3.752 \(\int \cot ^{\frac{5}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=140 \[ -\frac{2 \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}-\frac{2 i \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 d}-\frac{(1-i) \sqrt{a} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

[Out]

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

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Rubi [A]  time = 0.335324, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {4241, 3561, 3598, 12, 3544, 205} \[ -\frac{2 \cot ^{\frac{3}{2}}(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}-\frac{2 i \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{3 d}-\frac{(1-i) \sqrt{a} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rule 3561

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(d*(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*(c^2 + d^2)*
(n + 1)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - a*c*(n + 1) + a*d*(m + n + 1)*T
an[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^
2 + d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || IntegersQ[2*m, 2*n])

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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3544

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

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 0.962706, size = 125, normalized size = 0.89 \[ -\frac{i e^{-i (c+d x)} \sqrt{\cot (c+d x)} \left (4 e^{3 i (c+d x)}-3 \left (-1+e^{2 i (c+d x)}\right )^{3/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{3 d \left (-1+e^{2 i (c+d x)}\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.421, size = 1041, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d*2^(1/2)*(4*I*sin(d*x+c)*cos(d*x+c)*2^(1/2)-3*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*ln(-(((cos(d*x+c)-1)/s
in(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*
x+c)-cos(d*x+c)-sin(d*x+c)+1))+3*I*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*ln(-(((cos(d*x+c)-1)/sin(d*x
+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-c
os(d*x+c)-sin(d*x+c)+1))+6*I*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))
^(1/2)*2^(1/2)-1)+6*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^
(1/2)+1)+6*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)+
3*cos(d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos
(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))+6*I*cos(
d*x+c)^2*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)+4*2^(1/2)*cos(d
*x+c)^2-6*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-6*I*((cos(d*
x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-2*I*2^(1/2)*sin(d*x+c)-2*2^(1/2)
*cos(d*x+c)-6*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-6*((cos(d*
x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-3*((cos(d*x+c)-1)/sin(d*x+c))^(1
/2)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))-2*2^(1/2))*(cos(d*x+c)/sin(d*x+c))^(5/2)*(a*(I*sin(d*
x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*sin(d*x+c)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^2

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Maxima [B]  time = 2.31448, size = 1289, normalized size = 9.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/36*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(((36*I - 36)*cos(3*d*x + 3*c) +
(12*I - 12)*cos(d*x + c) - (36*I + 36)*sin(3*d*x + 3*c) - (12*I + 12)*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x
+ 2*c), cos(2*d*x + 2*c) - 1)) + ((36*I + 36)*cos(3*d*x + 3*c) + (12*I + 12)*cos(d*x + c) + (36*I - 36)*sin(3
*d*x + 3*c) + (12*I - 12)*sin(d*x + c))*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a) + ((
(36*I + 36)*cos(2*d*x + 2*c)^2 + (36*I + 36)*sin(2*d*x + 2*c)^2 - (72*I + 72)*cos(2*d*x + 2*c) + 36*I + 36)*ar
ctan2(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2
*c), cos(2*d*x + 2*c) - 1)) + 2*sin(d*x + c), 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c)
+ 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*cos(d*x + c)) + (-(18*I - 18)*cos(2*d*
x + 2*c)^2 - (18*I - 18)*sin(2*d*x + 2*c)^2 + (36*I - 36)*cos(2*d*x + 2*c) - 18*I + 18)*log(4*cos(d*x + c)^2 +
4*sin(d*x + c)^2 + 4*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(
sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))^2) + 8*(
cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*
x + 2*c), cos(2*d*x + 2*c) - 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))))*(c
os(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + ((((12*I - 12)*cos(d*x + c) -
(12*I + 12)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((12*I - 12)*cos(d*x + c) - (12*I + 12)*sin(d*x + c))*sin(2*d*
x + 2*c)^2 + (-(24*I - 24)*cos(d*x + c) + (24*I + 24)*sin(d*x + c))*cos(2*d*x + 2*c) + (12*I - 12)*cos(d*x + c
) - (12*I + 12)*sin(d*x + c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (((12*I + 12)*cos(d*x
+ c) + (12*I - 12)*sin(d*x + c))*cos(2*d*x + 2*c)^2 + ((12*I + 12)*cos(d*x + c) + (12*I - 12)*sin(d*x + c))*s
in(2*d*x + 2*c)^2 + (-(24*I + 24)*cos(d*x + c) - (24*I - 24)*sin(d*x + c))*cos(2*d*x + 2*c) + (12*I + 12)*cos(
d*x + c) + (12*I - 12)*sin(d*x + c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a))/((cos(
2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*d)

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Fricas [B]  time = 1.4566, size = 932, normalized size = 6.66 \begin{align*} \frac{-8 i \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} e^{\left (3 i \, d x + 3 i \, c\right )} - 3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{-\frac{2 i \, a}{d^{2}}} \log \left ({\left (\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (i \, d x + i \, c\right )} + i \, d \sqrt{-\frac{2 i \, a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + 3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt{-\frac{2 i \, a}{d^{2}}} \log \left ({\left (\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt{\frac{i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} - 1\right )} e^{\left (i \, d x + i \, c\right )} - i \, d \sqrt{-\frac{2 i \, a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right )}{6 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/6*(-8*I*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)
)*e^(3*I*d*x + 3*I*c) - 3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-2*I*a/d^2)*log((sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c
) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(e^(2*I*d*x + 2*I*c) - 1)*e^(I*d*x + I*c)
+ I*d*sqrt(-2*I*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)) + 3*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt(-2*I*a/
d^2)*log((sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)
)*(e^(2*I*d*x + 2*I*c) - 1)*e^(I*d*x + I*c) - I*d*sqrt(-2*I*a/d^2)*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)))
/(d*e^(2*I*d*x + 2*I*c) - d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**(5/2)*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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