Integrand size = 23, antiderivative size = 214 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\sqrt {b} \left (15 a^2+70 a b+63 b^2\right ) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{11/2} f}-\frac {\left (3 a^2+14 a b+13 b^2\right ) \cos (e+f x)}{2 a^5 f}+\frac {(a+3 b) (3 a+5 b) \cos ^3(e+f x)}{12 a^4 b f}-\frac {\cos ^5(e+f x)}{5 a^3 f}-\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b f \left (b+a \cos ^2(e+f x)\right )^2}-\frac {b (a+b) (3 a+11 b) \cos (e+f x)}{8 a^5 f \left (b+a \cos ^2(e+f x)\right )} \]
-1/2*(3*a^2+14*a*b+13*b^2)*cos(f*x+e)/a^5/f+1/12*(a+3*b)*(3*a+5*b)*cos(f*x +e)^3/a^4/b/f-1/5*cos(f*x+e)^5/a^3/f-1/4*(a+b)^2*cos(f*x+e)^7/a^2/b/f/(b+a *cos(f*x+e)^2)^2-1/8*b*(a+b)*(3*a+11*b)*cos(f*x+e)/a^5/f/(b+a*cos(f*x+e)^2 )+1/8*(15*a^2+70*a*b+63*b^2)*arctan(cos(f*x+e)*a^(1/2)/b^(1/2))*b^(1/2)/a^ (11/2)/f
Result contains complex when optimal does not.
Time = 9.36 (sec) , antiderivative size = 1641, normalized size of antiderivative = 7.67 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*(-900*a^(11/2)*b^(3/2)*Cos[ e + f*x] - 109000*a^(9/2)*b^(5/2)*Cos[e + f*x] - 936000*a^(7/2)*b^(7/2)*Co s[e + f*x] - 2803072*a^(5/2)*b^(9/2)*Cos[e + f*x] - 3763200*a^(3/2)*b^(11/ 2)*Cos[e + f*x] - 1935360*Sqrt[a]*b^(13/2)*Cos[e + f*x] - 900*a^(11/2)*b^( 3/2)*Cos[e + f*x]*Cos[2*(e + f*x)] + 900*a^(9/2)*b^(3/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) + 24000*a^(7/2)*b^(5/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) + 43200*a^(5/2)*b^(7/2)*Cos[e + f*x]*(a + 2*b + a*Cos [2*(e + f*x)]) + 225*a^5*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Co s[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)]) ^2 + 115200*a^2*b^3*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin [e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 537600*a*b^4*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2] )*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin [e])^2]*Tan[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 483840* b^5*ArcTan[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]* Tan[(f*x)/2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*T an[(f*x)/2]))/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 + 225*a^5*ArcTan[( (-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)...
Time = 0.51 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 4621, 366, 25, 360, 2341, 2009}
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.
| \(\displaystyle \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^5}{\left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4621 |
| \(\displaystyle -\frac {\int \frac {\cos ^6(e+f x) \left (1-\cos ^2(e+f x)\right )^2}{\left (a \cos ^2(e+f x)+b\right )^3}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 366 |
| \(\displaystyle -\frac {\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b \left (a \cos ^2(e+f x)+b\right )^2}-\frac {\int -\frac {\cos ^6(e+f x) \left (4 a^2+4 b \cos ^2(e+f x) a-7 (a+b)^2\right )}{\left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{4 a^2 b}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {\int \frac {\cos ^6(e+f x) \left (4 a^2+4 b \cos ^2(e+f x) a-7 (a+b)^2\right )}{\left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{4 a^2 