d*x+c))^(1/2)*co s(d*x+c)*sin(d*x+c)-I*ln((cot(d*x+c)-csc(d*x+c))^(1/2)-I)*cos(d*x+c)+2*cos (d*x+c)^2*2^(1/2)*arctan((1/2+1/2*I)*(cot(d*x+c)-csc(d*x+c))^(1/2)*2^(1/2) )+I*ln((cot(d*x+c)-csc(d*x+c))^(1/2)+1)*cos(d*x+c)^2+I*ln((cot(d*x+c)-csc( d*x+c))^(1/2)+I)*cos(d*x+c)-4*I*(cot(d*x+c)-csc(d*x+c))^(1/2)*cos(d*x+c)+I *ln((cot(d*x+c)-csc(d*x+c))^(1/2)+I)*cos(d*x+c)^2+2*I*2^(1/2)*arctan((1/2+ 1/2*I)*(cot(d*x+c)-csc(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)*sin(d*x+c)+I*ln(( cot(d*x+c)-csc(d*x+c))^(1/2)+1)*cos(d*x+c)+2*cos(d*x+c)*2^(1/2)*arctan((1/ 2+1/2*I)*(cot(d*x+c)-csc(d*x+c))^(1/2)*2^(1/2))-4*(cot(d*x+c)-csc(d*x+c))^ (1/2)*cos(d*x+c)^2+2*sin(d*x+c)*cos(d*x+c)*(cot(d*x+c)-csc(d*x+c))^(1/2)+s in(d*x+c)*cos(d*x+c)*ln((cot(d*x+c)-csc(d*x+c))^(1/2)-1)+sin(d*x+c)*cos(d* x+c)*ln((cot(d*x+c)-csc(d*x+c))^(1/2)-I)-sin(d*x+c)*cos(d*x+c)*ln((cot(d*x +c)-csc(d*x+c))^(1/2)+1)-sin(d*x+c)*cos(d*x+c)*ln((cot(d*x+c)-csc(d*x+c))^ (1/2)+I)-4*(cot(d*x+c)-csc(d*x+c))^(1/2)*cos(d*x+c)+2*sin(d*x+c)*(cot(d*x+ c)-csc(d*x+c))^(1/2))*cos(d*x+c)/(cos(d*x+c)-1)/(cos(d*x+c)+1)^2/(cot(d*x+ c)-csc(d*x+c))^(1/2)/(a*(1+I*tan(d*x+c)))^(1/2)/cot(d*x+c)^(5/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 703 vs. \(2 (163) = 326\).
Time = 0.27 (sec) , antiderivative size = 703, normalized size of antiderivative = 3.24 \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (3 \, e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} - {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {2 i}{a d^{2}}} \log \left (2 \, {\left (\sqrt {2} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {2 i}{a d^{2}}} + 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {2 i}{a d^{2}}} \log \left (-2 \, {\left (\sqrt {2} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} - a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {2 i}{a d^{2}}} - 2 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {i}{a d^{2}}} \log \left (-16 \, {\left (2 \, \sqrt {2} {\left (i \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a^{2} 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 {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{a d^{2}}} + 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) - {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {i}{a d^{2}}} \log \left (-16 \, {\left (2 \, \sqrt {2} {\left (-i \, a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, a^{2} 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 {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i}{a d^{2}}} + 3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right )}{4 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]
-1/4*(2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I *c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(3*e^(4*I*d*x + 4*I*c) - 2*e^(2*I*d*x + 2*I*c) - 1) - (a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(2*I/( a*d^2))*log(2*(sqrt(2)*(a*d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) *sqrt(2*I/(a*d^2)) + 2*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + (a*d*e^(3* I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(2*I/(a*d^2))*log(-2*(sqrt(2)*(a *d*e^(2*I*d*x + 2*I*c) - a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^ (2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(2*I/(a*d^2)) - 2*I* a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + (a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I *d*x + I*c))*sqrt(I/(a*d^2))*log(-16*(2*sqrt(2)*(I*a^2*d*e^(3*I*d*x + 3*I* c) - I*a^2*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^ (2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(I/(a*d^2)) + 3*a^2* e^(2*I*d*x + 2*I*c) - a^2)*e^(-2*I*d*x - 2*I*c)) - (a*d*e^(3*I*d*x + 3*I*c ) + a*d*e^(I*d*x + I*c))*sqrt(I/(a*d^2))*log(-16*(2*sqrt(2)*(-I*a^2*d*e^(3 *I*d*x + 3*I*c) + I*a^2*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1 ))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(I/(a*d ^2)) + 3*a^2*e^(2*I*d*x + 2*I*c) - a^2)*e^(-2*I*d*x - 2*I*c)))/(a*d*e^(3*I *d*x + 3*I*c) + a*d*e^(I*d*x + I*c))
\[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \cot ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
\[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\sqrt {i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\sqrt {i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{\cot ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{5/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]