3.3.27 \(\int \frac {\cot (e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\) [227]

Optimal. Leaf size=103 \[ \frac {\log (\cos (e+f x))}{(a-b)^2 f}+\frac {\log (\tan (e+f x))}{a^2 f}+\frac {(2 a-b) b \log \left (a+b \tan ^2(e+f x)\right )}{2 a^2 (a-b)^2 f}-\frac {b}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )} \]

[Out]

ln(cos(f*x+e))/(a-b)^2/f+ln(tan(f*x+e))/a^2/f+1/2*(2*a-b)*b*ln(a+b*tan(f*x+e)^2)/a^2/(a-b)^2/f-1/2*b/a/(a-b)/f
/(a+b*tan(f*x+e)^2)

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Rubi [A]
time = 0.08, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3751, 457, 84} \begin {gather*} \frac {b (2 a-b) \log \left (a+b \tan ^2(e+f x)\right )}{2 a^2 f (a-b)^2}+\frac {\log (\tan (e+f x))}{a^2 f}-\frac {b}{2 a f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac {\log (\cos (e+f x))}{f (a-b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]/(a + b*Tan[e + f*x]^2)^2,x]

[Out]

Log[Cos[e + f*x]]/((a - b)^2*f) + Log[Tan[e + f*x]]/(a^2*f) + ((2*a - b)*b*Log[a + b*Tan[e + f*x]^2])/(2*a^2*(
a - b)^2*f) - b/(2*a*(a - b)*f*(a + b*Tan[e + f*x]^2))

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3751

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

\begin {align*} \int \frac {\cot (e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x \left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\text {Subst}\left (\int \frac {1}{x (1+x) (a+b x)^2} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\text {Subst}\left (\int \left (\frac {1}{a^2 x}-\frac {1}{(a-b)^2 (1+x)}+\frac {b^2}{a (a-b) (a+b x)^2}+\frac {(2 a-b) b^2}{a^2 (a-b)^2 (a+b x)}\right ) \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\log (\cos (e+f x))}{(a-b)^2 f}+\frac {\log (\tan (e+f x))}{a^2 f}+\frac {(2 a-b) b \log \left (a+b \tan ^2(e+f x)\right )}{2 a^2 (a-b)^2 f}-\frac {b}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]
time = 1.42, size = 90, normalized size = 0.87 \begin {gather*} \frac {\frac {2 \log (\cos (e+f x))}{(a-b)^2}+\frac {2 \log (\tan (e+f x))+\frac {b \left ((2 a-b) \log \left (a+b \tan ^2(e+f x)\right )+\frac {a (-a+b)}{a+b \tan ^2(e+f x)}\right )}{(a-b)^2}}{a^2}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]/(a + b*Tan[e + f*x]^2)^2,x]

[Out]

((2*Log[Cos[e + f*x]])/(a - b)^2 + (2*Log[Tan[e + f*x]] + (b*((2*a - b)*Log[a + b*Tan[e + f*x]^2] + (a*(-a + b
))/(a + b*Tan[e + f*x]^2)))/(a - b)^2)/a^2)/(2*f)

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Maple [A]
time = 0.30, size = 115, normalized size = 1.12

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (\cos \left (f x +e \right )+1\right )}{2 a^{2}}+\frac {\ln \left (\cos \left (f x +e \right )-1\right )}{2 a^{2}}+\frac {b \left (\frac {a b}{\left (a -b \right )^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {\left (2 a -b \right ) \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{\left (a -b \right )^{2}}\right )}{2 a^{2}}}{f}\)	\(115\)
default	\(\frac {\frac {\ln \left (\cos \left (f x +e \right )+1\right )}{2 a^{2}}+\frac {\ln \left (\cos \left (f x +e \right )-1\right )}{2 a^{2}}+\frac {b \left (\frac {a b}{\left (a -b \right )^{2} \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}+\frac {\left (2 a -b \right ) \ln \left (a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b \right )}{\left (a -b \right )^{2}}\right )}{2 a^{2}}}{f}\)	\(115\)
norman	\(\frac {b^{2} \left (\tan ^{2}\left (f x +e \right )\right )}{2 a^{2} f \left (a -b \right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {\ln \left (\tan \left (f x +e \right )\right )}{a^{2} f}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \left (a^{2}-2 a b +b^{2}\right )}+\frac {b \left (2 a -b \right ) \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 a^{2} f \left (a^{2}-2 a b +b^{2}\right )}\)	\(127\)
risch	\(\frac {i x}{a^{2}-2 a b +b^{2}}-\frac {2 i x}{a^{2}}-\frac {2 i e}{a^{2} f}-\frac {4 i b x}{a \left (a^{2}-2 a b +b^{2}\right )}-\frac {4 i b e}{a f \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 i b^{2} x}{a^{2} \left (a^{2}-2 a b +b^{2}\right )}+\frac {2 i b^{2} e}{a^{2} f \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{a f \left (-a +b \right )^{2} \left (-a \,{\mathrm e}^{4 i \left (f x +e \right )}+b \,{\mathrm e}^{4 i \left (f x +e \right )}-2 a \,{\mathrm e}^{2 i \left (f x +e \right )}-2 b \,{\mathrm e}^{2 i \left (f x +e \right )}-a +b \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{a^{2} f}+\frac {b \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{a f \left (a^{2}-2 a b +b^{2}\right )}-\frac {b^{2} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{2 a^{2} f \left (a^{2}-2 a b +b^{2}\right )}\)	\(341\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)/(a+b*tan(f*x+e)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(1/2/a^2*ln(cos(f*x+e)+1)+1/2/a^2*ln(cos(f*x+e)-1)+1/2*b/a^2*(a*b/(a-b)^2/(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b
)+(2*a-b)/(a-b)^2*ln(a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)))

