\(\int \frac {\tan ^2(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [371]

Optimal result

Integrand size = 23, antiderivative size = 138 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {x}{a^3}+\frac {\left (3 a^2+12 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 \sqrt {b} (a+b)^{3/2} f}+\frac {\tan (e+f x)}{4 a f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(3 a+4 b) \tan (e+f x)}{8 a^2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

-x/a^3+1/8*(3*a^2+12*a*b+8*b^2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a^3 
/b^(1/2)/(a+b)^(3/2)/f+1/4*tan(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)^2+1/8*(3*a+ 
4*b)*tan(f*x+e)/a^2/(a+b)/f/(a+b+b*tan(f*x+e)^2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.91 (sec) , antiderivative size = 1334, normalized size of antiderivative = 9.67 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

Integrate[Tan[e + f*x]^2/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*((-6*a*(a + 2*b)*ArcTan[( 
Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(5/2) + (4*(3*a^2 + 8*a*b + 8* 
b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(5/2) - (4*a*Sqrt 
[b]*(3*a^2 + 16*a*b + 16*b^2 + 3*a*(a + 2*b)*Cos[2*(e + f*x)])*Sin[2*(e + 
f*x)])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2) + (2*Sqrt[b]*(3*a^3 + 
14*a^2*b + 24*a*b^2 + 16*b^3 + a*(3*a^2 + 4*a*b + 4*b^2)*Cos[2*(e + f*x)]) 
*Sin[2*(e + f*x)])/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2) - (Sqrt[b] 
*((2*(3*a^5 - 10*a^4*b + 80*a^3*b^2 + 480*a^2*b^3 + 640*a*b^4 + 256*b^5)*A 
rcTan[(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]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (Sec[2*e]*(256*b^2*( 
a + b)^2*(3*a^2 + 8*a*b + 8*b^2)*f*x*Cos[2*e] + 512*a*b^2*(a + b)^2*(a + 2 
*b)*f*x*Cos[2*f*x] + 128*a^4*b^2*f*x*Cos[2*(e + 2*f*x)] + 256*a^3*b^3*f*x* 
Cos[2*(e + 2*f*x)] + 128*a^2*b^4*f*x*Cos[2*(e + 2*f*x)] + 512*a^4*b^2*f*x* 
Cos[4*e + 2*f*x] + 2048*a^3*b^3*f*x*Cos[4*e + 2*f*x] + 2560*a^2*b^4*f*x*Co 
s[4*e + 2*f*x] + 1024*a*b^5*f*x*Cos[4*e + 2*f*x] + 128*a^4*b^2*f*x*Cos[6*e 
 + 4*f*x] + 256*a^3*b^3*f*x*Cos[6*e + 4*f*x] + 128*a^2*b^4*f*x*Cos[6*e + 4 
*f*x] - 9*a^6*Sin[2*e] + 12*a^5*b*Sin[2*e] + 684*a^4*b^2*Sin[2*e] + 2880*a 
^3*b^3*Sin[2*e] + 5280*a^2*b^4*Sin[2*e] + 4608*a*b^5*Sin[2*e] + 1536*b^6*S 
in[2*e] + 9*a^6*Sin[2*f*x] - 14*a^5*b*Sin[2*f*x] - 608*a^4*b^2*Sin[2*f*...
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.17, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4629, 2075, 373, 402, 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.

\(\displaystyle \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\tan (e+f x)^2}{\left (a+b \sec (e+f x)^2\right )^3}dx\)

\(\Big \downarrow \) 4629

\(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (a+b \left (\tan ^2(e+f x)+1\right )\right )^3}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2075

\(\displaystyle \frac {\int \frac {\tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 373

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\int \frac {1-3 \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 a}}{f}\)

\(\Big \downarrow \) 402

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\int \frac {-\left ((3 a+4 b) \tan ^2(e+f x)\right )+5 a+4 b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a (a+b)}-\frac {(3 a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}}{f}\)

\(\Big \downarrow \) 397

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {8 (a+b) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {\left (3 a^2+12 a b+8 b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a (a+b)}-\frac {(3 a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}}{f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {8 (a+b) \arctan (\tan (e+f x))}{a}-\frac {\left (3 a^2+12 a b+8 b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a (a+b)}-\frac {(3 a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\tan (e+f x)}{4 a \left (a+b \tan ^2(e+f x)+b\right )^2}-\frac {\frac {\frac {8 (a+b) \arctan (\tan (e+f x))}{a}-\frac {\left (3 a^2+12 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {b} \sqrt {a+b}}}{2 a (a+b)}-\frac {(3 a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a}}{f}\)

