3.1214 \(\int \frac{1}{(a+b \tan (e+f x))^2 (c+d \tan (e+f x))} \, dx\)

Optimal. Leaf size=183 \[ \frac{x \left (a^2 c-2 a b d-b^2 c\right )}{\left (a^2+b^2\right )^2 \left (c^2+d^2\right )}-\frac{b^2}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac{b^2 \left (-3 a^2 d+2 a b c-b^2 d\right ) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )^2 (b c-a d)^2}+\frac{d^3 \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right ) (b c-a d)^2} \]

[Out]

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

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Rubi [A]  time = 0.463219, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3569, 3651, 3530} \[ \frac{x \left (a^2 c-2 a b d-b^2 c\right )}{\left (a^2+b^2\right )^2 \left (c^2+d^2\right )}-\frac{b^2}{f \left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac{b^2 \left (-3 a^2 d+2 a b c-b^2 d\right ) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )^2 (b c-a d)^2}+\frac{d^3 \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right ) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])),x]

[Out]

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

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3651

Int[((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/(((a_) + (b_.)*tan[(e_.) + (f_.)
*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[((a*(A*c - c*C + B*d) + b*(B*c - A*d + C*d
))*x)/((a^2 + b^2)*(c^2 + d^2)), x] + (Dist[(A*b^2 - a*b*B + a^2*C)/((b*c - a*d)*(a^2 + b^2)), Int[(b - a*Tan[
e + f*x])/(a + b*Tan[e + f*x]), x], x] - Dist[(c^2*C - B*c*d + A*d^2)/((b*c - a*d)*(c^2 + d^2)), Int[(d - c*Ta
n[e + f*x])/(c + d*Tan[e + f*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]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d,
0] && NeQ[a^2 + b^2, 0] && EqQ[a*c + b*d, 0]

Rubi steps

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

Mathematica [A]  time = 3.32136, size = 302, normalized size = 1.65 \[ -\frac{\frac{\left (\frac{\sqrt{-b^2} \left (a^2 c-2 a b d-b^2 c\right )}{b}+a^2 d+2 a b c-b^2 d\right ) \log \left (\sqrt{-b^2}-b \tan (e+f x)\right )}{\left (a^2+b^2\right )^2 \left (c^2+d^2\right )}+\frac{\left (\frac{\sqrt{-b^2} \left (a^2 (-c)+2 a b d+b^2 c\right )}{b}+a^2 d+2 a b c-b^2 d\right ) \log \left (\sqrt{-b^2}+b \tan (e+f x)\right )}{\left (a^2+b^2\right )^2 \left (c^2+d^2\right )}+\frac{2 b^2}{\left (a^2+b^2\right ) (b c-a d) (a+b \tan (e+f x))}+\frac{2 b^2 \left (3 a^2 d-2 a b c+b^2 d\right ) \log (a+b \tan (e+f x))}{\left (a^2+b^2\right )^2 (b c-a d)^2}-\frac{2 d^3 \log (c+d \tan (e+f x))}{\left (c^2+d^2\right ) (b c-a d)^2}}{2 f} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + b*Tan[e + f*x])^2*(c + d*Tan[e + f*x])),x]

[Out]

