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

Optimal. Leaf size=97 \[ -\frac {\sqrt {b} (3 a-b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{3/2} f (a-b)^2}-\frac {b \tan (e+f x)}{2 a f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac {x}{(a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {3661, 414, 522, 203, 205} \[ -\frac {\sqrt {b} (3 a-b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{3/2} f (a-b)^2}-\frac {b \tan (e+f x)}{2 a f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac {x}{(a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[e + f*x]^2)^(-2),x]

[Out]

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 414

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 3661

Int[((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[(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, e, f, n, p}, x] && (IntegersQ[n, p] || IGtQ[p, 0] || EqQ[n^2, 4] || EqQ[n^2, 16])

Rubi steps

\begin {align*} \int \frac {1}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right ) \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac {b \tan (e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {2 a-b-b x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a (a-b) f}\\ &=-\frac {b \tan (e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}+\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b)^2 f}-\frac {((3 a-b) b) \operatorname {Subst}\left (\int \frac {1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a (a-b)^2 f}\\ &=\frac {x}{(a-b)^2}-\frac {(3 a-b) \sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 a^{3/2} (a-b)^2 f}-\frac {b \tan (e+f x)}{2 a (a-b) f \left (a+b \tan ^2(e+f x)\right )}\\ \end {align*}

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Mathematica [A]  time = 1.11, size = 88, normalized size = 0.91 \[ \frac {\frac {\sqrt {b} (b-3 a) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{3/2}}+\frac {b (b-a) \tan (e+f x)}{a \left (a+b \tan ^2(e+f x)\right )}+2 \tan ^{-1}(\tan (e+f x))}{2 f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTan[Tan[e + f*x]] + (Sqrt[b]*(-3*a + b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/a^(3/2) + (b*(-a + b)*Ta
n[e + f*x])/(a*(a + b*Tan[e + f*x]^2)))/(2*(a - b)^2*f)

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fricas [A]  time = 0.50, size = 390, normalized size = 4.02 \[ \left [\frac {8 \, a b f x \tan \left (f x + e\right )^{2} + 8 \, a^{2} f x - {\left ({\left (3 \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} - a b\right )} \sqrt {-\frac {b}{a}} \log \left (\frac {b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} + 4 \, {\left (a b \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right )\right )} \sqrt {-\frac {b}{a}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right ) - 4 \, {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{8 \, {\left ({\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}, \frac {4 \, a b f x \tan \left (f x + e\right )^{2} + 4 \, a^{2} f x - {\left ({\left (3 \, a b - b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} - a b\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt {\frac {b}{a}}}{2 \, b \tan \left (f x + e\right )}\right ) - 2 \, {\left (a b - b^{2}\right )} \tan \left (f x + e\right )}{4 \, {\left ({\left (a^{3} b - 2 \, a^{2} b^{2} + a b^{3}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{4} - 2 \, a^{3} b + a^{2} b^{2}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(8*a*b*f*x*tan(f*x + e)^2 + 8*a^2*f*x - ((3*a*b - b^2)*tan(f*x + e)^2 + 3*a^2 - a*b)*sqrt(-b/a)*log((b^2*
tan(f*x + e)^4 - 6*a*b*tan(f*x + e)^2 + a^2 + 4*(a*b*tan(f*x + e)^3 - a^2*tan(f*x + e))*sqrt(-b/a))/(b^2*tan(f
*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2)) - 4*(a*b - b^2)*tan(f*x + e))/((a^3*b - 2*a^2*b^2 + a*b^3)*f*tan(f*x
+ e)^2 + (a^4 - 2*a^3*b + a^2*b^2)*f), 1/4*(4*a*b*f*x*tan(f*x + e)^2 + 4*a^2*f*x - ((3*a*b - b^2)*tan(f*x + e)
^2 + 3*a^2 - a*b)*sqrt(b/a)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt(b/a)/(b*tan(f*x + e))) - 2*(a*b - b^2)*tan(
f*x + e))/((a^3*b - 2*a^2*b^2 + a*b^3)*f*tan(f*x + e)^2 + (a^4 - 2*a^3*b + a^2*b^2)*f)]

