\(\int \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^{3/2}} \, dx\) [855]

Optimal result

Integrand size = 38, antiderivative size = 147 \[ \int \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^{3/2}} \, dx=-\frac {(i A+B) (c-i d) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(i A-B) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {B (c+3 i d)+A (i c+d)}{2 a f \sqrt {a+i a \tan (e+f x)}} \]

[Out]

-1/4*(I*A+B)*(c-I*d)*arctanh(1/2*(a+I*a*tan(f*x+e))^(1/2)*2^(1/2)/a^(1/2))/a^(3/2)/f*2^(1/2)+1/2*(B*(c+3*I*d)+
A*(I*c+d))/a/f/(a+I*a*tan(f*x+e))^(1/2)+1/3*(I*A-B)*(c+I*d)/f/(a+I*a*tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {3671, 3607, 3561, 212} \[ \int \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^{3/2}} \, dx=-\frac {(B+i A) (c-i d) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(-B+i A) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {A (d+i c)+B (c+3 i d)}{2 a f \sqrt {a+i a \tan (e+f x)}} \]

[In]

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

[Out]

-1/2*((I*A + B)*(c - I*d)*ArcTanh[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*a^(3/2)*f) + ((I*A -
B)*(c + I*d))/(3*f*(a + I*a*Tan[e + f*x])^(3/2)) + (B*(c + (3*I)*d) + A*(I*c + d))/(2*a*f*Sqrt[a + I*a*Tan[e
+ f*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]

Rubi steps \begin{align*} \text {integral}& = \frac {(i A-B) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}-\frac {i \int \frac {a (B (c+i d)+A (i c+d))+2 a B d \tan (e+f x)}{\sqrt {a+i a \tan (e+f x)}} \, dx}{2 a^2} \\ & = \frac {(i A-B) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {B (c+3 i d)+A (i c+d)}{2 a f \sqrt {a+i a \tan (e+f x)}}+\frac {((A-i B) (c-i d)) \int \sqrt {a+i a \tan (e+f x)} \, dx}{4 a^2} \\ & = \frac {(i A-B) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {B (c+3 i d)+A (i c+d)}{2 a f \sqrt {a+i a \tan (e+f x)}}-\frac {(i (A-i B) (c-i d)) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (e+f x)}\right )}{2 a f} \\ & = -\frac {(i A+B) (c-i d) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} a^{3/2} f}+\frac {(i A-B) (c+i d)}{3 f (a+i a \tan (e+f x))^{3/2}}+\frac {B (c+3 i d)+A (i c+d)}{2 a f \sqrt {a+i a \tan (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.52 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.96 \[ \int \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^{3/2}} \, dx=-\frac {i \left (\frac {3 \sqrt {2} (A-i B) (c-i d) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {a}}-\frac {4 a (A+i B) (c+i d)}{(a+i a \tan (e+f x))^{3/2}}-\frac {6 (A c-i B c-i A d+3 B d)}{\sqrt {a+i a \tan (e+f x)}}\right )}{12 a f} \]

[In]

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

[Out]

((-1/12*I)*((3*Sqrt[2]*(A - I*B)*(c - I*d)*ArcTanh[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])])/Sqrt[a] - (4
*a*(A + I*B)*(c + I*d))/(a + I*a*Tan[e + f*x])^(3/2) - (6*(A*c - I*B*c - I*A*d + 3*B*d))/Sqrt[a + I*a*Tan[e +
f*x]]))/(a*f)

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {2 i \left (-\frac {\frac {1}{4} i A d +\frac {1}{4} i B c -\frac {1}{4} c A -\frac {3}{4} B d}{\sqrt {a +i a \tan \left (f x +e \right )}}+\frac {a \left (i A d +i B c +c A -B d \right )}{6 \left (a +i a \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\left (-\frac {1}{4} i A d -\frac {1}{4} i B c +\frac {1}{4} c A -\frac {1}{4} B d \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}\right )}{f a}\)	\(131\)
default	\(\frac {2 i \left (-\frac {\frac {1}{4} i A d +\frac {1}{4} i B c -\frac {1}{4} c A -\frac {3}{4} B d}{\sqrt {a +i a \tan \left (f x +e \right )}}+\frac {a \left (i A d +i B c +c A -B d \right )}{6 \left (a +i a \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\left (-\frac {1}{4} i A d -\frac {1}{4} i B c +\frac {1}{4} c A -\frac {1}{4} B d \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{2 \sqrt {a}}\right )}{f a}\)	\(131\)
parts	\(\frac {2 i c A a \left (\frac {1}{4 a^{2} \sqrt {a +i a \tan \left (f x +e \right )}}+\frac {1}{6 a \left (a +i a \tan \left (f x +e \right )\right )^{\frac {3}{2}}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 a^{\frac {5}{2}}}\right )}{f}+\frac {\left (A d +c B \right ) \left (-\frac {1}{3 \left (a +i a \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {1}{2 a \sqrt {a +i a \tan \left (f x +e \right )}}-\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4 a^{\frac {3}{2}}}\right )}{f}+\frac {2 i B d \left (\frac {3}{4 \sqrt {a +i a \tan \left (f x +e \right )}}-\frac {a}{6 \left (a +i a \tan \left (f x +e \right )\right )^{\frac {3}{2}}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{8 \sqrt {a}}\right )}{f a}\)	\(235\)

