3.261 \(\int \frac{\sec (e+f x) (c+d \sec (e+f x))^2}{(a+b \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=198 \[ \frac{2 \left (b^2 c^2-a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b^2 f \sqrt{a-b} \sqrt{a+b}}-\frac{(b c-a d)^2 \sin (e+f x)}{b f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac{2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a f (a-b)^{3/2} (a+b)^{3/2}}+\frac{d^2 \tanh ^{-1}(\sin (e+f x))}{b^2 f} \]

[Out]

(d^2*ArcTanh[Sin[e + f*x]])/(b^2*f) + (2*(b*c - a*d)^2*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(a
*(a - b)^(3/2)*(a + b)^(3/2)*f) + (2*(b^2*c^2 - a^2*d^2)*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/
(a*Sqrt[a - b]*b^2*Sqrt[a + b]*f) - ((b*c - a*d)^2*Sin[e + f*x])/(b*(a^2 - b^2)*f*(b + a*Cos[e + f*x]))

________________________________________________________________________________________

Rubi [A]  time = 0.366955, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used = {3988, 2952, 2664, 12, 2659, 208, 3770} \[ \frac{2 \left (b^2 c^2-a^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a b^2 f \sqrt{a-b} \sqrt{a+b}}-\frac{(b c-a d)^2 \sin (e+f x)}{b f \left (a^2-b^2\right ) (a \cos (e+f x)+b)}+\frac{2 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a+b}}\right )}{a f (a-b)^{3/2} (a+b)^{3/2}}+\frac{d^2 \tanh ^{-1}(\sin (e+f x))}{b^2 f} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c + d*Sec[e + f*x])^2)/(a + b*Sec[e + f*x])^2,x]

[Out]

(d^2*ArcTanh[Sin[e + f*x]])/(b^2*f) + (2*(b*c - a*d)^2*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(a
*(a - b)^(3/2)*(a + b)^(3/2)*f) + (2*(b^2*c^2 - a^2*d^2)*ArcTanh[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/
(a*Sqrt[a - b]*b^2*Sqrt[a + b]*f) - ((b*c - a*d)^2*Sin[e + f*x])/(b*(a^2 - b^2)*f*(b + a*Cos[e + f*x]))

Rule 3988

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]
*(d_.) + (c_))^(n_), x_Symbol] :> Dist[1/g^(m + n), Int[(g*Csc[e + f*x])^(m + n + p)*(b + a*Sin[e + f*x])^m*(d
+ c*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c - a*d, 0] && IntegerQ[m] && Inte
gerQ[n]

Rule 2952

Int[((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) +
(f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandTrig[(g*sin[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])
^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[b*c - a*d, 0] && (IntegersQ[m, n] || IntegersQ[m, p
] || IntegersQ[n, p]) && NeQ[p, 2]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.669054, size = 180, normalized size = 0.91 \[ \frac{\frac{2 \left (a^3 d^2-a b^2 \left (c^2+2 d^2\right )+2 b^3 c d\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (e+f x)\right )}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac{b (b c-a d)^2 \sin (e+f x)}{(b-a) (a+b) (a \cos (e+f x)+b)}+d^2 \left (-\log \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )\right )+d^2 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{b^2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x])^2)/(a + b*Sec[e + f*x])^2,x]

[Out]

((2*(2*b^3*c*d + a^3*d^2 - a*b^2*(c^2 + 2*d^2))*ArcTanh[((-a + b)*Tan[(e + f*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b
^2)^(3/2) - d^2*Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + d^2*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (b*(
b*c - a*d)^2*Sin[e + f*x])/((-a + b)*(a + b)*(b + a*Cos[e + f*x])))/(b^2*f)

________________________________________________________________________________________

Maple [B]  time = 0.088, size = 486, normalized size = 2.5 \begin{align*} 2\,{\frac{\tan \left ( 1/2\,fx+e/2 \right ){a}^{2}{d}^{2}}{fb \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}b-a-b \right ) }}-4\,{\frac{\tan \left ( 1/2\,fx+e/2 \right ) acd}{f \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}b-a-b \right ) }}+2\,{\frac{b\tan \left ( 1/2\,fx+e/2 \right ){c}^{2}}{f \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}b-a-b \right ) }}-2\,{\frac{{a}^{3}{d}^{2}}{f{b}^{2} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+2\,{\frac{{c}^{2}a}{f \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+4\,{\frac{a{d}^{2}}{f \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-4\,{\frac{bcd}{f \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,fx+e/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }+{\frac{{d}^{2}}{f{b}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) +1 \right ) }-{\frac{{d}^{2}}{f{b}^{2}}\ln \left ( \tan \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) -1 \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+b*sec(f*x+e))^2,x)

[Out]

2/f/b/(a^2-b^2)*tan(1/2*f*x+1/2*e)/(tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*a^2*d^2-4/f/(a^2-b^2)*t
an(1/2*f*x+1/2*e)/(tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*a*c*d+2/f*b/(a^2-b^2)*tan(1/2*f*x+1/2*e)
/(tan(1/2*f*x+1/2*e)^2*a-tan(1/2*f*x+1/2*e)^2*b-a-b)*c^2-2/f/b^2/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)
*tan(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2))*a^3*d^2+2/f/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*f*x
+1/2*e)/((a+b)*(a-b))^(1/2))*c^2*a+4/f/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*f*x+1/2*e)/((a+b)
*(a-b))^(1/2))*a*d^2-4/f*b/(a+b)/(a-b)/((a+b)*(a-b))^(1/2)*arctanh((a-b)*tan(1/2*f*x+1/2*e)/((a+b)*(a-b))^(1/2
))*c*d+1/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)+1)-1/f*d^2/b^2*ln(tan(1/2*f*x+1/2*e)-1)