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giac [A]  time = 1.80, size = 127, normalized size = 1.31 \[ -\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} {\left (3 \, a b - b^{2}\right )}}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b}} - \frac {2 \, {\left (f x + e\right )}}{a^{2} - 2 \, a b + b^{2}} + \frac {b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )} {\left (a^{2} - a b\right )}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*(3*a*b - b^2)/((a^3 - 2*a^2*b +
a*b^2)*sqrt(a*b)) - 2*(f*x + e)/(a^2 - 2*a*b + b^2) + b*tan(f*x + e)/((b*tan(f*x + e)^2 + a)*(a^2 - a*b)))/f

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maple [A]  time = 0.18, size = 160, normalized size = 1.65 \[ -\frac {b \tan \left (f x +e \right )}{2 \left (a -b \right )^{2} f \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {b^{2} \tan \left (f x +e \right )}{2 f \left (a -b \right )^{2} a \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}-\frac {3 b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \left (a -b \right )^{2} \sqrt {a b}}+\frac {b^{2} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {a b}}\right )}{2 f \left (a -b \right )^{2} a \sqrt {a b}}+\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \left (a -b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.58, size = 114, normalized size = 1.18 \[ -\frac {\frac {b \tan \left (f x + e\right )}{a^{3} - a^{2} b + {\left (a^{2} b - a b^{2}\right )} \tan \left (f x + e\right )^{2}} + \frac {{\left (3 \, a b - b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{3} - 2 \, a^{2} b + a b^{2}\right )} \sqrt {a b}} - \frac {2 \, {\left (f x + e\right )}}{a^{2} - 2 \, a b + b^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 13.53, size = 2489, normalized size = 25.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*tan(e + f*x)^2)^2,x)

[Out]

