3.1226 \(\int \frac{a+b \tan (e+f x)}{(c+d \tan (e+f x))^3} \, dx\)

Optimal. Leaf size=177 \[ \frac{b c-a d}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}-\frac{2 a c d-b \left (c^2-d^2\right )}{f \left (c^2+d^2\right )^2 (c+d \tan (e+f x))}+\frac{\left (a d \left (3 c^2-d^2\right )-b \left (c^3-3 c d^2\right )\right ) \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right )^3}+\frac{x \left (a c^3-3 a c d^2+3 b c^2 d-b d^3\right )}{\left (c^2+d^2\right )^3} \]

[Out]

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

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Rubi [A]  time = 0.290842, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3529, 3531, 3530} \[ \frac{b c-a d}{2 f \left (c^2+d^2\right ) (c+d \tan (e+f x))^2}-\frac{2 a c d-b \left (c^2-d^2\right )}{f \left (c^2+d^2\right )^2 (c+d \tan (e+f x))}+\frac{\left (a d \left (3 c^2-d^2\right )-b \left (c^3-3 c d^2\right )\right ) \log (c \cos (e+f x)+d \sin (e+f x))}{f \left (c^2+d^2\right )^3}+\frac{x \left (a c^3-3 a c d^2+3 b c^2 d-b d^3\right )}{\left (c^2+d^2\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 4.13314, size = 244, normalized size = 1.38 \[ \frac{(b c-a d) \left (\frac{d \left (\frac{\left (c^2+d^2\right ) \left (5 c^2+4 c d \tan (e+f x)+d^2\right )}{(c+d \tan (e+f x))^2}+\left (2 d^2-6 c^2\right ) \log (c+d \tan (e+f x))\right )}{\left (c^2+d^2\right )^3}+\frac{i \log (-\tan (e+f x)+i)}{(c+i d)^3}-\frac{\log (\tan (e+f x)+i)}{(d+i c)^3}\right )-b \left (\frac{2 d \left (\frac{c^2+d^2}{c+d \tan (e+f x)}-2 c \log (c+d \tan (e+f x))\right )}{\left (c^2+d^2\right )^2}+\frac{i \log (-\tan (e+f x)+i)}{(c+i d)^2}-\frac{i \log (\tan (e+f x)+i)}{(c-i d)^2}\right )}{2 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*((I*Log[I - Tan[e + f*x]])/(c + I*d)^2 - (I*Log[I + Tan[e + f*x]])/(c - I*d)^2 + (2*d*(-2*c*Log[c + d*Tan
[e + f*x]] + (c^2 + d^2)/(c + d*Tan[e + f*x])))/(c^2 + d^2)^2)) + (b*c - a*d)*((I*Log[I - Tan[e + f*x]])/(c +
I*d)^3 - Log[I + Tan[e + f*x]]/(I*c + d)^3 + (d*((-6*c^2 + 2*d^2)*Log[c + d*Tan[e + f*x]] + ((c^2 + d^2)*(5*c^
2 + d^2 + 4*c*d*Tan[e + f*x]))/(c + d*Tan[e + f*x])^2))/(c^2 + d^2)^3))/(2*d*f)

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Maple [B]  time = 0.035, size = 483, normalized size = 2.7 \begin{align*} -{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) a{c}^{2}d}{2\,f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) a{d}^{3}}{2\,f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) b{c}^{3}}{2\,f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}-{\frac{3\,\ln \left ( 1+ \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) bc{d}^{2}}{2\,f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}+{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) a{c}^{3}}{f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}-3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) ac{d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}+3\,{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) b{c}^{2}d}{f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) b{d}^{3}}{f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}+3\,{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ) a{c}^{2}d}{f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}-{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ) a{d}^{3}}{f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}-{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ) b{c}^{3}}{f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}+3\,{\frac{\ln \left ( c+d\tan \left ( fx+e \right ) \right ) bc{d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{3}}}-{\frac{ad}{2\,f \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) ^{2}}}+{\frac{bc}{2\,f \left ({c}^{2}+{d}^{2} \right ) \left ( c+d\tan \left ( fx+e \right ) \right ) ^{2}}}-2\,{\frac{acd}{f \left ({c}^{2}+{d}^{2} \right ) ^{2} \left ( c+d\tan \left ( fx+e \right ) \right ) }}+{\frac{b{c}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2} \left ( c+d\tan \left ( fx+e \right ) \right ) }}-{\frac{b{d}^{2}}{f \left ({c}^{2}+{d}^{2} \right ) ^{2} \left ( c+d\tan \left ( fx+e \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.71772, size = 433, normalized size = 2.45 \begin{align*} \frac{\frac{2 \,{\left (a c^{3} + 3 \, b c^{2} d - 3 \, a c d^{2} - b d^{3}\right )}{\left (f x + e\right )}}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} - \frac{2 \,{\left (b c^{3} - 3 \, a c^{2} d - 3 \, b c d^{2} + a d^{3}\right )} \log \left (d \tan \left (f x + e\right ) + c\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac{{\left (b c^{3} - 3 \, a c^{2} d - 3 \, b c d^{2} + a d^{3}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac{3 \, b c^{3} - 5 \, a c^{2} d - b c d^{2} - a d^{3} + 2 \,{\left (b c^{2} d - 2 \, a c d^{2} - b d^{3}\right )} \tan \left (f x + e\right )}{c^{6} + 2 \, c^{4} d^{2} + c^{2} d^{4} +{\left (c^{4} d^{2} + 2 \, c^{2} d^{4} + d^{6}\right )} \tan \left (f x + e\right )^{2} + 2 \,{\left (c^{5} d + 2 \, c^{3} d^{3} + c d^{5}\right )} \tan \left (f x + e\right )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.54457, size = 1038, normalized size = 5.86 \begin{align*} \frac{5 \, b c^{3} d^{2} - 7 \, a c^{2} d^{3} - b c d^{4} - a d^{5} + 2 \,{\left (a c^{5} + 3 \, b c^{4} d - 3 \, a c^{3} d^{2} - b c^{2} d^{3}\right )} f x -{\left (3 \, b c^{3} d^{2} - 5 \, a c^{2} d^{3} - 3 \, b c d^{4} + a d^{5} - 2 \,{\left (a c^{3} d^{2} + 3 \, b c^{2} d^{3} - 3 \, a c d^{4} - b d^{5}\right )} f x\right )} \tan \left (f x + e\right )^{2} -{\left (b c^{5} - 3 \, a c^{4} d - 3 \, b c^{3} d^{2} + a c^{2} d^{3} +{\left (b c^{3} d^{2} - 3 \, a c^{2} d^{3} - 3 \, b c d^{4} + a d^{5}\right )} \tan \left (f x + e\right )^{2} + 2 \,{\left (b c^{4} d - 3 \, a c^{3} d^{2} - 3 \, b c^{2} d^{3} + a c d^{4}\right )} \tan \left (f x + e\right )\right )} \log \left (\frac{d^{2} \tan \left (f x + e\right )^{2} + 2 \, c d \tan \left (f x + e\right ) + c^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) - 2 \,{\left (2 \, b c^{4} d - 3 \, a c^{3} d^{2} - 3 \, b c^{2} d^{3} + 3 \, a c d^{4} + b d^{5} - 2 \,{\left (a c^{4} d + 3 \, b c^{3} d^{2} - 3 \, a c^{2} d^{3} - b c d^{4}\right )} f x\right )} \tan \left (f x + e\right )}{2 \,{\left ({\left (c^{6} d^{2} + 3 \, c^{4} d^{4} + 3 \, c^{2} d^{6} + d^{8}\right )} f \tan \left (f x + e\right )^{2} + 2 \,{\left (c^{7} d + 3 \, c^{5} d^{3} + 3 \, c^{3} d^{5} + c d^{7}\right )} f \tan \left (f x + e\right ) +{\left (c^{8} + 3 \, c^{6} d^{2} + 3 \, c^{4} d^{4} + c^{2} d^{6}\right )} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: AttributeError

