3.2.33 \(\int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [133]

Optimal. Leaf size=214 \[ \frac {5 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {21 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d} \]

[Out]

5*I*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/a^(5/2)/d+1/8*I*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)+41/12*cot(d*x+c)/a^2/d/(a+I*a*tan(d*x+c))^(1/2)-21/4*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/
2)/a^3/d+1/5*cot(d*x+c)/d/(a+I*a*tan(d*x+c))^(5/2)+19/30*cot(d*x+c)/a/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.48, antiderivative size = 214, 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 = {3640, 3677, 3679, 3681, 3561, 212, 3680, 65, 214} \begin {gather*} \frac {5 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {21 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}+\frac {41 \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {19 \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*I)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) + ((I/4)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(S
qrt[2]*Sqrt[a])])/(Sqrt[2]*a^(5/2)*d) + Cot[c + d*x]/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (19*Cot[c + d*x])/(3
0*a*d*(a + I*a*Tan[c + d*x])^(3/2)) + (41*Cot[c + d*x])/(12*a^2*d*Sqrt[a + I*a*Tan[c + d*x]]) - (21*Cot[c + d*
x]*Sqrt[a + I*a*Tan[c + d*x]])/(4*a^3*d)

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 3679

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 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*} \int \frac {\cot ^2(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {\cot ^2(c+d x) \left (6 a-\frac {7}{2} i a \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\cot ^2(c+d x) \left (\frac {55 a^2}{2}-\frac {95}{4} i a^2 \tan (c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \cot (c+d x)}{12 a^2 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 {315 a^3}{4}-\frac {615}{8} i a^3 \tan (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {21 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {75 i a^4}{2}-\frac {315}{8} a^4 \tan (c+d x)\right ) \, dx}{15 a^7}\\ &=\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {21 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {(5 i) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{2 a^4}-\frac {\int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {21 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}+\frac {i \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 {(5 i) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{2 a^2 d}\\ &=\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {21 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}-\frac {5 \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 {5 i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{a^{5/2} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {\cot (c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {19 \cot (c+d x)}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {41 \cot (c+d x)}{12 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {21 \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a^3 d}\\ \end {align*}

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Mathematica [A]
time = 3.60, size = 263, normalized size = 1.23 \begin {gather*} -\frac {i \left (3+31 e^{2 i (c+d x)}+280 e^{4 i (c+d x)}-151 e^{6 i (c+d x)}-403 e^{8 i (c+d x)}+15 e^{5 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+300 \sqrt {2} e^{5 i (c+d x)} \left (-1+e^{2 i (c+d x)}\right ) \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 )\right )}{15 a^2 d \left (-1+e^{2 i (c+d x)}\right ) \left (1+e^{2 i (c+d x)}\right )^3 (-i+\tan (c+d x))^2 \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/15*I)*(3 + 31*E^((2*I)*(c + d*x)) + 280*E^((4*I)*(c + d*x)) - 151*E^((6*I)*(c + d*x)) - 403*E^((8*I)*(c +
d*x)) + 15*E^((5*I)*(c + d*x))*(-1 + E^((2*I)*(c + d*x)))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x
))] + 300*Sqrt[2]*E^((5*I)*(c + d*x))*(-1 + E^((2*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^2*d*(-1 + E^((2*I)*(c + d*x)))*(1 + E^((2*I)*(c + d*x))
)^3*(-I + Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1425 vs. \(2 (173 ) = 346\).
time = 0.91, size = 1426, normalized size = 6.66

method	result	size
default	\(\text {Expression too large to display}\)	\(1426\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-160*sin(d*x+c)*cos(d*x+c)^5-300*cos(d*x+c)^3*(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+300
*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c
)+1)/sin(d*x+c))+1260*sin(d*x+c)*cos(d*x+c)-300*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(((-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))-300*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a
rctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+300*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(((-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))-192*I*cos(d*x+c)^8-64*I*cos(d*x+c)^6-564*I*cos(d*x+c)^4
+820*I*cos(d*x+c)^2-300*(-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))*si
n(d*x+c)+300*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*
x+c)+1)/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)+15*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-I*cos(d*x+c
)+sin(d*x+c)+I)/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)*cos(d*x+c)^3+15*I*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-I*cos(d*x+c)+sin(d*x+c)+I)/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*2^(1/2))*2^(1/2)*cos(d*x+c)^2-15*I*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-I*
cos(d*x+c)+sin(d*x+c)+I)/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+300*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))-15*sin(d*x+c)*2^(1/2)*(-2*
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-I*cos(d*x+c)+sin(d*x+c)+I)/sin(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)*2^(1/2))-300*I*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(((-2*cos(d*x+c)/(1+cos(d*x+c)))^(
1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+300*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)^3+300*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)^2-300*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+c
os(d*x+c)))^(1/2))-15*I*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-I*cos(d*x+c)+sin(d*x+c)+I)/s
in(d*x+c)/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-668*sin(d*x+c)*cos(d*x+c)^3+15*cos(d*x+c)^2*sin(d*x+c)
*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-I*cos(d*x+c)+sin(d*x+c)+I)/sin(d*x+c)/(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-192*sin(d*x+c)*cos(d*x+c)^7)/(cos(d*x+c)^2-1)/a^3

