\(\int \frac {\tan ^6(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [369]

Optimal result

Integrand size = 23, antiderivative size = 147 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {x}{a^3}+\frac {\sqrt {a+b} \left (3 a^2-4 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b+b \tan ^2(e+f x)\right )} \]

[Out]

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {4226, 2000, 481, 592, 536, 209, 211} \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {x}{a^3}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\sqrt {a+b} \left (3 a^2-4 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 b^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b \tan ^2(e+f x)+b\right )^2} \]

[In]

Int[Tan[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

-(x/a^3) + (Sqrt[a + b]*(3*a^2 - 4*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(8*a^3*b^(5/2)*f)
- ((a + b)*Tan[e + f*x]^3)/(4*a*b*f*(a + b + b*Tan[e + f*x]^2)^2) - ((3*a - 4*b)*(a + b)*Tan[e + f*x])/(8*a^2*
b^2*f*(a + b + b*Tan[e + f*x]^2))

Rule 209

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

Rule 211

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

Rule 481

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-a)*e^(
2*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c - a*d)*(p + 1))), x] + Dist[e^
(2*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1)
+ (a*d*(m - n + n*q + 1) + b*c*n*(p + 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] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 536

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 592

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_)*((e_) + (f_.)*(x_)^(n_)), x
_Symbol] :> Simp[g^(n - 1)*(b*e - a*f)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(b*n*(b*c -
a*d)*(p + 1))), x] - Dist[g^n/(b*n*(b*c - a*d)*(p + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*S
imp[c*(b*e - a*f)*(m - n + 1) + (d*(b*e - a*f)*(m + n*q + 1) - b*n*(c*f - d*e)*(p + 1))*x^n, x], x], x] /; Fre
eQ[{a, b, c, d, e, f, g, q}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, 0]

Rule 2000

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 4226

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*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b \left (1+x^2\right )\right )^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \frac {x^6}{\left (1+x^2\right ) \left (a+b+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {\text {Subst}\left (\int \frac {x^2 \left (3 (a+b)+(3 a-b) x^2\right )}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a b f} \\ & = -\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {-((3 a-4 b) (a+b))+\left (-3 a^2+a b-4 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 b^2 f} \\ & = -\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{a^3 f}+\frac {\left ((a+b) \left (3 a^2-4 a b+8 b^2\right )\right ) \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 b^2 f} \\ & = -\frac {x}{a^3}+\frac {\sqrt {a+b} \left (3 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^{5/2} f}-\frac {(a+b) \tan ^3(e+f x)}{4 a b f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {(3 a-4 b) (a+b) \tan (e+f x)}{8 a^2 b^2 f \left (a+b+b \tan ^2(e+f x)\right )} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 7.61 (sec) , antiderivative size = 760, normalized size of antiderivative = 5.17 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\left (-3 a^3+a^2 b-4 a b^2-8 b^3\right ) (a+2 b+a \cos (2 e+2 f x))^3 \sec ^6(e+f x) \left (\frac {\arctan \left (\sec (f x) \left (\frac {\cos (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {i \sin (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \cos (2 e)}{64 a^3 b^2 \sqrt {a+b} f \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {i \arctan \left (\sec (f x) \left (\frac {\cos (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}-\frac {i \sin (2 e)}{2 \sqrt {a+b} \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right ) (-a \sin (f x)-2 b \sin (f x)+a \sin (2 e+f x))\right ) \sin (2 e)}{64 a^3 b^2 \sqrt {a+b} f \sqrt {b \cos (4 e)-i b \sin (4 e)}}\right )}{\left (a+b \sec ^2(e+f x)\right )^3}+\frac {(a+2 b+a \cos (2 e+2 f x)) \sec (2 e) \sec ^6(e+f x) \left (-24 a^2 b^2 f x \cos (2 e)-64 a b^3 f x \cos (2 e)-64 b^4 f x \cos (2 e)-16 a^2 b^2 f x \cos (2 f x)-32 a b^3 f x \cos (2 f x)-16 a^2 b^2 f x \cos (4 e+2 f x)-32 a b^3 f x \cos (4 e+2 f x)-4 a^2 b^2 f x \cos (2 e+4 f x)-4 a^2 b^2 f x \cos (6 e+4 f x)+9 a^4 \sin (2 e)+15 a^3 b \sin (2 e)-18 a^2 b^2 \sin (2 e)-72 a b^3 \sin (2 e)-48 b^4 \sin (2 e)-9 a^4 \sin (2 f x)-13 a^3 b \sin (2 f x)+28 a^2 b^2 \sin (2 f x)+32 a b^3 \sin (2 f x)+3 a^4 \sin (4 e+2 f x)-a^3 b \sin (4 e+2 f x)-20 a^2 b^2 \sin (4 e+2 f x)-16 a b^3 \sin (4 e+2 f x)-3 a^4 \sin (2 e+4 f x)+3 a^3 b \sin (2 e+4 f x)+6 a^2 b^2 \sin (2 e+4 f x)\right )}{128 a^3 b^2 f \left (a+b \sec ^2(e+f x)\right )^3} \]

