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

Optimal. Leaf size=85 \[ \frac {(a+2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 \sqrt {b} f \sqrt {a+b}}-\frac {x}{a^2}+\frac {\tan (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4141, 1975, 471, 522, 203, 205} \[ \frac {(a+2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 \sqrt {b} f \sqrt {a+b}}-\frac {x}{a^2}+\frac {\tan (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a^2) + ((a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(2*a^2*Sqrt[b]*Sqrt[a + b]*f) + Tan[e + f*x]
/(2*a*f*(a + b + 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 471

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(n*(b*c - a*d)*(p + 1)), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GeQ[n
, m - n + 1] && GtQ[m - n + 1, 0] && IntBinomialQ[a, b, c, d, e, m, 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 1975

Int[(u_)^(p_.)*(v_)^(q_.)*((e_.)*(x_))^(m_.), x_Symbol] :> Int[(e*x)^m*ExpandToSum[u, x]^p*ExpandToSum[v, x]^q
, x] /; FreeQ[{e, m, p, q}, x] && BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0]
&&  !BinomialMatchQ[{u, v}, x]

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps

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

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Mathematica [C]  time = 7.68, size = 346, normalized size = 4.07 \[ \frac {\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b)^2 \left (\frac {\frac {(a+2 b) \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {a \sqrt {b} \sin (2 (e+f x))}{(a+b) (a \cos (2 (e+f x))+a+2 b)}}{b^{3/2} f}-\frac {\frac {\left (a^2+8 a b+8 b^2\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b f (a+b) (\cos (e)-\sin (e)) (\sin (e)+\cos (e)) (a \cos (2 (e+f x))+a+2 b)}+\frac {\left (-a^3+6 a^2 b+24 a b^2+16 b^3\right ) (\cos (2 e)-i \sin (2 e)) \tan ^{-1}\left (\frac {(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right )}{b f (a+b)^{3/2} \sqrt {b (\cos (e)-i \sin (e))^4}}+16 x}{a^2}\right )}{64 \left (a+b \sec ^2(e+f x)\right )^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])^2*Sec[e + f*x]^4*(-((16*x + ((-a^3 + 6*a^2*b + 24*a*b^2 + 16*b^3)*ArcTan[(Sec[
f*x]*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin
[e])^4])]*(Cos[2*e] - I*Sin[2*e]))/(b*(a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + ((a^2 + 8*a*b + 8*b^2)*
((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(b*(a + b)*f*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[e] - Sin[e])*(Cos[e] + S
in[e])))/a^2) + (((a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(3/2) - (a*Sqrt[b]*Sin[2*(e +
f*x)])/((a + b)*(a + 2*b + a*Cos[2*(e + f*x)])))/(b^(3/2)*f)))/(64*(a + b*Sec[e + f*x]^2)^2)

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fricas [B]  time = 0.68, size = 458, normalized size = 5.39 \[ \left [-\frac {8 \, {\left (a^{2} b + a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 8 \, {\left (a b^{2} + b^{3}\right )} f x - 4 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{8 \, {\left ({\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}, -\frac {4 \, {\left (a^{2} b + a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 4 \, {\left (a b^{2} + b^{3}\right )} f x - 2 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{4 \, {\left ({\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.47, size = 99, normalized size = 1.16 \[ \frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (a + 2 \, b\right )}}{\sqrt {a b + b^{2}} a^{2}} - \frac {2 \, {\left (f x + e\right )}}{a^{2}} + \frac {\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^2/(a+b*sec(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 + b^2)))*(a + 2*b)/(sqrt(a*b + b^2)
*a^2) - 2*(f*x + e)/a^2 + tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)*a))/f

