3.517 \(\int \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=148 \[ -\frac{2 (5 A-9 i B) \left (a^3 \cot (c+d x)+i a^3\right )}{15 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{16 a^3 (5 A-6 i B)}{15 d \sqrt{\cot (c+d x)}}+\frac{8 \sqrt [4]{-1} a^3 (B+i A) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{2 i a B (a \cot (c+d x)+i a)^2}{5 d \cot ^{\frac{5}{2}}(c+d x)} \]

[Out]

(8*(-1)^(1/4)*a^3*(I*A + B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (16*a^3*(5*A - (6*I)*B))/(15*d*Sqrt[Co
t[c + d*x]]) + (((2*I)/5)*a*B*(I*a + a*Cot[c + d*x])^2)/(d*Cot[c + d*x]^(5/2)) - (2*(5*A - (9*I)*B)*(I*a^3 + a
^3*Cot[c + d*x]))/(15*d*Cot[c + d*x]^(3/2))

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Rubi [A]  time = 0.491551, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.139, Rules used = {3581, 3593, 3591, 3533, 208} \[ -\frac{2 (5 A-9 i B) \left (a^3 \cot (c+d x)+i a^3\right )}{15 d \cot ^{\frac{3}{2}}(c+d x)}-\frac{16 a^3 (5 A-6 i B)}{15 d \sqrt{\cot (c+d x)}}+\frac{8 \sqrt [4]{-1} a^3 (B+i A) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{2 i a B (a \cot (c+d x)+i a)^2}{5 d \cot ^{\frac{5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(8*(-1)^(1/4)*a^3*(I*A + B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (16*a^3*(5*A - (6*I)*B))/(15*d*Sqrt[Co
t[c + d*x]]) + (((2*I)/5)*a*B*(I*a + a*Cot[c + d*x])^2)/(d*Cot[c + d*x]^(5/2)) - (2*(5*A - (9*I)*B)*(I*a^3 + a
^3*Cot[c + d*x]))/(15*d*Cot[c + d*x]^(3/2))

Rule 3581

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3593

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^2*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^
(n + 1))/(d*f*(b*c + a*d)*(n + 1)), x] - Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m -
1) + b*d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ
[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]

Rule 3591

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b*c - a*d)*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2
+ b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rule 3533

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

Rubi steps

\begin{align*} \int \sqrt{\cot (c+d x)} (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\int \frac{(i a+a \cot (c+d x))^3 (B+A \cot (c+d x))}{\cot ^{\frac{7}{2}}(c+d x)} \, dx\\ &=\frac{2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}+\frac{2}{5} \int \frac{(i a+a \cot (c+d x))^2 \left (\frac{1}{2} a (5 i A+9 B)+\frac{1}{2} a (5 A-i B) \cot (c+d x)\right )}{\cot ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}-\frac{2 (5 A-9 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{15 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4}{15} \int \frac{(i a+a \cot (c+d x)) \left (2 a^2 (5 i A+6 B)+a^2 (5 A-3 i B) \cot (c+d x)\right )}{\cot ^{\frac{3}{2}}(c+d x)} \, dx\\ &=-\frac{16 a^3 (5 A-6 i B)}{15 d \sqrt{\cot (c+d x)}}+\frac{2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}-\frac{2 (5 A-9 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{15 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{4}{15} \int \frac{15 a^3 (i A+B)+15 a^3 (A-i B) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{16 a^3 (5 A-6 i B)}{15 d \sqrt{\cot (c+d x)}}+\frac{2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}-\frac{2 (5 A-9 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{15 d \cot ^{\frac{3}{2}}(c+d x)}+\frac{\left (120 a^6 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-15 a^3 (i A+B)+15 a^3 (A-i B) x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=\frac{8 \sqrt [4]{-1} a^3 (i A+B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{16 a^3 (5 A-6 i B)}{15 d \sqrt{\cot (c+d x)}}+\frac{2 i a B (i a+a \cot (c+d x))^2}{5 d \cot ^{\frac{5}{2}}(c+d x)}-\frac{2 (5 A-9 i B) \left (i a^3+a^3 \cot (c+d x)\right )}{15 d \cot ^{\frac{3}{2}}(c+d x)}\\ \end{align*}

