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

Optimal. Leaf size=220 \[ \frac {23 \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {11 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {21 i \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^2 d}+\frac {17 \cot ^2(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

23/4*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)+17/6*cot(d*x+c)^2/a/d/(a+I*a*tan(d*x+c))^(1/2)+21/4*I*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/
2)/a^2/d-11/3*cot(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/a^2/d+1/3*cot(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(3/2)

________________________________________________________________________________________

Rubi [A]  time = 0.74, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ \frac {23 \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 a^{3/2} d}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {11 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {21 i \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^2 d}+\frac {17 \cot ^2(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(23*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*a^(3/2)*d) - ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*S
qrt[a])]/(2*Sqrt[2]*a^(3/2)*d) + Cot[c + d*x]^2/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (17*Cot[c + d*x]^2)/(6*a*
d*Sqrt[a + I*a*Tan[c + d*x]]) + (((21*I)/4)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/(a^2*d) - (11*Cot[c + d*x
]^2*Sqrt[a + I*a*Tan[c + d*x]])/(3*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 ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\cot ^3(c+d x) \left (5 a-\frac {7}{2} i a \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2}\\ &=\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {17 \cot ^2(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {\int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \left (22 a^2-\frac {85}{4} i a^2 \tan (c+d x)\right ) \, dx}{3 a^4}\\ &=\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {17 \cot ^2(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}-\frac {11 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {\int \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {63 i a^3}{2}-33 a^3 \tan (c+d x)\right ) \, dx}{6 a^5}\\ &=\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {17 \cot ^2(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^2 d}-\frac {11 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {69 a^4}{4}+\frac {63}{4} i a^4 \tan (c+d x)\right ) \, dx}{6 a^6}\\ &=\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {17 \cot ^2(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^2 d}-\frac {11 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}-\frac {23 \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3}-\frac {i \int \sqrt {a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {17 \cot ^2(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^2 d}-\frac {11 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}-\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 {23 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 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 {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {17 \cot ^2(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^2 d}-\frac {11 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}+\frac {(23 i) \operatorname {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 a^2 d}\\ &=\frac {23 \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 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 {\cot ^2(c+d x)}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {17 \cot ^2(c+d x)}{6 a d \sqrt {a+i a \tan (c+d x)}}+\frac {21 i \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^2 d}-\frac {11 \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 a^2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 4.06, size = 214, normalized size = 0.97 \[ \frac {\sec ^{\frac {3}{2}}(c+d x) \left (\sqrt {2} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^{3/2} \left (23 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {1+e^{2 i (c+d x)}}}\right )-2 \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )-\frac {1}{3} \csc ^2(c+d x) \sqrt {\sec (c+d x)} (27 i \sin (2 (c+d x))-18 i \sin (4 (c+d x))+6 \cos (2 (c+d x))-19 \cos (4 (c+d x))+25)\right )}{8 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]^(3/2)*(Sqrt[2]*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^(3/2)
*(-2*ArcSinh[E^(I*(c + d*x))] + 23*Sqrt[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]]) -
(Csc[c + d*x]^2*Sqrt[Sec[c + d*x]]*(25 + 6*Cos[2*(c + d*x)] - 19*Cos[4*(c + d*x)] + (27*I)*Sin[2*(c + d*x)] -
(18*I)*Sin[4*(c + d*x)]))/3))/(8*d*(a + I*a*Tan[c + d*x])^(3/2))

