3.95 \(\int \frac{\sqrt{c+d \tan (e+f x)} (A+B \tan (e+f x)+C \tan ^2(e+f x))}{(a+b \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=317 \[ -\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}-\frac{\left (-a^2 b^2 (3 A d+2 B c-5 C d)+a^3 b B d+a^4 C d+a b^3 (4 A c-3 B d-4 c C)+b^4 (A d+2 B c)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{3/2} f \left (a^2+b^2\right )^2 \sqrt{b c-a d}}-\frac{\sqrt{c-i d} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2}-\frac{\sqrt{c+i d} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2} \]

[Out]

-(((I*A + B - I*C)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)^2*f)) - ((B - I*(
A - C))*Sqrt[c + I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*f) - ((a^3*b*B*d + a^4*C*d
+ b^4*(2*B*c + A*d) + a*b^3*(4*A*c - 4*c*C - 3*B*d) - a^2*b^2*(2*B*c + 3*A*d - 5*C*d))*ArcTanh[(Sqrt[b]*Sqrt[
c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(b^(3/2)*(a^2 + b^2)^2*Sqrt[b*c - a*d]*f) - ((A*b^2 - a*(b*B - a*C))*Sq
rt[c + d*Tan[e + f*x]])/(b*(a^2 + b^2)*f*(a + b*Tan[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.43942, antiderivative size = 317, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 7, integrand size = 47, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.149, Rules used = {3645, 3653, 3539, 3537, 63, 208, 3634} \[ -\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{c+d \tan (e+f x)}}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}-\frac{\left (-a^2 b^2 (3 A d+2 B c-5 C d)+a^3 b B d+a^4 C d+a b^3 (4 A c-3 B d-4 c C)+b^4 (A d+2 B c)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{3/2} f \left (a^2+b^2\right )^2 \sqrt{b c-a d}}-\frac{\sqrt{c-i d} (i A+B-i C) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (a-i b)^2}-\frac{\sqrt{c+i d} (B-i (A-C)) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (a+i b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((I*A + B - I*C)*Sqrt[c - I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((a - I*b)^2*f)) - ((B - I*(
A - C))*Sqrt[c + I*d]*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((a + I*b)^2*f) - ((a^3*b*B*d + a^4*C*d
+ b^4*(2*B*c + A*d) + a*b^3*(4*A*c - 4*c*C - 3*B*d) - a^2*b^2*(2*B*c + 3*A*d - 5*C*d))*ArcTanh[(Sqrt[b]*Sqrt[
c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(b^(3/2)*(a^2 + b^2)^2*Sqrt[b*c - a*d]*f) - ((A*b^2 - a*(b*B - a*C))*Sq
rt[c + d*Tan[e + f*x]])/(b*(a^2 + b^2)*f*(a + b*Tan[e + f*x]))

Rule 3645

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*d^2 + c*(c*C - B*d))*(a + b*T
an[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3653

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

Rule 3539

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

Rule 3537

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

Rule 63

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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 3634

