\(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\) [87]

Optimal result

Integrand size = 36, antiderivative size = 173 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {a^{5/2} (23 A-20 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {4 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 (7 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d} \]

[Out]

1/4*a^(5/2)*(23*A-20*I*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d-4*a^(5/2)*(A-I*B)*arctanh(1/2*(a+I*a*tan
(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-1/4*a^2*(7*I*A+4*B)*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-1/2*a*A*co
t(d*x+c)^2*(a+I*a*tan(d*x+c))^(3/2)/d

Rubi [A] (verified)

Time = 0.95 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3674, 3681, 3561, 212, 3680, 65, 214} \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {a^{5/2} (23 A-20 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {4 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 (4 B+7 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d} \]

[In]

Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

(a^(5/2)*(23*A - (20*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (4*Sqrt[2]*a^(5/2)*(A - I*B)*A
rcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (a^2*((7*I)*A + 4*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c
+ d*x]])/(4*d) - (a*A*Cot[c + d*x]^2*(a + I*a*Tan[c + d*x])^(3/2))/(2*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 3674

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] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && 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*} \text {integral}& = -\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac {1}{2} \int \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \left (\frac {1}{2} a (7 i A+4 B)-\frac {1}{2} a (A-4 i B) \tan (c+d x)\right ) \, dx \\ & = -\frac {a^2 (7 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac {1}{2} \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (23 A-20 i B)-\frac {3}{4} a^2 (3 i A+4 B) \tan (c+d x)\right ) \, dx \\ & = -\frac {a^2 (7 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac {1}{8} (a (23 A-20 i B)) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx-\left (4 a^2 (i A+B)\right ) \int \sqrt {a+i a \tan (c+d x)} \, dx \\ & = -\frac {a^2 (7 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}-\frac {\left (8 a^3 (A-i B)\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}-\frac {\left (a^3 (23 A-20 i B)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d} \\ & = -\frac {4 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 (7 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d}+\frac {\left (a^2 (23 i A+20 B)\right ) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 d} \\ & = \frac {a^{5/2} (23 A-20 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {4 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a^2 (7 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) (a+i a \tan (c+d x))^{3/2}}{2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.86 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.82 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {a^{5/2} (23 A-20 i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )-16 \sqrt {2} a^{5/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )-a^2 \cot (c+d x) (9 i A+4 B+2 A \cot (c+d x)) \sqrt {a+i a \tan (c+d x)}}{4 d} \]

[In]

Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]

[Out]

(a^(5/2)*(23*A - (20*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] - 16*Sqrt[2]*a^(5/2)*(A - I*B)*ArcTanh[
Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])] - a^2*Cot[c + d*x]*((9*I)*A + 4*B + 2*A*Cot[c + d*x])*Sqrt[a + I
*a*Tan[c + d*x]])/(4*d)

Maple [A] (verified)

Time = 0.28 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {2 a^{3} \left (-\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\left (-\frac {i B}{2}+\frac {9 A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B -\frac {7}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (-20 i B +23 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )}{d}\)	\(143\)
default	\(\frac {2 a^{3} \left (-\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}-\frac {\left (-\frac {i B}{2}+\frac {9 A}{8}\right ) \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}+\left (\frac {1}{2} i a B -\frac {7}{8} a A \right ) \sqrt {a +i a \tan \left (d x +c \right )}}{a^{2} \tan \left (d x +c \right )^{2}}+\frac {\left (-20 i B +23 A \right ) \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{8 \sqrt {a}}\right )}{d}\)	\(143\)

[In]

int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

2/d*a^3*(-1/2*(-4*I*B+4*A)*2^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-((-1/2*I*B+9/
8*A)*(a+I*a*tan(d*x+c))^(3/2)+(1/2*I*a*B-7/8*a*A)*(a+I*a*tan(d*x+c))^(1/2))/a^2/tan(d*x+c)^2+1/8*(23*A-20*I*B)
/a^(1/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2)))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 774 vs. \(2 (134) = 268\).

