Integrand size = 23, antiderivative size = 196 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {3 \left (a^2+6 a b+b^2\right ) x}{8 (a-b)^4}-\frac {3 \sqrt {a} \sqrt {b} (a+b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{2 (a-b)^4 f}-\frac {(5 a+b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )}-\frac {3 b (3 a+b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )} \]
3/8*(a^2+6*a*b+b^2)*x/(a-b)^4-3/2*(a+b)*arctan(b^(1/2)*tan(f*x+e)/a^(1/2)) *a^(1/2)*b^(1/2)/(a-b)^4/f-1/8*(5*a+b)*cos(f*x+e)*sin(f*x+e)/(a-b)^2/f/(a+ b*tan(f*x+e)^2)+1/4*cos(f*x+e)^3*sin(f*x+e)/(a-b)/f/(a+b*tan(f*x+e)^2)-3/8 *b*(3*a+b)*tan(f*x+e)/(a-b)^3/f/(a+b*tan(f*x+e)^2)
Time = 2.20 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.69 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {12 \left (a^2+6 a b+b^2\right ) (e+f x)-48 \sqrt {a} \sqrt {b} (a+b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )-8 (a-b) (a+b) \sin (2 (e+f x))-\frac {16 a (a-b) b \sin (2 (e+f x))}{a+b+(a-b) \cos (2 (e+f x))}+(a-b)^2 \sin (4 (e+f x))}{32 (a-b)^4 f} \]
(12*(a^2 + 6*a*b + b^2)*(e + f*x) - 48*Sqrt[a]*Sqrt[b]*(a + b)*ArcTan[(Sqr t[b]*Tan[e + f*x])/Sqrt[a]] - 8*(a - b)*(a + b)*Sin[2*(e + f*x)] - (16*a*( a - b)*b*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2*(e + f*x)]) + (a - b)^2* Sin[4*(e + f*x)])/(32*(a - b)^4*f)
Time = 0.41 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.17, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {3042, 4146, 372, 402, 27, 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.
| \(\displaystyle \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^4}{\left (a+b \tan (e+f x)^2\right )^2}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a\right )^2}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\int \frac {a-(4 a+b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a\right )^2}d\tan (e+f x)}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\frac {(5 a+b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )}-\frac {\int \frac {3 \left (a (a+b)-b (5 a+b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^2}d\tan (e+f x)}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\frac {(5 a+b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )}-\frac {3 \int \frac {a (a+b)-b (5 a+b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^2}d\tan (e+f x)}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\frac {(5 a+b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )}-\frac {3 \left (\frac {\int \frac {2 a \left (a (a+3 b)-b (3 a+b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{2 a (a-b)}-\frac {b (3 a+b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )}\right )}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\frac {(5 a+b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )}-\frac {3 \left (\frac {\int \frac {a (a+3 b)-b (3 a+b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{a-b}-\frac {b (3 a+b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )}\right )}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\frac {(5 a+b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )}-\frac {3 \left (\frac {\frac {\left (a^2+6 a b+b^2\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a-b}-\frac {4 a b (a+b) \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a-b}}{a-b}-\frac {b (3 a+b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )}\right )}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\frac {(5 a+b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )}-\frac {3 \left (\frac {\frac {\left (a^2+6 a b+b^2\right ) \arctan (\tan (e+f x))}{a-b}-\frac {4 a b (a+b) \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a-b}}{a-b}-\frac {b (3 a+b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )}\right )}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )}-\frac {\frac {(5 a+b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )}-\frac {3 \left (\frac {\frac {\left (a^2+6 a b+b^2\right ) \arctan (\tan (e+f x))}{a-b}-\frac {4 \sqrt {a} \sqrt {b} (a+b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a-b}}{a-b}-\frac {b (3 a+b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )}\right )}{2 (a-b)}}{4 (a-b)}}{f}\) |
(Tan[e + f*x]/(4*(a - b)*(1 + Tan[e + f*x]^2)^2*(a + b*Tan[e + f*x]^2)) - (((5*a + b)*Tan[e + f*x])/(2*(a - b)*(1 + Tan[e + f*x]^2)*(a + b*Tan[e + f *x]^2)) - (3*((((a^2 + 6*a*b + b^2)*ArcTan[Tan[e + f*x]])/(a - b) - (4*Sqr t[a]*Sqrt[b]*(a + b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a - b))/(a - b) - (b*(3*a + b)*Tan[e + f*x])/((a - b)*(a + b*Tan[e + f*x]^2))))/(2*(a - b)))/(4*(a - b)))/f
3.1.74.3.1 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[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 = 15.25 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.81
