\(\int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [78]

Optimal result

Integrand size = 34, antiderivative size = 113 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {2} a^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 i a B \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

-2*a^(3/2)*A*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d+2*a^(3/2)*(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+2*I*a*B*(a+I*a*tan(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3675, 3681, 3561, 212, 3680, 65, 214} \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {2 \sqrt {2} a^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 i a B \sqrt {a+i a \tan (c+d x)}}{d} \]

[In]

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

[Out]

(-2*a^(3/2)*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + (2*Sqrt[2]*a^(3/2)*(A - I*B)*ArcTanh[Sqrt[a + I
*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((2*I)*a*B*Sqrt[a + I*a*Tan[c + d*x]])/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 3675

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*B*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*
(m + n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)
+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*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] && 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 {2 i a B \sqrt {a+i a \tan (c+d x)}}{d}+2 \int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {a A}{2}+\frac {1}{2} a (i A+2 B) \tan (c+d x)\right ) \, dx \\ & = \frac {2 i a B \sqrt {a+i a \tan (c+d x)}}{d}+A \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx+(2 a (i A+B)) \int \sqrt {a+i a \tan (c+d x)} \, dx \\ & = \frac {2 i a B \sqrt {a+i a \tan (c+d x)}}{d}+\frac {\left (a^2 A\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}+\frac {\left (4 a^2 (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 {2 \sqrt {2} a^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 i a B \sqrt {a+i a \tan (c+d x)}}{d}-\frac {(2 i a A) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d} \\ & = -\frac {2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {2} a^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {2 i a B \sqrt {a+i a \tan (c+d x)}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.64 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {-2 a^{3/2} A \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )+2 \sqrt {2} a^{3/2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )+2 i a B \sqrt {a+i a \tan (c+d x)}}{d} \]

[In]

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

[Out]

(-2*a^(3/2)*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] + 2*Sqrt[2]*a^(3/2)*(A - I*B)*ArcTanh[Sqrt[a + I*a*T
an[c + d*x]]/(Sqrt[2]*Sqrt[a])] + (2*I)*a*B*Sqrt[a + I*a*Tan[c + d*x]])/d

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {2 a \left (i B \sqrt {a +i a \tan \left (d x +c \right )}+\sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-\sqrt {a}\, A \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )\right )}{d}\)	\(87\)
default	\(\frac {2 a \left (i B \sqrt {a +i a \tan \left (d x +c \right )}+\sqrt {a}\, \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )-\sqrt {a}\, A \,\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )\right )}{d}\)	\(87\)

[In]

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

[Out]

2/d*a*(I*B*(a+I*a*tan(d*x+c))^(1/2)+a^(1/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/
2))-a^(1/2)*A*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. 514 vs. \(2 (86) = 172\).

Time = 0.26 (sec) , antiderivative size = 514, normalized size of antiderivative = 4.55 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {-4 i \, \sqrt {2} B a \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )} - 2 \, \sqrt {2} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} d \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a}\right ) + 2 \, \sqrt {2} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} d \log \left (\frac {4 \, {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a}\right ) + \sqrt {\frac {A^{2} a^{3}}{d^{2}}} d \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} + 2 \, \sqrt {2} \sqrt {\frac {A^{2} a^{3}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right ) - \sqrt {\frac {A^{2} a^{3}}{d^{2}}} d \log \left (\frac {16 \, {\left (3 \, A a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + A a^{2} - 2 \, \sqrt {2} \sqrt {\frac {A^{2} a^{3}}{d^{2}}} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{A}\right )}{2 \, d} \]

[In]

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

[Out]

-1/2*(-4*I*sqrt(2)*B*a*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 2*sqrt(2)*sqrt((A^2 - 2*I*A*B - B^2
)*a^3/d^2)*d*log(4*((-I*A - B)*a^2*e^(I*d*x + I*c) - sqrt((A^2 - 2*I*A*B - B^2)*a^3/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(2)*sqrt((A^2 - 2*I*A*B
- B^2)*a^3/d^2)*d*log(4*((-I*A - B)*a^2*e^(I*d*x + I*c) - sqrt((A^2 - 2*I*A*B - B^2)*a^3/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)) + sqrt(A^2*a^3/d^2)*d*lo
g(16*(3*A*a^2*e^(2*I*d*x + 2*I*c) + A*a^2 + 2*sqrt(2)*sqrt(A^2*a^3/d^2)*(d*e^(3*I*d*x + 3*I*c) + 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)/A) - sqrt(A^2*a^3/d^2)*d*log(16*(3*A*a^2*e^(2*I*
d*x + 2*I*c) + A*a^2 - 2*sqrt(2)*sqrt(A^2*a^3/d^2)*(d*e^(3*I*d*x + 3*I*c) + 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)/A))/d

