\(\int \frac {\tan ^4(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [370]

Optimal result

Integrand size = 23, antiderivative size = 137 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {x}{a^3}+\frac {\left (a^2-4 a b-8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 b^{3/2} \sqrt {a+b} f}-\frac {(a+b) \tan (e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(a-4 b) \tan (e+f x)}{8 a^2 b f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 9.95 (sec) , antiderivative size = 1333, normalized size of antiderivative = 9.73 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

Integrate[Tan[e + f*x]^4/(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[(S 
qrt[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 + 1 
4*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)*Ar 
cTan[(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*C 
os[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*C 
os[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*Cos 
[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*Si 
n[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*x...
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4629, 2075, 372, 402, 27, 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.

\(\displaystyle \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4629

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

\(\Big \downarrow \) 2075

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

\(\Big \downarrow \) 372

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

\(\Big \downarrow \) 402

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 397

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 218

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

Input:

Int[Tan[e + f*x]^4/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(-1/4*((a + b)*Tan[e + f*x])/(a*b*(a + b + b*Tan[e + f*x]^2)^2) + (((8*b*A 
rcTan[Tan[e + f*x]])/a + ((a^2 - 4*a*b - 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f* 
x])/Sqrt[a + b]])/(a*Sqrt[b]*Sqrt[a + b]))/(2*a) + ((a - 4*b)*Tan[e + f*x] 
)/(2*a*(a + b + b*Tan[e + f*x]^2)))/(4*a*b))/f
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 372

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ 
), x_Symbol] :> Simp[(-a)*e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2 
)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] + Simp[e^4/(2*b*(b*c - a*d)*(p + 1 
))   Int[(e*x)^(m - 4)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[a*c*(m - 3) + 
 (a*d*(m + 2*q - 1) + 2*b*c*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, 
e, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && GtQ[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 = 9.06 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}+\frac {\frac {\left (\frac {1}{8} a^{2}-\frac {1}{2} a b \right ) \tan \left (f x +e \right )^{3}-\frac {a \left (a^{2}+5 a b +4 b^{2}\right ) \tan \left (f x +e \right )}{8 b}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (a^{2}-4 a b -8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 b \sqrt {\left (a +b \right ) b}}}{a^{3}}}{f}\)	\(124\)
default	\(\frac {\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}+\frac {\frac {\left (\frac {1}{8} a^{2}-\frac {1}{2} a b \right ) \tan \left (f x +e \right )^{3}-\frac {a \left (a^{2}+5 a b +4 b^{2}\right ) \tan \left (f x +e \right )}{8 b}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (a^{2}-4 a b -8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 b \sqrt {\left (a +b \right ) b}}}{a^{3}}}{f}\)	\(124\)
risch	\(\frac {x}{a^{3}}-\frac {i \left (a^{3} {\mathrm e}^{6 i \left (f x +e \right )}+12 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}+16 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+3 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+26 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+56 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+48 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+3 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+20 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{2} b +32 \,{\mathrm e}^{2 i \left (f x +e \right )} a \,b^{2}+a^{3}+6 a^{2} b \right )}{4 a^{3} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2} b}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{16 \sqrt {-a b -b^{2}}\, f b a}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, f \,a^{2}}+\frac {b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, f \,a^{3}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{16 \sqrt {-a b -b^{2}}\, f b a}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, f \,a^{2}}-\frac {b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{2 \sqrt {-a b -b^{2}}\, f \,a^{3}}\)	\(749\)

Input:

int(tan(f*x+e)^4/(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^2-1/2*a*b)*tan(f*x+e)^3-1/8*a 
*(a^2+5*a*b+4*b^2)/b*tan(f*x+e))/(a+b+b*tan(f*x+e)^2)^2+1/8*(a^2-4*a*b-8*b 
^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. 342 vs. \(2 (123) = 246\).

Time = 0.13 (sec) , antiderivative size = 763, normalized size of antiderivative = 5.57 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(tan(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\int \frac {\tan ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \] Input:

integrate(tan(f*x+e)**4/(a+b*sec(f*x+e)**2)**3,x)
 

Output:

Integral(tan(e + f*x)**4/(a + b*sec(e + f*x)**2)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.19 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {{\left (a b - 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} - {\left (a^{2} + 5 \, a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )}{a^{2} b^{3} \tan \left (f x + e\right )^{4} + a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3} + 2 \, {\left (a^{3} b^{2} + a^{2} b^{3}\right )} \tan \left (f x + e\right )^{2}} + \frac {8 \, {\left (f x + e\right )}}{a^{3}} + \frac {{\left (a^{2} - 4 \, a b - 8 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3} b}}{8 \, f} \] Input:

integrate(tan(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.52 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.07 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {f x + e}{a^{3} f} + \frac {{\left (a^{2} - 4 \, a b - 8 \, b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )}{8 \, \sqrt {a b + b^{2}} a^{3} b f} + \frac {a b \tan \left (f x + e\right )^{3} - 4 \, b^{2} \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right ) - 5 \, a b \tan \left (f x + e\right ) - 4 \, b^{2} \tan \left (f x + e\right )}{8 \, {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2} a^{2} b f} \] Input:

integrate(tan(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 15.69 (sec) , antiderivative size = 1117, normalized size of antiderivative = 8.15 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \] Input:

int(tan(e + f*x)^4/(a + b/cos(e + f*x)^2)^3,x)
 

Output:

((tan(e + f*x)^3*(a - 4*b))/(8*a^2) - (tan(e + f*x)*(a + b)*(a + 4*b))/(8* 
a^2*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(tan(e + f*x)/(32*(b/(4*a) - 1/32)) + tan(e + f*x)/(4*(a 
/(32*b) - 1/4)))/(a^3*f) + (atan(-(((((-b^3*(a + b))^(1/2)*((2*a^6*b^3 + ( 
a^7*b^2)/2)/(a^6*b) - (tan(e + f*x)*(512*a^6*b^4 + 256*a^7*b^3)*(-b^3*(a + 
 b))^(1/2)*(4*a*b - a^2 + 8*b^2))/(512*a^4*b*(a^3*b^4 + a^4*b^3)))*(4*a*b 
- a^2 + 8*b^2))/(16*(a^3*b^4 + a^4*b^3)) - (tan(e + f*x)*(64*a*b^3 - 8*a^3 
*b + a^4 + 128*b^4))/(32*a^4*b))*(-b^3*(a + b))^(1/2)*(4*a*b - a^2 + 8*b^2 
)*1i)/(16*(a^3*b^4 + a^4*b^3)) - ((((-b^3*(a + b))^(1/2)*((2*a^6*b^3 + (a^ 
7*b^2)/2)/(a^6*b) + (tan(e + f*x)*(512*a^6*b^4 + 256*a^7*b^3)*(-b^3*(a + b 
))^(1/2)*(4*a*b - a^2 + 8*b^2))/(512*a^4*b*(a^3*b^4 + a^4*b^3)))*(4*a*b - 
a^2 + 8*b^2))/(16*(a^3*b^4 + a^4*b^3)) + (tan(e + f*x)*(64*a*b^3 - 8*a^3*b 
 + a^4 + 128*b^4))/(32*a^4*b))*(-b^3*(a + b))^(1/2)*(4*a*b - a^2 + 8*b^2)* 
1i)/(16*(a^3*b^4 + a^4*b^3)))/(((a*b^2)/4 - (a^2*b)/4 + a^3/32 + b^3)/(a^6 
*b) + ((((-b^3*(a + b))^(1/2)*((2*a^6*b^3 + (a^7*b^2)/2)/(a^6*b) - (tan(e 
+ f*x)*(512*a^6*b^4 + 256*a^7*b^3)*(-b^3*(a + b))^(1/2)*(4*a*b - a^2 + 8*b 
^2))/(512*a^4*b*(a^3*b^4 + a^4*b^3)))*(4*a*b - a^2 + 8*b^2))/(16*(a^3*b^4 
+ a^4*b^3)) - (tan(e + f*x)*(64*a*b^3 - 8*a^3*b + a^4 + 128*b^4))/(32*a^4* 
b))*(-b^3*(a + b))^(1/2)*(4*a*b - a^2 + 8*b^2))/(16*(a^3*b^4 + a^4*b^3)) + 
 ((((-b^3*(a + b))^(1/2)*((2*a^6*b^3 + (a^7*b^2)/2)/(a^6*b) + (tan(e + ...
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1337, normalized size of antiderivative = 9.76 \[ \int \frac {\tan ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(tan(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x)
 

Output:

(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 - 4*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 - 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**4 + 6*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 + 24*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*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 + sqrt(b)*sqrt(a + b)*ata 
n((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**4 - 2*sqrt(b)*sqrt( 
a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**3*b - 15* 
sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b)) 
*a**2*b**2 - 20*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - s 
qrt(a))/sqrt(b))*a*b**3 - 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + 
 f*x)/2) - sqrt(a))/sqrt(b))*b**4 + 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 - 4*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) + s 
qrt(a))/sqrt(b))*sin(e + f*x)**4*a**2*b**2 - 2*sqrt(b)*sqrt(a + b)*atan...