| method | result | size |
| derivativedivides | \(\frac {-\frac {a b \left (\frac {\left (\frac {a}{2}-\frac {b}{2}\right ) \tan \left (f x +e \right )}{a +b \tan \left (f x +e \right )^{2}}+\frac {3 \left (a +b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a -b \right )^{4}}+\frac {\frac {\left (-\frac {5}{8} a^{2}+\frac {1}{4} a b +\frac {3}{8} b^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}+\frac {5}{8} b^{2}-\frac {1}{4} a b \right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {3 \left (a^{2}+6 a b +b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{\left (a -b \right )^{4}}}{f}\) | \(159\) |
| default | \(\frac {-\frac {a b \left (\frac {\left (\frac {a}{2}-\frac {b}{2}\right ) \tan \left (f x +e \right )}{a +b \tan \left (f x +e \right )^{2}}+\frac {3 \left (a +b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{2 \sqrt {a b}}\right )}{\left (a -b \right )^{4}}+\frac {\frac {\left (-\frac {5}{8} a^{2}+\frac {1}{4} a b +\frac {3}{8} b^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}+\frac {5}{8} b^{2}-\frac {1}{4} a b \right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {3 \left (a^{2}+6 a b +b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{\left (a -b \right )^{4}}}{f}\) | \(159\) |
| risch | \(\frac {3 x \,a^{2}}{8 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}+\frac {9 x a b}{4 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}+\frac {3 x \,b^{2}}{8 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2}}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )}}{64 \left (a -b \right )^{2} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} a}{8 \left (a -b \right )^{3} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{8 \left (a -b \right )^{3} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} a}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )}}{64 \left (a^{2}-2 a b +b^{2}\right ) f}-\frac {i a b \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}{f \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) a}{4 \left (a -b \right )^{4} f}+\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right ) b}{4 \left (a -b \right )^{4} f}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) a}{4 \left (a -b \right )^{4} f}-\frac {3 \sqrt {-a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right ) b}{4 \left (a -b \right )^{4} f}\) | \(572\) |
1/f*(-a*b/(a-b)^4*((1/2*a-1/2*b)*tan(f*x+e)/(a+b*tan(f*x+e)^2)+3/2*(a+b)/( a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2)))+1/(a-b)^4*(((-5/8*a^2+1/4*a*b +3/8*b^2)*tan(f*x+e)^3+(-3/8*a^2+5/8*b^2-1/4*a*b)*tan(f*x+e))/(1+tan(f*x+e )^2)^2+3/8*(a^2+6*a*b+b^2)*arctan(tan(f*x+e))))
Time = 0.38 (sec) , antiderivative size = 705, normalized size of antiderivative = 3.60 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\left [\frac {3 \, {\left (a^{3} + 5 \, a^{2} b - 5 \, a b^{2} - b^{3}\right )} f x \cos \left (f x + e\right )^{2} + 3 \, {\left (a^{2} b + 6 \, a b^{2} + b^{3}\right )} f x + 3 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \sqrt {-a b} \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 ) + {\left (2 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{3} - 9 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (3 \, a^{2} b - 2 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\right )}}, \frac {3 \, {\left (a^{3} + 5 \, a^{2} b - 5 \, a b^{2} - b^{3}\right )} f x \cos \left (f x + e\right )^{2} + 3 \, {\left (a^{2} b + 6 \, a b^{2} + b^{3}\right )} f x + 6 \, {\left ({\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} + a b + b^{2}\right )} \sqrt {a b} \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 ) + {\left (2 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{5} - {\left (5 \, a^{3} - 9 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - 3 \, {\left (3 \, a^{2} b - 2 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b - 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} - 4 \, a b^{4} + b^{5}\right )} f\right )}}\right ] \]
[1/8*(3*(a^3 + 5*a^2*b - 5*a*b^2 - b^3)*f*x*cos(f*x + e)^2 + 3*(a^2*b + 6* a*b^2 + b^3)*f*x + 3*((a^2 - b^2)*cos(f*x + e)^2 + a*b + b^2)*sqrt(-a*b)*l og(((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) ) + (2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^5 - (5*a^3 - 9*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^3 - 3*(3*a^2*b - 2*a*b^2 - b^3)*cos(f*x + e)) *sin(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)* f*cos(f*x + e)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f), 1/8 *(3*(a^3 + 5*a^2*b - 5*a*b^2 - b^3)*f*x*cos(f*x + e)^2 + 3*(a^2*b + 6*a*b^ 2 + b^3)*f*x + 6*((a^2 - b^2)*cos(f*x + e)^2 + a*b + b^2)*sqrt(a*b)*arctan (1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(a*b)/(a*b*cos(f*x + e)*sin(f*x + e) )) + (2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^5 - (5*a^3 - 9*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e)^3 - 3*(3*a^2*b - 2*a*b^2 - b^3)*cos(f*x + e) )*sin(f*x + e))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5) *f*cos(f*x + e)^2 + (a^4*b - 4*a^3*b^2 + 6*a^2*b^3 - 4*a*b^4 + b^5)*f)]
