Integrand size = 23, antiderivative size = 250 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {3 \left (a^2+10 a b+5 b^2\right ) x}{8 (a-b)^5}-\frac {3 \sqrt {b} \left (5 a^2+10 a b+b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{8 \sqrt {a} (a-b)^5 f}-\frac {(5 a+3 b) \cos (e+f x) \sin (e+f x)}{8 (a-b)^2 f \left (a+b \tan ^2(e+f x)\right )^2}+\frac {\cos ^3(e+f x) \sin (e+f x)}{4 (a-b) f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+5 b) \tan (e+f x)}{8 (a-b)^3 f \left (a+b \tan ^2(e+f x)\right )^2}-\frac {3 b (a+b) \tan (e+f x)}{2 (a-b)^4 f \left (a+b \tan ^2(e+f x)\right )} \]
3/8*(a^2+10*a*b+5*b^2)*x/(a-b)^5-3/8*(5*a^2+10*a*b+b^2)*arctan(b^(1/2)*tan (f*x+e)/a^(1/2))*b^(1/2)/(a-b)^5/f/a^(1/2)-1/8*(5*a+3*b)*cos(f*x+e)*sin(f* x+e)/(a-b)^2/f/(a+b*tan(f*x+e)^2)^2+1/4*cos(f*x+e)^3*sin(f*x+e)/(a-b)/f/(a +b*tan(f*x+e)^2)^2-1/8*b*(7*a+5*b)*tan(f*x+e)/(a-b)^3/f/(a+b*tan(f*x+e)^2) ^2-3/2*b*(a+b)*tan(f*x+e)/(a-b)^4/f/(a+b*tan(f*x+e)^2)
Time = 1.61 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {12 \left (a^2+10 a b+5 b^2\right ) (e+f x)-\frac {12 \sqrt {b} \left (5 a^2+10 a b+b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{\sqrt {a}}-8 (a-b) (a+2 b) \sin (2 (e+f x))+\frac {16 a (a-b) b^2 \sin (2 (e+f x))}{(a+b+(a-b) \cos (2 (e+f x)))^2}-\frac {4 (a-b) b (9 a+5 b) \sin (2 (e+f x))}{a+b+(a-b) \cos (2 (e+f x))}+(a-b)^2 \sin (4 (e+f x))}{32 (a-b)^5 f} \]
(12*(a^2 + 10*a*b + 5*b^2)*(e + f*x) - (12*Sqrt[b]*(5*a^2 + 10*a*b + b^2)* ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/Sqrt[a] - 8*(a - b)*(a + 2*b)*Sin[ 2*(e + f*x)] + (16*a*(a - b)*b^2*Sin[2*(e + f*x)])/(a + b + (a - b)*Cos[2* (e + f*x)])^2 - (4*(a - b)*b*(9*a + 5*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)^5*f)
Time = 0.50 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.16, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4146, 372, 402, 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.
| \(\displaystyle \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^4}{\left (a+b \tan (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tan ^4(e+f x)}{\left (\tan ^2(e+f x)+1\right )^3 \left (b \tan ^2(e+f x)+a\right )^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 372 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\int \frac {a-(4 a+3 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^2 \left (b \tan ^2(e+f x)+a\right )^3}d\tan (e+f x)}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {(5 a+3 b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\int \frac {a (3 a+5 b)-5 b (5 a+3 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^3}d\tan (e+f x)}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {(5 a+3 b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {\int \frac {12 a \left (a (a+3 b)-b (7 a+5 b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^2}d\tan (e+f x)}{4 a (a-b)}-\frac {b (7 a+5 b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )^2}}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {(5 a+3 b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {3 \int \frac {a (a+3 b)-b (7 a+5 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )^2}d\tan (e+f x)}{a-b}-\frac {b (7 a+5 b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )^2}}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {(5 a+3 b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {3 \left (\frac {\int \frac {2 a \left (a^2+6 b a+b^2-4 b (a+b) \tan ^2(e+f x)\right )}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{2 a (a-b)}-\frac {4 b (a+b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )}\right )}{a-b}-\frac {b (7 a+5 b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )^2}}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {(5 a+3 b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {3 \left (\frac {\int \frac {a^2+6 b a+b^2-4 b (a+b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a\right )}d\tan (e+f x)}{a-b}-\frac {4 b (a+b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )}\right )}{a-b}-\frac {b (7 a+5 b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )^2}}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {(5 