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

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Maple [B]  time = 0.058, size = 411, normalized size = 2.3 \begin{align*} -{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){a}^{2}d}{2\,f \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ({c}^{2}+{d}^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) abc}{f \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ){b}^{2}d}{2\,f \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){a}^{2}c}{f \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ({c}^{2}+{d}^{2} \right ) }}-2\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) abd}{f \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ({c}^{2}+{d}^{2} \right ) }}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ){b}^{2}c}{f \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{{d}^{3}\ln \left ( c+d\tan \left ( fx+e \right ) \right ) }{f \left ( ad-bc \right ) ^{2} \left ({c}^{2}+{d}^{2} \right ) }}+{\frac{{b}^{2}}{f \left ({a}^{2}+{b}^{2} \right ) \left ( ad-bc \right ) \left ( a+b\tan \left ( fx+e \right ) \right ) }}-3\,{\frac{{b}^{2}\ln \left ( a+b\tan \left ( fx+e \right ) \right ){a}^{2}d}{f \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( ad-bc \right ) ^{2}}}+2\,{\frac{{b}^{3}\ln \left ( a+b\tan \left ( fx+e \right ) \right ) ac}{f \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( ad-bc \right ) ^{2}}}-{\frac{{b}^{4}\ln \left ( a+b\tan \left ( fx+e \right ) \right ) d}{f \left ({a}^{2}+{b}^{2} \right ) ^{2} \left ( ad-bc \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f/(a^2+b^2)^2/(c^2+d^2)*ln(1+tan(f*x+e)^2)*a^2*d-1/f/(a^2+b^2)^2/(c^2+d^2)*ln(1+tan(f*x+e)^2)*a*b*c+1/2/f
/(a^2+b^2)^2/(c^2+d^2)*ln(1+tan(f*x+e)^2)*b^2*d+1/f/(a^2+b^2)^2/(c^2+d^2)*arctan(tan(f*x+e))*a^2*c-2/f/(a^2+b^
2)^2/(c^2+d^2)*arctan(tan(f*x+e))*a*b*d-1/f/(a^2+b^2)^2/(c^2+d^2)*arctan(tan(f*x+e))*b^2*c+1/f*d^3/(a*d-b*c)^2
/(c^2+d^2)*ln(c+d*tan(f*x+e))+1/f*b^2/(a^2+b^2)/(a*d-b*c)/(a+b*tan(f*x+e))-3/f*b^2/(a^2+b^2)^2/(a*d-b*c)^2*ln(
a+b*tan(f*x+e))*a^2*d+2/f*b^3/(a^2+b^2)^2/(a*d-b*c)^2*ln(a+b*tan(f*x+e))*a*c-1/f*b^4/(a^2+b^2)^2/(a*d-b*c)^2*l
n(a+b*tan(f*x+e))*d

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Maxima [B]  time = 1.76249, size = 518, normalized size = 2.83 \begin{align*} \frac{\frac{2 \, d^{3} \log \left (d \tan \left (f x + e\right ) + c\right )}{b^{2} c^{4} - 2 \, a b c^{3} d - 2 \, a b c d^{3} + a^{2} d^{4} +{\left (a^{2} + b^{2}\right )} c^{2} d^{2}} - \frac{2 \,{\left (2 \, a b d -{\left (a^{2} - b^{2}\right )} c\right )}{\left (f x + e\right )}}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} c^{2} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{2}} - \frac{2 \, b^{2}}{{\left (a^{3} b + a b^{3}\right )} c -{\left (a^{4} + a^{2} b^{2}\right )} d +{\left ({\left (a^{2} b^{2} + b^{4}\right )} c -{\left (a^{3} b + a b^{3}\right )} d\right )} \tan \left (f x + e\right )} + \frac{2 \,{\left (2 \, a b^{3} c -{\left (3 \, a^{2} b^{2} + b^{4}\right )} d\right )} \log \left (b \tan \left (f x + e\right ) + a\right )}{{\left (a^{4} b^{2} + 2 \, a^{2} b^{4} + b^{6}\right )} c^{2} - 2 \,{\left (a^{5} b + 2 \, a^{3} b^{3} + a b^{5}\right )} c d +{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d^{2}} - \frac{{\left (2 \, a b c +{\left (a^{2} - b^{2}\right )} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} c^{2} +{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} d^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*d^3*log(d*tan(f*x + e) + c)/(b^2*c^4 - 2*a*b*c^3*d - 2*a*b*c*d^3 + a^2*d^4 + (a^2 + b^2)*c^2*d^2) - 2*(
2*a*b*d - (a^2 - b^2)*c)*(f*x + e)/((a^4 + 2*a^2*b^2 + b^4)*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2) - 2*b^2/((a^3*b
+ a*b^3)*c - (a^4 + a^2*b^2)*d + ((a^2*b^2 + b^4)*c - (a^3*b + a*b^3)*d)*tan(f*x + e)) + 2*(2*a*b^3*c - (3*a^
2*b^2 + b^4)*d)*log(b*tan(f*x + e) + a)/((a^4*b^2 + 2*a^2*b^4 + b^6)*c^2 - 2*(a^5*b + 2*a^3*b^3 + a*b^5)*c*d +
(a^6 + 2*a^4*b^2 + a^2*b^4)*d^2) - (2*a*b*c + (a^2 - b^2)*d)*log(tan(f*x + e)^2 + 1)/((a^4 + 2*a^2*b^2 + b^4)
*c^2 + (a^4 + 2*a^2*b^2 + b^4)*d^2))/f

