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

Optimal. Leaf size=132 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {3}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{3 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.39, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3559, 3596, 3600, 3480, 206, 3599, 63, 208} \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {3}{2 a d \sqrt {a+i a \tan (c+d x)}}+\frac {1}{3 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 1.99, size = 196, normalized size = 1.48 \[ -\frac {i \left (11 e^{2 i (c+d x)}+10 e^{4 i (c+d x)}+3 e^{3 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-12 \sqrt {2} e^{3 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )+1\right )}{3 a d \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/3*I)*(1 + 11*E^((2*I)*(c + d*x)) + 10*E^((4*I)*(c + d*x)) + 3*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c +
d*x))]*ArcSinh[E^(I*(c + d*x))] - 12*Sqrt[2]*E^((3*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[(Sqrt[2
]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]]))/(a*d*(1 + E^((2*I)*(c + d*x)))^2*(-I + Tan[c + d*x])*Sqrt[
a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.44, size = 500, normalized size = 3.79 \[ \frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \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}} a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \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 ) - 6 \, a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 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^{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 ) + 6 \, a^{2} d \sqrt {\frac {1}{a^{3} d^{2}}} e^{\left (3 i \, d x + 3 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^{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 (10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 11 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*sqrt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*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*sq
rt(1/2)*a^2*d*sqrt(1/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*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)) - 6*a^2*d*sqrt(
1/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*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)) + 6*a^2*
d*sqrt(1/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*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))*(10*e^(4*I*d*x + 4*I*c) + 11*e^(2*I*d*x + 2*I*c) + 1))*e^(-3*I*d*x
- 3*I*c)/(a^2*d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.56, size = 703, normalized size = 5.33 \[ -\frac {\sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-3 i \sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+16 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \sqrt {2}\, \cos \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-12 i \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sin \left (d x +c \right )+12 i \cos \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )-16 \left (\cos ^{4}\left (d x +c \right )\right )+3 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\left (i \cos \left (d x +c \right )-i-\sin \left (d x +c \right )\right ) \sqrt {2}}{2 \sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+12 \cos \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )+12 \ln \left (-\frac {-\sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+12 i \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \ln \left (-\frac {-\sin \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )+36 i \cos \left (d x +c \right ) \sin \left (d x +c \right )+12 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right )-28 \left (\cos ^{2}\left (d x +c \right )\right )\right )}{24 d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)/(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)*(-3*I*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arct
an(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))*sin(d*x+c)+16*I*co
s(d*x+c)^3*sin(d*x+c)+3*2^(1/2)*cos(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))-12*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ar
ctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)+12*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)-16*cos(d*x+c)^4+3*2^(1/2)*(-2*c
os(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))+12*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))+12*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))*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*sin(d*x+c)+12*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))+36*I*sin(d*x+c)*cos(d*x+c)+12*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arct
an(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-28*cos(d*x+c)^2)/a^2

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maxima [A]  time = 0.69, size = 139, normalized size = 1.05 \[ -\frac {\frac {3 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {3}{2}}} - \frac {24 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {3}{2}}} - \frac {4 \, {\left (9 i \, a \tan \left (d x + c\right ) + 11 \, a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(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^(3/2) - 24*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/a^(3/2)
- 4*(9*I*a*tan(d*x + c) + 11*a)/((I*a*tan(d*x + c) + a)^(3/2)*a))/d

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mupad [B]  time = 4.05, size = 109, normalized size = 0.83 \[ -\frac {2\,\mathrm {atanh}\left (\frac {a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a^3}}\right )}{d\,\sqrt {a^3}}+\frac {\frac {3\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{2\,a}+\frac {1}{3}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a^3}}\right )}{4\,d\,\sqrt {a^3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3*(a + a*tan(c + d*x)*1i))/(2*a) + 1/3)/(d*(a + a*tan(c + d*x)*1i)^(3/2)) - (2*atanh((a*(a + a*tan(c + d*x)*
1i)^(1/2))/(a^3)^(1/2)))/(d*(a^3)^(1/2)) + (2^(1/2)*atanh((2^(1/2)*a*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(a^3)^(
1/2))))/(4*d*(a^3)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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