\(\int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\) [106]

Optimal result

Integrand size = 36, antiderivative size = 167 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A-13 B}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i A-31 B}{20 a^2 d \sqrt {a+i a \tan (c+d x)}} \]

[Out]

1/8*(I*A+B)*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)+1/20*(-I*A+31*B)/a^2/d/(a+
I*a*tan(d*x+c))^(1/2)+1/5*(I*A-B)*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(5/2)+1/30*(3*I*A-13*B)/a/d/(a+I*a*tan(d*x
+c))^(3/2)

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3676, 3671, 3607, 3561, 212} \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}-\frac {-31 B+i A}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(-B+i A) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {-13 B+3 i A}{30 a d (a+i a \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*a^(5/2)*d) + ((I*A - B)*Tan[c + d
*x]^2)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + ((3*I)*A - 13*B)/(30*a*d*(a + I*a*Tan[c + d*x])^(3/2)) - (I*A - 31
*B)/(20*a^2*d*Sqrt[a + I*a*Tan[c + d*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 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 3607

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

Rule 3671

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(A*b - a*B))*(a*c + b*d)*((a + b*Tan[e + f*x])^m/(2*a^2*f*m)), x] + Dis
t[1/(2*a*b), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[A*b*c + a*B*c + a*A*d + b*B*d + 2*a*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] && EqQ[a^2 + b^2, 0]

Rule 3676

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

Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}-\frac {\int \frac {\tan (c+d x) \left (2 a (i A-B)-\frac {1}{2} a (A-9 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2} \\ & = \frac {(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A-13 B}{30 a d (a+i a \tan (c+d x))^{3/2}}+\frac {i \int \frac {\frac {1}{2} a^2 (3 i A-13 B)-a^2 (A-9 i B) \tan (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx}{10 a^4} \\ & = \frac {(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A-13 B}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i A-31 B}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}-\frac {(A-i B) \int \sqrt {a+i a \tan (c+d x)} \, dx}{8 a^3} \\ & = \frac {(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A-13 B}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i A-31 B}{20 a^2 d \sqrt {a+i a \tan (c+d x)}}+\frac {(i A+B) \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 {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{4 \sqrt {2} a^{5/2} d}+\frac {(i A-B) \tan ^2(c+d x)}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A-13 B}{30 a d (a+i a \tan (c+d x))^{3/2}}-\frac {i A-31 B}{20 a^2 d \sqrt {a+i a \tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.22 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.06 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {2 B \tan ^2(c+d x)}{d (a+i a \tan (c+d x))^{5/2}}+\frac {i \left (\frac {15 \sqrt {2} (A-i B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{a^{3/2}}-\frac {24 a (A-9 i B)}{(a+i a \tan (c+d x))^{5/2}}+\frac {20 (3 A-19 i B)}{(a+i a \tan (c+d x))^{3/2}}-\frac {30 (A-i B)}{a \sqrt {a+i a \tan (c+d x)}}\right )}{120 a d} \]

[In]

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

[Out]

(-2*B*Tan[c + d*x]^2)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + ((I/120)*((15*Sqrt[2]*(A - I*B)*ArcTanh[Sqrt[a + I*a*
Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/a^(3/2) - (24*a*(A - (9*I)*B))/(a + I*a*Tan[c + d*x])^(5/2) + (20*(3*A - (19
*I)*B))/(a + I*a*Tan[c + d*x])^(3/2) - (30*(A - I*B))/(a*Sqrt[a + I*a*Tan[c + d*x]])))/(a*d)

Maple [A] (verified)

