\(\int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx\) [769]

Optimal result

Integrand size = 43, antiderivative size = 184 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {(5 i A-B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a c^{3/2} f}-\frac {5 i A-B}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {5 i A-B}{8 a c f \sqrt {c-i c \tan (e+f x)}} \]

[Out]

1/16*(5*I*A-B)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2))/a/c^(3/2)/f*2^(1/2)+1/8*(-5*I*A+B)/a/c/f/
(c-I*c*tan(f*x+e))^(1/2)+1/12*(-5*I*A+B)/a/f/(c-I*c*tan(f*x+e))^(3/2)+1/2*(I*A-B)/a/f/(1+I*tan(f*x+e))/(c-I*c*
tan(f*x+e))^(3/2)

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.116, Rules used = {3669, 79, 53, 65, 214} \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {(-B+5 i A) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a c^{3/2} f}+\frac {-B+i A}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {-B+5 i A}{8 a c f \sqrt {c-i c \tan (e+f x)}}-\frac {-B+5 i A}{12 a f (c-i c \tan (e+f x))^{3/2}} \]

[In]

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

[Out]

(((5*I)*A - B)*ArcTanh[Sqrt[c - I*c*Tan[e + f*x]]/(Sqrt[2]*Sqrt[c])])/(8*Sqrt[2]*a*c^(3/2)*f) - ((5*I)*A - B)/
(12*a*f*(c - I*c*Tan[e + f*x])^(3/2)) + (I*A - B)/(2*a*f*(1 + I*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2)) -
((5*I)*A - B)/(8*a*c*f*Sqrt[c - I*c*Tan[e + f*x]])

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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 3669

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

Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int \frac {A+B x}{(a+i a x)^2 (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {((5 A+i B) c) \text {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{5/2}} \, dx,x,\tan (e+f x)\right )}{4 f} \\ & = -\frac {5 i A-B}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}+\frac {(5 A+i B) \text {Subst}\left (\int \frac {1}{(a+i a x) (c-i c x)^{3/2}} \, dx,x,\tan (e+f x)\right )}{8 f} \\ & = -\frac {5 i A-B}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {5 i A-B}{8 a c f \sqrt {c-i c \tan (e+f x)}}+\frac {(5 A+i B) \text {Subst}\left (\int \frac {1}{(a+i a x) \sqrt {c-i c x}} \, dx,x,\tan (e+f x)\right )}{16 c f} \\ & = -\frac {5 i A-B}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {5 i A-B}{8 a c f \sqrt {c-i c \tan (e+f x)}}+\frac {(5 i A-B) \text {Subst}\left (\int \frac {1}{2 a-\frac {a x^2}{c}} \, dx,x,\sqrt {c-i c \tan (e+f x)}\right )}{8 c^2 f} \\ & = \frac {(5 i A-B) \text {arctanh}\left (\frac {\sqrt {c-i c \tan (e+f x)}}{\sqrt {2} \sqrt {c}}\right )}{8 \sqrt {2} a c^{3/2} f}-\frac {5 i A-B}{12 a f (c-i c \tan (e+f x))^{3/2}}+\frac {i A-B}{2 a f (1+i \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}-\frac {5 i A-B}{8 a c f \sqrt {c-i c \tan (e+f x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 4.71 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.56 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {3 (-5 i A+B) \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},-\frac {1}{2} i (i+\tan (e+f x))\right )+2 i \cos (e+f x) ((A+5 i B) \cos (e+f x)+(-5 i A+B) \sin (e+f x))}{24 a c f \sqrt {c-i c \tan (e+f x)}} \]

[In]

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

[Out]

(3*((-5*I)*A + B)*Hypergeometric2F1[-1/2, 1, 1/2, (-1/2*I)*(I + Tan[e + f*x])] + (2*I)*Cos[e + f*x]*((A + (5*I
)*B)*Cos[e + f*x] + ((-5*I)*A + B)*Sin[e + f*x]))/(24*a*c*f*Sqrt[c - I*c*Tan[e + f*x]])