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

Rubi steps

\begin{align*} \int \frac{\sqrt{c+d \tan (e+f x)} \left (A+B \tan (e+f x)+C \tan ^2(e+f x)\right )}{(a+b \tan (e+f x))^2} \, dx &=-\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{c+d \tan (e+f x)}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{\int \frac{\frac{1}{2} \left (2 (b B-a C) \left (b c-\frac{a d}{2}\right )+2 A b \left (a c+\frac{b d}{2}\right )\right )-b ((A-C) (b c-a d)-B (a c+b d)) \tan (e+f x)-\frac{1}{2} \left (A b^2-a b B-a^2 C-2 b^2 C\right ) d \tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{b \left (a^2+b^2\right )}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{c+d \tan (e+f x)}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{\int \frac{b \left (a^2 (A c-c C-B d)-b^2 (A c-c C-B d)+2 a b (B c+(A-C) d)\right )-b \left (2 a b (A c-c C-B d)-a^2 (B c+(A-C) d)+b^2 (B c+(A-C) d)\right ) \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{b \left (a^2+b^2\right )^2}+\frac{\left (a^3 b B d+a^4 C d+b^4 (2 B c+A d)+a b^3 (4 A c-4 c C-3 B d)-a^2 b^2 (2 B c+3 A d-5 C d)\right ) \int \frac{1+\tan ^2(e+f x)}{(a+b \tan (e+f x)) \sqrt{c+d \tan (e+f x)}} \, dx}{2 b \left (a^2+b^2\right )^2}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{c+d \tan (e+f x)}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}+\frac{((A-i B-C) (c-i d)) \int \frac{1+i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a-i b)^2}+\frac{((A+i B-C) (c+i d)) \int \frac{1-i \tan (e+f x)}{\sqrt{c+d \tan (e+f x)}} \, dx}{2 (a+i b)^2}+\frac{\left (a^3 b B d+a^4 C d+b^4 (2 B c+A d)+a b^3 (4 A c-4 c C-3 B d)-a^2 b^2 (2 B c+3 A d-5 C d)\right ) \operatorname{Subst}\left (\int \frac{1}{(a+b x) \sqrt{c+d x}} \, dx,x,\tan (e+f x)\right )}{2 b \left (a^2+b^2\right )^2 f}\\ &=-\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{c+d \tan (e+f x)}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac{(i (A+i B-C) (c+i d)) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c+i d x}} \, dx,x,-i \tan (e+f x)\right )}{2 (a+i b)^2 f}+\frac{((A-i B-C) (i c+d)) \operatorname{Subst}\left (\int \frac{1}{(-1+x) \sqrt{c-i d x}} \, dx,x,i \tan (e+f x)\right )}{2 (a-i b)^2 f}+\frac{\left (a^3 b B d+a^4 C d+b^4 (2 B c+A d)+a b^3 (4 A c-4 c C-3 B d)-a^2 b^2 (2 B c+3 A d-5 C d)\right ) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b c}{d}+\frac{b x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{b \left (a^2+b^2\right )^2 d f}\\ &=-\frac{\left (a^3 b B d+a^4 C d+b^4 (2 B c+A d)+a b^3 (4 A c-4 c C-3 B d)-a^2 b^2 (2 B c+3 A d-5 C d)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right )^2 \sqrt{b c-a d} f}-\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{c+d \tan (e+f x)}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}-\frac{((A+i B-C) (c+i d)) \operatorname{Subst}\left (\int \frac{1}{-1+\frac{i c}{d}-\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(a+i b)^2 d f}+\frac{((i A+B-i C) (i c+d)) \operatorname{Subst}\left (\int \frac{1}{-1-\frac{i c}{d}+\frac{i x^2}{d}} \, dx,x,\sqrt{c+d \tan (e+f x)}\right )}{(a-i b)^2 d f}\\ &=-\frac{(B+i (A-C)) \sqrt{c-i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{(a-i b)^2 f}-\frac{(B-i (A-C)) \sqrt{c+i d} \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{(a+i b)^2 f}-\frac{\left (a^3 b B d+a^4 C d+b^4 (2 B c+A d)+a b^3 (4 A c-4 c C-3 B d)-a^2 b^2 (2 B c+3 A d-5 C d)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{b^{3/2} \left (a^2+b^2\right )^2 \sqrt{b c-a d} f}-\frac{\left (A b^2-a (b B-a C)\right ) \sqrt{c+d \tan (e+f x)}}{b \left (a^2+b^2\right ) f (a+b \tan (e+f x))}\\ \end{align*}