a+3 b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {3 \left (\frac {\frac {\left (a^2+10 a b+5 b^2\right ) \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a-b}-\frac {b \left (5 a^2+10 a b+b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a-b}}{a-b}-\frac {4 b (a+b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )}\right )}{a-b}-\frac {b (7 a+5 b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )^2}}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {(5 a+3 b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {3 \left (\frac {\frac {\left (a^2+10 a b+5 b^2\right ) \arctan (\tan (e+f x))}{a-b}-\frac {b \left (5 a^2+10 a b+b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a}d\tan (e+f x)}{a-b}}{a-b}-\frac {4 b (a+b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )}\right )}{a-b}-\frac {b (7 a+5 b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )^2}}{2 (a-b)}}{4 (a-b)}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\tan (e+f x)}{4 (a-b) \left (\tan ^2(e+f x)+1\right )^2 \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {(5 a+3 b) \tan (e+f x)}{2 (a-b) \left (\tan ^2(e+f x)+1\right ) \left (a+b \tan ^2(e+f x)\right )^2}-\frac {\frac {3 \left (\frac {\frac {\left (a^2+10 a b+5 b^2\right ) \arctan (\tan (e+f x))}{a-b}-\frac {\sqrt {b} \left (5 a^2+10 a b+b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{\sqrt {a} (a-b)}}{a-b}-\frac {4 b (a+b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )}\right )}{a-b}-\frac {b (7 a+5 b) \tan (e+f x)}{(a-b) \left (a+b \tan ^2(e+f x)\right )^2}}{2 (a-b)}}{4 (a-b)}}{f}\) |
(Tan[e + f*x]/(4*(a - b)*(1 + Tan[e + f*x]^2)^2*(a + b*Tan[e + f*x]^2)^2) - (((5*a + 3*b)*Tan[e + f*x])/(2*(a - b)*(1 + Tan[e + f*x]^2)*(a + b*Tan[e + f*x]^2)^2) - (-((b*(7*a + 5*b)*Tan[e + f*x])/((a - b)*(a + b*Tan[e + f* x]^2)^2)) + (3*((((a^2 + 10*a*b + 5*b^2)*ArcTan[Tan[e + f*x]])/(a - b) - ( Sqrt[b]*(5*a^2 + 10*a*b + b^2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(Sq rt[a]*(a - b)))/(a - b) - (4*b*(a + b)*Tan[e + f*x])/((a - b)*(a + b*Tan[e + f*x]^2))))/(a - b))/(2*(a - b)))/(4*(a - b)))/f
3.1.86.3.1 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 = 42.12 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.84
| method | result | size |
| derivativedivides | \(\frac {-\frac {b \left (\frac {\left (\frac {7}{8} a^{2} b -\frac {1}{4} a \,b^{2}-\frac {5}{8} b^{3}\right ) \tan \left (f x +e \right )^{3}+\frac {3 a \left (3 a^{2}-2 a b -b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {3 \left (5 a^{2}+10 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{\left (a -b \right )^{5}}+\frac {\frac {\left (-\frac {1}{4} a b +\frac {7}{8} b^{2}-\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}-\frac {3}{4} a b +\frac {9}{8} b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {3 \left (a^{2}+10 a b +5 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{\left (a -b \right )^{5}}}{f}\) | \(209\) |
| default | \(\frac {-\frac {b \left (\frac {\left (\frac {7}{8} a^{2} b -\frac {1}{4} a \,b^{2}-\frac {5}{8} b^{3}\right ) \tan \left (f x +e \right )^{3}+\frac {3 a \left (3 a^{2}-2 a b -b^{2}\right ) \tan \left (f x +e \right )}{8}}{\left (a +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {3 \left (5 a^{2}+10 a b +b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{\left (a -b \right )^{5}}+\frac {\frac {\left (-\frac {1}{4} a b +\frac {7}{8} b^{2}-\frac {5}{8} a^{2}\right ) \tan \left (f x +e \right )^{3}+\left (-\frac {3}{8} a^{2}-\frac {3}{4} a b +\frac {9}{8} b^{2}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{2}}+\frac {3 \left (a^{2}+10 a b +5 b^{2}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{8}}{\left (a -b \right )^{5}}}{f}\) | \(209\) |