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Maxima [A]
time = 0.29, size = 127, normalized size = 1.23 \begin {gather*} \frac {\frac {b^{2}}{a^{4} - 2 \, a^{3} b + a^{2} b^{2} - {\left (a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {{\left (2 \, a b - b^{2}\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{4} - 2 \, a^{3} b + a^{2} b^{2}} + \frac {\log \left (\sin \left (f x + e\right )^{2}\right )}{a^{2}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*tan(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

1/2*(b^2/(a^4 - 2*a^3*b + a^2*b^2 - (a^4 - 3*a^3*b + 3*a^2*b^2 - a*b^3)*sin(f*x + e)^2) + (2*a*b - b^2)*log(-(
a - b)*sin(f*x + e)^2 + a)/(a^4 - 2*a^3*b + a^2*b^2) + log(sin(f*x + e)^2)/a^2)/f

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Fricas [A]
time = 1.19, size = 205, normalized size = 1.99 \begin {gather*} \frac {a b^{2} \tan \left (f x + e\right )^{2} + a b^{2} + {\left (a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac {\tan \left (f x + e\right )^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) + {\left (2 \, a^{2} b - a b^{2} + {\left (2 \, a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left ({\left (a^{4} b - 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*tan(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/2*(a*b^2*tan(f*x + e)^2 + a*b^2 + (a^3 - 2*a^2*b + a*b^2 + (a^2*b - 2*a*b^2 + b^3)*tan(f*x + e)^2)*log(tan(f
*x + e)^2/(tan(f*x + e)^2 + 1)) + (2*a^2*b - a*b^2 + (2*a*b^2 - b^3)*tan(f*x + e)^2)*log((b*tan(f*x + e)^2 + a
)/(tan(f*x + e)^2 + 1)))/((a^4*b - 2*a^3*b^2 + a^2*b^3)*f*tan(f*x + e)^2 + (a^5 - 2*a^4*b + a^3*b^2)*f)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 2341 vs. \(2 (80) = 160\).
time = 104.09, size = 2341, normalized size = 22.73 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Piecewise((zoo*x*cot(e)/tan(e)**4, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((-log(tan(e + f*x)**2 + 1)/(2*f) + log(ta
n(e + f*x))/f)/a**2, Eq(b, 0)), ((-log(tan(e + f*x)**2 + 1)/(2*f) + log(tan(e + f*x))/f + 1/(2*f*tan(e + f*x)*
*2) - 1/(4*f*tan(e + f*x)**4))/b**2, Eq(a, 0)), (-2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**4/(4*b**2*f*tan(e +
f*x)**4 + 8*b**2*f*tan(e + f*x)**2 + 4*b**2*f) - 4*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2/(4*b**2*f*tan(e +
f*x)**4 + 8*b**2*f*tan(e + f*x)**2 + 4*b**2*f) - 2*log(tan(e + f*x)**2 + 1)/(4*b**2*f*tan(e + f*x)**4 + 8*b**
2*f*tan(e + f*x)**2 + 4*b**2*f) + 4*log(tan(e + f*x))*tan(e + f*x)**4/(4*b**2*f*tan(e + f*x)**4 + 8*b**2*f*tan
(e + f*x)**2 + 4*b**2*f) + 8*log(tan(e + f*x))*tan(e + f*x)**2/(4*b**2*f*tan(e + f*x)**4 + 8*b**2*f*tan(e + f*
x)**2 + 4*b**2*f) + 4*log(tan(e + f*x))/(4*b**2*f*tan(e + f*x)**4 + 8*b**2*f*tan(e + f*x)**2 + 4*b**2*f) + 2*t
an(e + f*x)**2/(4*b**2*f*tan(e + f*x)**4 + 8*b**2*f*tan(e + f*x)**2 + 4*b**2*f) + 3/(4*b**2*f*tan(e + f*x)**4
+ 8*b**2*f*tan(e + f*x)**2 + 4*b**2*f), Eq(a, b)), (x*cot(e)/(a + b*tan(e)**2)**2, Eq(f, 0)), (-a**3*log(tan(e
+ f*x)**2 + 1)/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b
**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) + 2*a**3*log(tan(e + f*x))/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a
**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) + 2*a**2*b*log(-sqrt(
-a/b) + tan(e + f*x))/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*
a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) + 2*a**2*b*log(sqrt(-a/b) + tan(e + f*x))/(2*a**5*f + 2*a**4*b*f*
tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2)
- a**2*b*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3
*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) + 2*a**2*b*log(tan(e + f*x))*tan(e +
f*x)**2/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f +
2*a**2*b**3*f*tan(e + f*x)**2) - 4*a**2*b*log(tan(e + f*x))/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*b*
f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) - a**2*b/(2*a**5*f + 2*a**4
*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)
**2) + 2*a*b**2*log(-sqrt(-a/b) + tan(e + f*x))*tan(e + f*x)**2/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**
4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) - a*b**2*log(-sqrt(-a/b
) + tan(e + f*x))/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3
*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) + 2*a*b**2*log(sqrt(-a/b) + tan(e + f*x))*tan(e + f*x)**2/(2*a**5*f +
2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e
+ f*x)**2) - a*b**2*log(sqrt(-a/b) + tan(e + f*x))/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a*
*3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) - 4*a*b**2*log(tan(e + f*x))*tan(e
+ f*x)**2/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f
+ 2*a**2*b**3*f*tan(e + f*x)**2) + 2*a*b**2*log(tan(e + f*x))/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*
b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) + a*b**2/(2*a**5*f + 2*a*
*4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*
x)**2) - b**3*log(-sqrt(-a/b) + tan(e + f*x))*tan(e + f*x)**2/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*
b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) - b**3*log(sqrt(-a/b) + t
an(e + f*x))*tan(e + f*x)**2/(2*a**5*f + 2*a**4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)*
*2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2) + 2*b**3*log(tan(e + f*x))*tan(e + f*x)**2/(2*a**5*f + 2*a
**4*b*f*tan(e + f*x)**2 - 4*a**4*b*f - 4*a**3*b**2*f*tan(e + f*x)**2 + 2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f
*x)**2), True))