Mathematica [B]  time = 6.39603, size = 764, normalized size = 2.41 \[ -\frac{2 C \sqrt{c+d \tan (e+f x)}}{b f (a+b \tan (e+f x))}-\frac{2 \left (-\frac{\sqrt{c+d \tan (e+f x)} \left (\frac{1}{2} b^2 (-a C d-A b c+2 b c C)-a \left (-\frac{1}{2} a (-a C d-b B d+b c C)-\frac{1}{2} b^2 (d (A-C)+B c)\right )\right )}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}-\frac{\frac{2 \sqrt{b c-a d} \left (-\frac{1}{4} a^2 d (b c-a d) \left (a^2 (-C)-a b B+A b^2-2 b^2 C\right )+\frac{1}{4} b^2 (b c-a d) \left (a^2 C d+a b (2 A c-B d-2 c C)+b^2 (A d+2 B c)\right )+\frac{1}{2} a b^2 (b c-a d) (-a A d-a B c+a C d+A b c-b B d-b c C)\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{c+d \tan (e+f x)}}{\sqrt{b c-a d}}\right )}{\sqrt{b} f \left (a^2+b^2\right ) (a d-b c)}+\frac{\frac{i \sqrt{c-i d} \left (\frac{1}{2} b (b c-a d) \left (a^2 (A c-B d-c C)+2 a b (d (A-C)+B c)-b^2 (A c-B d-c C)\right )+\frac{1}{2} i b (b c-a d) \left (a^2 (-(d (A-C)+B c))+2 a b (A c-B d-c C)+b^2 (d (A-C)+B c)\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c-i d}}\right )}{f (-c+i d)}-\frac{i \sqrt{c+i d} \left (\frac{1}{2} b (b c-a d) \left (a^2 (A c-B d-c C)+2 a b (d (A-C)+B c)-b^2 (A c-B d-c C)\right )-\frac{1}{2} i b (b c-a d) \left (a^2 (-(d (A-C)+B c))+2 a b (A c-B d-c C)+b^2 (d (A-C)+B c)\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{c+d \tan (e+f x)}}{\sqrt{c+i d}}\right )}{f (-c-i d)}}{a^2+b^2}}{\left (a^2+b^2\right ) (b c-a d)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*C*Sqrt[c + d*Tan[e + f*x]])/(b*f*(a + b*Tan[e + f*x])) - (2*(-((((I*Sqrt[c - I*d]*((b*(b*c - a*d)*(a^2*(A*
c - c*C - B*d) - b^2*(A*c - c*C - B*d) + 2*a*b*(B*c + (A - C)*d)))/2 + (I/2)*b*(b*c - a*d)*(2*a*b*(A*c - c*C -
B*d) - a^2*(B*c + (A - C)*d) + b^2*(B*c + (A - C)*d)))*ArcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c - I*d]])/((-c
+ I*d)*f) - (I*Sqrt[c + I*d]*((b*(b*c - a*d)*(a^2*(A*c - c*C - B*d) - b^2*(A*c - c*C - B*d) + 2*a*b*(B*c + (A
- C)*d)))/2 - (I/2)*b*(b*c - a*d)*(2*a*b*(A*c - c*C - B*d) - a^2*(B*c + (A - C)*d) + b^2*(B*c + (A - C)*d)))*A
rcTanh[Sqrt[c + d*Tan[e + f*x]]/Sqrt[c + I*d]])/((-c - I*d)*f))/(a^2 + b^2) + (2*Sqrt[b*c - a*d]*(-(a^2*(A*b^2
- a*b*B - a^2*C - 2*b^2*C)*d*(b*c - a*d))/4 + (a*b^2*(b*c - a*d)*(A*b*c - a*B*c - b*c*C - a*A*d - b*B*d + a*C
*d))/2 + (b^2*(b*c - a*d)*(a^2*C*d + b^2*(2*B*c + A*d) + a*b*(2*A*c - 2*c*C - B*d)))/4)*ArcTanh[(Sqrt[b]*Sqrt[
c + d*Tan[e + f*x]])/Sqrt[b*c - a*d]])/(Sqrt[b]*(a^2 + b^2)*(-(b*c) + a*d)*f))/((a^2 + b^2)*(b*c - a*d))) - ((
(b^2*(-(A*b*c) + 2*b*c*C - a*C*d))/2 - a*(-(b^2*(B*c + (A - C)*d))/2 - (a*(b*c*C - b*B*d - a*C*d))/2))*Sqrt[c
+ d*Tan[e + f*x]])/((a^2 + b^2)*(b*c - a*d)*f*(a + b*Tan[e + f*x]))))/b

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Maple [B]  time = 0.214, size = 5778, normalized size = 18.2 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^2,x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c + d \tan{\left (e + f x \right )}} \left (A + B \tan{\left (e + f x \right )} + C \tan ^{2}{\left (e + f x \right )}\right )}{\left (a + b \tan{\left (e + f x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)**2)/(a+b*tan(f*x+e))**2,x)

[Out]

Integral(sqrt(c + d*tan(e + f*x))*(A + B*tan(e + f*x) + C*tan(e + f*x)**2)/(a + b*tan(e + f*x))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \tan \left (f x + e\right )^{2} + B \tan \left (f x + e\right ) + A\right )} \sqrt{d \tan \left (f x + e\right ) + c}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^(1/2)*(A+B*tan(f*x+e)+C*tan(f*x+e)^2)/(a+b*tan(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((C*tan(f*x + e)^2 + B*tan(f*x + e) + A)*sqrt(d*tan(f*x + e) + c)/(b*tan(f*x + e) + a)^2, x)