Timed out. \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Timed out} \]
Time = 0.32 (sec) , antiderivative size = 312, normalized size of antiderivative = 1.59 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {3 \, {\left (a^{2} + 6 \, a b + b^{2}\right )} {\left (f x + e\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {12 \, {\left (a^{2} b + a b^{2}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a b}} - \frac {3 \, {\left (3 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{5} + {\left (5 \, a^{2} + 14 \, a b + 5 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{2} + 3 \, a b\right )} \tan \left (f x + e\right )}{{\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} \tan \left (f x + e\right )^{6} + {\left (a^{4} - a^{3} b - 3 \, a^{2} b^{2} + 5 \, a b^{3} - 2 \, b^{4}\right )} \tan \left (f x + e\right )^{4} + a^{4} - 3 \, a^{3} b + 3 \, a^{2} b^{2} - a b^{3} + {\left (2 \, a^{4} - 5 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3} - b^{4}\right )} \tan \left (f x + e\right )^{2}}}{8 \, f} \]
1/8*(3*(a^2 + 6*a*b + b^2)*(f*x + e)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - 12*(a^2*b + a*b^2)*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*(3*a*b + b^2)*tan(f*x + e) ^5 + (5*a^2 + 14*a*b + 5*b^2)*tan(f*x + e)^3 + 3*(a^2 + 3*a*b)*tan(f*x + e ))/((a^3*b - 3*a^2*b^2 + 3*a*b^3 - b^4)*tan(f*x + e)^6 + (a^4 - a^3*b - 3* a^2*b^2 + 5*a*b^3 - 2*b^4)*tan(f*x + e)^4 + a^4 - 3*a^3*b + 3*a^2*b^2 - a* b^3 + (2*a^4 - 5*a^3*b + 3*a^2*b^2 + a*b^3 - b^4)*tan(f*x + e)^2))/f
Time = 0.64 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.31 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\frac {\frac {3 \, {\left (a^{2} + 6 \, a b + b^{2}\right )} {\left (f x + e\right )}}{a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}} - \frac {4 \, a b \tan \left (f x + e\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (b \tan \left (f x + e\right )^{2} + a\right )}} - \frac {12 \, {\left (a^{2} b + a b^{2}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a b}} - \frac {5 \, a \tan \left (f x + e\right )^{3} + 3 \, b \tan \left (f x + e\right )^{3} + 3 \, a \tan \left (f x + e\right ) + 5 \, b \tan \left (f x + e\right )}{{\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} {\left (\tan \left (f x + e\right )^{2} + 1\right )}^{2}}}{8 \, f} \]
1/8*(3*(a^2 + 6*a*b + b^2)*(f*x + e)/(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4) - 4*a*b*tan(f*x + e)/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(b*tan(f*x + e)^2 + a)) - 12*(a^2*b + a*b^2)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arc tan(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)) - (5*a*tan(f*x + e)^3 + 3*b*tan(f*x + e)^3 + 3*a*tan(f*x + e) + 5*b*tan(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*(tan(f*x + e)^2 + 1) ^2))/f
Time = 15.29 (sec) , antiderivative size = 4616, normalized size of antiderivative = 23.55 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \]
(atan(((((tan(e + f*x)*(108*a*b^6 + 9*b^7 + 486*a^2*b^5 + 396*a^3*b^4 + 15 3*a^4*b^3))/(32*(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)) - (3*(((9*a*b^11)/2 - (69*a^2*b^10)/2 + 114*a^3*b^9 - 210*a^ 4*b^8 + 231*a^5*b^7 - 147*a^6*b^6 + 42*a^7*b^5 + 6*a^8*b^4 - (15*a^9*b^3)/ 2 + (3*a^10*b^2)/2)/(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) - (3*tan(e + f*x )*(a*b*6i + a^2*1i + b^2*1i)*(256*b^11 - 1792*a*b^10 + 5120*a^2*b^9 - 7168 *a^3*b^8 + 3584*a^4*b^7 + 3584*a^5*b^6 - 7168*a^6*b^5 + 5120*a^7*b^4 - 179 2*a^8*b^3 + 256*a^9*b^2))/(512*(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*b*6i + a^2*1i + b^2*1i))/(16*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2 )))*(a*b*6i + a^2*1i + b^2*1i)*3i)/(16*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6* a^2*b^2)) + (((tan(e + f*x)*(108*a*b^6 + 9*b^7 + 486*a^2*b^5 + 396*a^3*b^4 + 153*a^4*b^3))/(32*(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)) + (3*(((9*a*b^11)/2 - (69*a^2*b^10)/2 + 114*a^3*b^9 - 2 10*a^4*b^8 + 231*a^5*b^7 - 147*a^6*b^6 + 42*a^7*b^5 + 6*a^8*b^4 - (15*a^9* b^3)/2 + (3*a^10*b^2)/2)/(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) + (3*tan(e + f*x)*(a*b*6i + a^2*1i + b^2*1i)*(256*b^11 - 1792*a*b^10 + 5120*a^2*b^9 - 7168*a^3*b^8 + 3584*a^4*b^7 + 3584*a^5*b^6 - 7168*a^6*b^5 + 5120*a^7*b...