Integrand size = 23, antiderivative size = 178 \[ \int \frac {\sin ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\frac {\left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) x}{16 (a-b)^4}-\frac {a^{5/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{(a-b)^4 f}-\frac {\left (11 a^2-4 a b+b^2\right ) \cos (e+f x) \sin (e+f x)}{16 (a-b)^3 f}+\frac {(3 a-b) \cos ^3(e+f x) \sin (e+f x)}{8 (a-b)^2 f}+\frac {\cos ^3(e+f x) \sin ^3(e+f x)}{6 (a-b) f} \]
1/16*(5*a^3+15*a^2*b-5*a*b^2+b^3)*x/(a-b)^4-1/16*(11*a^2-4*a*b+b^2)*cos(f* x+e)*sin(f*x+e)/(a-b)^3/f+1/8*(3*a-b)*cos(f*x+e)^3*sin(f*x+e)/(a-b)^2/f+1/ 6*cos(f*x+e)^3*sin(f*x+e)^3/(a-b)/f-a^(5/2)*arctan(b^(1/2)*tan(f*x+e)/a^(1 /2))*b^(1/2)/(a-b)^4/f
Time = 1.09 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.79 \[ \int \frac {\sin ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {-12 \left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) (e+f x)+192 a^{5/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )+3 (a-b) (5 a-b) (3 a+b) \sin (2 (e+f x))-3 (a-b)^2 (3 a-b) \sin (4 (e+f x))+(a-b)^3 \sin (6 (e+f x))}{192 (a-b)^4 f} \]
-1/192*(-12*(5*a^3 + 15*a^2*b - 5*a*b^2 + b^3)*(e + f*x) + 192*a^(5/2)*Sqr t[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] + 3*(a - b)*(5*a - b)*(3*a + b )*Sin[2*(e + f*x)] - 3*(a - b)^2*(3*a - b)*Sin[4*(e + f*x)] + (a - b)^3*Si n[6*(e + f*x)])/((a - b)^4*f)
Time = 0.44 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.25, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4146, 372, 27, 440, 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.
| \(\displaystyle \int \frac {\sin ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^6}{a+b \tan (e+f x)^2}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tan ^6(e+f x)}{\left (\tan ^2(e+f x)+1\right )^4 \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 (a-b) \left (\tan ^2(e+f x)+1\right )^3}-\frac {\int \frac {3 \tan ^2(e+f x) \left (a-(2 a-b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{6 (a-b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 (a-b) \left (\tan ^2(e+f x)+1\right )^3}-\frac {\int \frac {\tan ^2(e+f x) \left (a-(2 a-b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{2 (a-b)}}{f}\) |
| \(\Big \downarrow \) 440 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 (a-b) \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\int \frac {a (3 a-b)-\left (8 a^2-3 b a+b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{4 (a-b)}-\frac {(3 a-b) \tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}}{2 (a-b)}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 (a-b) \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (11 a^2-4 a b+b^2\right ) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {\int \frac {a (5 a-b) (a+b)-b \left (11 a^2-4 b a+b^2\right ) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{2 (a-b)}}{4 (a-b)}-\frac {(3 a-b) \tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}}{2 (a-b)}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 (a-b) \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (11 a^2-4 a b+b^2\right ) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a-b}-\frac {16 a^3 b \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a-b}}{2 (a-b)}}{4 (a-b)}-\frac {(3 a-b) \tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}}{2 (a-b)}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 (a-b) \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (11 a^2-4 a b+b^2\right ) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) \arctan (\tan (e+f x))}{a-b}-\frac {16 a^3 b \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a-b}}{2 (a-b)}}{4 (a-b)}-\frac {(3 a-b) \tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}}{2 (a-b)}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\tan ^3(e+f x)}{6 (a-b) \left (\tan ^2(e+f x)+1\right )^3}-\frac {\frac {\frac {\left (11 a^2-4 a b+b^2\right ) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right )}-\frac {\frac {\left (5 a^3+15 a^2 b-5 a b^2+b^3\right ) \arctan (\tan (e+f x))}{a-b}-\frac {16 a^{5/2} \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a-b}}{2 (a-b)}}{4 (a-b)}-\frac {(3 a-b) \tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2}}{2 (a-b)}}{f}\) |