[In]

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

[Out]

2*I/f/a*(-(1/4*I*A*d+1/4*I*B*c-1/4*c*A-3/4*B*d)/(a+I*a*tan(f*x+e))^(1/2)+1/6*a*(-B*d+I*A*d+I*B*c+c*A)/(a+I*a*t
an(f*x+e))^(3/2)-1/2*(-1/4*I*A*d-1/4*I*B*c+1/4*c*A-1/4*B*d)*2^(1/2)/a^(1/2)*arctanh(1/2*(a+I*a*tan(f*x+e))^(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. 619 vs. \(2 (108) = 216\).

Time = 0.25 (sec) , antiderivative size = 619, normalized size of antiderivative = 4.21 \[ \int \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} c^{2} + 2 \, {\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} c d - {\left (A^{2} - 2 i \, A B - B^{2}\right )} d^{2}}{a^{3} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, a^{2} f\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} c^{2} + 2 \, {\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} c d - {\left (A^{2} - 2 i \, A B - B^{2}\right )} d^{2}}{a^{3} f^{2}}} + {\left ({\left (A - i \, B\right )} a c + {\left (-i \, A - B\right )} a d\right )} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{{\left (A - i \, B\right )} c - {\left (i \, A + B\right )} d}\right ) - 3 \, \sqrt {\frac {1}{2}} a^{2} f \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} c^{2} + 2 \, {\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} c d - {\left (A^{2} - 2 i \, A B - B^{2}\right )} d^{2}}{a^{3} f^{2}}} e^{\left (3 i \, f x + 3 i \, e\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, a^{2} f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, a^{2} f\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} c^{2} + 2 \, {\left (-i \, A^{2} - 2 \, A B + i \, B^{2}\right )} c d - {\left (A^{2} - 2 i \, A B - B^{2}\right )} d^{2}}{a^{3} f^{2}}} + {\left ({\left (A - i \, B\right )} a c + {\left (-i \, A - B\right )} a d\right )} e^{\left (i \, f x + i \, e\right )}\right )} e^{\left (-i \, f x - i \, e\right )}}{{\left (A - i \, B\right )} c - {\left (i \, A + B\right )} d}\right ) + \sqrt {2} {\left ({\left (i \, A - B\right )} c - {\left (A + i \, B\right )} d - 2 \, {\left ({\left (-2 i \, A - B\right )} c - {\left (A + 4 i \, B\right )} d\right )} e^{\left (4 i \, f x + 4 i \, e\right )} + {\left ({\left (5 i \, A + B\right )} c + {\left (A + 7 i \, B\right )} d\right )} e^{\left (2 i \, f x + 2 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-3 i \, f x - 3 i \, e\right )}}{12 \, a^{2} f} \]

[In]

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

[Out]