b}+\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b \left (a \cos ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle -\frac {\frac {\frac {b^2 (a+b) (3 a+11 b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\int \frac {-8 a^4 b \cos ^6(e+f x)+2 a^3 (a+b) (3 a+11 b) \cos ^4(e+f x)-2 a^2 b (a+b) (3 a+11 b) \cos ^2(e+f x)+a b^2 (a+b) (3 a+11 b)}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{2 a^4}}{4 a^2 b}+\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b \left (a \cos ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle -\frac {\frac {\frac {b^2 (a+b) (3 a+11 b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {\int \left (-8 a^3 b \cos ^4(e+f x)+2 a^2 (a+3 b) (3 a+5 b) \cos ^2(e+f x)-4 a b \left (3 a^2+14 b a+13 b^2\right )+\frac {63 a b^4+70 a^2 b^3+15 a^3 b^2}{a \cos ^2(e+f x)+b}\right )d\cos (e+f x)}{2 a^4}}{4 a^2 b}+\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b \left (a \cos ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {(a+b)^2 \cos ^7(e+f x)}{4 a^2 b \left (a \cos ^2(e+f x)+b\right )^2}+\frac {\frac {b^2 (a+b) (3 a+11 b) \cos (e+f x)}{2 a^3 \left (a \cos ^2(e+f x)+b\right )}-\frac {-\frac {8}{5} a^3 b \cos ^5(e+f x)+\sqrt {a} b^{3/2} \left (15 a^2+70 a b+63 b^2\right ) \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )-4 a b \left (3 a^2+14 a b+13 b^2\right ) \cos (e+f x)+\frac {2}{3} a^2 (a+3 b) (3 a+5 b) \cos ^3(e+f x)}{2 a^4}}{4 a^2 b}}{f}\) |
-((((a + b)^2*Cos[e + f*x]^7)/(4*a^2*b*(b + a*Cos[e + f*x]^2)^2) + ((b^2*( a + b)*(3*a + 11*b)*Cos[e + f*x])/(2*a^3*(b + a*Cos[e + f*x]^2)) - (Sqrt[a ]*b^(3/2)*(15*a^2 + 70*a*b + 63*b^2)*ArcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b] ] - 4*a*b*(3*a^2 + 14*a*b + 13*b^2)*Cos[e + f*x] + (2*a^2*(a + 3*b)*(3*a + 5*b)*Cos[e + f*x]^3)/3 - (8*a^3*b*Cos[e + f*x]^5)/5)/(2*a^4))/(4*a^2*b))/ f)
3.1.54.3.1 Defintions of rubi rules used
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(2*a* b^2*e*(p + 1))), x] + Simp[1/(2*a*b^2*(p + 1)) Int[(e*x)^m*(a + b*x^2)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + 2*b^2*c^2*(p + 1) + 2*a*b*d^2*(p + 1)*x^ 2, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && LtQ[p , -1]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*((b + a*(ff*x)^n)^p/(ff*x)^(n*p)), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/ 2] && IntegerQ[n] && IntegerQ[p]
Time = 10.37 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.89
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {\cos \left (f x +e \right )^{5} a^{2}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}-a \cos \left (f x +e \right )^{3} b +a^{2} \cos \left (f x +e \right )+6 a b \cos \left (f x +e \right )+6 b^{2} \cos \left (f x +e \right )}{a^{5}}+\frac {b \left (\frac {\left (-\frac {9}{8} a^{3}-\frac {13}{4} a^{2} b -\frac {17}{8} a \,b^{2}\right ) \cos \left (f x +e \right )^{3}-\frac {b \left (7 a^{2}+22 a b +15 b^{2}\right ) \cos \left (f x +e \right )}{8}}{\left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}+70 a b +63 b^{2}\right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}}{f}\) | \(190\) |