Input:

Int[Tan[e + f*x]^2/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(Tan[e + f*x]/(4*a*(a + b + b*Tan[e + f*x]^2)^2) - (((8*(a + b)*ArcTan[Tan 
[e + f*x]])/a - ((3*a^2 + 12*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sq 
rt[a + b]])/(a*Sqrt[b]*Sqrt[a + b]))/(2*a*(a + b)) - ((3*a + 4*b)*Tan[e + 
f*x])/(2*a*(a + b)*(a + b + b*Tan[e + f*x]^2)))/(4*a))/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 402

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x 
_)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ 
(q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) 
 Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) 
*(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b 
, c, d, e, f, q}, x] && LtQ[p, -1]
 

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 = 6.91 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}+\frac {\frac {\frac {a b \left (3 a +4 b \right ) \tan \left (f x +e \right )^{3}}{8 a +8 b}+\frac {\left (5 a +4 b \right ) a \tan \left (f x +e \right )}{8}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+12 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}}{a^{3}}}{f}\)	\(125\)
default	\(\frac {-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}+\frac {\frac {\frac {a b \left (3 a +4 b \right ) \tan \left (f x +e \right )^{3}}{8 a +8 b}+\frac {\left (5 a +4 b \right ) a \tan \left (f x +e \right )}{8}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}+12 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}}{a^{3}}}{f}\)	\(125\)
risch	\(-\frac {x}{a^{3}}+\frac {i \left (5 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}+20 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}+16 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+15 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+58 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+88 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+48 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+15 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+44 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{2} b +32 \,{\mathrm e}^{2 i \left (f x +e \right )} a \,b^{2}+5 a^{3}+6 a^{2} b \right )}{4 a^{3} \left (a +b \right ) 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 )^{2}}+\frac {3 \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 )}{16 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f a}+\frac {3 \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}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) 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 ) b^{2}}{2 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f \,a^{3}}-\frac {3 \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 )}{16 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f a}-\frac {3 \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}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) 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 ) b^{2}}{2 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f \,a^{3}}\)	\(791\)

Input:

int(tan(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/f*(-1/a^3*arctan(tan(f*x+e))+1/a^3*((1/8*a*b*(3*a+4*b)/(a+b)*tan(f*x+e)^ 
3+1/8*(5*a+4*b)*a*tan(f*x+e))/(a+b+b*tan(f*x+e)^2)^2+1/8*(3*a^2+12*a*b+8*b 
^2)/(a+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. 390 vs. \(2 (124) = 248\).

Time = 0.14 (sec) , antiderivative size = 860, normalized size of antiderivative = 6.23 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(tan(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
 

Output:

[-1/32*(32*(a^4*b + 2*a^3*b^2 + a^2*b^3)*f*x*cos(f*x + e)^4 + 64*(a^3*b^2 
+ 2*a^2*b^3 + a*b^4)*f*x*cos(f*x + e)^2 + 32*(a^2*b^3 + 2*a*b^4 + b^5)*f*x 
 + ((3*a^4 + 12*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 + 3*a^2*b^2 + 12*a*b^3 + 
 8*b^4 + 2*(3*a^3*b + 12*a^2*b^2 + 8*a*b^3)*cos(f*x + e)^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)) - 4*(( 
5*a^4*b + 11*a^3*b^2 + 6*a^2*b^3)*cos(f*x + e)^3 + (3*a^3*b^2 + 7*a^2*b^3 
+ 4*a*b^4)*cos(f*x + e))*sin(f*x + e))/((a^7*b + 2*a^6*b^2 + a^5*b^3)*f*co 
s(f*x + e)^4 + 2*(a^6*b^2 + 2*a^5*b^3 + a^4*b^4)*f*cos(f*x + e)^2 + (a^5*b 
^3 + 2*a^4*b^4 + a^3*b^5)*f), -1/16*(16*(a^4*b + 2*a^3*b^2 + a^2*b^3)*f*x* 
cos(f*x + e)^4 + 32*(a^3*b^2 + 2*a^2*b^3 + a*b^4)*f*x*cos(f*x + e)^2 + 16* 
(a^2*b^3 + 2*a*b^4 + b^5)*f*x + ((3*a^4 + 12*a^3*b + 8*a^2*b^2)*cos(f*x + 
e)^4 + 3*a^2*b^2 + 12*a*b^3 + 8*b^4 + 2*(3*a^3*b + 12*a^2*b^2 + 8*a*b^3)*c 
os(f*x + e)^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))) - 2*((5*a^4*b + 11*a^3*b^2 + 6 
*a^2*b^3)*cos(f*x + e)^3 + (3*a^3*b^2 + 7*a^2*b^3 + 4*a*b^4)*cos(f*x + e)) 
*sin(f*x + e))/((a^7*b + 2*a^6*b^2 + a^5*b^3)*f*cos(f*x + e)^4 + 2*(a^6*b^ 
2 + 2*a^5*b^3 + a^4*b^4)*f*cos(f*x + e)^2 + (a^5*b^3 + 2*a^4*b^4 + a^3*b^5 
)*f)]
 