________________________________________________________________________________________

fricas [B]  time = 0.47, size = 678, normalized size = 3.08 \[ -\frac {12 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - 2 \, 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 )} \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 ) - 12 \, \sqrt {\frac {1}{2}} {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - 2 \, 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 )} \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 ) - 69 \, {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - 2 \, 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 )} \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 ) + 69 \, {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - 2 \, 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 )} \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 ) + 4 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (37 \, e^{\left (8 i \, d x + 8 i \, c\right )} - 33 \, e^{\left (6 i \, d x + 6 i \, c\right )} - 50 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 21 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}}{48 \, {\left (a^{2} d e^{\left (7 i \, d x + 7 i \, c\right )} - 2 \, 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)^3/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/48*(12*sqrt(1/2)*(a^2*d*e^(7*I*d*x + 7*I*c) - 2*a^2*d*e^(5*I*d*x + 5*I*c) + 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))*s
qrt(1/(a^3*d^2)) + a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 12*sqrt(1/2)*(a^2*d*e^(7*I*d*x + 7*I*c) - 2*a^2*d*e^
(5*I*d*x + 5*I*c) + 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)) -
69*(a^2*d*e^(7*I*d*x + 7*I*c) - 2*a^2*d*e^(5*I*d*x + 5*I*c) + 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)) + 69*(a^2*d*e^(7*I*d*x + 7*I*c) - 2*a^2*d*e^
(5*I*d*x + 5*I*c) + 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)) + 4*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(37*e^(8*I*d*x + 8*I*c) - 33*e^(6*I*d*x
+ 6*I*c) - 50*e^(4*I*d*x + 4*I*c) + 21*e^(2*I*d*x + 2*I*c) + 1))/(a^2*d*e^(7*I*d*x + 7*I*c) - 2*a^2*d*e^(5*I*d
*x + 5*I*c) + a^2*d*e^(3*I*d*x + 3*I*c))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 1.42, size = 1409, normalized size = 6.40 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-6*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)-136*I*c
os(d*x+c)^3*sin(d*x+c)+6*I*2^(1/2)*(-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))*cos(d*x+c)^2*sin(d*x+c)-69*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)^2-6*(
-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))*cos(d*x+c)^3*2^(1/2)+32*cos(d*x+c)^6+69*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)+252*I*sin(d*x+c)*cos(d*x+c)-
69*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))*sin(d*x+c)-6*(-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))*cos(d*x+c)^2*2^(1/2)-69*cos(d*x+c)^2*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))-69*(-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)^3-32*I*cos(d*x+c)^5*sin(d*x+c)+69*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*sin(d*x+c)+6*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+co
s(d*x+c)))^(1/2)*2^(1/2))-69*cos(d*x+c)^2*(-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))+120*cos(d*x+c)^4+69*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))-69*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-sin(d*x+c)*(-2*co
s(d*x+c)/(1+cos(d*x+c)))^(1/2)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^3+6*2^(1/2)*(-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))+69*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)+69*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)
)+69*(-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))-176*cos(d*x+c)^2)/(co
s(d*x+c)^2-1)/a^2

________________________________________________________________________________________

maxima [A]  time = 0.56, size = 221, normalized size = 1.00 \[ -\frac {a^{2} {\left (\frac {2 \, {\left (63 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} - 107 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a + 34 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} + 4 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4} + {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{5}} - \frac {3 \, \sqrt {2} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {7}{2}}} + \frac {69 \, \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {7}{2}}}\right )}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*a^2*(2*(63*(I*a*tan(d*x + c) + a)^3 - 107*(I*a*tan(d*x + c) + a)^2*a + 34*(I*a*tan(d*x + c) + a)*a^2 + 4
*a^3)/((I*a*tan(d*x + c) + a)^(7/2)*a^3 - 2*(I*a*tan(d*x + c) + a)^(5/2)*a^4 + (I*a*tan(d*x + c) + a)^(3/2)*a^
5) - 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^(7/2) + 69*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/a^(7/2))/
d

________________________________________________________________________________________

mupad [B]  time = 3.94, size = 186, normalized size = 0.85 \[ -\frac {\frac {a^2}{3}+\frac {21\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{4\,a}+\frac {17\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6}-\frac {107\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{12}}{d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}-2\,a\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}+a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {23\,\mathrm {atanh}\left (\frac {a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{\sqrt {a^3}}\right )}{4\,d\,\sqrt {a^3}}-\frac {\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a^3}}\right )}{4\,d\,\sqrt {a^3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(23*atanh((a*(a + a*tan(c + d*x)*1i)^(1/2))/(a^3)^(1/2)))/(4*d*(a^3)^(1/2)) - ((21*(a + a*tan(c + d*x)*1i)^3)/
(4*a) - (107*(a + a*tan(c + d*x)*1i)^2)/12 + (17*a*(a + a*tan(c + d*x)*1i))/6 + a^2/3)/(d*(a + a*tan(c + d*x)*
1i)^(7/2) - 2*a*d*(a + a*tan(c + d*x)*1i)^(5/2) + a^2*d*(a + a*tan(c + d*x)*1i)^(3/2)) - (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))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________