| default | \(\frac {-\frac {\frac {\cos \left (f x +e \right )^{5} a^{2}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}-a \cos \left (f x +e \right )^{3} b +a^{2} \cos \left (f x +e \right )+6 a b \cos \left (f x +e \right )+6 b^{2} \cos \left (f x +e \right )}{a^{5}}+\frac {b \left (\frac {\left (-\frac {9}{8} a^{3}-\frac {13}{4} a^{2} b -\frac {17}{8} a \,b^{2}\right ) \cos \left (f x +e \right )^{3}-\frac {b \left (7 a^{2}+22 a b +15 b^{2}\right ) \cos \left (f x +e \right )}{8}}{\left (b +a \cos \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}+70 a b +63 b^{2}\right ) \arctan \left (\frac {a \cos \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{5}}}{f}\) | \(190\) |
| risch | \(-\frac {{\mathrm e}^{5 i \left (f x +e \right )}}{160 a^{3} f}+\frac {5 \,{\mathrm e}^{3 i \left (f x +e \right )}}{96 a^{3} f}+\frac {{\mathrm e}^{3 i \left (f x +e \right )} b}{8 a^{4} f}-\frac {5 \,{\mathrm e}^{i \left (f x +e \right )}}{16 a^{3} f}-\frac {21 \,{\mathrm e}^{i \left (f x +e \right )} b}{8 a^{4} f}-\frac {3 \,{\mathrm e}^{i \left (f x +e \right )} b^{2}}{a^{5} f}-\frac {5 \,{\mathrm e}^{-i \left (f x +e \right )}}{16 a^{3} f}-\frac {21 \,{\mathrm e}^{-i \left (f x +e \right )} b}{8 a^{4} f}-\frac {3 \,{\mathrm e}^{-i \left (f x +e \right )} b^{2}}{a^{5} f}+\frac {5 \,{\mathrm e}^{-3 i \left (f x +e \right )}}{96 a^{3} f}+\frac {{\mathrm e}^{-3 i \left (f x +e \right )} b}{8 a^{4} f}-\frac {{\mathrm e}^{-5 i \left (f x +e \right )}}{160 a^{3} f}-\frac {b \left (9 a^{3} {\mathrm e}^{7 i \left (f x +e \right )}+26 a^{2} b \,{\mathrm e}^{7 i \left (f x +e \right )}+17 a \,b^{2} {\mathrm e}^{7 i \left (f x +e \right )}+27 a^{3} {\mathrm e}^{5 i \left (f x +e \right )}+106 a^{2} b \,{\mathrm e}^{5 i \left (f x +e \right )}+139 a \,b^{2} {\mathrm e}^{5 i \left (f x +e \right )}+60 b^{3} {\mathrm e}^{5 i \left (f x +e \right )}+27 a^{3} {\mathrm e}^{3 i \left (f x +e \right )}+106 a^{2} b \,{\mathrm e}^{3 i \left (f x +e \right )}+139 a \,b^{2} {\mathrm e}^{3 i \left (f x +e \right )}+60 b^{3} {\mathrm e}^{3 i \left (f x +e \right )}+9 a^{3} {\mathrm e}^{i \left (f x +e \right )}+26 a^{2} b \,{\mathrm e}^{i \left (f x +e \right )}+17 a \,b^{2} {\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{5} 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 {63 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b^{2}}{16 a^{6} f}+\frac {35 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{8 a^{5} f}-\frac {63 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b^{2}}{16 a^{6} f}+\frac {15 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{16 a^{4} f}-\frac {35 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right ) b}{8 a^{5} f}-\frac {15 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{16 a^{4} f}\) | \(753\) |
1/f*(-1/a^5*(1/5*cos(f*x+e)^5*a^2-2/3*a^2*cos(f*x+e)^3-a*cos(f*x+e)^3*b+a^ 2*cos(f*x+e)+6*a*b*cos(f*x+e)+6*b^2*cos(f*x+e))+b/a^5*(((-9/8*a^3-13/4*a^2 *b-17/8*a*b^2)*cos(f*x+e)^3-1/8*b*(7*a^2+22*a*b+15*b^2)*cos(f*x+e))/(b+a*c os(f*x+e)^2)^2+1/8*(15*a^2+70*a*b+63*b^2)/(a*b)^(1/2)*arctan(a*cos(f*x+e)/ (a*b)^(1/2))))
