\(\int \frac {\tan ^2(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [358]

Optimal result

Integrand size = 23, antiderivative size = 85 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=-\frac {x}{a^2}+\frac {(a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^2 \sqrt {b} \sqrt {a+b} f}+\frac {\tan (e+f x)}{2 a f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 5.38 (sec) , antiderivative size = 346, normalized size of antiderivative = 4.07 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^2 \sec ^4(e+f x) \left (-\frac {16 x+\frac {\left (-a^3+6 a^2 b+24 a b^2+16 b^3\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))}{b (a+b)^{3/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {\left (a^2+8 a b+8 b^2\right ) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{b (a+b) f (a+2 b+a \cos (2 (e+f x))) (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}}{a^2}+\frac {\frac {(a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{3/2}}-\frac {a \sqrt {b} \sin (2 (e+f x))}{(a+b) (a+2 b+a \cos (2 (e+f x)))}}{b^{3/2} f}\right )}{64 \left (a+b \sec ^2(e+f x)\right )^2} \] Input:

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

Output:

((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] + Sin[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)
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4629, 2075, 373, 397, 216, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)^2}{\left (a+b \sec (e+f x)^2\right )^2}dx\)

\(\Big \downarrow \) 4629

\(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^2}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2075

\(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 373

\(\displaystyle \frac {\frac {\tan (e+f x)}{2 a \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\int \frac {1-\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a}}{f}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\tan (e+f x)}{2 a \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\frac {2 \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {(a+2 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\tan (e+f x)}{2 a \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\frac {2 \arctan (\tan (e+f x))}{a}-\frac {(a+2 b) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\tan (e+f x)}{2 a \left (a+b \tan ^2(e+f x)+b\right )}-\frac {\frac {2 \arctan (\tan (e+f x))}{a}-\frac {(a+2 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {b} \sqrt {a+b}}}{2 a}}{f}\)

Input:

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

Output:

(-1/2*((2*ArcTan[Tan[e + f*x]])/a - ((a + 2*b)*ArcTan[(Sqrt[b]*Tan[e + f*x 
])/Sqrt[a + b]])/(a*Sqrt[b]*Sqrt[a + b]))/a + Tan[e + f*x]/(2*a*(a + b + b 
*Tan[e + f*x]^2)))/f
 

Defintions of rubi rules used

rule 216

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

rule 218

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

rule 373

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[e*(e*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 
1)/(2*(b*c - a*d)*(p + 1))), x] - Simp[e^2/(2*(b*c - a*d)*(p + 1))   Int[(e 
*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(m - 1) + d*(m + 2*p + 
 2*q + 3)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 
 0] && LtQ[p, -1] && GtQ[m, 1] && LeQ[m, 3] && IntBinomialQ[a, b, c, d, e, 
m, 2, p, q, x]
 

rule 397

Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ 
Symbol] :> Simp[(b*e - a*f)/(b*c - a*d)   Int[1/(a + b*x^2), x], x] - Simp[ 
(d*e - c*f)/(b*c - a*d)   Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e 
, f}, x]
 

rule 2075

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4629

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]}, Sim 
p[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] && Inte 
gerQ[n/2] && (IntegerQ[m/2] || EqQ[n, 2])
 

Maple [A] (verified)

Time = 1.68 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{2}}+\frac {\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {\left (a +2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}}{a^{2}}}{f}\)	\(77\)
default	\(\frac {-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{2}}+\frac {\frac {a \tan \left (f x +e \right )}{2 a +2 b +2 b \tan \left (f x +e \right )^{2}}+\frac {\left (a +2 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \sqrt {\left (a +b \right ) b}}}{a^{2}}}{f}\)	\(77\)
risch	\(-\frac {x}{a^{2}}+\frac {i \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}{a^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, f a}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right ) b}{2 \sqrt {-a b -b^{2}}\, f \,a^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, f a}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right ) b}{2 \sqrt {-a b -b^{2}}\, f \,a^{2}}\)	\(435\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (73) = 146\).

Time = 0.11 (sec) , antiderivative size = 458, normalized size of antiderivative = 5.39 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [-\frac {8 \, {\left (a^{2} b + a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 8 \, {\left (a b^{2} + b^{3}\right )} f x - 4 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{8 \, {\left ({\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}, -\frac {4 \, {\left (a^{2} b + a b^{2}\right )} f x \cos \left (f x + e\right )^{2} + 4 \, {\left (a b^{2} + b^{3}\right )} f x - 2 \, {\left (a^{2} b + a b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) + {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} + a b + 2 \, b^{2}\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{4 \, {\left ({\left (a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{2} + a^{2} b^{3}\right )} f\right )}}\right ] \] Input:

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

Output:

[-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)]
 

Sympy [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.88 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {\tan \left (f x + e\right )}{a b \tan \left (f x + e\right )^{2} + a^{2} + a b} + \frac {{\left (a + 2 \, b\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{2}} - \frac {2 \, {\left (f x + e\right )}}{a^{2}}}{2 \, f} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.98 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {{\left (a + 2 \, b\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )}{2 \, \sqrt {a b + b^{2}} a^{2} f} - \frac {f x + e}{a^{2} f} + \frac {\tan \left (f x + e\right )}{2 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )} a f} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 15.57 (sec) , antiderivative size = 711, normalized size of antiderivative = 8.36 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx =\text {Too large to display} \] Input:

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

Output:

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))
 

Reduce [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 519, normalized size of antiderivative = 6.11 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) \sin \left (f x +e \right )^{2} a^{2}+2 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) \sin \left (f x +e \right )^{2} a b -\sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) a^{2}-3 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) a b -2 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) b^{2}+\sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) \sin \left (f x +e \right )^{2} a^{2}+2 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) \sin \left (f x +e \right )^{2} a b -\sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) a^{2}-3 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) a b -2 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) b^{2}-\cos \left (f x +e \right ) \sin \left (f x +e \right ) a^{2} b -\cos \left (f x +e \right ) \sin \left (f x +e \right ) a \,b^{2}-2 \sin \left (f x +e \right )^{2} a^{2} b f x -2 \sin \left (f x +e \right )^{2} a \,b^{2} f x +2 a^{2} b f x +4 a \,b^{2} f x +2 b^{3} f x}{2 a^{2} b f \left (\sin \left (f x +e \right )^{2} a^{2}+\sin \left (f x +e \right )^{2} a b -a^{2}-2 a b -b^{2}\right )} \] Input:

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

Output:

(sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b) 
)*sin(e + f*x)**2*a**2 + 2*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + 
f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b - sqrt(b)*sqrt(a + b)*atan 
((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**2 - 3*sqrt(b)*sqrt(a 
 + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a*b - 2*sqrt( 
b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*b**2 
 + sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt( 
b))*sin(e + f*x)**2*a**2 + 2*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e 
+ f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b - sqrt(b)*sqrt(a + b)*at 
an((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a**2 - 3*sqrt(b)*sqrt 
(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*a*b - 2*sqr 
t(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*b* 
*2 - cos(e + f*x)*sin(e + f*x)*a**2*b - cos(e + f*x)*sin(e + f*x)*a*b**2 - 
 2*sin(e + f*x)**2*a**2*b*f*x - 2*sin(e + f*x)**2*a*b**2*f*x + 2*a**2*b*f* 
x + 4*a*b**2*f*x + 2*b**3*f*x)/(2*a**2*b*f*(sin(e + f*x)**2*a**2 + sin(e + 
 f*x)**2*a*b - a**2 - 2*a*b - b**2))