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Giac [B]  time = 1.42768, size = 570, normalized size = 3.22 \begin{align*} \frac{\frac{2 \,{\left (a c^{3} + 3 \, b c^{2} d - 3 \, a c d^{2} - b d^{3}\right )}{\left (f x + e\right )}}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} + \frac{{\left (b c^{3} - 3 \, a c^{2} d - 3 \, b c d^{2} + a d^{3}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}} - \frac{2 \,{\left (b c^{3} d - 3 \, a c^{2} d^{2} - 3 \, b c d^{3} + a d^{4}\right )} \log \left ({\left | d \tan \left (f x + e\right ) + c \right |}\right )}{c^{6} d + 3 \, c^{4} d^{3} + 3 \, c^{2} d^{5} + d^{7}} + \frac{3 \, b c^{3} d^{2} \tan \left (f x + e\right )^{2} - 9 \, a c^{2} d^{3} \tan \left (f x + e\right )^{2} - 9 \, b c d^{4} \tan \left (f x + e\right )^{2} + 3 \, a d^{5} \tan \left (f x + e\right )^{2} + 8 \, b c^{4} d \tan \left (f x + e\right ) - 22 \, a c^{3} d^{2} \tan \left (f x + e\right ) - 18 \, b c^{2} d^{3} \tan \left (f x + e\right ) + 2 \, a c d^{4} \tan \left (f x + e\right ) - 2 \, b d^{5} \tan \left (f x + e\right ) + 6 \, b c^{5} - 14 \, a c^{4} d - 7 \, b c^{3} d^{2} - 3 \, a c^{2} d^{3} - b c d^{4} - a d^{5}}{{\left (c^{6} + 3 \, c^{4} d^{2} + 3 \, c^{2} d^{4} + d^{6}\right )}{\left (d \tan \left (f x + e\right ) + c\right )}^{2}}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(a*c^3 + 3*b*c^2*d - 3*a*c*d^2 - b*d^3)*(f*x + e)/(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6) + (b*c^3 - 3*a*c^
2*d - 3*b*c*d^2 + a*d^3)*log(tan(f*x + e)^2 + 1)/(c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6) - 2*(b*c^3*d - 3*a*c^2*d^
2 - 3*b*c*d^3 + a*d^4)*log(abs(d*tan(f*x + e) + c))/(c^6*d + 3*c^4*d^3 + 3*c^2*d^5 + d^7) + (3*b*c^3*d^2*tan(f
*x + e)^2 - 9*a*c^2*d^3*tan(f*x + e)^2 - 9*b*c*d^4*tan(f*x + e)^2 + 3*a*d^5*tan(f*x + e)^2 + 8*b*c^4*d*tan(f*x
+ e) - 22*a*c^3*d^2*tan(f*x + e) - 18*b*c^2*d^3*tan(f*x + e) + 2*a*c*d^4*tan(f*x + e) - 2*b*d^5*tan(f*x + e)
+ 6*b*c^5 - 14*a*c^4*d - 7*b*c^3*d^2 - 3*a*c^2*d^3 - b*c*d^4 - a*d^5)/((c^6 + 3*c^4*d^2 + 3*c^2*d^4 + d^6)*(d*
tan(f*x + e) + c)^2))/f