Time = 0.32 (sec) , antiderivative size = 579, normalized size of antiderivative = 2.71 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [-\frac {48 \, a^{4} \cos \left (f x + e\right )^{9} - 16 \, {\left (10 \, a^{4} + 9 \, a^{3} b\right )} \cos \left (f x + e\right )^{7} + 16 \, {\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{5} + 50 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 30 \, {\left (15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4}\right )} \cos \left (f x + e\right )}{240 \, {\left (a^{7} f \cos \left (f x + e\right )^{4} + 2 \, a^{6} b f \cos \left (f x + e\right )^{2} + a^{5} b^{2} f\right )}}, -\frac {24 \, a^{4} \cos \left (f x + e\right )^{9} - 8 \, {\left (10 \, a^{4} + 9 \, a^{3} b\right )} \cos \left (f x + e\right )^{7} + 8 \, {\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{5} + 25 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (15 \, a^{4} + 70 \, a^{3} b + 63 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 70 \, a^{2} b^{2} + 63 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right ) + 15 \, {\left (15 \, a^{2} b^{2} + 70 \, a b^{3} + 63 \, b^{4}\right )} \cos \left (f x + e\right )}{120 \, {\left (a^{7} f \cos \left (f x + e\right )^{4} + 2 \, a^{6} b f \cos \left (f x + e\right )^{2} + a^{5} b^{2} f\right )}}\right ] \]
[-1/240*(48*a^4*cos(f*x + e)^9 - 16*(10*a^4 + 9*a^3*b)*cos(f*x + e)^7 + 16 *(15*a^4 + 70*a^3*b + 63*a^2*b^2)*cos(f*x + e)^5 + 50*(15*a^3*b + 70*a^2*b ^2 + 63*a*b^3)*cos(f*x + e)^3 - 15*((15*a^4 + 70*a^3*b + 63*a^2*b^2)*cos(f *x + e)^4 + 15*a^2*b^2 + 70*a*b^3 + 63*b^4 + 2*(15*a^3*b + 70*a^2*b^2 + 63 *a*b^3)*cos(f*x + e)^2)*sqrt(-b/a)*log(-(a*cos(f*x + e)^2 + 2*a*sqrt(-b/a) *cos(f*x + e) - b)/(a*cos(f*x + e)^2 + b)) + 30*(15*a^2*b^2 + 70*a*b^3 + 6 3*b^4)*cos(f*x + e))/(a^7*f*cos(f*x + e)^4 + 2*a^6*b*f*cos(f*x + e)^2 + a^ 5*b^2*f), -1/120*(24*a^4*cos(f*x + e)^9 - 8*(10*a^4 + 9*a^3*b)*cos(f*x + e )^7 + 8*(15*a^4 + 70*a^3*b + 63*a^2*b^2)*cos(f*x + e)^5 + 25*(15*a^3*b + 7 0*a^2*b^2 + 63*a*b^3)*cos(f*x + e)^3 - 15*((15*a^4 + 70*a^3*b + 63*a^2*b^2 )*cos(f*x + e)^4 + 15*a^2*b^2 + 70*a*b^3 + 63*b^4 + 2*(15*a^3*b + 70*a^2*b ^2 + 63*a*b^3)*cos(f*x + e)^2)*sqrt(b/a)*arctan(a*sqrt(b/a)*cos(f*x + e)/b ) + 15*(15*a^2*b^2 + 70*a*b^3 + 63*b^4)*cos(f*x + e))/(a^7*f*cos(f*x + e)^ 4 + 2*a^6*b*f*cos(f*x + e)^2 + a^5*b^2*f)]
Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 204, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {15 \, {\left ({\left (9 \, a^{3} b + 26 \, a^{2} b^{2} + 17 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (7 \, a^{2} b^{2} + 22 \, a b^{3} + 15 \, b^{4}\right )} \cos \left (f x + e\right )\right )}}{a^{7} \cos \left (f x + e\right )^{4} + 2 \, a^{6} b \cos \left (f x + e\right )^{2} + a^{5} b^{2}} - \frac {15 \, {\left (15 \, a^{2} b + 70 \, a b^{2} + 63 \, b^{3}\right )} \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{5}} + \frac {8 \, {\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - 5 \, {\left (2 \, a^{2} + 3 \, a b\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 6 \, a b + 6 \, b^{2}\right )} \cos \left (f x + e\right )\right )}}{a^{5}}}{120 \, f} \]
-1/120*(15*((9*a^3*b + 26*a^2*b^2 + 17*a*b^3)*cos(f*x + e)^3 + (7*a^2*b^2 + 22*a*b^3 + 15*b^4)*cos(f*x + e))/(a^7*cos(f*x + e)^4 + 2*a^6*b*cos(f*x + e)^2 + a^5*b^2) - 15*(15*a^2*b + 70*a*b^2 + 63*b^3)*arctan(a*cos(f*x + e) /sqrt(a*b))/(sqrt(a*b)*a^5) + 8*(3*a^2*cos(f*x + e)^5 - 5*(2*a^2 + 3*a*b)* cos(f*x + e)^3 + 15*(a^2 + 6*a*b + 6*b^2)*cos(f*x + e))/a^5)/f
Leaf count of result is larger than twice the leaf count of optimal. 781 vs. \(2 (194) = 388\).
