3.1201 \(\int \frac {(c+d \tan (e+f x))^2}{(a+b \tan (e+f x))^2} \, dx\)

Optimal. Leaf size=126 \[ -\frac {(b c-a d)^2}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac {2 (a c+b d) (b c-a d) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )^2}-\frac {x (b (c-d)-a (c+d)) (a (c-d)+b (c+d))}{\left (a^2+b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.23, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {3542, 3531, 3530} \[ -\frac {(b c-a d)^2}{b f \left (a^2+b^2\right ) (a+b \tan (e+f x))}+\frac {2 (a c+b d) (b c-a d) \log (a \cos (e+f x)+b \sin (e+f x))}{f \left (a^2+b^2\right )^2}-\frac {x (b (c-d)-a (c+d)) (a (c-d)+b (c+d))}{\left (a^2+b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 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]

Rubi steps

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

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Mathematica [C]  time = 2.20, size = 321, normalized size = 2.55 \[ \frac {(c+d \tan (e+f x))^2 (a \cos (e+f x)+b \sin (e+f x)) \left (-2 i (e+f x) \left (a^2 c d+a b \left (d^2-c^2\right )-b^2 c d\right ) (a \cos (e+f x)+b \sin (e+f x))-\left (a^2 c d+a b \left (d^2-c^2\right )-b^2 c d\right ) (a \cos (e+f x)+b \sin (e+f x)) \log \left ((a \cos (e+f x)+b \sin (e+f x))^2\right )+2 i \left (a^2 c d+a b \left (d^2-c^2\right )-b^2 c d\right ) \tan ^{-1}(\tan (e+f x)) (a \cos (e+f x)+b \sin (e+f x))+\frac {\left (a^2+b^2\right ) (b c-a d)^2 \sin (e+f x)}{a}+(e+f x) (a (c+d)+b (d-c)) (a (c-d)+b (c+d)) (a \cos (e+f x)+b \sin (e+f x))\right )}{f \left (a^2+b^2\right )^2 (a+b \tan (e+f x))^2 (c \cos (e+f x)+d \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.49, size = 304, normalized size = 2.41 \[ -\frac {b^{3} c^{2} - 2 \, a b^{2} c d + a^{2} b d^{2} - {\left (4 \, a^{2} b c d + {\left (a^{3} - a b^{2}\right )} c^{2} - {\left (a^{3} - a b^{2}\right )} d^{2}\right )} f x - {\left (a^{2} b c^{2} - a^{2} b d^{2} - {\left (a^{3} - a b^{2}\right )} c d + {\left (a b^{2} c^{2} - a b^{2} d^{2} - {\left (a^{2} b - b^{3}\right )} c d\right )} \tan \left (f x + e\right )\right )} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{2} + 2 \, a b \tan \left (f x + e\right ) + a^{2}}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (a b^{2} c^{2} - 2 \, a^{2} b c d + a^{3} d^{2} + {\left (4 \, a b^{2} c d + {\left (a^{2} b - b^{3}\right )} c^{2} - {\left (a^{2} b - b^{3}\right )} d^{2}\right )} f x\right )} \tan \left (f x + e\right )}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} f \tan \left (f x + e\right ) + {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} f} \]

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))^2,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 1.46, size = 332, normalized size = 2.63 \[ \frac {\frac {{\left (a^{2} c^{2} - b^{2} c^{2} + 4 \, a b c d - a^{2} d^{2} + b^{2} d^{2}\right )} {\left (f x + e\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a b c^{2} - a^{2} c d + b^{2} c d - a b d^{2}\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (a b^{2} c^{2} - a^{2} b c d + b^{3} c d - a b^{2} d^{2}\right )} \log \left ({\left | b \tan \left (f x + e\right ) + a \right |}\right )}{a^{4} b + 2 \, a^{2} b^{3} + b^{5}} - \frac {2 \, a b^{3} c^{2} \tan \left (f x + e\right ) - 2 \, a^{2} b^{2} c d \tan \left (f x + e\right ) + 2 \, b^{4} c d \tan \left (f x + e\right ) - 2 \, a b^{3} d^{2} \tan \left (f x + e\right ) + 3 \, a^{2} b^{2} c^{2} + b^{4} c^{2} - 4 \, a^{3} b c d + a^{4} d^{2} - a^{2} b^{2} d^{2}}{{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} {\left (b \tan \left (f x + e\right ) + a\right )}}}{f} \]