(2*atan((((((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)*1i)/(3*a^4*b - a^5 + a^2
*b^3 - 3*a^3*b^2) - (tan(e + f*x)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2
))/(2*(a^4 - 2*a^3*b + a^2*b^2)*(2*a^2 - 4*a*b + 2*b^2)))/(2*a^2 - 4*a*b + 2*b^2) + (tan(e + f*x)*(b^5 - 6*a*b
^4 + 13*a^2*b^3))/(2*(a^4 - 2*a^3*b + a^2*b^2)))/(2*a^2 - 4*a*b + 2*b^2) - ((((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b
^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)*1i)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) + (tan(e + f*x)*(16*a^2*b^
7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2))/(2*(a^4 - 2*a^3*b + a^2*b^2)*(2*a^2 - 4*a
*b + 2*b^2)))/(2*a^2 - 4*a*b + 2*b^2) - (tan(e + f*x)*(b^5 - 6*a*b^4 + 13*a^2*b^3))/(2*(a^4 - 2*a^3*b + a^2*b^
2)))/(2*a^2 - 4*a*b + 2*b^2))/((((((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)*1
i)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) - (tan(e + f*x)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 -
48*a^6*b^3 + 16*a^7*b^2))/(2*(a^4 - 2*a^3*b + a^2*b^2)*(2*a^2 - 4*a*b + 2*b^2)))*1i)/(2*a^2 - 4*a*b + 2*b^2) +
(tan(e + f*x)*(b^5 - 6*a*b^4 + 13*a^2*b^3)*1i)/(2*(a^4 - 2*a^3*b + a^2*b^2)))/(2*a^2 - 4*a*b + 2*b^2) + (((((
2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)*1i)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*
b^2) + (tan(e + f*x)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2))/(2*(a^4 -
2*a^3*b + a^2*b^2)*(2*a^2 - 4*a*b + 2*b^2)))*1i)/(2*a^2 - 4*a*b + 2*b^2) - (tan(e + f*x)*(b^5 - 6*a*b^4 + 13*a
^2*b^3)*1i)/(2*(a^4 - 2*a^3*b + a^2*b^2)))/(2*a^2 - 4*a*b + 2*b^2) - ((3*a*b^3)/2 - b^4/2)/(3*a^4*b - a^5 + a^
2*b^3 - 3*a^3*b^2))))/(f*(2*a^2 - 4*a*b + 2*b^2)) - (atan((((-a^3*b)^(1/2)*((tan(e + f*x)*(b^5 - 6*a*b^4 + 13*
a^2*b^3))/(2*(a^4 - 2*a^3*b + a^2*b^2)) - (((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a
^6*b^2)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) - (tan(e + f*x)*(-a^3*b)^(1/2)*(3*a - b)*(16*a^2*b^7 - 48*a^3*b^
6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2))/(8*(a^4 - 2*a^3*b + a^2*b^2)*(a^5 - 2*a^4*b + a^3*b^2)
))*(-a^3*b)^(1/2)*(3*a - b))/(4*(a^5 - 2*a^4*b + a^3*b^2)))*(3*a - b)*1i)/(4*(a^5 - 2*a^4*b + a^3*b^2)) + ((-a
^3*b)^(1/2)*((tan(e + f*x)*(b^5 - 6*a*b^4 + 13*a^2*b^3))/(2*(a^4 - 2*a^3*b + a^2*b^2)) + (((2*a*b^7 - 12*a^2*b
^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) + (tan(e + f*x)*(
-a^3*b)^(1/2)*(3*a - b)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2))/(8*(a^4
- 2*a^3*b + a^2*b^2)*(a^5 - 2*a^4*b + a^3*b^2)))*(-a^3*b)^(1/2)*(3*a - b))/(4*(a^5 - 2*a^4*b + a^3*b^2)))*(3*
a - b)*1i)/(4*(a^5 - 2*a^4*b + a^3*b^2)))/(((3*a*b^3)/2 - b^4/2)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) + ((-a^
3*b)^(1/2)*((tan(e + f*x)*(b^5 - 6*a*b^4 + 13*a^2*b^3))/(2*(a^4 - 2*a^3*b + a^2*b^2)) - (((2*a*b^7 - 12*a^2*b^
6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)/(3*a^4*b - a^5 + a^2*b^3 - 3*a^3*b^2) - (tan(e + f*x)*(-
a^3*b)^(1/2)*(3*a - b)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2))/(8*(a^4
- 2*a^3*b + a^2*b^2)*(a^5 - 2*a^4*b + a^3*b^2)))*(-a^3*b)^(1/2)*(3*a - b))/(4*(a^5 - 2*a^4*b + a^3*b^2)))*(3*a
- b))/(4*(a^5 - 2*a^4*b + a^3*b^2)) - ((-a^3*b)^(1/2)*((tan(e + f*x)*(b^5 - 6*a*b^4 + 13*a^2*b^3))/(2*(a^4 -
2*a^3*b + a^2*b^2)) + (((2*a*b^7 - 12*a^2*b^6 + 28*a^3*b^5 - 32*a^4*b^4 + 18*a^5*b^3 - 4*a^6*b^2)/(3*a^4*b - a
^5 + a^2*b^3 - 3*a^3*b^2) + (tan(e + f*x)*(-a^3*b)^(1/2)*(3*a - b)*(16*a^2*b^7 - 48*a^3*b^6 + 32*a^4*b^5 + 32*
a^5*b^4 - 48*a^6*b^3 + 16*a^7*b^2))/(8*(a^4 - 2*a^3*b + a^2*b^2)*(a^5 - 2*a^4*b + a^3*b^2)))*(-a^3*b)^(1/2)*(3
*a - b))/(4*(a^5 - 2*a^4*b + a^3*b^2)))*(3*a - b))/(4*(a^5 - 2*a^4*b + a^3*b^2))))*(-a^3*b)^(1/2)*(3*a - b)*1i
)/(2*f*(a^5 - 2*a^4*b + a^3*b^2)) - (b*tan(e + f*x))/(2*a*f*(a + b*tan(e + f*x)^2)*(a - b))

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sympy [A]  time = 28.54, size = 2322, normalized size = 23.94 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*tan(f*x+e)**2)**2,x)

[Out]