| risch | \(\frac {3 x \,a^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a -b \right )^{2}}+\frac {15 x a b}{4 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a -b \right )^{2}}+\frac {15 x \,b^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a -b \right )^{2}}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{4 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a -b \right ) f}-\frac {i {\mathrm e}^{4 i \left (f x +e \right )}}{64 \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 ) \left (a -b \right ) f}+\frac {i {\mathrm e}^{-4 i \left (f x +e \right )}}{64 \left (a^{2}-2 a b +b^{2}\right ) f \left (a -b \right )}-\frac {i \left (-9 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}-9 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}+13 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+5 b^{3} {\mathrm e}^{6 i \left (f x +e \right )}-27 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}-33 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}-37 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-15 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}-27 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}-11 a^{2} b \,{\mathrm e}^{2 i \left (f x +e \right )}+23 a \,b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+15 b^{3} {\mathrm e}^{2 i \left (f x +e \right )}-9 a^{3}+13 a^{2} b +a \,b^{2}-5 b^{3}\right ) b}{4 \left (-a^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) \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 )^{2} \left (a^{2}-2 a b +b^{2}\right ) f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{4 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} a}{8 \left (a^{2}-2 a b +b^{2}\right ) \left (a -b \right )^{2} f}+\frac {15 a \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 )}{16 \left (a -b \right )^{5} f}+\frac {15 \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}{8 \left (a -b \right )^{5} 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^{2}}{16 a \left (a -b \right )^{5} f}-\frac {15 a \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 )}{16 \left (a -b \right )^{5} f}-\frac {15 \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}{8 \left (a -b \right )^{5} 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^{2}}{16 a \left (a -b \right )^{5} f}\) | \(948\) |
1/f*(-b/(a-b)^5*(((7/8*a^2*b-1/4*a*b^2-5/8*b^3)*tan(f*x+e)^3+3/8*a*(3*a^2- 2*a*b-b^2)*tan(f*x+e))/(a+b*tan(f*x+e)^2)^2+3/8*(5*a^2+10*a*b+b^2)/(a*b)^( 1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2)))+1/(a-b)^5*(((-1/4*a*b+7/8*b^2-5/8*a ^2)*tan(f*x+e)^3+(-3/8*a^2-3/4*a*b+9/8*b^2)*tan(f*x+e))/(1+tan(f*x+e)^2)^2 +3/8*(a^2+10*a*b+5*b^2)*arctan(tan(f*x+e))))
Leaf count of result is larger than twice the leaf count of optimal. 551 vs. \(2 (230) = 460\).
Time = 0.43 (sec) , antiderivative size = 1191, normalized size of antiderivative = 4.76 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
[1/32*(12*(a^4 + 8*a^3*b - 14*a^2*b^2 + 5*b^4)*f*x*cos(f*x + e)^4 + 24*(a^ 3*b + 9*a^2*b^2 - 5*a*b^3 - 5*b^4)*f*x*cos(f*x + e)^2 + 12*(a^2*b^2 + 10*a *b^3 + 5*b^4)*f*x - 3*((5*a^4 - 14*a^2*b^2 + 8*a*b^3 + b^4)*cos(f*x + e)^4 + 5*a^2*b^2 + 10*a*b^3 + b^4 + 2*(5*a^3*b + 5*a^2*b^2 - 9*a*b^3 - b^4)*co s(f*x + e)^2)*sqrt(-b/a)*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^2 + a*b)*cos(f*x + e)^3 - a*b*cos(f*x + e) )*sqrt(-b/a)*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)) + 4*(2*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a* b^3 + b^4)*cos(f*x + e)^7 - (5*a^4 - 12*a^3*b + 6*a^2*b^2 + 4*a*b^3 - 3*b^ 4)*cos(f*x + e)^5 - (19*a^3*b - 21*a^2*b^2 - 15*a*b^3 + 17*b^4)*cos(f*x + e)^3 - 12*(a^2*b^2 - b^4)*cos(f*x + e))*sin(f*x + e))/((a^7 - 7*a^6*b + 21 *a^5*b^2 - 35*a^4*b^3 + 35*a^3*b^4 - 21*a^2*b^5 + 7*a*b^6 - b^7)*f*cos(f*x + e)^4 + 2*(a^6*b - 6*a^5*b^2 + 15*a^4*b^3 - 20*a^3*b^4 + 15*a^2*b^5 - 6* a*b^6 + b^7)*f*cos(f*x + e)^2 + (a^5*b^2 - 5*a^4*b^3 + 10*a^3*b^4 - 10*a^2 *b^5 + 5*a*b^6 - b^7)*f), 1/16*(6*(a^4 + 8*a^3*b - 14*a^2*b^2 + 5*b^4)*f*x *cos(f*x + e)^4 + 12*(a^3*b + 9*a^2*b^2 - 5*a*b^3 - 5*b^4)*f*x*cos(f*x + e )^2 + 6*(a^2*b^2 + 10*a*b^3 + 5*b^4)*f*x + 3*((5*a^4 - 14*a^2*b^2 + 8*a*b^ 3 + b^4)*cos(f*x + e)^4 + 5*a^2*b^2 + 10*a*b^3 + b^4 + 2*(5*a^3*b + 5*a^2* b^2 - 9*a*b^3 - b^4)*cos(f*x + e)^2)*sqrt(b/a)*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(b/a)/(b*cos(f*x + e)*sin(f*x + e))) + 2*(2*(a^4 - 4*a...