Time = 0.13 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.74

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {\frac {A}{8}+\frac {7 i B}{8}}{\sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a \left (5 i B +3 A \right )}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{2} \left (i B +A \right )}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\left (-\frac {A}{8}+\frac {i B}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}\right )}{d \,a^{2}}\)	\(124\)
default	\(\frac {2 i \left (-\frac {\frac {A}{8}+\frac {7 i B}{8}}{\sqrt {a +i a \tan \left (d x +c \right )}}+\frac {a \left (5 i B +3 A \right )}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {a^{2} \left (i B +A \right )}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}-\frac {\left (-\frac {A}{8}+\frac {i B}{8}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}\right )}{d \,a^{2}}\)	\(124\)
parts	\(\frac {2 i A \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 a^{\frac {3}{2}}}+\frac {1}{4 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}-\frac {1}{8 a \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {a}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d a}+\frac {2 B \left (\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{16 \sqrt {a}}+\frac {7}{8 \sqrt {a +i a \tan \left (d x +c \right )}}-\frac {5 a}{12 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}+\frac {a^{2}}{10 \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}\right )}{d \,a^{2}}\)	\(189\)

[In]

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

[Out]

2*I/d/a^2*(-(1/8*A+7/8*I*B)/(a+I*a*tan(d*x+c))^(1/2)+1/12*a*(3*A+5*I*B)/(a+I*a*tan(d*x+c))^(3/2)-1/10*a^2*(A+I
*B)/(a+I*a*tan(d*x+c))^(5/2)-1/2*(-1/8*A+1/8*I*B)*2^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(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. 393 vs. \(2 (128) = 256\).

Time = 0.26 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.35 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} + {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a^{5} d^{2}}} - {\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left ({\left (-3 i \, A + 83 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} - 2 \, {\left (-3 i \, A - 32 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (-3 i \, A + 8 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]

[In]

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

[Out]

-1/120*(15*sqrt(1/2)*a^3*d*sqrt(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*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(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2))
+ (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 15*sqrt(1/2)*a^3*d*sqrt(-(A^2 - 2*I*A*B - B^2)/
(a^5*d^2))*e^(5*I*d*x + 5*I*c)*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(-(A^2 - 2*I*A*B - B^2)/(a^5*d^2)) - (-I*A - B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A +
B)) - sqrt(2)*((-3*I*A + 83*B)*e^(6*I*d*x + 6*I*c) - 2*(-3*I*A - 32*B)*e^(4*I*d*x + 4*I*c) - 2*(-3*I*A + 8*B)
*e^(2*I*d*x + 2*I*c) - 3*I*A + 3*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-5*I*d*x - 5*I*c)/(a^3*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=-\frac {i \, {\left (15 \, \sqrt {2} {\left (A - i \, B\right )} \sqrt {a} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} {\left (A + 7 i \, B\right )} a - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} {\left (3 \, A + 5 i \, B\right )} a^{2} + 12 \, {\left (A + i \, B\right )} a^{3}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\right )}}{240 \, a^{3} d} \]

[In]

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

[Out]

-1/240*I*(15*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))) + 4*(15*(I*a*tan(d*x + c) + a)^2*(A + 7*I*B)*a - 10*(I*a*tan(d*x + c) + a)*(3*A +
5*I*B)*a^2 + 12*(A + I*B)*a^3)/(I*a*tan(d*x + c) + a)^(5/2))/(a^3*d)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 7.95 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.12 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx=\frac {A\,1{}\mathrm {i}}{20\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {A\,{\mathrm {tan}\left (c+d\,x\right )}^2\,1{}\mathrm {i}}{4\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\frac {B\,a^2}{5}+\frac {7\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{4}-\frac {5\,B\,a\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}{6}}{a^2\,d\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{5/2}\,d}+\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}}\right )}{8\,a^{5/2}\,d} \]

[In]

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

[Out]

(A*1i)/(20*d*(a + a*tan(c + d*x)*1i)^(5/2)) + (A*tan(c + d*x)^2*1i)/(4*d*(a + a*tan(c + d*x)*1i)^(5/2)) + ((B*
a^2)/5 + (7*B*(a + a*tan(c + d*x)*1i)^2)/4 - (5*B*a*(a + a*tan(c + d*x)*1i))/6)/(a^2*d*(a + a*tan(c + d*x)*1i)
^(5/2)) - (2^(1/2)*A*atan((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*(-a)^(1/2)))*1i)/(8*(-a)^(5/2)*d) + (2^(1
/2)*B*atanh((2^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^(1/2))))/(8*a^(5/2)*d)