Piecewise((zoo*x/tan(e)**4, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (x/a**2, Eq(b, 0)), ((x + 1/(f*tan(e + f*x)) - 1/
(3*f*tan(e + f*x)**3))/b**2, Eq(a, 0)), (3*f*x*tan(e + f*x)**4/(8*b**2*f*tan(e + f*x)**4 + 16*b**2*f*tan(e + f
*x)**2 + 8*b**2*f) + 6*f*x*tan(e + f*x)**2/(8*b**2*f*tan(e + f*x)**4 + 16*b**2*f*tan(e + f*x)**2 + 8*b**2*f) +
3*f*x/(8*b**2*f*tan(e + f*x)**4 + 16*b**2*f*tan(e + f*x)**2 + 8*b**2*f) + 3*tan(e + f*x)**3/(8*b**2*f*tan(e +
f*x)**4 + 16*b**2*f*tan(e + f*x)**2 + 8*b**2*f) + 5*tan(e + f*x)/(8*b**2*f*tan(e + f*x)**4 + 16*b**2*f*tan(e
+ f*x)**2 + 8*b**2*f), Eq(a, b)), (x/(a + b*tan(e)**2)**2, Eq(f, 0)), (4*I*a**(5/2)*f*x*sqrt(1/b)/(4*I*a**(9/2
)*f*sqrt(1/b) + 4*I*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(7/2)*b*f*sqrt(1/b) - 8*I*a**(5/2)*b**2*f*
sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1/b) + 4*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2) + 4
*I*a**(3/2)*b*f*x*sqrt(1/b)*tan(e + f*x)**2/(4*I*a**(9/2)*f*sqrt(1/b) + 4*I*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x
)**2 - 8*I*a**(7/2)*b*f*sqrt(1/b) - 8*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1
/b) + 4*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2) - 2*I*a**(3/2)*b*sqrt(1/b)*tan(e + f*x)/(4*I*a**(9/2)*f*s
qrt(1/b) + 4*I*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(7/2)*b*f*sqrt(1/b) - 8*I*a**(5/2)*b**2*f*sqrt(
1/b)*tan(e + f*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1/b) + 4*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2) + 2*I*sq
rt(a)*b**2*sqrt(1/b)*tan(e + f*x)/(4*I*a**(9/2)*f*sqrt(1/b) + 4*I*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I
*a**(7/2)*b*f*sqrt(1/b) - 8*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1/b) + 4*I*
a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2) - 3*a**2*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(9/2)*f*s
qrt(1/b) + 4*I*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(7/2)*b*f*sqrt(1/b) - 8*I*a**(5/2)*b**2*f*sqrt(
1/b)*tan(e + f*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1/b) + 4*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2) + 3*a**2
*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(9/2)*f*sqrt(1/b) + 4*I*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x)**
2 - 8*I*a**(7/2)*b*f*sqrt(1/b) - 8*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1/b)
+ 4*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2) - 3*a*b*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x
)**2/(4*I*a**(9/2)*f*sqrt(1/b) + 4*I*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(7/2)*b*f*sqrt(1/b) - 8*I
*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1/b) + 4*I*a**(3/2)*b**3*f*sqrt(1/b)*tan
(e + f*x)**2) + a*b*log(-I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(9/2)*f*sqrt(1/b) + 4*I*a**(7/2)*b*f*sqrt
(1/b)*tan(e + f*x)**2 - 8*I*a**(7/2)*b*f*sqrt(1/b) - 8*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(5
/2)*b**2*f*sqrt(1/b) + 4*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2) + 3*a*b*log(I*sqrt(a)*sqrt(1/b) + tan(e
+ f*x))*tan(e + f*x)**2/(4*I*a**(9/2)*f*sqrt(1/b) + 4*I*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(7/2)*
b*f*sqrt(1/b) - 8*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1/b) + 4*I*a**(3/2)*b
**3*f*sqrt(1/b)*tan(e + f*x)**2) - a*b*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))/(4*I*a**(9/2)*f*sqrt(1/b) + 4*I
*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(7/2)*b*f*sqrt(1/b) - 8*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f
*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1/b) + 4*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2) + b**2*log(-I*sqrt(a)*
sqrt(1/b) + tan(e + f*x))*tan(e + f*x)**2/(4*I*a**(9/2)*f*sqrt(1/b) + 4*I*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x)*
*2 - 8*I*a**(7/2)*b*f*sqrt(1/b) - 8*I*a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1/b
) + 4*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(e + f*x)**2) - b**2*log(I*sqrt(a)*sqrt(1/b) + tan(e + f*x))*tan(e + f*x)
**2/(4*I*a**(9/2)*f*sqrt(1/b) + 4*I*a**(7/2)*b*f*sqrt(1/b)*tan(e + f*x)**2 - 8*I*a**(7/2)*b*f*sqrt(1/b) - 8*I*
a**(5/2)*b**2*f*sqrt(1/b)*tan(e + f*x)**2 + 4*I*a**(5/2)*b**2*f*sqrt(1/b) + 4*I*a**(3/2)*b**3*f*sqrt(1/b)*tan(
e + f*x)**2), True))

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