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Fricas [B]  time = 3.35744, size = 1542, normalized size = 8.43 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*b^5*c^3 - 2*a*b^4*c^2*d + 2*b^5*c*d^2 - 2*a*b^4*d^3 + 2*(2*a^4*b*c^2*d + 2*a^4*b*d^3 - (a^3*b^2 - a*b^
4)*c^3 - (a^5 + 3*a^3*b^2)*c*d^2)*f*x - (2*a^2*b^3*c^3 + 2*a^2*b^3*c*d^2 - (3*a^3*b^2 + a*b^4)*c^2*d - (3*a^3*
b^2 + a*b^4)*d^3 + (2*a*b^4*c^3 + 2*a*b^4*c*d^2 - (3*a^2*b^3 + b^5)*c^2*d - (3*a^2*b^3 + b^5)*d^3)*tan(f*x + e
))*log((b^2*tan(f*x + e)^2 + 2*a*b*tan(f*x + e) + a^2)/(tan(f*x + e)^2 + 1)) - ((a^4*b + 2*a^2*b^3 + b^5)*d^3*
tan(f*x + e) + (a^5 + 2*a^3*b^2 + a*b^4)*d^3)*log((d^2*tan(f*x + e)^2 + 2*c*d*tan(f*x + e) + c^2)/(tan(f*x + e
)^2 + 1)) - 2*(a*b^4*c^3 - a^2*b^3*c^2*d + a*b^4*c*d^2 - a^2*b^3*d^3 - (2*a^3*b^2*c^2*d + 2*a^3*b^2*d^3 - (a^2
*b^3 - b^5)*c^3 - (a^4*b + 3*a^2*b^3)*c*d^2)*f*x)*tan(f*x + e))/(((a^4*b^3 + 2*a^2*b^5 + b^7)*c^4 - 2*(a^5*b^2
+ 2*a^3*b^4 + a*b^6)*c^3*d + (a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*c^2*d^2 - 2*(a^5*b^2 + 2*a^3*b^4 + a*b^6)*
c*d^3 + (a^6*b + 2*a^4*b^3 + a^2*b^5)*d^4)*f*tan(f*x + e) + ((a^5*b^2 + 2*a^3*b^4 + a*b^6)*c^4 - 2*(a^6*b + 2*
a^4*b^3 + a^2*b^5)*c^3*d + (a^7 + 3*a^5*b^2 + 3*a^3*b^4 + a*b^6)*c^2*d^2 - 2*(a^6*b + 2*a^4*b^3 + a^2*b^5)*c*d
^3 + (a^7 + 2*a^5*b^2 + a^3*b^4)*d^4)*f)

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Sympy [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(1/(a+b*tan(f*x+e))**2/(c+d*tan(f*x+e)),x)

[Out]