Mathematica [A]  time = 6.33981, size = 140, normalized size = 0.95 \[ \frac{a^3 \sec ^2(c+d x) \left (-5 (3 B+i A) \sin (2 (c+d x))-9 (5 A-7 i B) \cos (2 (c+d x))+\frac{120 (A-i B) \cos ^2(c+d x) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt{i \tan (c+d x)}}-45 A+57 i B\right )}{15 d \sqrt{\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(a^3*Sec[c + d*x]^2*(-45*A + (57*I)*B - 9*(5*A - (7*I)*B)*Cos[2*(c + d*x)] - 5*(I*A + 3*B)*Sin[2*(c + d*x)] +
(120*(A - I*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Cos[c + d*x]^2)/Sqrt[I*Tan[
c + d*x]]))/(15*d*Sqrt[Cot[c + d*x]])

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Maple [C]  time = 0.61, size = 973, normalized size = 6.6 \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)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

-1/15*a^3/d*2^(1/2)*(cos(d*x+c)-1)*(-3*I*B*2^(1/2)-63*I*B*cos(d*x+c)^3*2^(1/2)-60*I*B*cos(d*x+c)^2*sin(d*x+c)*
EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((cos(d*x+c)-1)/sin(d*x+c))^(1
/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)+60*A*((cos(d*x+
c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(
1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*sin(d*x+c)-6
0*A*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))
/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*sin(d*x+c
)-60*B*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+
c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1
/2))*sin(d*x+c)+63*I*B*cos(d*x+c)^2*2^(1/2)+5*I*A*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+45*A*cos(d*x+c)^3*2^(1/2)+3*
I*B*2^(1/2)*cos(d*x+c)+15*B*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)-60*I*A*cos(d*x+c)^2*sin(d*x+c)*EllipticPi((-(cos(d
*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+
sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-45*A*2^(1/2)*cos(d*x+c)^2-15*B*2^(
1/2)*cos(d*x+c)*sin(d*x+c)+60*I*B*cos(d*x+c)^2*sin(d*x+c)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1
/2),1/2*2^(1/2))*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-
1-sin(d*x+c))/sin(d*x+c))^(1/2)-5*I*A*cos(d*x+c)*sin(d*x+c)*2^(1/2))*(cos(d*x+c)+1)^2*(cos(d*x+c)/sin(d*x+c))^
(1/2)/cos(d*x+c)^3/sin(d*x+c)^3

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Maxima [A]  time = 1.59765, size = 270, normalized size = 1.82 \begin{align*} -\frac{15 \,{\left (\sqrt{2}{\left (\left (2 i + 2\right ) \, A - \left (2 i - 2\right ) \, B\right )} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + \sqrt{2}{\left (\left (2 i + 2\right ) \, A - \left (2 i - 2\right ) \, B\right )} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt{2}{\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} + 2 \,{\left (3 i \, B a^{3} - \frac{5 \,{\left (-i \, A - 3 \, B\right )} a^{3}}{\tan \left (d x + c\right )} + \frac{{\left (45 \, A - 60 i \, B\right )} a^{3}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac{5}{2}}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/15*(15*(sqrt(2)*((2*I + 2)*A - (2*I - 2)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + sqrt(2)*
((2*I + 2)*A - (2*I - 2)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + sqrt(2)*((I - 1)*A + (I +
1)*B)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - sqrt(2)*((I - 1)*A + (I + 1)*B)*log(-sqrt(2)/sqrt
(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 + 2*(3*I*B*a^3 - 5*(-I*A - 3*B)*a^3/tan(d*x + c) + (45*A - 60*I*B)*a
^3/tan(d*x + c)^2)*tan(d*x + c)^(5/2))/d

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Fricas [B]  time = 1.5634, size = 1369, normalized size = 9.25 \begin{align*} \frac{15 \, \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac{{\left (8 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 15 \, \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac{{\left (8 \,{\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) +{\left ({\left (400 i \, A + 624 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (320 i \, A + 288 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (-400 i \, A - 528 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (-320 i \, A - 384 \, B\right )} a^{3}\right )} \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}}}{60 \,{\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, 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)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/60*(15*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d
*e^(2*I*d*x + 2*I*c) + d)*log(-(8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^
6/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*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)/((4*I*A + 4*B)*a^3)) - 15*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e
^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(-(8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt((-64*I*A^2
- 128*A*B + 64*I*B^2)*a^6/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*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)/((4*I*A + 4*B)*a^3)) + ((400*I*A + 624*B)*a^3*e^(6*I*d*x + 6*I*c) + (320*I*A +
288*B)*a^3*e^(4*I*d*x + 4*I*c) + (-400*I*A - 528*B)*a^3*e^(2*I*d*x + 2*I*c) + (-320*I*A - 384*B)*a^3)*sqrt((I*
e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*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)**(1/2)*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \sqrt{\cot \left (d x + c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3*sqrt(cot(d*x + c)), x)