3.1202 \(\int \frac{(c+d \tan (e+f x))^2}{(a+b \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=214 \[ -\frac{\left (-3 a^2 b \left (c^2-d^2\right )+2 a^3 c d-6 a b^2 c d+b^3 \left (c^2-d^2\right )\right ) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )^3}+\frac{x \left (6 a^2 b c d+a^3 \left (c^2-d^2\right )-3 a b^2 \left (c^2-d^2\right )-2 b^3 c d\right )}{\left (a^2+b^2\right )^3}-\frac{(b c-a d)^2}{2 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}-\frac{2 (a c+b d) (b c-a d)}{f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))} \]

[Out]

((6*a^2*b*c*d - 2*b^3*c*d + a^3*(c^2 - d^2) - 3*a*b^2*(c^2 - d^2))*x)/(a^2 + b^2)^3 - ((2*a^3*c*d - 6*a*b^2*c*
d - 3*a^2*b*(c^2 - d^2) + b^3*(c^2 - d^2))*Log[a*Cos[e + f*x] + b*Sin[e + f*x]])/((a^2 + b^2)^3*f) - (b*c - a*
d)^2/(2*b*(a^2 + b^2)*f*(a + b*Tan[e + f*x])^2) - (2*(b*c - a*d)*(a*c + b*d))/((a^2 + b^2)^2*f*(a + b*Tan[e +
f*x]))

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Rubi [A]  time = 0.357269, antiderivative size = 214, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {3542, 3529, 3531, 3530} \[ -\frac{\left (-3 a^2 b \left (c^2-d^2\right )+2 a^3 c d-6 a b^2 c d+b^3 \left (c^2-d^2\right )\right ) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )^3}+\frac{x \left (6 a^2 b c d+a^3 \left (c^2-d^2\right )-3 a b^2 \left (c^2-d^2\right )-2 b^3 c d\right )}{\left (a^2+b^2\right )^3}-\frac{(b c-a d)^2}{2 b f \left (a^2+b^2\right ) (a+b \tan (e+f x))^2}-\frac{2 (a c+b d) (b c-a d)}{f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Tan[e + f*x])^2/(a + b*Tan[e + f*x])^3,x]

[Out]

((6*a^2*b*c*d - 2*b^3*c*d + a^3*(c^2 - d^2) - 3*a*b^2*(c^2 - d^2))*x)/(a^2 + b^2)^3 - ((2*a^3*c*d - 6*a*b^2*c*
d - 3*a^2*b*(c^2 - d^2) + b^3*(c^2 - d^2))*Log[a*Cos[e + f*x] + b*Sin[e + f*x]])/((a^2 + b^2)^3*f) - (b*c - a*
d)^2/(2*b*(a^2 + b^2)*f*(a + b*Tan[e + f*x])^2) - (2*(b*c - a*d)*(a*c + b*d))/((a^2 + b^2)^2*f*(a + b*Tan[e +
f*x]))

Rule 3542

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

Rule 3529

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

Rule 3531

Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((a*c +
b*d)*x)/(a^2 + b^2), x] + Dist[(b*c - a*d)/(a^2 + b^2), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[a*c + b*d, 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{(c+d \tan (e+f x))^2}{(a+b \tan (e+f x))^3} \, dx &=-\frac{(b c-a d)^2}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}+\frac{\int \frac{2 b c d+a \left (c^2-d^2\right )+\left (2 a c d-b \left (c^2-d^2\right )\right ) \tan (e+f x)}{(a+b \tan (e+f x))^2} \, dx}{a^2+b^2}\\ &=-\frac{(b c-a d)^2}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac{2 (b c-a d) (a c+b d)}{\left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}+\frac{\int \frac{(a c+b c-a d+b d) (a c-b c+a d+b d)-2 (b c-a d) (a c+b d) \tan (e+f x)}{a+b \tan (e+f x)} \, dx}{\left (a^2+b^2\right )^2}\\ &=\frac{\left (6 a^2 b c d-2 b^3 c d+a^3 \left (c^2-d^2\right )-3 a b^2 \left (c^2-d^2\right )\right ) x}{\left (a^2+b^2\right )^3}-\frac{(b c-a d)^2}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac{2 (b c-a d) (a c+b d)}{\left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}-\frac{\left (2 a^3 c d-6 a b^2 c d-3 a^2 b \left (c^2-d^2\right )+b^3 \left (c^2-d^2\right )\right ) \int \frac{b-a \tan (e+f x)}{a+b \tan (e+f x)} \, dx}{\left (a^2+b^2\right )^3}\\ &=\frac{\left (6 a^2 b c d-2 b^3 c d+a^3 \left (c^2-d^2\right )-3 a b^2 \left (c^2-d^2\right )\right ) x}{\left (a^2+b^2\right )^3}-\frac{\left (2 a^3 c d-6 a b^2 c d-3 a^2 b \left (c^2-d^2\right )+b^3 \left (c^2-d^2\right )\right ) \log (a \cos (e+f x)+b \sin (e+f x))}{\left (a^2+b^2\right )^3 f}-\frac{(b c-a d)^2}{2 b \left (a^2+b^2\right ) f (a+b \tan (e+f x))^2}-\frac{2 (b c-a d) (a c+b d)}{\left (a^2+b^2\right )^2 f (a+b \tan (e+f x))}\\ \end{align*}

