3.124 \(\int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=181 \[ \frac {3 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {7 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a^2 d}+\frac {13 \cot (c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

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)

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Rubi [A]  time = 0.58, antiderivative size = 181, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3559, 3596, 3598, 3600, 3480, 206, 3599, 63, 208} \[ \frac {3 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {7 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{2 a^2 d}+\frac {13 \cot (c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^2/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((3*I)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(a^(3/2)*d) + ((I/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(S
qrt[2]*Sqrt[a])])/(Sqrt[2]*a^(3/2)*d) + Cot[c + d*x]/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (13*Cot[c + d*x])/(6
*a*d*Sqrt[a + I*a*Tan[c + d*x]]) - (7*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/(2*a^2*d)

Rule 63

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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3480

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 3559

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 3596

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 3598

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*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] - Dist[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] /; FreeQ[{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]

Rule 3599

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 3600

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*} \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {\cot (c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\cot ^2(c+d x) \left (4 a-\frac {5}{2} i a \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\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 {\int \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {21 a^2}{2}-\frac {39}{4} i a^2 \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=\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}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {9 i a^3}{2}-\frac {21}{4} a^3 \tan (c+d x)\right ) \, dx}{3 a^5}\\ &=\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}-\frac {(3 i) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{2 a^3}-\frac {\int \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\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}+\frac {i \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{2 a d}-\frac {(3 i) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 a d}\\ &=\frac {i \tanh ^{-1}\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}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{a^2 d}\\ &=\frac {3 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {i \tanh ^{-1}\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}\\ \end {align*}

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Mathematica [A]  time = 1.90, size = 214, normalized size = 1.18 \[ -\frac {e^{-4 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \csc (2 (c+d x)) \left (\sqrt {1+e^{2 i (c+d x)}} \left (-15 e^{2 i (c+d x)}+28 e^{4 i (c+d x)}-1\right )-3 e^{3 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right ) \sinh ^{-1}\left (e^{i (c+d x)}\right )-18 \sqrt {2} e^{3 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right ) \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )\right )}{12 a d \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^2/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

-1/12*(Sqrt[1 + E^((2*I)*(c + d*x))]*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-1 - 15*E^((2*I)*(c + d*x)) + 28*E^((4*I)
*(c + d*x))) - 3*E^((3*I)*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))*ArcSinh[E^(I*(c + d*x))] - 18*Sqrt[2]*E^((3*I)
*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]])*Csc[2
*(c + d*x)])/(a*d*E^((4*I)*(c + d*x))*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.46, size = 596, normalized size = 3.29 \[ \frac {\sqrt {\frac {1}{2}} {\left (3 i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - 3 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 ) + \sqrt {\frac {1}{2}} {\left (-3 i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 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 ) + {\left (9 i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - 9 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 ) + {\left (-9 i \, a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} + 9 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 )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/12*(sqrt(1/2)*(3*I*a^2*d*e^(5*I*d*x + 5*I*c) - 3*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)) + sqrt(1/2)*(-3*I*a^2*d*e^(5*I*d*x + 5*I*c) + 3*I*a^2*d*e^(3*I*d*x + 3*I*c))*sqr
t(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) - 9*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) + 9*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)
)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot \left (d x + c\right )^{2}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(cot(d*x + c)^2/(I*a*tan(d*x + c) + a)^(3/2), x)

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maple [B]  time = 1.35, size = 1411, normalized size = 7.80 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

1/24/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-36*I*cos(d*x+c)^4+18*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)-
16*I*cos(d*x+c)^6+52*I*cos(d*x+c)^2-18*I*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2))-3*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I
*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-16*cos(d*x+c)^5*sin(d*x+c)+
18*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2-18*I*sin
(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-
1)/sin(d*x+c))-18*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+18*cos
(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-18*co
s(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+
c)-1)/sin(d*x+c))+3*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(
-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+3*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*co
s(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)*2^(1/2)+18*I*(-2*co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^3+3*2^(1/2)*sin(d*x+c
)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*2^(1/2))-18*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c))-44*cos(d*x+c)^3*sin(d*x+c)-3*I*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*2^(1/2))-3*I*2^(1/2)*cos(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(I*cos(d*x+c)-I-sin(d*x+
c))/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-18*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)+18*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(
d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c))+18*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
ln(-(-sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c))+84*cos(d*x+c)*sin(d*x+c))/(cos
(d*x+c)^2-1)/a^2

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maxima [A]  time = 0.54, size = 184, normalized size = 1.02 \[ -\frac {i \, a {\left (\frac {4 \, {\left (21 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 13 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a - 2 \, a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3}} + \frac {3 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} + \frac {36 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}}\right )}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^2/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-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*t
an(d*x + c) + a) + sqrt(a)))/a^(5/2))/d

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mupad [B]  time = 4.02, size = 178, normalized size = 0.98 \[ -\frac {\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,13{}\mathrm {i}}{6\,d}+\frac {a\,1{}\mathrm {i}}{3\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,7{}\mathrm {i}}{2\,a\,d}}{a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\mathrm {atan}\left (\frac {\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2}\right )\,\sqrt {-a^3}\,3{}\mathrm {i}}{a^3\,d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^2}\right )\,\sqrt {-a^3}\,1{}\mathrm {i}}{4\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^2/(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

- (((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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**2/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(cot(c + d*x)**2/(I*a*(tan(c + d*x) - I))**(3/2), x)

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