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.77

method	result	size
derivativedivides	\(\frac {2 i c \left (\frac {\frac {\left (\frac {i B}{8}+\frac {A}{8}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {i B}{2}+\frac {5 A}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{4 c^{2}}-\frac {A}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +A}{12 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f a}\)	\(141\)
default	\(\frac {2 i c \left (\frac {\frac {\left (\frac {i B}{8}+\frac {A}{8}\right ) \sqrt {c -i c \tan \left (f x +e \right )}}{\frac {c}{2}+\frac {i c \tan \left (f x +e \right )}{2}}+\frac {\left (\frac {i B}{2}+\frac {5 A}{2}\right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )}{4 \sqrt {c}}}{4 c^{2}}-\frac {A}{4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}}-\frac {-i B +A}{12 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\right )}{f a}\)	\(141\)

[In]

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

[Out]

2*I/f/a*c*(1/4/c^2*((1/8*I*B+1/8*A)*(c-I*c*tan(f*x+e))^(1/2)/(1/2*c+1/2*I*c*tan(f*x+e))+1/4*(1/2*I*B+5/2*A)*2^
(1/2)/c^(1/2)*arctanh(1/2*(c-I*c*tan(f*x+e))^(1/2)*2^(1/2)/c^(1/2)))-1/4/c^2*A/(c-I*c*tan(f*x+e))^(1/2)-1/12/c
*(A-I*B)/(c-I*c*tan(f*x+e))^(3/2))

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. 388 vs. \(2 (139) = 278\).

Time = 0.26 (sec) , antiderivative size = 388, normalized size of antiderivative = 2.11 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (3 \, \sqrt {\frac {1}{2}} a c^{2} f \sqrt {-\frac {25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a c f e^{\left (2 i \, f x + 2 i \, e\right )} + a c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} + 5 i \, A - B\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a c f}\right ) - 3 \, \sqrt {\frac {1}{2}} a c^{2} f \sqrt {-\frac {25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} e^{\left (2 i \, f x + 2 i \, e\right )} \log \left (-\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a c f e^{\left (2 i \, f x + 2 i \, e\right )} + a c f\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {-\frac {25 \, A^{2} + 10 i \, A B - B^{2}}{a^{2} c^{3} f^{2}}} - 5 i \, A + B\right )} e^{\left (-i \, f x - i \, e\right )}}{4 \, a c f}\right ) - \sqrt {2} {\left (2 \, {\left (i \, A + B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 4 \, {\left (4 i \, A + B\right )} e^{\left (4 i \, f x + 4 i \, e\right )} - {\left (-11 i \, A - 5 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}\right )} e^{\left (-2 i \, f x - 2 i \, e\right )}}{48 \, a c^{2} f} \]

[In]

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

[Out]

1/48*(3*sqrt(1/2)*a*c^2*f*sqrt(-(25*A^2 + 10*I*A*B - B^2)/(a^2*c^3*f^2))*e^(2*I*f*x + 2*I*e)*log(1/4*(sqrt(2)*
sqrt(1/2)*(a*c*f*e^(2*I*f*x + 2*I*e) + a*c*f)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(-(25*A^2 + 10*I*A*B - B^2
)/(a^2*c^3*f^2)) + 5*I*A - B)*e^(-I*f*x - I*e)/(a*c*f)) - 3*sqrt(1/2)*a*c^2*f*sqrt(-(25*A^2 + 10*I*A*B - B^2)/
(a^2*c^3*f^2))*e^(2*I*f*x + 2*I*e)*log(-1/4*(sqrt(2)*sqrt(1/2)*(a*c*f*e^(2*I*f*x + 2*I*e) + a*c*f)*sqrt(c/(e^(
2*I*f*x + 2*I*e) + 1))*sqrt(-(25*A^2 + 10*I*A*B - B^2)/(a^2*c^3*f^2)) - 5*I*A + B)*e^(-I*f*x - I*e)/(a*c*f)) -
sqrt(2)*(2*(I*A + B)*e^(6*I*f*x + 6*I*e) + 4*(4*I*A + B)*e^(4*I*f*x + 4*I*e) - (-11*I*A - 5*B)*e^(2*I*f*x + 2
*I*e) - 3*I*A + 3*B)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-2*I*f*x - 2*I*e)/(a*c^2*f)