Timed out. \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Time = 0.32 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.84 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\frac {3 \, {\left (a^{2} + 10 \, a b + 5 \, b^{2}\right )} {\left (f x + e\right )}}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} - \frac {3 \, {\left (5 \, a^{2} b + 10 \, a b^{2} + b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} \sqrt {a b}} - \frac {12 \, {\left (a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{7} + {\left (19 \, a^{2} b + 34 \, a b^{2} + 19 \, b^{3}\right )} \tan \left (f x + e\right )^{5} + {\left (5 \, a^{3} + 31 \, a^{2} b + 31 \, a b^{2} + 5 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (a^{3} + 6 \, a^{2} b + a b^{2}\right )} \tan \left (f x + e\right )}{{\left (a^{4} b^{2} - 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} - 4 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{8} + 2 \, {\left (a^{5} b - 3 \, a^{4} b^{2} + 2 \, a^{3} b^{3} + 2 \, a^{2} b^{4} - 3 \, a b^{5} + b^{6}\right )} \tan \left (f x + e\right )^{6} + a^{6} - 4 \, a^{5} b + 6 \, a^{4} b^{2} - 4 \, a^{3} b^{3} + a^{2} b^{4} + {\left (a^{6} - 9 \, a^{4} b^{2} + 16 \, a^{3} b^{3} - 9 \, a^{2} b^{4} + b^{6}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{6} - 3 \, a^{5} b + 2 \, a^{4} b^{2} + 2 \, a^{3} b^{3} - 3 \, a^{2} b^{4} + a b^{5}\right )} \tan \left (f x + e\right )^{2}}}{8 \, f} \]
1/8*(3*(a^2 + 10*a*b + 5*b^2)*(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) - 3*(5*a^2*b + 10*a*b^2 + b^3)*arctan(b*tan(f*x + e)/sqrt(a*b))/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*s qrt(a*b)) - (12*(a*b^2 + b^3)*tan(f*x + e)^7 + (19*a^2*b + 34*a*b^2 + 19*b ^3)*tan(f*x + e)^5 + (5*a^3 + 31*a^2*b + 31*a*b^2 + 5*b^3)*tan(f*x + e)^3 + 3*(a^3 + 6*a^2*b + a*b^2)*tan(f*x + e))/((a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^ 4 - 4*a*b^5 + b^6)*tan(f*x + e)^8 + 2*(a^5*b - 3*a^4*b^2 + 2*a^3*b^3 + 2*a ^2*b^4 - 3*a*b^5 + b^6)*tan(f*x + e)^6 + a^6 - 4*a^5*b + 6*a^4*b^2 - 4*a^3 *b^3 + a^2*b^4 + (a^6 - 9*a^4*b^2 + 16*a^3*b^3 - 9*a^2*b^4 + b^6)*tan(f*x + e)^4 + 2*(a^6 - 3*a^5*b + 2*a^4*b^2 + 2*a^3*b^3 - 3*a^2*b^4 + a*b^5)*tan (f*x + e)^2))/f
Time = 1.00 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.52 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\frac {3 \, {\left (a^{2} + 10 \, a b + 5 \, b^{2}\right )} {\left (f x + e\right )}}{a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}} - \frac {3 \, {\left (5 \, a^{2} b + 10 \, a b^{2} + b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )\right )}}{{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )} \sqrt {a b}} - \frac {12 \, a b^{2} \tan \left (f x + e\right )^{7} + 12 \, b^{3} \tan \left (f x + e\right )^{7} + 19 \, a^{2} b \tan \left (f x + e\right )^{5} + 34 \, a b^{2} \tan \left (f x + e\right )^{5} + 19 \, b^{3} \tan \left (f x + e\right )^{5} + 5 \, a^{3} \tan \left (f x + e\right )^{3} + 31 \, a^{2} b \tan \left (f x + e\right )^{3} + 31 \, a b^{2} \tan \left (f x + e\right )^{3} + 5 \, b^{3} \tan \left (f x + e\right )^{3} + 3 \, a^{3} \tan \left (f x + e\right ) + 18 \, a^{2} b \tan \left (f x + e\right ) + 3 \, a b^{2} \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{4} + a \tan \left (f x + e\right )^{2} + b \tan \left (f x + e\right )^{2} + a\right )}^{2} {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )}}}{8 \, f} \]
1/8*(3*(a^2 + 10*a*b + 5*b^2)*(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) - 3*(5*a^2*b + 10*a*b^2 + b^3)*(pi*floor((f*x + e) /pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))/((a^5 - 5*a^4*b + 10 *a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*sqrt(a*b)) - (12*a*b^2*tan(f*x + e) ^7 + 12*b^3*tan(f*x + e)^7 + 19*a^2*b*tan(f*x + e)^5 + 34*a*b^2*tan(f*x + e)^5 + 19*b^3*tan(f*x + e)^5 + 5*a^3*tan(f*x + e)^3 + 31*a^2*b*tan(f*x + e )^3 + 31*a*b^2*tan(f*x + e)^3 + 5*b^3*tan(f*x + e)^3 + 3*a^3*tan(f*x + e) + 18*a^2*b*tan(f*x + e) + 3*a*b^2*tan(f*x + e))/((b*tan(f*x + e)^4 + a*tan (f*x + e)^2 + b*tan(f*x + e)^2 + a)^2*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)))/f
Time = 15.65 (sec) , antiderivative size = 5965, normalized size of antiderivative = 23.86 \[ \int \frac {\sin ^4(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
(atan(((((tan(e + f*x)*(540*a*b^6 + 117*b^7 + 990*a^2*b^5 + 540*a^3*b^4 + 117*a^4*b^3))/(16*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28*a^6*b^2)) + (3*((6*a*b^13 - (3*b^14)/2 + 2 1*a^2*b^12 - 210*a^3*b^11 + (1395*a^4*b^10)/2 - 1332*a^5*b^9 + 1638*a^6*b^ 8 - 1332*a^7*b^7 + (1395*a^8*b^6)/2 - 210*a^9*b^5 + 21*a^10*b^4 + 6*a^11*b ^3 - (3*a^12*b^2)/2)/(a^12 - 12*a^11*b - 12*a*b^11 + b^12 + 66*a^2*b^10 - 220*a^3*b^9 + 495*a^4*b^8 - 792*a^5*b^7 + 924*a^6*b^6 - 792*a^7*b^5 + 495* a^8*b^4 - 220*a^9*b^3 + 66*a^10*b^2) - (3*tan(e + f*x)*(10*a*b + a^2 + 5*b ^2)*(1152*a*b^12 - 128*b^13 - 4480*a^2*b^11 + 9600*a^3*b^10 - 11520*a^4*b^ 9 + 5376*a^5*b^8 + 5376*a^6*b^7 - 11520*a^7*b^6 + 9600*a^8*b^5 - 4480*a^9* b^4 + 1152*a^10*b^3 - 128*a^11*b^2))/(256*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^ 2*b^6 - 56*a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28*a^6*b^2)))*(10*a*b + a^2 + 5*b^2))/(16*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3* b^2*10i)))*(10*a*b + a^2 + 5*b^2)*3i)/(16*(a*b^4*5i - a^4*b*5i + a^5*1i - b^5*1i - a^2*b^3*10i + a^3*b^2*10i)) + (((tan(e + f*x)*(540*a*b^6 + 117*b^ 7 + 990*a^2*b^5 + 540*a^3*b^4 + 117*a^4*b^3))/(16*(a^8 - 8*a^7*b - 8*a*b^7 + b^8 + 28*a^2*b^6 - 56*a^3*b^5 + 70*a^4*b^4 - 56*a^5*b^3 + 28*a^6*b^2)) - (3*((6*a*b^13 - (3*b^14)/2 + 21*a^2*b^12 - 210*a^3*b^11 + (1395*a^4*b^10 )/2 - 1332*a^5*b^9 + 1638*a^6*b^8 - 1332*a^7*b^7 + (1395*a^8*b^6)/2 - 2...