Sympy [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \] Input:

integrate(tan(f*x+e)**2/(a+b*sec(f*x+e)**2)**3,x)
 

Output:

Integral(tan(e + f*x)**2/(a + b*sec(e + f*x)**2)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.38 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{4} + a^{3} b\right )} \sqrt {{\left (a + b\right )} b}} + \frac {{\left (3 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + {\left (5 \, a^{2} + 9 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )}{a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3} + {\left (a^{3} b^{2} + a^{2} b^{3}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} \tan \left (f x + e\right )^{2}} - \frac {8 \, {\left (f x + e\right )}}{a^{3}}}{8 \, f} \] Input:

integrate(tan(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
 

Output:

1/8*((3*a^2 + 12*a*b + 8*b^2)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^4 
 + a^3*b)*sqrt((a + b)*b)) + ((3*a*b + 4*b^2)*tan(f*x + e)^3 + (5*a^2 + 9* 
a*b + 4*b^2)*tan(f*x + e))/(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3 + (a^3*b^2 
 + a^2*b^3)*tan(f*x + e)^4 + 2*(a^4*b + 2*a^3*b^2 + a^2*b^3)*tan(f*x + e)^ 
2) - 8*(f*x + e)/a^3)/f
 

Giac [A] (verification not implemented)

Time = 0.39 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.17 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {{\left (3 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )}{8 \, {\left (a^{4} f + a^{3} b f\right )} \sqrt {a b + b^{2}}} + \frac {3 \, a b \tan \left (f x + e\right )^{3} + 4 \, b^{2} \tan \left (f x + e\right )^{3} + 5 \, a^{2} \tan \left (f x + e\right ) + 9 \, a b \tan \left (f x + e\right ) + 4 \, b^{2} \tan \left (f x + e\right )}{8 \, {\left (a^{3} f + a^{2} b f\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} - \frac {f x + e}{a^{3} f} \] Input:

integrate(tan(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
 

Output:

1/8*(3*a^2 + 12*a*b + 8*b^2)*arctan(b*tan(f*x + e)/sqrt(a*b + b^2))/((a^4* 
f + a^3*b*f)*sqrt(a*b + b^2)) + 1/8*(3*a*b*tan(f*x + e)^3 + 4*b^2*tan(f*x 
+ e)^3 + 5*a^2*tan(f*x + e) + 9*a*b*tan(f*x + e) + 4*b^2*tan(f*x + e))/((a 
^3*f + a^2*b*f)*(b*tan(f*x + e)^2 + a + b)^2) - (f*x + e)/(a^3*f)
 

Mupad [B] (verification not implemented)

Time = 18.13 (sec) , antiderivative size = 2405, normalized size of antiderivative = 17.43 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \] Input:

int(tan(e + f*x)^2/(a + b/cos(e + f*x)^2)^3,x)
 

Output:

((tan(e + f*x)*(5*a + 4*b))/(8*a^2) + (tan(e + f*x)^3*(3*a*b + 4*b^2))/(8* 
a^2*(a + b)))/(f*(2*a*b + a^2 + b^2 + tan(e + f*x)^2*(2*a*b + 2*b^2) + b^2 
*tan(e + f*x)^4)) - atan((((((2*a^6*b^4 + (9*a^7*b^3)/2 + (5*a^8*b^2)/2)*1 
i)/(2*(2*a^7*b + a^8 + a^6*b^2)) - (tan(e + f*x)*(512*a^6*b^5 + 1280*a^7*b 
^4 + 1024*a^8*b^3 + 256*a^9*b^2))/(128*a^3*(2*a^5*b + a^6 + a^4*b^2)))/(2* 
a^3) + (tan(e + f*x)*(320*a*b^4 + 9*a^4*b + 128*b^5 + 256*a^2*b^3 + 72*a^3 
*b^2))/(64*(2*a^5*b + a^6 + a^4*b^2)))/a^3 - ((((2*a^6*b^4 + (9*a^7*b^3)/2 
 + (5*a^8*b^2)/2)*1i)/(2*(2*a^7*b + a^8 + a^6*b^2)) + (tan(e + f*x)*(512*a 
^6*b^5 + 1280*a^7*b^4 + 1024*a^8*b^3 + 256*a^9*b^2))/(128*a^3*(2*a^5*b + a 
^6 + a^4*b^2)))/(2*a^3) - (tan(e + f*x)*(320*a*b^4 + 9*a^4*b + 128*b^5 + 2 
56*a^2*b^3 + 72*a^3*b^2))/(64*(2*a^5*b + a^6 + a^4*b^2)))/a^3)/(((9*a*b^3) 
/4 + (9*a^3*b)/32 + b^4 + (3*a^2*b^2)/2)/(2*a^7*b + a^8 + a^6*b^2) + ((((( 
2*a^6*b^4 + (9*a^7*b^3)/2 + (5*a^8*b^2)/2)*1i)/(2*(2*a^7*b + a^8 + a^6*b^2 
)) - (tan(e + f*x)*(512*a^6*b^5 + 1280*a^7*b^4 + 1024*a^8*b^3 + 256*a^9*b^ 
2))/(128*a^3*(2*a^5*b + a^6 + a^4*b^2)))*1i)/(2*a^3) + (tan(e + f*x)*(320* 
a*b^4 + 9*a^4*b + 128*b^5 + 256*a^2*b^3 + 72*a^3*b^2)*1i)/(64*(2*a^5*b + a 
^6 + a^4*b^2)))/a^3 + (((((2*a^6*b^4 + (9*a^7*b^3)/2 + (5*a^8*b^2)/2)*1i)/ 
(2*(2*a^7*b + a^8 + a^6*b^2)) + (tan(e + f*x)*(512*a^6*b^5 + 1280*a^7*b^4 
+ 1024*a^8*b^3 + 256*a^9*b^2))/(128*a^3*(2*a^5*b + a^6 + a^4*b^2)))*1i)/(2 
*a^3) - (tan(e + f*x)*(320*a*b^4 + 9*a^4*b + 128*b^5 + 256*a^2*b^3 + 72...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1422, normalized size of antiderivative = 10.30 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(tan(f*x+e)^2/(a+b*sec(f*x+e)^2)^3,x)
 

Output:

(3*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt( 
b))*sin(e + f*x)**4*a**4 + 12*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e 
 + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b + 8*sqrt(b)*sqrt(a + 
 b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4 
*a**2*b**2 - 6*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sq 
rt(a))/sqrt(b))*sin(e + f*x)**2*a**4 - 30*sqrt(b)*sqrt(a + b)*atan((sqrt(a 
 + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**3*b - 40*sqr 
t(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*si 
n(e + f*x)**2*a**2*b**2 - 16*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**3 + 3*sqrt(b)*sqrt(a + 
b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**4 + 18*sqrt(b 
)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**3* 
b + 35*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/s 
qrt(b))*a**2*b**2 + 28*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x) 
/2) - sqrt(a))/sqrt(b))*a*b**3 + 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*t 
an((e + f*x)/2) - sqrt(a))/sqrt(b))*b**4 + 3*sqrt(b)*sqrt(a + b)*atan((sqr 
t(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**4 + 12*sq 
rt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*s 
in(e + f*x)**4*a**3*b + 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f 
*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**2*b**2 - 6*sqrt(b)*sqrt(a...