(Tan[e + f*x]^3/(6*(a - b)*(1 + Tan[e + f*x]^2)^3) - (-1/4*((3*a - b)*Tan[ e + f*x])/((a - b)*(1 + Tan[e + f*x]^2)^2) + (-1/2*(((5*a^3 + 15*a^2*b - 5 *a*b^2 + b^3)*ArcTan[Tan[e + f*x]])/(a - b) - (16*a^(5/2)*Sqrt[b]*ArcTan[( Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a - b))/(a - b) + ((11*a^2 - 4*a*b + b^2) *Tan[e + f*x])/(2*(a - b)*(1 + Tan[e + f*x]^2)))/(4*(a - b)))/(2*(a - b))) /f
3.1.61.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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])
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]
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]
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]
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]
Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ )*((e_) + (f_.)*(x_)^2), x_Symbol] :> Simp[g*(b*e - a*f)*(g*x)^(m - 1)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*b*(b*c - a*d)*(p + 1))), x] - Simp[ g^2/(2*b*(b*c - a*d)*(p + 1)) Int[(g*x)^(m - 2)*(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f)*(m - 1) + (d*(b*e - a*f)*(m + 2*q + 1) - b*2*(c *f - d*e)*(p + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, q}, x] && LtQ[p, -1] && GtQ[m, 1]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Time = 21.62 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.04
| method | result | size |
| derivativedivides | \(\frac {-\frac {a^{3} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{4} \sqrt {a b}}+\frac {\frac {\left (-\frac {11}{16} a^{3}+\frac {15}{16} a^{2} b -\frac {5}{16} a \,b^{2}+\frac {1}{16} b^{3}\right ) \tan \left (f x +e \right )^{5}+\left (-\frac {5}{6} a^{3}+\frac {1}{2} a^{2} b +\frac {1}{2} a \,b^{2}-\frac {1}{6} b^{3}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {5}{16} a^{3}+\frac {1}{16} a^{2} b +\frac {5}{16} a \,b^{2}-\frac {1}{16} b^{3}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3}}+\frac {\left (5 a^{3}+15 a^{2} b -5 a \,b^{2}+b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{\left (a -b \right )^{4}}}{f}\) | \(185\) |
| default | \(\frac {-\frac {a^{3} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{\left (a -b \right )^{4} \sqrt {a b}}+\frac {\frac {\left (-\frac {11}{16} a^{3}+\frac {15}{16} a^{2} b -\frac {5}{16} a \,b^{2}+\frac {1}{16} b^{3}\right ) \tan \left (f x +e \right )^{5}+\left (-\frac {5}{6} a^{3}+\frac {1}{2} a^{2} b +\frac {1}{2} a \,b^{2}-\frac {1}{6} b^{3}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {5}{16} a^{3}+\frac {1}{16} a^{2} b +\frac {5}{16} a \,b^{2}-\frac {1}{16} b^{3}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3}}+\frac {\left (5 a^{3}+15 a^{2} b -5 a \,b^{2}+b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{\left (a -b \right )^{4}}}{f}\) | \(185\) |
| risch | \(\frac {5 x \,a^{3}}{16 \left (a -b \right )^{4}}+\frac {15 x \,a^{2} b}{16 \left (a -b \right )^{4}}-\frac {5 x a \,b^{2}}{16 \left (a -b \right )^{4}}+\frac {x \,b^{3}}{16 \left (a -b \right )^{4}}+\frac {15 i {\mathrm e}^{2 i \left (f x +e \right )} a^{2}}{128 \left (a -b \right )^{3} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} a b}{64 \left (a -b \right )^{3} f}-\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b^{2}}{128 \left (a -b \right )^{3} f}-\frac {15 i {\mathrm e}^{-2 i \left (f x +e \right )} a^{2}}{128 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right ) f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} a