Mathematica [C]  time = 3.34698, size = 291, normalized size = 1.36 \[ \frac{(b c-a d) \left (-\frac{2 \left (3 a^2 b \left (d^2-c^2\right )+2 a^3 c d-6 a b^2 c d+b^3 \left (c^2-d^2\right )\right ) \log (a+b \tan (e+f x))}{\left (a^2+b^2\right )^2}-\frac{2 (b c-a d) \left (a^2 (-d)+2 a b c+b^2 d\right )}{b \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac{(b+i a)^3 (c+i d)^2 \log (-\tan (e+f x)+i)}{\left (a^2+b^2\right )^2}+\frac{i (a+i b) (c-i d)^2 \log (\tan (e+f x)+i)}{(a-i b)^2}\right )-\frac{b^2 (c+d \tan (e+f x))^3}{(a+b \tan (e+f x))^2}+\frac{b d (c+d \tan (e+f x))^2}{a+b \tan (e+f x)}}{2 f \left (a^2+b^2\right ) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*Tan[e + f*x])^2/(a + b*Tan[e + f*x])^3,x]

[Out]

((b*d*(c + d*Tan[e + f*x])^2)/(a + b*Tan[e + f*x]) - (b^2*(c + d*Tan[e + f*x])^3)/(a + b*Tan[e + f*x])^2 + (b*
c - a*d)*(((I*a + b)^3*(c + I*d)^2*Log[I - Tan[e + f*x]])/(a^2 + b^2)^2 + (I*(a + I*b)*(c - I*d)^2*Log[I + Tan
[e + f*x]])/(a - I*b)^2 - (2*(2*a^3*c*d - 6*a*b^2*c*d + b^3*(c^2 - d^2) + 3*a^2*b*(-c^2 + d^2))*Log[a + b*Tan[
e + f*x]])/(a^2 + b^2)^2 - (2*(b*c - a*d)*(2*a*b*c - a^2*d + b^2*d))/(b*(a^2 + b^2)*(a + b*Tan[e + f*x]))))/(2
*(a^2 + b^2)*(b*c - a*d)*f)

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Maple [B]  time = 0.04, size = 753, normalized size = 3.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*tan(f*x+e))^2/(a+b*tan(f*x+e))^3,x)

[Out]

-2/f/(a^2+b^2)^3*ln(a+b*tan(f*x+e))*a^3*c*d+2/f/(a^2+b^2)^2/(a+b*tan(f*x+e))*a^2*c*d-3/2/f/(a^2+b^2)^3*ln(1+ta
n(f*x+e)^2)*a^2*b*c^2-2/f/(a^2+b^2)^2/(a+b*tan(f*x+e))*b^2*c*d-2/f/(a^2+b^2)^3*arctan(tan(f*x+e))*b^3*c*d+1/f/
(a^2+b^2)/(a+b*tan(f*x+e))^2*a*c*d-3/f/(a^2+b^2)^3*arctan(tan(f*x+e))*a*b^2*c^2+3/f/(a^2+b^2)^3*ln(a+b*tan(f*x
+e))*a^2*b*c^2-3/f/(a^2+b^2)^3*ln(a+b*tan(f*x+e))*a^2*b*d^2+1/f/(a^2+b^2)^3*ln(1+tan(f*x+e)^2)*a^3*c*d-1/2/f/(
a^2+b^2)^3*ln(1+tan(f*x+e)^2)*b^3*d^2+1/f/(a^2+b^2)^3*arctan(tan(f*x+e))*a^3*c^2-1/f/(a^2+b^2)^3*arctan(tan(f*
x+e))*a^3*d^2-1/2/f/(a^2+b^2)*b/(a+b*tan(f*x+e))^2*c^2-1/f/(a^2+b^2)^3*ln(a+b*tan(f*x+e))*b^3*c^2+1/f/(a^2+b^2
)^3*ln(a+b*tan(f*x+e))*b^3*d^2+2/f/(a^2+b^2)^2/(a+b*tan(f*x+e))*a*b*d^2-1/2/f/(a^2+b^2)/b/(a+b*tan(f*x+e))^2*a
^2*d^2+6/f/(a^2+b^2)^3*arctan(tan(f*x+e))*a^2*b*c*d-3/f/(a^2+b^2)^3*ln(1+tan(f*x+e)^2)*a*b^2*c*d+1/2/f/(a^2+b^
2)^3*ln(1+tan(f*x+e)^2)*b^3*c^2-2/f/(a^2+b^2)^2/(a+b*tan(f*x+e))*a*b*c^2+3/f/(a^2+b^2)^3*arctan(tan(f*x+e))*a*
b^2*d^2+3/2/f/(a^2+b^2)^3*ln(1+tan(f*x+e)^2)*a^2*b*d^2+6/f/(a^2+b^2)^3*ln(a+b*tan(f*x+e))*a*b^2*c*d