________________________________________________________________________________________

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(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+b*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 21.6519, size = 1719, normalized size = 8.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+b*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

[-1/2*((a*b^3*c^2 - 2*b^4*c*d - (a^3*b - 2*a*b^3)*d^2 + (a^2*b^2*c^2 - 2*a*b^3*c*d - (a^4 - 2*a^2*b^2)*d^2)*co
s(f*x + e))*sqrt(a^2 - b^2)*log((2*a*b*cos(f*x + e) - (a^2 - 2*b^2)*cos(f*x + e)^2 - 2*sqrt(a^2 - b^2)*(b*cos(
f*x + e) + a)*sin(f*x + e) + 2*a^2 - b^2)/(a^2*cos(f*x + e)^2 + 2*a*b*cos(f*x + e) + b^2)) - ((a^5 - 2*a^3*b^2
+ a*b^4)*d^2*cos(f*x + e) + (a^4*b - 2*a^2*b^3 + b^5)*d^2)*log(sin(f*x + e) + 1) + ((a^5 - 2*a^3*b^2 + a*b^4)
*d^2*cos(f*x + e) + (a^4*b - 2*a^2*b^3 + b^5)*d^2)*log(-sin(f*x + e) + 1) + 2*((a^2*b^3 - b^5)*c^2 - 2*(a^3*b^
2 - a*b^4)*c*d + (a^4*b - a^2*b^3)*d^2)*sin(f*x + e))/((a^5*b^2 - 2*a^3*b^4 + a*b^6)*f*cos(f*x + e) + (a^4*b^3
- 2*a^2*b^5 + b^7)*f), 1/2*(2*(a*b^3*c^2 - 2*b^4*c*d - (a^3*b - 2*a*b^3)*d^2 + (a^2*b^2*c^2 - 2*a*b^3*c*d - (
a^4 - 2*a^2*b^2)*d^2)*cos(f*x + e))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(f*x + e) + a)/((a^2 - b^2
)*sin(f*x + e))) + ((a^5 - 2*a^3*b^2 + a*b^4)*d^2*cos(f*x + e) + (a^4*b - 2*a^2*b^3 + b^5)*d^2)*log(sin(f*x +
e) + 1) - ((a^5 - 2*a^3*b^2 + a*b^4)*d^2*cos(f*x + e) + (a^4*b - 2*a^2*b^3 + b^5)*d^2)*log(-sin(f*x + e) + 1)
- 2*((a^2*b^3 - b^5)*c^2 - 2*(a^3*b^2 - a*b^4)*c*d + (a^4*b - a^2*b^3)*d^2)*sin(f*x + e))/((a^5*b^2 - 2*a^3*b^
4 + a*b^6)*f*cos(f*x + e) + (a^4*b^3 - 2*a^2*b^5 + b^7)*f)]

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))**2/(a+b*sec(f*x+e))**2,x)

[Out]

Integral((c + d*sec(e + f*x))**2*sec(e + f*x)/(a + b*sec(e + f*x))**2, x)

________________________________________________________________________________________

Giac [A]  time = 1.44071, size = 374, normalized size = 1.89 \begin{align*} \frac{\frac{d^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1 \right |}\right )}{b^{2}} - \frac{d^{2} \log \left ({\left | \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 1 \right |}\right )}{b^{2}} - \frac{2 \,{\left (a b^{2} c^{2} - 2 \, b^{3} c d - a^{3} d^{2} + 2 \, a b^{2} d^{2}\right )}{\left (\pi \left \lfloor \frac{f x + e}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (2 \, a - 2 \, b\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\sqrt{-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{2} b^{2} - b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{2 \,{\left (b^{2} c^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 2 \, a b c d \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + a^{2} d^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )\right )}}{{\left (a^{2} b - b^{3}\right )}{\left (a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a - b\right )}}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c+d*sec(f*x+e))^2/(a+b*sec(f*x+e))^2,x, algorithm="giac")

[Out]

(d^2*log(abs(tan(1/2*f*x + 1/2*e) + 1))/b^2 - d^2*log(abs(tan(1/2*f*x + 1/2*e) - 1))/b^2 - 2*(a*b^2*c^2 - 2*b^
3*c*d - a^3*d^2 + 2*a*b^2*d^2)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*a - 2*b) + arctan((a*tan(1/2*f*x + 1/2*
e) - b*tan(1/2*f*x + 1/2*e))/sqrt(-a^2 + b^2)))/((a^2*b^2 - b^4)*sqrt(-a^2 + b^2)) + 2*(b^2*c^2*tan(1/2*f*x +
1/2*e) - 2*a*b*c*d*tan(1/2*f*x + 1/2*e) + a^2*d^2*tan(1/2*f*x + 1/2*e))/((a^2*b - b^3)*(a*tan(1/2*f*x + 1/2*e)
^2 - b*tan(1/2*f*x + 1/2*e)^2 - a - b)))/f