b}{64 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right ) f}+\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b^{2}}{128 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right ) f}-\frac {\sqrt {-a b}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 \left (a -b \right )^{4} f}+\frac {\sqrt {-a b}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 \left (a -b \right )^{4} f}-\frac {\sin \left (6 f x +6 e \right )}{192 \left (a -b \right ) f}+\frac {3 \sin \left (4 f x +4 e \right ) a}{64 \left (a -b \right )^{2} f}-\frac {\sin \left (4 f x +4 e \right ) b}{64 \left (a -b \right )^{2} f}\) | \(417\) |
1/f*(-a^3*b/(a-b)^4/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))+1/(a-b)^4 *(((-11/16*a^3+15/16*a^2*b-5/16*a*b^2+1/16*b^3)*tan(f*x+e)^5+(-5/6*a^3+1/2 *a^2*b+1/2*a*b^2-1/6*b^3)*tan(f*x+e)^3+(-5/16*a^3+1/16*a^2*b+5/16*a*b^2-1/ 16*b^3)*tan(f*x+e))/(1+tan(f*x+e)^2)^3+1/16*(5*a^3+15*a^2*b-5*a*b^2+b^3)*a rctan(tan(f*x+e))))
Time = 0.34 (sec) , antiderivative size = 521, normalized size of antiderivative = 2.93 \[ \int \frac {\sin ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\left [\frac {12 \, \sqrt {-a b} a^{2} \log \left (\frac {{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \, {\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 3 \, {\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} f x - {\left (8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (13 \, a^{3} - 33 \, a^{2} b + 27 \, a b^{2} - 7 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{3} - 15 \, a^{2} b + 5 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f}, \frac {24 \, \sqrt {a b} a^{2} \arctan \left (\frac {{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {a b}}{2 \, a b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) + 3 \, {\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} f x - {\left (8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{5} - 2 \, {\left (13 \, a^{3} - 33 \, a^{2} b + 27 \, a b^{2} - 7 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{3} - 15 \, a^{2} b + 5 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} f}\right ] \]
[1/48*(12*sqrt(-a*b)*a^2*log(((a^2 + 6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a* b + b^2)*cos(f*x + e)^2 + 4*((a + b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt (-a*b)*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2)) + 3*(5*a^3 + 15*a^2*b - 5*a*b^2 + b^3)*f*x - ( 8*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^5 - 2*(13*a^3 - 33*a^2*b + 27*a*b^2 - 7*b^3)*cos(f*x + e)^3 + 3*(11*a^3 - 15*a^2*b + 5*a*b^2 - b^3)*c os(f*x + e))*sin(f*x + e))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*f) , 1/48*(24*sqrt(a*b)*a^2*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(a*b) /(a*b*cos(f*x + e)*sin(f*x + e))) + 3*(5*a^3 + 15*a^2*b - 5*a*b^2 + b^3)*f *x - (8*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^5 - 2*(13*a^3 - 33*a^ 2*b + 27*a*b^2 - 7*b^3)*cos(f*x + e)^3 + 3*(11*a^3 - 15*a^2*b + 5*a*b^2 - b^3)*cos(f*x + e))*sin(f*x + e))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b ^4)*f)]
Timed out. \[ \int \frac {\sin ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]
Time = 0.50 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.71 \[ \int \frac {\sin ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {48 \, a^{3} b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a b}} - \frac {3 \, {\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} {\left (f x + e\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {3 \, {\left (11 \, a^{2} - 4 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a^{2} + 2 \, a b - b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{2} + 4 \, a b - b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{6} + 3 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{4} + a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3} + 3 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \tan \left (f x + e\right )^{2}}}{48 \, f} \]