[In]

Integrate[Tan[e + f*x]^6/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

((-3*a^3 + a^2*b - 4*a*b^2 - 8*b^3)*(a + 2*b + a*Cos[2*e + 2*f*x])^3*Sec[e + f*x]^6*((ArcTan[Sec[f*x]*(Cos[2*e
]/(2*Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*Sin[2*e])/(Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*
e]]))*(-(a*Sin[f*x]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Cos[2*e])/(64*a^3*b^2*Sqrt[a + b]*f*Sqrt[b*Cos[4*e] -
I*b*Sin[4*e]]) - ((I/64)*ArcTan[Sec[f*x]*(Cos[2*e]/(2*Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]) - ((I/2)*S
in[2*e])/(Sqrt[a + b]*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]]))*(-(a*Sin[f*x]) - 2*b*Sin[f*x] + a*Sin[2*e + f*x])]*Sin
[2*e])/(a^3*b^2*Sqrt[a + b]*f*Sqrt[b*Cos[4*e] - I*b*Sin[4*e]])))/(a + b*Sec[e + f*x]^2)^3 + ((a + 2*b + a*Cos[
2*e + 2*f*x])*Sec[2*e]*Sec[e + f*x]^6*(-24*a^2*b^2*f*x*Cos[2*e] - 64*a*b^3*f*x*Cos[2*e] - 64*b^4*f*x*Cos[2*e]
- 16*a^2*b^2*f*x*Cos[2*f*x] - 32*a*b^3*f*x*Cos[2*f*x] - 16*a^2*b^2*f*x*Cos[4*e + 2*f*x] - 32*a*b^3*f*x*Cos[4*e
+ 2*f*x] - 4*a^2*b^2*f*x*Cos[2*e + 4*f*x] - 4*a^2*b^2*f*x*Cos[6*e + 4*f*x] + 9*a^4*Sin[2*e] + 15*a^3*b*Sin[2*
e] - 18*a^2*b^2*Sin[2*e] - 72*a*b^3*Sin[2*e] - 48*b^4*Sin[2*e] - 9*a^4*Sin[2*f*x] - 13*a^3*b*Sin[2*f*x] + 28*a
^2*b^2*Sin[2*f*x] + 32*a*b^3*Sin[2*f*x] + 3*a^4*Sin[4*e + 2*f*x] - a^3*b*Sin[4*e + 2*f*x] - 20*a^2*b^2*Sin[4*e
+ 2*f*x] - 16*a*b^3*Sin[4*e + 2*f*x] - 3*a^4*Sin[2*e + 4*f*x] + 3*a^3*b*Sin[2*e + 4*f*x] + 6*a^2*b^2*Sin[2*e
+ 4*f*x]))/(128*a^3*b^2*f*(a + b*Sec[e + f*x]^2)^3)

Maple [A] (verified)

Time = 27.17 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.91

method	result	size
derivativedivides	\(\frac {\frac {\left (a +b \right ) \left (\frac {-\frac {a \left (5 a -4 b \right ) \tan \left (f x +e \right )^{3}}{8 b}-\frac {a \left (3 a^{2}-a b -4 b^{2}\right ) \tan \left (f x +e \right )}{8 b^{2}}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 b^{2} \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}}{f}\)	\(134\)
default	\(\frac {\frac {\left (a +b \right ) \left (\frac {-\frac {a \left (5 a -4 b \right ) \tan \left (f x +e \right )^{3}}{8 b}-\frac {a \left (3 a^{2}-a b -4 b^{2}\right ) \tan \left (f x +e \right )}{8 b^{2}}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (3 a^{2}-4 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 b^{2} \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}}{f}\)	\(134\)
risch	\(-\frac {x}{a^{3}}+\frac {i \left (-3 a^{4} {\mathrm e}^{6 i \left (f x +e \right )}+a^{3} b \,{\mathrm e}^{6 i \left (f x +e \right )}+20 a^{2} b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+16 a \,b^{3} {\mathrm e}^{6 i \left (f x +e \right )}-9 a^{4} {\mathrm e}^{4 i \left (f x +e \right )}-15 a^{3} b \,{\mathrm e}^{4 i \left (f x +e \right )}+18 a^{2} b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+72 a \,b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+48 b^{4} {\mathrm e}^{4 i \left (f x +e \right )}-9 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{4}-13 a^{3} b \,{\mathrm e}^{2 i \left (f x +e \right )}+28 a^{2} b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+32 a \,b^{3} {\mathrm e}^{2 i \left (f x +e \right )}-3 a^{4}+3 a^{3} b +6 a^{2} b^{2}\right )}{4 a^{3} b^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}-\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{16 b^{3} f a}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 b^{2} f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 b f \,a^{3}}+\frac {3 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{16 b^{3} f a}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 b^{2} f \,a^{2}}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 b f \,a^{3}}\)	\(584\)

[In]

int(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (133) = 266\).