Time = 0.27 (sec) , antiderivative size = 774, normalized size of antiderivative = 4.47 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(32*sqrt(2)*sqrt((A^2 - 2*I*A*B - B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*lo
g(4*((-I*A - B)*a^3*e^(I*d*x + I*c) - sqrt((A^2 - 2*I*A*B - B^2)*a^5/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a^2)) - 32*sqrt(2)*sqrt((A^2 - 2*I*A*B - B^2)*a^5/
d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(4*((-I*A - B)*a^3*e^(I*d*x + I*c) - sqrt((A^2 -
2*I*A*B - B^2)*a^5/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/
((-I*A - B)*a^2)) + sqrt((529*A^2 - 920*I*A*B - 400*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*
I*c) + d)*log(-16*(3*(-23*I*A - 20*B)*a^3*e^(2*I*d*x + 2*I*c) + (-23*I*A - 20*B)*a^3 + 2*sqrt(2)*sqrt((529*A^2
- 920*I*A*B - 400*B^2)*a^5/d^2)*(I*d*e^(3*I*d*x + 3*I*c) + I*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1)))*e^(-2*I*d*x - 2*I*c)/((23*I*A + 20*B)*a)) - sqrt((529*A^2 - 920*I*A*B - 400*B^2)*a^5/d^2)*(d*e^(4*I*d*x
+ 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-16*(3*(-23*I*A - 20*B)*a^3*e^(2*I*d*x + 2*I*c) + (-23*I*A - 20*B)
*a^3 + 2*sqrt(2)*sqrt((529*A^2 - 920*I*A*B - 400*B^2)*a^5/d^2)*(-I*d*e^(3*I*d*x + 3*I*c) - I*d*e^(I*d*x + I*c)
)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/((23*I*A + 20*B)*a)) - 4*sqrt(2)*((11*A - 4*I*B)*a^2
*e^(5*I*d*x + 5*I*c) + 4*A*a^2*e^(3*I*d*x + 3*I*c) - (7*A - 4*I*B)*a^2*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2
*I*c) + 1)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F(-1)]

Timed out. \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Timed out} \]

[In]

integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.19 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\frac {{\left (16 \, \sqrt {2} {\left (A - i \, B\right )} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - {\left (23 \, A - 20 i \, B\right )} \sqrt {a} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) + \frac {2 \, {\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (9 \, A - 4 i \, B\right )} a - \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (7 \, A - 4 i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + a^{2}}\right )} a^{2}}{8 \, d} \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/8*(16*sqrt(2)*(A - I*B)*sqrt(a)*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))) - (23*A - 20*I*B)*sqrt(a)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x
+ c) + a) + sqrt(a))) + 2*((I*a*tan(d*x + c) + a)^(3/2)*(9*A - 4*I*B)*a - sqrt(I*a*tan(d*x + c) + a)*(7*A - 4
*I*B)*a^2)/((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2))*a^2/d

Giac [F]

\[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\int { {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^3, x)

Mupad [B] (verification not implemented)

Time = 9.47 (sec) , antiderivative size = 2991, normalized size of antiderivative = 17.29 \[ \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx=\text {Too large to display} \]

[In]