Sympy [F]

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

[In]

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

[Out]

-I*(Integral(A/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - I*c*sqrt(-I*c*tan(e + f*x) + c)), x) + Inte
gral(B*tan(e + f*x)/(-I*c*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2 - I*c*sqrt(-I*c*tan(e + f*x) + c)), x))/
a

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.91 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i \, {\left (\frac {3 \, \sqrt {2} {\left (5 \, A + i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {c} - \sqrt {-i \, c \tan \left (f x + e\right ) + c}}{\sqrt {2} \sqrt {c} + \sqrt {-i \, c \tan \left (f x + e\right ) + c}}\right )}{a \sqrt {c}} + \frac {4 \, {\left (3 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{2} {\left (5 \, A + i \, B\right )} - 4 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )} {\left (5 \, A + i \, B\right )} c - 8 \, {\left (A - i \, B\right )} c^{2}\right )}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a - 2 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} a c}\right )}}{96 \, c f} \]

[In]

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

[Out]

-1/96*I*(3*sqrt(2)*(5*A + I*B)*log(-(sqrt(2)*sqrt(c) - sqrt(-I*c*tan(f*x + e) + c))/(sqrt(2)*sqrt(c) + sqrt(-I
*c*tan(f*x + e) + c)))/(a*sqrt(c)) + 4*(3*(-I*c*tan(f*x + e) + c)^2*(5*A + I*B) - 4*(-I*c*tan(f*x + e) + c)*(5
*A + I*B)*c - 8*(A - I*B)*c^2)/((-I*c*tan(f*x + e) + c)^(5/2)*a - 2*(-I*c*tan(f*x + e) + c)^(3/2)*a*c))/(c*f)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.70 (sec) , antiderivative size = 261, normalized size of antiderivative = 1.42 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\frac {B\,c}{3}-\frac {B\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}{6}+\frac {B\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{8\,c}}{a\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}-2\,a\,c\,f\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}-\frac {\frac {A\,\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )\,5{}\mathrm {i}}{6\,a\,f}+\frac {A\,c\,1{}\mathrm {i}}{3\,a\,f}-\frac {A\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2\,5{}\mathrm {i}}{8\,a\,c\,f}}{2\,c\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2}-{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}+\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-c}}\right )\,5{}\mathrm {i}}{16\,a\,{\left (-c\right )}^{3/2}\,f}-\frac {\sqrt {2}\,B\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {c}}\right )}{16\,a\,c^{3/2}\,f} \]

[In]

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

[Out]

((B*c)/3 - (B*(c - c*tan(e + f*x)*1i))/6 + (B*(c - c*tan(e + f*x)*1i)^2)/(8*c))/(a*f*(c - c*tan(e + f*x)*1i)^(
5/2) - 2*a*c*f*(c - c*tan(e + f*x)*1i)^(3/2)) - ((A*(c - c*tan(e + f*x)*1i)*5i)/(6*a*f) + (A*c*1i)/(3*a*f) - (
A*(c - c*tan(e + f*x)*1i)^2*5i)/(8*a*c*f))/(2*c*(c - c*tan(e + f*x)*1i)^(3/2) - (c - c*tan(e + f*x)*1i)^(5/2))
+ (2^(1/2)*A*atan((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*(-c)^(1/2)))*5i)/(16*a*(-c)^(3/2)*f) - (2^(1/2)*
B*atanh((2^(1/2)*(c - c*tan(e + f*x)*1i)^(1/2))/(2*c^(1/2))))/(16*a*c^(3/2)*f)