\(\int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [132]

Optimal result

Integrand size = 24, antiderivative size = 159 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

-2*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d+1/8*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)+7/4/a^2/d/(a+I*a*tan(d*x+c))^(1/2)+1/5/d/(a+I*a*tan(d*x+c))^(5/2)+1/2/a/d/(a+I*a*tan(d*x+
c))^(3/2)

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3640, 3677, 3681, 3561, 212, 3680, 65, 214} \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {7}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {1}{5 d (a+i a \tan (c+d x))^{5/2}} \]

[In]

Int[Cot[c + d*x]/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(-2*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) + ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqr
t[a])]/(4*Sqrt[2]*a^(5/2)*d) + 1/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + 1/(2*a*d*(a + I*a*Tan[c + d*x])^(3/2)) +
7/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[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] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

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]

Rule 3561

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3640

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Dist[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[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])

Rule 3677

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(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] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[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
] /; FreeQ[{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]

Rule 3680

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[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]

Rule 3681

Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(A*b + a*B)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m, x], x] - Dist[(B*c
- A*d)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m*((a - b*Tan[e + f*x])/(c + d*Tan[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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {\cot (c+d x) \left (5 a-\frac {5}{2} i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2} \\ & = \frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\cot (c+d x) \left (15 a^2-\frac {45}{4} i a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4} \\ & = \frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (15 a^3-\frac {105}{8} i a^3 \tan (c+d x)\right ) \, dx}{15 a^6} \\ & = \frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{a^4}+\frac {i \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = \frac {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {\text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 a^2 d}+\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{a^2 d} \\ & = \frac {\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 {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(2 i) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{a^3 d} \\ & = -\frac {2 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {\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 {1}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {1}{2 a d (a+i a \tan (c+d x))^{3/2}}+\frac {7}{4 a^2 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.37 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.90 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {-\frac {80 \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {5 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{a^{5/2}}+\frac {8}{(a+i a \tan (c+d x))^{5/2}}+\frac {20}{a (a+i a \tan (c+d x))^{3/2}}+\frac {70}{a^2 \sqrt {a+i a \tan (c+d x)}}}{40 d} \]

[In]

Integrate[Cot[c + d*x]/(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-80*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/a^(5/2) + (5*Sqrt[2]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sq
rt[2]*Sqrt[a])])/a^(5/2) + 8/(a + I*a*Tan[c + d*x])^(5/2) + 20/(a*(a + I*a*Tan[c + d*x])^(3/2)) + 70/(a^2*Sqrt
[a + I*a*Tan[c + d*x]]))/(40*d)

Maple [A] (verified)

Time = 1.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {2 a^{2} \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {9}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {9}{2}}}+\frac {7}{8 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{4 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{10 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d}\)	\(122\)
default	\(\frac {2 a^{2} \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{a^{\frac {9}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {9}{2}}}+\frac {7}{8 a^{4} \sqrt {a +i a \tan \left (d x +c \right )}}+\frac {1}{4 a^{3} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {1}{10 a^{2} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d}\)	\(122\)

[In]

int(cot(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

2/d*a^2*(-1/a^(9/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))+1/16/a^(9/2)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+
c))^(1/2)*2^(1/2)/a^(1/2))+7/8/a^4/(a+I*a*tan(d*x+c))^(1/2)+1/4/a^3/(a+I*a*tan(d*x+c))^(3/2)+1/10/a^2/(a+I*a*t
an(d*x+c))^(5/2))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 511 vs. \(2 (120) = 240\).

Time = 0.26 (sec) , antiderivative size = 511, normalized size of antiderivative = 3.21 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {{\left (5 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} + a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 5 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a^{5} d^{2}}} - a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 20 \, a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} 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^{5} d^{2}}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 20 \, a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} 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^{5} 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 (41 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 48 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 8 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{40 \, a^{3} d} \]

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/40*(5*sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*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(1/(a^5*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 5*sq
rt(1/2)*a^3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*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(1/(a^5*d^2)) - a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 20*a^3*d*sqrt
(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) + 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(1/(a^5*d^2)) + a^2)*e^(-2*I*d*x - 2*I*c)) + 20*a^
3*d*sqrt(1/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) - 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(1/(a^5*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))*(41*e^(6*I*d*x + 6*I*c) + 48*e^(4*I*d*x + 4*I*c) + 8*e^(2*I*d*x +
2*I*c) + 1))*e^(-5*I*d*x - 5*I*c)/(a^3*d)

Sympy [F]

\[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int \frac {\cot {\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Integral(cot(c + d*x)/(I*a*(tan(c + d*x) - I))**(5/2), x)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.01 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {\frac {5 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} - \frac {80 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {4 \, {\left (35 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} + 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + 4 \, a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2}}}{80 \, d} \]

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

-1/80*(5*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) - 80*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/a^(5/2)
- 4*(35*(I*a*tan(d*x + c) + a)^2 + 10*(I*a*tan(d*x + c) + a)*a + 4*a^2)/((I*a*tan(d*x + c) + a)^(5/2)*a^2))/d

Giac [F]

\[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=\int { \frac {\cot \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(cot(d*x+c)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(cot(d*x + c)/(I*a*tan(d*x + c) + a)^(5/2), x)

Mupad [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.83 \[ \int \frac {\cot (c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a^5}}\right )}{d\,\sqrt {a^5}}+\frac {\frac {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}{2\,a}+\frac {7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4\,a^2}+\frac {1}{5}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,a^2\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a^5}}\right )}{8\,d\,\sqrt {a^5}} \]

[In]

int(cot(c + d*x)/(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

((a + a*tan(c + d*x)*1i)/(2*a) + (7*(a + a*tan(c + d*x)*1i)^2)/(4*a^2) + 1/5)/(d*(a + a*tan(c + d*x)*1i)^(5/2)
) - (2*atanh((a^2*(a + a*tan(c + d*x)*1i)^(1/2))/(a^5)^(1/2)))/(d*(a^5)^(1/2)) + (2^(1/2)*atanh((2^(1/2)*a^2*(
a + a*tan(c + d*x)*1i)^(1/2))/(2*(a^5)^(1/2))))/(8*d*(a^5)^(1/2))