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Maxima [B]  time = 1.71763, size = 582, normalized size = 2.72 \begin{align*} \frac{\frac{2 \,{\left ({\left (a^{3} - 3 \, a b^{2}\right )} c^{2} + 2 \,{\left (3 \, a^{2} b - b^{3}\right )} c d -{\left (a^{3} - 3 \, a b^{2}\right )} d^{2}\right )}{\left (f x + e\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left ({\left (3 \, a^{2} b - b^{3}\right )} c^{2} - 2 \,{\left (a^{3} - 3 \, a b^{2}\right )} c d -{\left (3 \, a^{2} b - b^{3}\right )} d^{2}\right )} \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left ({\left (3 \, a^{2} b - b^{3}\right )} c^{2} - 2 \,{\left (a^{3} - 3 \, a b^{2}\right )} c d -{\left (3 \, a^{2} b - b^{3}\right )} d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (5 \, a^{2} b^{2} + b^{4}\right )} c^{2} - 2 \,{\left (3 \, a^{3} b - a b^{3}\right )} c d +{\left (a^{4} - 3 \, a^{2} b^{2}\right )} d^{2} + 4 \,{\left (a b^{3} c^{2} - a b^{3} d^{2} -{\left (a^{2} b^{2} - b^{4}\right )} c d\right )} \tan \left (f x + e\right )}{a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5} +{\left (a^{4} b^{3} + 2 \, a^{2} b^{5} + b^{7}\right )} \tan \left (f x + e\right )^{2} + 2 \,{\left (a^{5} b^{2} + 2 \, a^{3} b^{4} + a b^{6}\right )} \tan \left (f x + e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^2/(a+b*tan(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.68775, size = 1423, normalized size = 6.65 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^2/(a+b*tan(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/2*((7*a^2*b^3 + b^5)*c^2 - 2*(5*a^3*b^2 - a*b^4)*c*d + 3*(a^4*b - a^2*b^3)*d^2 - 2*((a^5 - 3*a^3*b^2)*c^2 +
2*(3*a^4*b - a^2*b^3)*c*d - (a^5 - 3*a^3*b^2)*d^2)*f*x - ((5*a^2*b^3 - b^5)*c^2 - 6*(a^3*b^2 - a*b^4)*c*d + (
a^4*b - 5*a^2*b^3)*d^2 + 2*((a^3*b^2 - 3*a*b^4)*c^2 + 2*(3*a^2*b^3 - b^5)*c*d - (a^3*b^2 - 3*a*b^4)*d^2)*f*x)*
tan(f*x + e)^2 - ((3*a^4*b - a^2*b^3)*c^2 - 2*(a^5 - 3*a^3*b^2)*c*d - (3*a^4*b - a^2*b^3)*d^2 + ((3*a^2*b^3 -
b^5)*c^2 - 2*(a^3*b^2 - 3*a*b^4)*c*d - (3*a^2*b^3 - b^5)*d^2)*tan(f*x + e)^2 + 2*((3*a^3*b^2 - a*b^4)*c^2 - 2*
(a^4*b - 3*a^2*b^3)*c*d - (3*a^3*b^2 - a*b^4)*d^2)*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)) - 2*(3*(a^3*b^2 - a*b^4)*c^2 - 2*(2*a^4*b - 3*a^2*b^3 + b^5)*c*d + (a^5 - 3*a^3*b
^2 + 2*a*b^4)*d^2 + 2*((a^4*b - 3*a^2*b^3)*c^2 + 2*(3*a^3*b^2 - a*b^4)*c*d - (a^4*b - 3*a^2*b^3)*d^2)*f*x)*tan
(f*x + e))/((a^6*b^2 + 3*a^4*b^4 + 3*a^2*b^6 + b^8)*f*tan(f*x + e)^2 + 2*(a^7*b + 3*a^5*b^3 + 3*a^3*b^5 + a*b^
7)*f*tan(f*x + e) + (a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*b^6)*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))**2/(a+b*tan(f*x+e))**3,x)