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.15 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=-\frac {\sqrt {2} {\left (A - i \, B\right )} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - A a^{\frac {3}{2}} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right ) - 2 i \, \sqrt {i \, a \tan \left (d x + c\right ) + a} B a}{d} \]

[In]

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

[Out]

-(sqrt(2)*(A - I*B)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*ta
n(d*x + c) + a))) - A*a^(3/2)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)
)) - 2*I*sqrt(I*a*tan(d*x + c) + a)*B*a)/d

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.01 (sec) , antiderivative size = 553, normalized size of antiderivative = 4.89 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx=\frac {B\,a\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{d}-\frac {2\,A\,\mathrm {atanh}\left (-\frac {32\,A^3\,a^6\,d\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{-32\,d\,A^3\,a^8+128{}\mathrm {i}\,d\,A^2\,B\,a^8+64\,d\,A\,B^2\,a^8}+\frac {64\,A\,B^2\,a^6\,d\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{-32\,d\,A^3\,a^8+128{}\mathrm {i}\,d\,A^2\,B\,a^8+64\,d\,A\,B^2\,a^8}+\frac {A^2\,B\,a^6\,d\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,128{}\mathrm {i}}{-32\,d\,A^3\,a^8+128{}\mathrm {i}\,d\,A^2\,B\,a^8+64\,d\,A\,B^2\,a^8}\right )\,\sqrt {a^3}}{d}+\frac {2\,\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,A^3\,a^6\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,16{}\mathrm {i}}{32\,d\,A^3\,a^8-160{}\mathrm {i}\,d\,A^2\,B\,a^8-192\,d\,A\,B^2\,a^8+64{}\mathrm {i}\,d\,B^3\,a^8}-\frac {32\,\sqrt {2}\,B^3\,a^6\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{32\,d\,A^3\,a^8-160{}\mathrm {i}\,d\,A^2\,B\,a^8-192\,d\,A\,B^2\,a^8+64{}\mathrm {i}\,d\,B^3\,a^8}-\frac {\sqrt {2}\,A\,B^2\,a^6\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,96{}\mathrm {i}}{32\,d\,A^3\,a^8-160{}\mathrm {i}\,d\,A^2\,B\,a^8-192\,d\,A\,B^2\,a^8+64{}\mathrm {i}\,d\,B^3\,a^8}+\frac {80\,\sqrt {2}\,A^2\,B\,a^6\,d\,\sqrt {-a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{32\,d\,A^3\,a^8-160{}\mathrm {i}\,d\,A^2\,B\,a^8-192\,d\,A\,B^2\,a^8+64{}\mathrm {i}\,d\,B^3\,a^8}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,\sqrt {-a^3}}{d} \]

[In]

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

[Out]

(B*a*(a + a*tan(c + d*x)*1i)^(1/2)*2i)/d - (2*A*atanh((64*A*B^2*a^6*d*(a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2
))/(64*A*B^2*a^8*d - 32*A^3*a^8*d + A^2*B*a^8*d*128i) - (32*A^3*a^6*d*(a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2
))/(64*A*B^2*a^8*d - 32*A^3*a^8*d + A^2*B*a^8*d*128i) + (A^2*B*a^6*d*(a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)
*128i)/(64*A*B^2*a^8*d - 32*A^3*a^8*d + A^2*B*a^8*d*128i))*(a^3)^(1/2))/d + (2*2^(1/2)*atanh((2^(1/2)*A^3*a^6*
d*(-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*16i)/(32*A^3*a^8*d + B^3*a^8*d*64i - 192*A*B^2*a^8*d - A^2*B*a^8*
d*160i) - (32*2^(1/2)*B^3*a^6*d*(-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(32*A^3*a^8*d + B^3*a^8*d*64i - 19
2*A*B^2*a^8*d - A^2*B*a^8*d*160i) - (2^(1/2)*A*B^2*a^6*d*(-a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2)*96i)/(32*A
^3*a^8*d + B^3*a^8*d*64i - 192*A*B^2*a^8*d - A^2*B*a^8*d*160i) + (80*2^(1/2)*A^2*B*a^6*d*(-a^3)^(1/2)*(a + a*t
an(c + d*x)*1i)^(1/2))/(32*A^3*a^8*d + B^3*a^8*d*64i - 192*A*B^2*a^8*d - A^2*B*a^8*d*160i))*(A*1i + B)*(-a^3)^
(1/2))/d