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))^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.24, size = 465, normalized size = 3.69 \[ -\frac {a^{2} d^{2}}{f \left (a^{2}+b^{2}\right ) b \left (a +b \tan \left (f x +e \right )\right )}+\frac {2 a c d}{f \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (f x +e \right )\right )}-\frac {b \,c^{2}}{f \left (a^{2}+b^{2}\right ) \left (a +b \tan \left (f x +e \right )\right )}-\frac {2 \ln \left (a +b \tan \left (f x +e \right )\right ) a^{2} c d}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {2 \ln \left (a +b \tan \left (f x +e \right )\right ) a b \,c^{2}}{f \left (a^{2}+b^{2}\right )^{2}}-\frac {2 \ln \left (a +b \tan \left (f x +e \right )\right ) a b \,d^{2}}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {2 \ln \left (a +b \tan \left (f x +e \right )\right ) b^{2} c d}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) a^{2} c^{2}}{f \left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tan \left (f x +e \right )\right ) a^{2} d^{2}}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {4 \arctan \left (\tan \left (f x +e \right )\right ) a b c d}{f \left (a^{2}+b^{2}\right )^{2}}-\frac {\arctan \left (\tan \left (f x +e \right )\right ) b^{2} c^{2}}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) b^{2} d^{2}}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a^{2} c d}{f \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a b \,c^{2}}{f \left (a^{2}+b^{2}\right )^{2}}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) a b \,d^{2}}{f \left (a^{2}+b^{2}\right )^{2}}-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right ) b^{2} c d}{f \left (a^{2}+b^{2}\right )^{2}} \]

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))^2,x)

[Out]

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

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maxima [A]  time = 0.83, size = 230, normalized size = 1.83 \[ \frac {\frac {{\left (4 \, a b c d + {\left (a^{2} - b^{2}\right )} c^{2} - {\left (a^{2} - b^{2}\right )} d^{2}\right )} {\left (f x + e\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac {2 \, {\left (a b c^{2} - a b d^{2} - {\left (a^{2} - b^{2}\right )} c d\right )} \log \left (b \tan \left (f x + e\right ) + a\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {{\left (a b c^{2} - a b d^{2} - {\left (a^{2} - b^{2}\right )} c d\right )} \log \left (\tan \left (f x + e\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac {b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}}{a^{3} b + a b^{3} + {\left (a^{2} b^{2} + b^{4}\right )} \tan \left (f x + e\right )}}{f} \]

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))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 7.12, size = 208, normalized size = 1.65 \[ \frac {\ln \left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )\,\left (-2\,c\,d\,a^2+\left (2\,c^2-2\,d^2\right )\,a\,b+2\,c\,d\,b^2\right )}{f\,\left (a^4+2\,a^2\,b^2+b^4\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,\left (c^2+c\,d\,2{}\mathrm {i}-d^2\right )}{2\,f\,\left (-a^2\,1{}\mathrm {i}+2\,a\,b+b^2\,1{}\mathrm {i}\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )\,\left (c^2\,1{}\mathrm {i}+2\,c\,d-d^2\,1{}\mathrm {i}\right )}{2\,f\,\left (-a^2+a\,b\,2{}\mathrm {i}+b^2\right )}-\frac {a^2\,d^2-2\,a\,b\,c\,d+b^2\,c^2}{b\,f\,\left (a^2+b^2\right )\,\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.14, size = 4258, normalized size = 33.79 \[ \text {result too large to display} \]

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))**2,x)

[Out]