1/12*(3*sqrt(1/2)*a^2*f*sqrt(-((A^2 - 2*I*A*B - B^2)*c^2 + 2*(-I*A^2 - 2*A*B + I*B^2)*c*d - (A^2 - 2*I*A*B - B
^2)*d^2)/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(4*(sqrt(2)*sqrt(1/2)*(I*a^2*f*e^(2*I*f*x + 2*I*e) + I*a^2*f)*sqrt(
a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-((A^2 - 2*I*A*B - B^2)*c^2 + 2*(-I*A^2 - 2*A*B + I*B^2)*c*d - (A^2 - 2*I*A*
B - B^2)*d^2)/(a^3*f^2)) + ((A - I*B)*a*c + (-I*A - B)*a*d)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)/((A - I*B)*c - (
I*A + B)*d)) - 3*sqrt(1/2)*a^2*f*sqrt(-((A^2 - 2*I*A*B - B^2)*c^2 + 2*(-I*A^2 - 2*A*B + I*B^2)*c*d - (A^2 - 2*
I*A*B - B^2)*d^2)/(a^3*f^2))*e^(3*I*f*x + 3*I*e)*log(4*(sqrt(2)*sqrt(1/2)*(-I*a^2*f*e^(2*I*f*x + 2*I*e) - I*a^
2*f)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-((A^2 - 2*I*A*B - B^2)*c^2 + 2*(-I*A^2 - 2*A*B + I*B^2)*c*d - (A^
2 - 2*I*A*B - B^2)*d^2)/(a^3*f^2)) + ((A - I*B)*a*c + (-I*A - B)*a*d)*e^(I*f*x + I*e))*e^(-I*f*x - I*e)/((A -
I*B)*c - (I*A + B)*d)) + sqrt(2)*((I*A - B)*c - (A + I*B)*d - 2*((-2*I*A - B)*c - (A + 4*I*B)*d)*e^(4*I*f*x +
4*I*e) + ((5*I*A + B)*c + (A + 7*I*B)*d)*e^(2*I*f*x + 2*I*e))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-3*I*f*x -
3*I*e)/(a^2*f)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (A - i \, B\right )} {\left (c - i \, d\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (f x + e\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (f x + e\right ) + a}}\right )}{\sqrt {a}} + \frac {4 \, {\left (2 \, {\left (A + i \, B\right )} a c + 2 \, {\left (i \, A - B\right )} a d + 3 \, {\left ({\left (A - i \, B\right )} c + {\left (-i \, A + 3 \, B\right )} d\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}\right )}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}\right )}}{24 \, a f} \]

[In]

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

[Out]

1/24*I*(3*sqrt(2)*(A - I*B)*(c - I*d)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(f*x + e) + a))/(sqrt(2)*sqrt(a) + s
qrt(I*a*tan(f*x + e) + a)))/sqrt(a) + 4*(2*(A + I*B)*a*c + 2*(I*A - B)*a*d + 3*((A - I*B)*c + (-I*A + 3*B)*d)*
(I*a*tan(f*x + e) + a))/(I*a*tan(f*x + e) + a)^(3/2))/(a*f)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.59 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.67 \[ \int \frac {(A+B \tan (e+f x)) (c+d \tan (e+f x))}{(a+i a \tan (e+f x))^{3/2}} \, dx=\frac {\frac {\left (A\,c+A\,d\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{3\,f}+\frac {\left (A\,c-A\,d\,1{}\mathrm {i}\right )\,\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,a\,f}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\frac {B\,c+B\,d\,1{}\mathrm {i}}{3\,f}-\frac {\left (B\,c+B\,d\,3{}\mathrm {i}\right )\,\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{2\,a\,f}}{{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}+\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,B\,\left (d+c\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}\,\left (B\,c-B\,d\,1{}\mathrm {i}\right )}\right )\,\left (d+c\,1{}\mathrm {i}\right )}{4\,{\left (-a\right )}^{3/2}\,f}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,A\,\left (d+c\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {a}\,\left (A\,c-A\,d\,1{}\mathrm {i}\right )}\right )\,\left (d+c\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,a^{3/2}\,f} \]

[In]

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

[Out]

(((A*c + A*d*1i)*1i)/(3*f) + ((A*c - A*d*1i)*(a + a*tan(e + f*x)*1i)*1i)/(2*a*f))/(a + a*tan(e + f*x)*1i)^(3/2
) - ((B*c + B*d*1i)/(3*f) - ((B*c + B*d*3i)*(a + a*tan(e + f*x)*1i))/(2*a*f))/(a + a*tan(e + f*x)*1i)^(3/2) +
(2^(1/2)*B*atanh((2^(1/2)*B*(c*1i + d)*(a + a*tan(e + f*x)*1i)^(1/2))/(2*(-a)^(1/2)*(B*c - B*d*1i)))*(c*1i + d
))/(4*(-a)^(3/2)*f) + (2^(1/2)*A*atan((2^(1/2)*A*(c*1i + d)*(a + a*tan(e + f*x)*1i)^(1/2))/(2*a^(1/2)*(A*c - A
*d*1i)))*(c*1i + d)*1i)/(4*a^(3/2)*f)