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Maxima [A]
time = 0.53, size = 202, normalized size = 0.94 \begin {gather*} -\frac {i \, a {\left (\frac {4 \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} - 205 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a - 38 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{2} - 12 \, a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{3} - {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{4}} + \frac {15 \, \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 {600 \, \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 )}}{240 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/240*I*a*(4*(315*(I*a*tan(d*x + c) + a)^3 - 205*(I*a*tan(d*x + c) + a)^2*a - 38*(I*a*tan(d*x + c) + a)*a^2 -
12*a^3)/((I*a*tan(d*x + c) + a)^(7/2)*a^3 - (I*a*tan(d*x + c) + a)^(5/2)*a^4) + 15*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) + 600*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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 612 vs. \(2 (165) = 330\).
time = 0.38, size = 612, normalized size = 2.86 \begin {gather*} -\frac {15 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + i \, a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \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 ) + 15 \, \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - i \, a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \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 ) + 150 \, {\left (-i \, a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} + i \, a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \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 ) + 150 \, {\left (i \, a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - i \, a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \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 (-403 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 151 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 280 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 31 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )}}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/120*(15*sqrt(1/2)*(-I*a^3*d*e^(7*I*d*x + 7*I*c) + I*a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(1/(a^5*d^2))*log(4*(sqr
t(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)) + 15*sqrt(1/2)*(I*a^3*d*e^(7*I*d*x + 7*I*c) - I*a^3*d*e^(5*I*d*x + 5*I*c))*sqr
t(1/(a^5*d^2))*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)) + 150*(-I*a^3*d*e^(7*I*d*x + 7*I*c) + I*a^3*d*e^(5*I
*d*x + 5*I*c))*sqrt(1/(a^5*d^2))*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)) + 150*(I
*a^3*d*e^(7*I*d*x + 7*I*c) - I*a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(1/(a^5*d^2))*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))*(-403*I*e^(8*I*d*x + 8*I*c) - 1
51*I*e^(6*I*d*x + 6*I*c) + 280*I*e^(4*I*d*x + 4*I*c) + 31*I*e^(2*I*d*x + 2*I*c) + 3*I))/(a^3*d*e^(7*I*d*x + 7*
I*c) - a^3*d*e^(5*I*d*x + 5*I*c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{2}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 4.06, size = 201, normalized size = 0.94 \begin {gather*} -\frac {\frac {\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )\,19{}\mathrm {i}}{30\,d}+\frac {a\,1{}\mathrm {i}}{5\,d}+\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2\,41{}\mathrm {i}}{12\,a\,d}-\frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3\,21{}\mathrm {i}}{4\,a^2\,d}}{a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{7/2}}-\frac {\mathrm {atan}\left (\frac {\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^3}\right )\,\sqrt {-a^5}\,5{}\mathrm {i}}{a^5\,d}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {-a^5}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^3}\right )\,\sqrt {-a^5}\,1{}\mathrm {i}}{8\,a^5\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (((a + a*tan(c + d*x)*1i)*19i)/(30*d) + (a*1i)/(5*d) + ((a + a*tan(c + d*x)*1i)^2*41i)/(12*a*d) - ((a + a*ta
n(c + d*x)*1i)^3*21i)/(4*a^2*d))/(a*(a + a*tan(c + d*x)*1i)^(5/2) - (a + a*tan(c + d*x)*1i)^(7/2)) - (atan(((-
a^5)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/a^3)*(-a^5)^(1/2)*5i)/(a^5*d) - (2^(1/2)*atan((2^(1/2)*(-a^5)^(1/2)*
(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^3))*(-a^5)^(1/2)*1i)/(8*a^5*d)

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