-1/48*(48*a^3*b*arctan(b*tan(f*x + e)/sqrt(a*b))/((a^4 - 4*a^3*b + 6*a^2*b ^2 - 4*a*b^3 + b^4)*sqrt(a*b)) - 3*(5*a^3 + 15*a^2*b - 5*a*b^2 + b^3)*(f*x + e)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) + (3*(11*a^2 - 4*a*b + b ^2)*tan(f*x + e)^5 + 8*(5*a^2 + 2*a*b - b^2)*tan(f*x + e)^3 + 3*(5*a^2 + 4 *a*b - b^2)*tan(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*tan(f*x + e)^6 + 3*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*tan(f*x + e)^4 + a^3 - 3*a^2*b + 3*a*b ^2 - b^3 + 3*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*tan(f*x + e)^2))/f
Time = 0.52 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.57 \[ \int \frac {\sin ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx=-\frac {\frac {48 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )} a^{3} b}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a b}} - \frac {3 \, {\left (5 \, a^{3} + 15 \, a^{2} b - 5 \, a b^{2} + b^{3}\right )} {\left (f x + e\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} + \frac {33 \, a^{2} \tan \left (f x + e\right )^{5} - 12 \, a b \tan \left (f x + e\right )^{5} + 3 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 16 \, a b \tan \left (f x + e\right )^{3} - 8 \, b^{2} \tan \left (f x + e\right )^{3} + 15 \, a^{2} \tan \left (f x + e\right ) + 12 \, a b \tan \left (f x + e\right ) - 3 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{48 \, f} \]
-1/48*(48*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqr t(a*b)))*a^3*b/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*sqrt(a*b)) - 3 *(5*a^3 + 15*a^2*b - 5*a*b^2 + b^3)*(f*x + e)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) + (33*a^2*tan(f*x + e)^5 - 12*a*b*tan(f*x + e)^5 + 3*b^2*t an(f*x + e)^5 + 40*a^2*tan(f*x + e)^3 + 16*a*b*tan(f*x + e)^3 - 8*b^2*tan( f*x + e)^3 + 15*a^2*tan(f*x + e) + 12*a*b*tan(f*x + e) - 3*b^2*tan(f*x + e ))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(tan(f*x + e)^2 + 1)^3))/f
Time = 15.19 (sec) , antiderivative size = 4910, normalized size of antiderivative = 27.58 \[ \int \frac {\sin ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx=\text {Too large to display} \]
(atan(-((((((3*a^2*b^11 - (a*b^12)/4 - (55*a^3*b^10)/4 + 32*a^4*b^9 - (77* a^5*b^8)/2 + 14*a^6*b^7 + (49*a^7*b^6)/2 - 40*a^8*b^5 + (107*a^9*b^4)/4 - 9*a^10*b^3 + (5*a^11*b^2)/4)/(9*a*b^8 - 9*a^8*b + a^9 - b^9 - 36*a^2*b^7 + 84*a^3*b^6 - 126*a^4*b^5 + 126*a^5*b^4 - 84*a^6*b^3 + 36*a^7*b^2) - (tan( e + f*x)*(a^2*b*15i - a*b^2*5i + a^3*5i + b^3*1i)*(1024*b^11 - 7168*a*b^10 + 20480*a^2*b^9 - 28672*a^3*b^8 + 14336*a^4*b^7 + 14336*a^5*b^6 - 28672*a ^6*b^5 + 20480*a^7*b^4 - 7168*a^8*b^3 + 1024*a^9*b^2))/(4096*(a^4 - 4*a^3* b - 4*a*b^3 + b^4 + 6*a^2*b^2)*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b^2)))*(a^2*b*15i - a*b^2*5i + a^3*5i + b^3*1i))/(3 2*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) - (tan(e + f*x)*(b^9 - 10*a *b^8 + 55*a^2*b^7 - 140*a^3*b^6 + 175*a^4*b^5 + 150*a^5*b^4 + 281*a^6*b^3) )/(128*(a^6 - 6*a^5*b - 6*a*b^5 + b^6 + 15*a^2*b^4 - 20*a^3*b^3 + 15*a^4*b ^2)))*(a^2*b*15i - a*b^2*5i + a^3*5i + b^3*1i)*1i)/(32*(a^4 - 4*a^3*b - 4* a*b^3 + b^4 + 6*a^2*b^2)) - (((((3*a^2*b^11 - (a*b^12)/4 - (55*a^3*b^10)/4 + 32*a^4*b^9 - (77*a^5*b^8)/2 + 14*a^6*b^7 + (49*a^7*b^6)/2 - 40*a^8*b^5 + (107*a^9*b^4)/4 - 9*a^10*b^3 + (5*a^11*b^2)/4)/(9*a*b^8 - 9*a^8*b + a^9 - b^9 - 36*a^2*b^7 + 84*a^3*b^6 - 126*a^4*b^5 + 126*a^5*b^4 - 84*a^6*b^3 + 36*a^7*b^2) + (tan(e + f*x)*(a^2*b*15i - a*b^2*5i + a^3*5i + b^3*1i)*(102 4*b^11 - 7168*a*b^10 + 20480*a^2*b^9 - 28672*a^3*b^8 + 14336*a^4*b^7 + 143 36*a^5*b^6 - 28672*a^6*b^5 + 20480*a^7*b^4 - 7168*a^8*b^3 + 1024*a^9*b^...