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.56893, size = 829, normalized size = 3.87 \begin{align*} \frac{\frac{2 \,{\left (a^{3} c^{2} - 3 \, a b^{2} c^{2} + 6 \, a^{2} b c d - 2 \, b^{3} c d - a^{3} d^{2} + 3 \, a b^{2} d^{2}\right )}{\left (f x + e\right )}}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac{{\left (3 \, a^{2} b c^{2} - b^{3} c^{2} - 2 \, a^{3} c d + 6 \, a b^{2} c d - 3 \, a^{2} b d^{2} + b^{3} d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} + \frac{2 \,{\left (3 \, a^{2} b^{2} c^{2} - b^{4} c^{2} - 2 \, a^{3} b c d + 6 \, a b^{3} c d - 3 \, a^{2} b^{2} d^{2} + b^{4} d^{2}\right )} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}} - \frac{9 \, a^{2} b^{4} c^{2} \tan \left (f x + e\right )^{2} - 3 \, b^{6} c^{2} \tan \left (f x + e\right )^{2} - 6 \, a^{3} b^{3} c d \tan \left (f x + e\right )^{2} + 18 \, a b^{5} c d \tan \left (f x + e\right )^{2} - 9 \, a^{2} b^{4} d^{2} \tan \left (f x + e\right )^{2} + 3 \, b^{6} d^{2} \tan \left (f x + e\right )^{2} + 22 \, a^{3} b^{3} c^{2} \tan \left (f x + e\right ) - 2 \, a b^{5} c^{2} \tan \left (f x + e\right ) - 16 \, a^{4} b^{2} c d \tan \left (f x + e\right ) + 36 \, a^{2} b^{4} c d \tan \left (f x + e\right ) + 4 \, b^{6} c d \tan \left (f x + e\right ) - 22 \, a^{3} b^{3} d^{2} \tan \left (f x + e\right ) + 2 \, a b^{5} d^{2} \tan \left (f x + e\right ) + 14 \, a^{4} b^{2} c^{2} + 3 \, a^{2} b^{4} c^{2} + b^{6} c^{2} - 12 \, a^{5} b c d + 14 \, a^{3} b^{3} c d + 2 \, a b^{5} c d + a^{6} d^{2} - 11 \, a^{4} b^{2} d^{2}}{{\left (a^{6} b + 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} + b^{7}\right )}{\left (b \tan \left (f x + e\right ) + a\right )}^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*tan(f*x+e))^2/(a+b*tan(f*x+e))^3,x, algorithm="giac")

[Out]

1/2*(2*(a^3*c^2 - 3*a*b^2*c^2 + 6*a^2*b*c*d - 2*b^3*c*d - a^3*d^2 + 3*a*b^2*d^2)*(f*x + e)/(a^6 + 3*a^4*b^2 +
3*a^2*b^4 + b^6) - (3*a^2*b*c^2 - b^3*c^2 - 2*a^3*c*d + 6*a*b^2*c*d - 3*a^2*b*d^2 + b^3*d^2)*log(tan(f*x + e)^
2 + 1)/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) + 2*(3*a^2*b^2*c^2 - b^4*c^2 - 2*a^3*b*c*d + 6*a*b^3*c*d - 3*a^2*b^
2*d^2 + b^4*d^2)*log(abs(b*tan(f*x + e) + a))/(a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7) - (9*a^2*b^4*c^2*tan(f*x +
e)^2 - 3*b^6*c^2*tan(f*x + e)^2 - 6*a^3*b^3*c*d*tan(f*x + e)^2 + 18*a*b^5*c*d*tan(f*x + e)^2 - 9*a^2*b^4*d^2*
tan(f*x + e)^2 + 3*b^6*d^2*tan(f*x + e)^2 + 22*a^3*b^3*c^2*tan(f*x + e) - 2*a*b^5*c^2*tan(f*x + e) - 16*a^4*b^
2*c*d*tan(f*x + e) + 36*a^2*b^4*c*d*tan(f*x + e) + 4*b^6*c*d*tan(f*x + e) - 22*a^3*b^3*d^2*tan(f*x + e) + 2*a*
b^5*d^2*tan(f*x + e) + 14*a^4*b^2*c^2 + 3*a^2*b^4*c^2 + b^6*c^2 - 12*a^5*b*c*d + 14*a^3*b^3*c*d + 2*a*b^5*c*d
+ a^6*d^2 - 11*a^4*b^2*d^2)/((a^6*b + 3*a^4*b^3 + 3*a^2*b^5 + b^7)*(b*tan(f*x + e) + a)^2))/f