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Giac [A]
time = 1.00, size = 152, normalized size = 1.48 \begin {gather*} \frac {\frac {{\left (2 \, a b - b^{2}\right )} \log \left ({\left | -a \sin \left (f x + e\right )^{2} + b \sin \left (f x + e\right )^{2} + a \right |}\right )}{a^{4} - 2 \, a^{3} b + a^{2} b^{2}} - \frac {2 \, a b \sin \left (f x + e\right )^{2} - b^{2} \sin \left (f x + e\right )^{2} - 2 \, a b}{{\left (a^{3} - a^{2} b\right )} {\left (a \sin \left (f x + e\right )^{2} - b \sin \left (f x + e\right )^{2} - a\right )}} + \frac {\log \left (\sin \left (f x + e\right )^{2}\right )}{a^{2}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)/(a+b*tan(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/2*((2*a*b - b^2)*log(abs(-a*sin(f*x + e)^2 + b*sin(f*x + e)^2 + a))/(a^4 - 2*a^3*b + a^2*b^2) - (2*a*b*sin(f
*x + e)^2 - b^2*sin(f*x + e)^2 - 2*a*b)/((a^3 - a^2*b)*(a*sin(f*x + e)^2 - b*sin(f*x + e)^2 - a)) + log(sin(f*
x + e)^2)/a^2)/f

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Mupad [B]
time = 11.73, size = 104, normalized size = 1.01 \begin {gather*} \frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )}{a^2\,f}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,{\left (a-b\right )}^2}-\frac {b}{2\,a\,f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (a-b\right )}+\frac {b\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (2\,a-b\right )}{2\,a^2\,f\,{\left (a-b\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)/(a + b*tan(e + f*x)^2)^2,x)

[Out]

log(tan(e + f*x))/(a^2*f) - log(tan(e + f*x)^2 + 1)/(2*f*(a - b)^2) - b/(2*a*f*(a + b*tan(e + f*x)^2)*(a - b))
+ (b*log(a + b*tan(e + f*x)^2)*(2*a - b))/(2*a^2*f*(a - b)^2)

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