int(cot(c + d*x)^3*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

2*atanh((17*A^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((1041*A^2*a^5)/(128*d^2) - ((289*A^4*a^22)/(4*d^4) + (3
136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)/(64*a^6) -
(57*B^2*a^5)/(8*d^2) - (A*B*a^5*243i)/(16*d^2))^(1/2))/(4*((663*A^3*a^11*d)/16 - B^3*a^11*d*252i - (7*A*d^3*(
(289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*
884i)/d^4)^(1/2))/8 + (B*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*
a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)*1i)/2 + 507*A*B^2*a^11*d + (A^2*B*a^11*d*861i)/4)) - (3*d^4*(a
+ a*tan(c + d*x)*1i)^(1/2)*((1041*A^2*a^5)/(128*d^2) - ((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A
^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)/(64*a^6) - (57*B^2*a^5)/(8*d^2) - (A*
B*a^5*243i)/(16*d^2))^(1/2)*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a
^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2))/(2*((663*A^3*a^14*d)/16 - B^3*a^14*d*252i + 507*A*B^2*a^14*d +
(A^2*B*a^14*d*861i)/4 - (7*A*a^3*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 +
(A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2))/8 + (B*a^3*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^2
2)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)*1i)/2)) + (28*B^2*a^8
*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((1041*A^2*a^5)/(128*d^2) - ((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 -
(1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)/(64*a^6) - (57*B^2*a^5)/(8*d^
2) - (A*B*a^5*243i)/(16*d^2))^(1/2))/((663*A^3*a^11*d)/16 - B^3*a^11*d*252i - (7*A*d^3*((289*A^4*a^22)/(4*d^4)
+ (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2))/8 +
(B*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3
*B*a^22*884i)/d^4)^(1/2)*1i)/2 + 507*A*B^2*a^11*d + (A^2*B*a^11*d*861i)/4) + (A*B*a^8*d^2*(a + a*tan(c + d*x)*
1i)^(1/2)*((1041*A^2*a^5)/(128*d^2) - ((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4
+ (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)/(64*a^6) - (57*B^2*a^5)/(8*d^2) - (A*B*a^5*243i)/(16*d
^2))^(1/2)*26i)/((663*A^3*a^11*d)/16 - B^3*a^11*d*252i - (7*A*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^
4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2))/8 + (B*d^3*((289*A^4*a^22
)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1
/2)*1i)/2 + 507*A*B^2*a^11*d + (A^2*B*a^11*d*861i)/4))*((1041*A^2*a^5)/(128*d^2) - ((289*A^4*a^22)/(4*d^4) + (
3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)/(64*a^6)
- (57*B^2*a^5)/(8*d^2) - (A*B*a^5*243i)/(16*d^2))^(1/2) + 2*atanh((3*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*(((289*
A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)
/d^4)^(1/2)/(64*a^6) + (1041*A^2*a^5)/(128*d^2) - (57*B^2*a^5)/(8*d^2) - (A*B*a^5*243i)/(16*d^2))^(1/2)*((289*
A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)
/d^4)^(1/2))/(2*((663*A^3*a^14*d)/16 - B^3*a^14*d*252i + 507*A*B^2*a^14*d + (A^2*B*a^14*d*861i)/4 + (7*A*a^3*d
^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a
^22*884i)/d^4)^(1/2))/8 - (B*a^3*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 +
(A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)*1i)/2)) + (17*A^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)
*(((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^
22*884i)/d^4)^(1/2)/(64*a^6) + (1041*A^2*a^5)/(128*d^2) - (57*B^2*a^5)/(8*d^2) - (A*B*a^5*243i)/(16*d^2))^(1/2
))/(4*((663*A^3*a^11*d)/16 - B^3*a^11*d*252i + (7*A*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*
A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2))/8 - (B*d^3*((289*A^4*a^22)/(4*d^4)
+ (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)*1i)/2
+ 507*A*B^2*a^11*d + (A^2*B*a^11*d*861i)/4)) + (28*B^2*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((289*A^4*a^22)/
(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2
)/(64*a^6) + (1041*A^2*a^5)/(128*d^2) - (57*B^2*a^5)/(8*d^2) - (A*B*a^5*243i)/(16*d^2))^(1/2))/((663*A^3*a^11*
d)/16 - B^3*a^11*d*252i + (7*A*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (
A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2))/8 - (B*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4
- (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)*1i)/2 + 507*A*B^2*a^11*d +
(A^2*B*a^11*d*861i)/4) + (A*B*a^8*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)
/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)/(64*a^6) + (1041*A^2*a^
5)/(128*d^2) - (57*B^2*a^5)/(8*d^2) - (A*B*a^5*243i)/(16*d^2))^(1/2)*26i)/((663*A^3*a^11*d)/16 - B^3*a^11*d*25
2i + (7*A*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4
+ (A^3*B*a^22*884i)/d^4)^(1/2))/8 - (B*d^3*((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22
)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*a^22*884i)/d^4)^(1/2)*1i)/2 + 507*A*B^2*a^11*d + (A^2*B*a^11*d*861i)/4
))*(((289*A^4*a^22)/(4*d^4) + (3136*B^4*a^22)/d^4 - (1752*A^2*B^2*a^22)/d^4 + (A*B^3*a^22*5824i)/d^4 + (A^3*B*
a^22*884i)/d^4)^(1/2)/(64*a^6) + (1041*A^2*a^5)/(128*d^2) - (57*B^2*a^5)/(8*d^2) - (A*B*a^5*243i)/(16*d^2))^(1
/2) - (((7*A*a^4 - B*a^4*4i)*(a + a*tan(c + d*x)*1i)^(1/2))/(4*d) - ((9*A*a^3 - B*a^3*4i)*(a + a*tan(c + d*x)*
1i)^(3/2))/(4*d))/((a + a*tan(c + d*x)*1i)^2 - 2*a*(a + a*tan(c + d*x)*1i) + a^2)