Time = 0.47 (sec) , antiderivative size = 781, normalized size of antiderivative = 3.65 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
-1/120*(15*(15*a^2*b + 70*a*b^2 + 63*b^3)*arctan(-(a*cos(f*x + e) - b)/(sq rt(a*b)*cos(f*x + e) + sqrt(a*b)))/(sqrt(a*b)*a^5) + 30*(9*a^3*b + 33*a^2* b^2 + 39*a*b^3 + 15*b^4 + 27*a^3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 49*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 23*a*b^3*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 45*b^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 27*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 27*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 3*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 45*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 9*a^3*b*(co s(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 11*a^2*b^2*(cos(f*x + e) - 1)^3/( cos(f*x + e) + 1)^3 - 13*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 15*b^4*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3)/((a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos (f*x + e) + 1)^2)^2*a^5) - 16*(8*a^2 + 75*a*b + 90*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 330*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1 ) - 360*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 80*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 480*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 540*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 270*a*b*(cos(f* x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 360*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 45*a*b*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 90*b^...
Time = 17.90 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.19 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {{\cos \left (e+f\,x\right )}^3\,\left (\frac {b}{a^4}+\frac {2}{3\,a^3}\right )}{f}-\frac {\left (\frac {9\,a^3\,b}{8}+\frac {13\,a^2\,b^2}{4}+\frac {17\,a\,b^3}{8}\right )\,{\cos \left (e+f\,x\right )}^3+\left (\frac {7\,a^2\,b^2}{8}+\frac {11\,a\,b^3}{4}+\frac {15\,b^4}{8}\right )\,\cos \left (e+f\,x\right )}{f\,\left (a^7\,{\cos \left (e+f\,x\right )}^4+2\,a^6\,b\,{\cos \left (e+f\,x\right )}^2+a^5\,b^2\right )}-\frac {{\cos \left (e+f\,x\right )}^5}{5\,a^3\,f}-\frac {\cos \left (e+f\,x\right )\,\left (\frac {1}{a^3}-\frac {3\,b^2}{a^5}+\frac {3\,b\,\left (\frac {3\,b}{a^4}+\frac {2}{a^3}\right )}{a}\right )}{f}+\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\sqrt {b}\,\cos \left (e+f\,x\right )\,\left (15\,a^2+70\,a\,b+63\,b^2\right )}{15\,a^2\,b+70\,a\,b^2+63\,b^3}\right )\,\left (15\,a^2+70\,a\,b+63\,b^2\right )}{8\,a^{11/2}\,f} \]
(cos(e + f*x)^3*(b/a^4 + 2/(3*a^3)))/f - (cos(e + f*x)^3*((17*a*b^3)/8 + ( 9*a^3*b)/8 + (13*a^2*b^2)/4) + cos(e + f*x)*((11*a*b^3)/4 + (15*b^4)/8 + ( 7*a^2*b^2)/8))/(f*(a^5*b^2 + a^7*cos(e + f*x)^4 + 2*a^6*b*cos(e + f*x)^2)) - cos(e + f*x)^5/(5*a^3*f) - (cos(e + f*x)*(1/a^3 - (3*b^2)/a^5 + (3*b*(( 3*b)/a^4 + 2/a^3))/a))/f + (b^(1/2)*atan((a^(1/2)*b^(1/2)*cos(e + f*x)*(70 *a*b + 15*a^2 + 63*b^2))/(70*a*b^2 + 15*a^2*b + 63*b^3))*(70*a*b + 15*a^2 + 63*b^2))/(8*a^(11/2)*f)