Piecewise((zoo*x*(c + d*tan(e))**2/tan(e)**2, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (c**2*f*x*tan(e + f*x)**2/(-4*b
**2*f*tan(e + f*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - 2*I*c**2*f*x*tan(e + f*x)/(-4*b**2*f*tan(e + f*x
)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - c**2*f*x/(-4*b**2*f*tan(e + f*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4
*b**2*f) + c**2*tan(e + f*x)/(-4*b**2*f*tan(e + f*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - 2*I*c**2/(-4*b
**2*f*tan(e + f*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - 2*I*c*d*f*x*tan(e + f*x)**2/(-4*b**2*f*tan(e + f
*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - 4*c*d*f*x*tan(e + f*x)/(-4*b**2*f*tan(e + f*x)**2 + 8*I*b**2*f*
tan(e + f*x) + 4*b**2*f) + 2*I*c*d*f*x/(-4*b**2*f*tan(e + f*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - 2*I*
c*d*tan(e + f*x)/(-4*b**2*f*tan(e + f*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - d**2*f*x*tan(e + f*x)**2/(
-4*b**2*f*tan(e + f*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) + 2*I*d**2*f*x*tan(e + f*x)/(-4*b**2*f*tan(e +
f*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) + d**2*f*x/(-4*b**2*f*tan(e + f*x)**2 + 8*I*b**2*f*tan(e + f*x)
+ 4*b**2*f) + 3*d**2*tan(e + f*x)/(-4*b**2*f*tan(e + f*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - 2*I*d**2
/(-4*b**2*f*tan(e + f*x)**2 + 8*I*b**2*f*tan(e + f*x) + 4*b**2*f), Eq(a, -I*b)), (c**2*f*x*tan(e + f*x)**2/(-4
*b**2*f*tan(e + f*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) + 2*I*c**2*f*x*tan(e + f*x)/(-4*b**2*f*tan(e + f
*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - c**2*f*x/(-4*b**2*f*tan(e + f*x)**2 - 8*I*b**2*f*tan(e + f*x) +
4*b**2*f) + c**2*tan(e + f*x)/(-4*b**2*f*tan(e + f*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) + 2*I*c**2/(-4
*b**2*f*tan(e + f*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) + 2*I*c*d*f*x*tan(e + f*x)**2/(-4*b**2*f*tan(e +
f*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - 4*c*d*f*x*tan(e + f*x)/(-4*b**2*f*tan(e + f*x)**2 - 8*I*b**2*
f*tan(e + f*x) + 4*b**2*f) - 2*I*c*d*f*x/(-4*b**2*f*tan(e + f*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) + 2*
I*c*d*tan(e + f*x)/(-4*b**2*f*tan(e + f*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - d**2*f*x*tan(e + f*x)**2
/(-4*b**2*f*tan(e + f*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) - 2*I*d**2*f*x*tan(e + f*x)/(-4*b**2*f*tan(e
+ f*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) + d**2*f*x/(-4*b**2*f*tan(e + f*x)**2 - 8*I*b**2*f*tan(e + f*
x) + 4*b**2*f) + 3*d**2*tan(e + f*x)/(-4*b**2*f*tan(e + f*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f) + 2*I*d*
*2/(-4*b**2*f*tan(e + f*x)**2 - 8*I*b**2*f*tan(e + f*x) + 4*b**2*f), Eq(a, I*b)), ((c**2*x + c*d*log(tan(e + f
*x)**2 + 1)/f - d**2*x + d**2*tan(e + f*x)/f)/a**2, Eq(b, 0)), (x*(c + d*tan(e))**2/(a + b*tan(e))**2, Eq(f, 0
)), (-a**4*d**2/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f +
b**6*f*tan(e + f*x)) + a**3*b*c**2*f*x/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*t
an(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - 2*a**3*b*c*d*log(a/b + tan(e + f*x))/(a**5*b*f + a**4*b**2*f*t
an(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) + a**3*b*c*d*log(ta
n(e + f*x)**2 + 1)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*
f + b**6*f*tan(e + f*x)) + 2*a**3*b*c*d/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*t
an(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - a**3*b*d**2*f*x/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*
b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) + a**2*b**2*c**2*f*x*tan(e + f*x)/(a**5*
b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x))
+ 2*a**2*b**2*c**2*log(a/b + tan(e + f*x))/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*
f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - a**2*b**2*c**2*log(tan(e + f*x)**2 + 1)/(a**5*b*f + a**4*b*
*2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - a**2*b**2*c
**2/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(
e + f*x)) + 4*a**2*b**2*c*d*f*x/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f
*x) + a*b**5*f + b**6*f*tan(e + f*x)) - 2*a**2*b**2*c*d*log(a/b + tan(e + f*x))*tan(e + f*x)/(a**5*b*f + a**4*
b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) + a**2*b**2
*c*d*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*
f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - a**2*b**2*d**2*f*x*tan(e + f*x)/(a**5*b*f + a**4*b**2*f*tan
(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - 2*a**2*b**2*d**2*lo
g(a/b + tan(e + f*x))/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b*
*5*f + b**6*f*tan(e + f*x)) + a**2*b**2*d**2*log(tan(e + f*x)**2 + 1)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2
*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - a**2*b**2*d**2/(a**5*b*f + a**4*
b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - a*b**3*c*
*2*f*x/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*t
an(e + f*x)) + 2*a*b**3*c**2*log(a/b + tan(e + f*x))*tan(e + f*x)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**
3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - a*b**3*c**2*log(tan(e + f*x)**2 + 1)
*tan(e + f*x)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b
**6*f*tan(e + f*x)) + 4*a*b**3*c*d*f*x*tan(e + f*x)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a
**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) + 2*a*b**3*c*d*log(a/b + tan(e + f*x))/(a**5*b*f + a
**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - a*b**
3*c*d*log(tan(e + f*x)**2 + 1)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*
x) + a*b**5*f + b**6*f*tan(e + f*x)) + 2*a*b**3*c*d/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a
**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) + a*b**3*d**2*f*x/(a**5*b*f + a**4*b**2*f*tan(e + f*
x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - 2*a*b**3*d**2*log(a/b + ta
n(e + f*x))*tan(e + f*x)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a
*b**5*f + b**6*f*tan(e + f*x)) + a*b**3*d**2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)/(a**5*b*f + a**4*b**2*f*tan
(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - b**4*c**2*f*x*tan(e
+ f*x)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*
tan(e + f*x)) - b**4*c**2/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) +
a*b**5*f + b**6*f*tan(e + f*x)) + 2*b**4*c*d*log(a/b + tan(e + f*x))*tan(e + f*x)/(a**5*b*f + a**4*b**2*f*tan(
e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)) - b**4*c*d*log(tan(e +
f*x)**2 + 1)*tan(e + f*x)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b**3*f + 2*a**2*b**4*f*tan(e + f*x) +
a*b**5*f + b**6*f*tan(e + f*x)) + b**4*d**2*f*x*tan(e + f*x)/(a**5*b*f + a**4*b**2*f*tan(e + f*x) + 2*a**3*b*
*3*f + 2*a**2*b**4*f*tan(e + f*x) + a*b**5*f + b**6*f*tan(e + f*x)), True))

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