Timed out

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Giac [B]  time = 1.33831, size = 732, normalized size = 4. \begin{align*} \frac{\frac{2 \, d^{4} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{b^{2} c^{4} d - 2 \, a b c^{3} d^{2} + a^{2} c^{2} d^{3} + b^{2} c^{2} d^{3} - 2 \, a b c d^{4} + a^{2} d^{5}} + \frac{2 \,{\left (a^{2} c - b^{2} c - 2 \, a b d\right )}{\left (f x + e\right )}}{a^{4} c^{2} + 2 \, a^{2} b^{2} c^{2} + b^{4} c^{2} + a^{4} d^{2} + 2 \, a^{2} b^{2} d^{2} + b^{4} d^{2}} - \frac{{\left (2 \, a b c + a^{2} d - b^{2} d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{4} c^{2} + 2 \, a^{2} b^{2} c^{2} + b^{4} c^{2} + a^{4} d^{2} + 2 \, a^{2} b^{2} d^{2} + b^{4} d^{2}} + \frac{2 \,{\left (2 \, a b^{4} c - 3 \, a^{2} b^{3} d - b^{5} d\right )} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{4} b^{3} c^{2} + 2 \, a^{2} b^{5} c^{2} + b^{7} c^{2} - 2 \, a^{5} b^{2} c d - 4 \, a^{3} b^{4} c d - 2 \, a b^{6} c d + a^{6} b d^{2} + 2 \, a^{4} b^{3} d^{2} + a^{2} b^{5} d^{2}} - \frac{2 \,{\left (2 \, a b^{4} c \tan \left (f x + e\right ) - 3 \, a^{2} b^{3} d \tan \left (f x + e\right ) - b^{5} d \tan \left (f x + e\right ) + 3 \, a^{2} b^{3} c + b^{5} c - 4 \, a^{3} b^{2} d - 2 \, a b^{4} d\right )}}{{\left (a^{4} b^{2} c^{2} + 2 \, a^{2} b^{4} c^{2} + b^{6} c^{2} - 2 \, a^{5} b c d - 4 \, a^{3} b^{3} c d - 2 \, a b^{5} c d + a^{6} d^{2} + 2 \, a^{4} b^{2} d^{2} + a^{2} b^{4} d^{2}\right )}{\left (b \tan \left (f x + e\right ) + a\right )}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*d^4*log(abs(d*tan(f*x + e) + c))/(b^2*c^4*d - 2*a*b*c^3*d^2 + a^2*c^2*d^3 + b^2*c^2*d^3 - 2*a*b*c*d^4 +
a^2*d^5) + 2*(a^2*c - b^2*c - 2*a*b*d)*(f*x + e)/(a^4*c^2 + 2*a^2*b^2*c^2 + b^4*c^2 + a^4*d^2 + 2*a^2*b^2*d^2
+ b^4*d^2) - (2*a*b*c + a^2*d - b^2*d)*log(tan(f*x + e)^2 + 1)/(a^4*c^2 + 2*a^2*b^2*c^2 + b^4*c^2 + a^4*d^2 +
2*a^2*b^2*d^2 + b^4*d^2) + 2*(2*a*b^4*c - 3*a^2*b^3*d - b^5*d)*log(abs(b*tan(f*x + e) + a))/(a^4*b^3*c^2 + 2*
a^2*b^5*c^2 + b^7*c^2 - 2*a^5*b^2*c*d - 4*a^3*b^4*c*d - 2*a*b^6*c*d + a^6*b*d^2 + 2*a^4*b^3*d^2 + a^2*b^5*d^2)
- 2*(2*a*b^4*c*tan(f*x + e) - 3*a^2*b^3*d*tan(f*x + e) - b^5*d*tan(f*x + e) + 3*a^2*b^3*c + b^5*c - 4*a^3*b^2
*d - 2*a*b^4*d)/((a^4*b^2*c^2 + 2*a^2*b^4*c^2 + b^6*c^2 - 2*a^5*b*c*d - 4*a^3*b^3*c*d - 2*a*b^5*c*d + a^6*d^2
+ 2*a^4*b^2*d^2 + a^2*b^4*d^2)*(b*tan(f*x + e) + a)))/f