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maple [A]  time = 0.95, size = 108, normalized size = 1.27 \[ \frac {\tan \left (f x +e \right )}{2 a f \left (a +b +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {\arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{2 f a \sqrt {\left (a +b \right ) b}}+\frac {b \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{2} \sqrt {\left (a +b \right ) b}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.42, size = 75, normalized size = 0.88 \[ \frac {\frac {\tan \left (f x + e\right )}{a b \tan \left (f x + e\right )^{2} + a^{2} + a b} + \frac {{\left (a + 2 \, b\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{2}} - \frac {2 \, {\left (f x + e\right )}}{a^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 5.00, size = 711, normalized size = 8.36 \[ \frac {\mathrm {tan}\left (e+f\,x\right )}{2\,a\,f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}-\frac {x}{a^2}-\frac {\mathrm {atan}\left (\frac {\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a^2\,b+4\,a\,b^2+8\,b^3\right )}{2\,a^2}-\frac {\left (2\,a\,b^2-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (16\,a^5\,b^2+32\,a^4\,b^3\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{8\,a^2\,\left (a^3\,b+a^2\,b^2\right )}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^3\,b+a^2\,b^2\right )}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{4\,\left (a^3\,b+a^2\,b^2\right )}+\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a^2\,b+4\,a\,b^2+8\,b^3\right )}{2\,a^2}+\frac {\left (2\,a\,b^2+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (16\,a^5\,b^2+32\,a^4\,b^3\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{8\,a^2\,\left (a^3\,b+a^2\,b^2\right )}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^3\,b+a^2\,b^2\right )}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{4\,\left (a^3\,b+a^2\,b^2\right )}}{\frac {b^2+\frac {a\,b}{2}}{a^3}-\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a^2\,b+4\,a\,b^2+8\,b^3\right )}{2\,a^2}-\frac {\left (2\,a\,b^2-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (16\,a^5\,b^2+32\,a^4\,b^3\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{8\,a^2\,\left (a^3\,b+a^2\,b^2\right )}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^3\,b+a^2\,b^2\right )}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^3\,b+a^2\,b^2\right )}+\frac {\left (\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a^2\,b+4\,a\,b^2+8\,b^3\right )}{2\,a^2}+\frac {\left (2\,a\,b^2+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (16\,a^5\,b^2+32\,a^4\,b^3\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{8\,a^2\,\left (a^3\,b+a^2\,b^2\right )}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^3\,b+a^2\,b^2\right )}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}}{4\,\left (a^3\,b+a^2\,b^2\right )}}\right )\,\left (a+2\,b\right )\,\sqrt {-b\,\left (a+b\right )}\,1{}\mathrm {i}}{2\,f\,\left (a^3\,b+a^2\,b^2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

tan(e + f*x)/(2*a*f*(a + b + b*tan(e + f*x)^2)) - x/a^2 - (atan(((((tan(e + f*x)*(4*a*b^2 + a^2*b + 8*b^3))/(2
*a^2) - ((2*a*b^2 - (tan(e + f*x)*(32*a^4*b^3 + 16*a^5*b^2)*(a + 2*b)*(-b*(a + b))^(1/2))/(8*a^2*(a^3*b + a^2*
b^2)))*(a + 2*b)*(-b*(a + b))^(1/2))/(4*(a^3*b + a^2*b^2)))*(a + 2*b)*(-b*(a + b))^(1/2)*1i)/(4*(a^3*b + a^2*b
^2)) + (((tan(e + f*x)*(4*a*b^2 + a^2*b + 8*b^3))/(2*a^2) + ((2*a*b^2 + (tan(e + f*x)*(32*a^4*b^3 + 16*a^5*b^2
)*(a + 2*b)*(-b*(a + b))^(1/2))/(8*a^2*(a^3*b + a^2*b^2)))*(a + 2*b)*(-b*(a + b))^(1/2))/(4*(a^3*b + a^2*b^2))
)*(a + 2*b)*(-b*(a + b))^(1/2)*1i)/(4*(a^3*b + a^2*b^2)))/(((a*b)/2 + b^2)/a^3 - (((tan(e + f*x)*(4*a*b^2 + a^
2*b + 8*b^3))/(2*a^2) - ((2*a*b^2 - (tan(e + f*x)*(32*a^4*b^3 + 16*a^5*b^2)*(a + 2*b)*(-b*(a + b))^(1/2))/(8*a
^2*(a^3*b + a^2*b^2)))*(a + 2*b)*(-b*(a + b))^(1/2))/(4*(a^3*b + a^2*b^2)))*(a + 2*b)*(-b*(a + b))^(1/2))/(4*(
a^3*b + a^2*b^2)) + (((tan(e + f*x)*(4*a*b^2 + a^2*b + 8*b^3))/(2*a^2) + ((2*a*b^2 + (tan(e + f*x)*(32*a^4*b^3
+ 16*a^5*b^2)*(a + 2*b)*(-b*(a + b))^(1/2))/(8*a^2*(a^3*b + a^2*b^2)))*(a + 2*b)*(-b*(a + b))^(1/2))/(4*(a^3*
b + a^2*b^2)))*(a + 2*b)*(-b*(a + b))^(1/2))/(4*(a^3*b + a^2*b^2))))*(a + 2*b)*(-b*(a + b))^(1/2)*1i)/(2*f*(a^
3*b + a^2*b^2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(e + f*x)**2/(a + b*sec(e + f*x)**2)**2, x)

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