Time = 0.30 (sec) , antiderivative size = 664, normalized size of antiderivative = 4.52 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [-\frac {32 \, a^{2} b^{2} f x \cos \left (f x + e\right )^{4} + 64 \, a b^{3} f x \cos \left (f x + e\right )^{2} + 32 \, b^{4} f x - {\left ({\left (3 \, a^{4} - 4 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} - 4 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (3 \, a^{3} b - 4 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {a + b}{b}} \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 b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - b^{2} \cos \left (f x + e\right )\right )} \sqrt {-\frac {a + b}{b}} \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 ) + 4 \, {\left (3 \, {\left (a^{4} - a^{3} b - 2 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{3} b + a^{2} b^{2} - 4 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left (a^{5} b^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b^{3} f \cos \left (f x + e\right )^{2} + a^{3} b^{4} f\right )}}, -\frac {16 \, a^{2} b^{2} f x \cos \left (f x + e\right )^{4} + 32 \, a b^{3} f x \cos \left (f x + e\right )^{2} + 16 \, b^{4} f x + {\left ({\left (3 \, a^{4} - 4 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b^{2} - 4 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (3 \, a^{3} b - 4 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {a + b}{b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a + b}{b}}}{2 \, {\left (a + b\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 2 \, {\left (3 \, {\left (a^{4} - a^{3} b - 2 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (5 \, a^{3} b + a^{2} b^{2} - 4 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left (a^{5} b^{2} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b^{3} f \cos \left (f x + e\right )^{2} + a^{3} b^{4} f\right )}}\right ] \]

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\int \frac {\tan ^{6}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]

[In]

integrate(tan(f*x+e)**6/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Integral(tan(e + f*x)**6/(a + b*sec(e + f*x)**2)**3, x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.31 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (5 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (3 \, a^{3} + 2 \, a^{2} b - 5 \, a b^{2} - 4 \, b^{3}\right )} \tan \left (f x + e\right )}{a^{2} b^{4} \tan \left (f x + e\right )^{4} + a^{4} b^{2} + 2 \, a^{3} b^{3} + a^{2} b^{4} + 2 \, {\left (a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (f x + e\right )^{2}} + \frac {8 \, {\left (f x + e\right )}}{a^{3}} - \frac {{\left (3 \, a^{3} - a^{2} b + 4 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{3} b^{2}}}{8 \, f} \]

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 2.54 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.36 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {8 \, {\left (f x + e\right )}}{a^{3}} - \frac {{\left (3 \, a^{3} - a^{2} b + 4 \, a b^{2} + 8 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{\sqrt {a b + b^{2}} a^{3} b^{2}} + \frac {5 \, a^{2} b \tan \left (f x + e\right )^{3} + a b^{2} \tan \left (f x + e\right )^{3} - 4 \, b^{3} \tan \left (f x + e\right )^{3} + 3 \, a^{3} \tan \left (f x + e\right ) + 2 \, a^{2} b \tan \left (f x + e\right ) - 5 \, a b^{2} \tan \left (f x + e\right ) - 4 \, b^{3} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2} a^{2} b^{2}}}{8 \, f} \]

[In]

integrate(tan(f*x+e)^6/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 20.95 (sec) , antiderivative size = 615, normalized size of antiderivative = 4.18 \[ \int \frac {\tan ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\mathrm {atan}\left (\frac {25\,\mathrm {tan}\left (e+f\,x\right )}{32\,\left (\frac {5\,b}{4\,a}-\frac {3\,a}{16\,b}+\frac {9\,a^2}{32\,b^2}+\frac {25}{32}\right )}-\frac {3\,\mathrm {tan}\left (e+f\,x\right )}{16\,\left (\frac {9\,a}{32\,b}+\frac {25\,b}{32\,a}+\frac {5\,b^2}{4\,a^2}-\frac {3}{16}\right )}+\frac {9\,\mathrm {tan}\left (e+f\,x\right )}{32\,\left (\frac {25\,b^2}{32\,a^2}-\frac {3\,b}{16\,a}+\frac {5\,b^3}{4\,a^3}+\frac {9}{32}\right )}+\frac {5\,\mathrm {tan}\left (e+f\,x\right )}{4\,\left (\frac {25\,a}{32\,b}-\frac {3\,a^2}{16\,b^2}+\frac {9\,a^3}{32\,b^3}+\frac {5}{4}\right )}\right )}{a^3\,f}-\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (5\,a^2+a\,b-4\,b^2\right )}{8\,a^2\,b}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a+b\right )\,\left (-3\,a^2+a\,b+4\,b^2\right )}{8\,a^2\,b^2}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}-\frac {\mathrm {atanh}\left (\frac {27\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^6-a\,b^5}}{256\,\left (\frac {27\,a\,b^2}{256}-\frac {27\,b^3}{128}+\frac {171\,b^4}{256\,a}-\frac {7\,b^5}{64\,a^2}+\frac {5\,b^6}{32\,a^3}+\frac {5\,b^7}{4\,a^4}\right )}-\frac {81\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^6-a\,b^5}}{256\,\left (\frac {27\,a^2\,b}{256}-\frac {27\,a\,b^2}{128}+\frac {171\,b^3}{256}-\frac {7\,b^4}{64\,a}+\frac {5\,b^5}{32\,a^2}+\frac {5\,b^6}{4\,a^3}\right )}-\frac {35\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^6-a\,b^5}}{32\,\left (\frac {171\,a^2\,b}{256}-\frac {7\,a\,b^2}{64}-\frac {27\,a^3}{128}+\frac {5\,b^3}{32}+\frac {5\,b^4}{4\,a}+\frac {27\,a^4}{256\,b}\right )}+\frac {5\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^6-a\,b^5}}{4\,\left (\frac {5\,a\,b^2}{32}-\frac {7\,a^2\,b}{64}+\frac {171\,a^3}{256}+\frac {5\,b^3}{4}-\frac {27\,a^4}{128\,b}+\frac {27\,a^5}{256\,b^2}\right )}+\frac {63\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-b^6-a\,b^5}}{64\,\left (\frac {171\,a\,b^2}{256}-\frac {27\,a^2\,b}{128}+\frac {27\,a^3}{256}-\frac {7\,b^3}{64}+\frac {5\,b^4}{32\,a}+\frac {5\,b^5}{4\,a^2}\right )}\right )\,\sqrt {-b^5\,\left (a+b\right )}\,\left (3\,a^2-4\,a\,b+8\,b^2\right )}{8\,a^3\,b^5\,f} \]

[In]

int(tan(e + f*x)^6/(a + b/cos(e + f*x)^2)^3,x)

[Out]

- atan((25*tan(e + f*x))/(32*((5*b)/(4*a) - (3*a)/(16*b) + (9*a^2)/(32*b^2) + 25/32)) - (3*tan(e + f*x))/(16*(
(9*a)/(32*b) + (25*b)/(32*a) + (5*b^2)/(4*a^2) - 3/16)) + (9*tan(e + f*x))/(32*((25*b^2)/(32*a^2) - (3*b)/(16*
a) + (5*b^3)/(4*a^3) + 9/32)) + (5*tan(e + f*x))/(4*((25*a)/(32*b) - (3*a^2)/(16*b^2) + (9*a^3)/(32*b^3) + 5/4
)))/(a^3*f) - ((tan(e + f*x)^3*(a*b + 5*a^2 - 4*b^2))/(8*a^2*b) - (tan(e + f*x)*(a + b)*(a*b - 3*a^2 + 4*b^2))
/(8*a^2*b^2))/(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)) - (atanh((27*tan(e
+ f*x)*(- a*b^5 - b^6)^(1/2))/(256*((27*a*b^2)/256 - (27*b^3)/128 + (171*b^4)/(256*a) - (7*b^5)/(64*a^2) + (5
*b^6)/(32*a^3) + (5*b^7)/(4*a^4))) - (81*tan(e + f*x)*(- a*b^5 - b^6)^(1/2))/(256*((27*a^2*b)/256 - (27*a*b^2)
/128 + (171*b^3)/256 - (7*b^4)/(64*a) + (5*b^5)/(32*a^2) + (5*b^6)/(4*a^3))) - (35*tan(e + f*x)*(- a*b^5 - b^6
)^(1/2))/(32*((171*a^2*b)/256 - (7*a*b^2)/64 - (27*a^3)/128 + (5*b^3)/32 + (5*b^4)/(4*a) + (27*a^4)/(256*b)))
+ (5*tan(e + f*x)*(- a*b^5 - b^6)^(1/2))/(4*((5*a*b^2)/32 - (7*a^2*b)/64 + (171*a^3)/256 + (5*b^3)/4 - (27*a^4
)/(128*b) + (27*a^5)/(256*b^2))) + (63*tan(e + f*x)*(- a*b^5 - b^6)^(1/2))/(64*((171*a*b^2)/256 - (27*a^2*b)/1
28 + (27*a^3)/256 - (7*b^3)/64 + (5*b^4)/(32*a) + (5*b^5)/(4*a^2))))*(-b^5*(a + b))^(1/2)*(3*a^2 - 4*a*b + 8*b
^2))/(8*a^3*b^5*f)