Integrand size = 23, antiderivative size = 264 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {\sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 (a-b)^{11/2} f}-\frac {\left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{5 (a-b)^5 f}+\frac {(10 a-b) \cos ^3(e+f x)}{15 (a-b)^4 f}-\frac {\cos ^5(e+f x)}{5 (a-b) f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{20 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b \left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{40 (a-b)^5 f \left (a-b+b \sec ^2(e+f x)\right )} \]
-1/5*(5*a^2+20*a*b+2*b^2)*cos(f*x+e)/(a-b)^5/f+1/15*(10*a-b)*cos(f*x+e)^3/ (a-b)^4/f-1/5*cos(f*x+e)^5/(a-b)/f/(a-b+b*sec(f*x+e)^2)^2-1/20*b*(5*a^2+4* b^2)*sec(f*x+e)/(a-b)^4/f/(a-b+b*sec(f*x+e)^2)^2-1/40*b*(35*a^2+40*a*b+24* b^2)*sec(f*x+e)/(a-b)^5/f/(a-b+b*sec(f*x+e)^2)-1/8*(15*a^2+40*a*b+8*b^2)*a rctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^(1/2)/(a-b)^(11/2)/f
Time = 6.42 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.05 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\frac {30 \sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{11/2}}+\frac {30 \sqrt {b} \left (15 a^2+40 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{11/2}}+\frac {-30 \cos (e+f x) \left (11 b^2+16 a b \left (2+\frac {b}{a+b+(a-b) \cos (2 (e+f x))}\right )+a^2 \left (5-\frac {8 b^2}{(a+b+(a-b) \cos (2 (e+f x)))^2}+\frac {18 b}{a+b+(a-b) \cos (2 (e+f x))}\right )\right )+(a-b) (5 (5 a+7 b) \cos (3 (e+f x))+3 (-a+b) \cos (5 (e+f x)))}{(a-b)^5}}{240 f} \]
((30*Sqrt[b]*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[( e + f*x)/2])/Sqrt[b]])/(a - b)^(11/2) + (30*Sqrt[b]*(15*a^2 + 40*a*b + 8*b ^2)*ArcTan[(Sqrt[a - b] + Sqrt[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(11/ 2) + (-30*Cos[e + f*x]*(11*b^2 + 16*a*b*(2 + b/(a + b + (a - b)*Cos[2*(e + f*x)])) + a^2*(5 - (8*b^2)/(a + b + (a - b)*Cos[2*(e + f*x)])^2 + (18*b)/ (a + b + (a - b)*Cos[2*(e + f*x)]))) + (a - b)*(5*(5*a + 7*b)*Cos[3*(e + f *x)] + 3*(-a + b)*Cos[5*(e + f*x)]))/(a - b)^5)/(240*f)
Time = 0.64 (sec) , antiderivative size = 273, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3042, 4147, 365, 25, 361, 25, 1582, 1584, 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.
| \(\displaystyle \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^5}{\left (a+b \tan (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int \frac {\cos ^6(e+f x) \left (1-\sec ^2(e+f x)\right )^2}{\left (b \sec ^2(e+f x)+a-b\right )^3}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 365 |
| \(\displaystyle \frac {\frac {\int -\frac {\cos ^4(e+f x) \left (-5 (a-b) \sec ^2(e+f x)+10 a-b\right )}{\left (b \sec ^2(e+f x)+a-b\right )^3}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\int \frac {\cos ^4(e+f x) \left (-5 (a-b) \sec ^2(e+f x)+10 a-b\right )}{\left (b \sec ^2(e+f x)+a-b\right )^3}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 361 |
| \(\displaystyle \frac {-\frac {\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{4 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}-\frac {1}{4} b \int -\frac {\cos ^4(e+f x) \left (\frac {3 \left (5 a^2+4 b^2\right ) \sec ^4(e+f x)}{(a-b)^3}-\frac {4 \left (5 a^2+4 b^2\right ) \sec ^2(e+f x)}{(a-b)^2 b}+\frac {4 (10 a-b)}{(a-b) b}\right )}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {\frac {1}{4} b \int \frac {\cos ^4(e+f x) \left (\frac {3 \left (5 a^2+4 b^2\right ) \sec ^4(e+f x)}{(a-b)^3}-\frac {4 \left (5 a^2+4 b^2\right ) \sec ^2(e+f x)}{(a-b)^2 b}+\frac {4 (10 a-b)}{(a-b) b}\right )}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)+\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{4 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {-\frac {\frac {1}{4} b \left (\frac {\int \frac {\cos ^4(e+f x) \left (\frac {b^2 \left (35 a^2+40 b a+24 b^2\right ) \sec ^4(e+f x)}{a-b}-8 b \left (5 a^2+10 b a+3 b^2\right ) \sec ^2(e+f x)+8 (a-b) (10 a-b) b\right )}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{2 b^2 (a-b)^3}+\frac {\left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{2 (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}\right )+\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{4 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {-\frac {\frac {1}{4} b \left (\frac {\int \left (8 (10 a-b) b \cos ^4(e+f x)-\frac {8 b \left (5 a^2+20 b a+2 b^2\right ) \cos ^2(e+f x)}{a-b}+\frac {5 b^2 \left (15 a^2+40 b a+8 b^2\right )}{(a-b) \left (b \sec ^2(e+f x)+a-b\right )}\right )d\sec (e+f x)}{2 b^2 (a-b)^3}+\frac {\left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{2 (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}\right )+\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{4 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {\frac {1}{4} b \left (\frac {\frac {5 b^{3/2} \left (15 a^2+40 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{(a-b)^{3/2}}+\frac {8 b \left (5 a^2+20 a b+2 b^2\right ) \cos (e+f x)}{a-b}-\frac {8}{3} b (10 a-b) \cos ^3(e+f x)}{2 b^2 (a-b)^3}+\frac {\left (35 a^2+40 a b+24 b^2\right ) \sec (e+f x)}{2 (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}\right )+\frac {b \left (5 a^2+4 b^2\right ) \sec (e+f x)}{4 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}}{5 (a-b)}-\frac {\cos ^5(e+f x)}{5 (a-b) \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
(-1/5*Cos[e + f*x]^5/((a - b)*(a - b + b*Sec[e + f*x]^2)^2) - ((b*(5*a^2 + 4*b^2)*Sec[e + f*x])/(4*(a - b)^3*(a - b + b*Sec[e + f*x]^2)^2) + (b*(((5 *b^(3/2)*(15*a^2 + 40*a*b + 8*b^2)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a - b)^(3/2) + (8*b*(5*a^2 + 20*a*b + 2*b^2)*Cos[e + f*x])/(a - b) - (8*(10*a - b)*b*Cos[e + f*x]^3)/3)/(2*(a - b)^3*b^2) + ((35*a^2 + 40*a*b + 24*b^2)*Sec[e + f*x])/(2*(a - b)^4*(a - b + b*Sec[e + f*x]^2))))/4)/(5*( a - b)))/f
3.1.80.3.1 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[x^m*(a + b*x^2)^(p + 1)*E xpandToSum[2*b*(p + 1)*Together[(b^(m/2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d)*x^(-m + 2))/(a + b*x^2)] - ((-a)^(m/2 - 1)*(b*c - a*d))/x^m, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ILtQ[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[c^2*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] - Simp[1/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)*(a + b*x^2)^p*Simp[2*b*c^2*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*d^2*(m + 1)*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1]
Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^ 4)^(p_.), x_Symbol] :> Simp[(-d)^(m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*((d + e*x^2)^(q + 1)/(2*e^(2*p + m/2)*(q + 1))), x] + Simp[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(2*(-d)^(-m/2 + 1)*e^(2*p)*(q + 1)*(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[1/(f*ff^ m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a - b + b*ff^2*x^2)^p/x^(m + 1 )), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[( m - 1)/2]
Time = 73.24 (sec) , antiderivative size = 282, normalized size of antiderivative = 1.07
| method | result | size |
| derivativedivides | \(\frac {-\frac {\frac {a^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a b \cos \left (f x +e \right )^{5}}{5}+\frac {b^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}+\frac {a b \cos \left (f x +e \right )^{3}}{3}+\frac {b^{2} \cos \left (f x +e \right )^{3}}{3}+a^{2} \cos \left (f x +e \right )+4 a b \cos \left (f x +e \right )+\cos \left (f x +e \right ) b^{2}}{\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {b \left (\frac {-\frac {a \left (9 a^{2}-a b -8 b^{2}\right ) \cos \left (f x +e \right )^{3}}{8}+\left (-\frac {7}{8} a^{2} b -a \,b^{2}\right ) \cos \left (f x +e \right )}{\left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )^{2}}+\frac {\left (15 a^{2}+40 a b +8 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{8 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{5}}}{f}\) | \(282\) |
| default | \(\frac {-\frac {\frac {a^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a b \cos \left (f x +e \right )^{5}}{5}+\frac {b^{2} \cos \left (f x +e \right )^{5}}{5}-\frac {2 a^{2} \cos \left (f x +e \right )^{3}}{3}+\frac {a b \cos \left (f x +e \right )^{3}}{3}+\frac {b^{2} \cos \left (f x +e \right )^{3}}{3}+a^{2} \cos \left (f x +e \right )+4 a b \cos \left (f x +e \right )+\cos \left (f x +e \right ) b^{2}}{\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a^{2}-2 a b +b^{2}\right )}+\frac {b \left (\frac {-\frac {a \left (9 a^{2}-a b -8 b^{2}\right ) \cos \left (f x +e \right )^{3}}{8}+\left (-\frac {7}{8} a^{2} b -a \,b^{2}\right ) \cos \left (f x +e \right )}{\left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )^{2}}+\frac {\left (15 a^{2}+40 a b +8 b^{2}\right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{8 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{5}}}{f}\) | \(282\) |
| risch | \(\text {Expression too large to display}\) | \(1175\) |
1/f*(-1/(a^3-3*a^2*b+3*a*b^2-b^3)/(a^2-2*a*b+b^2)*(1/5*a^2*cos(f*x+e)^5-2/ 5*a*b*cos(f*x+e)^5+1/5*b^2*cos(f*x+e)^5-2/3*a^2*cos(f*x+e)^3+1/3*a*b*cos(f *x+e)^3+1/3*b^2*cos(f*x+e)^3+a^2*cos(f*x+e)+4*a*b*cos(f*x+e)+cos(f*x+e)*b^ 2)+b/(a-b)^5*((-1/8*a*(9*a^2-a*b-8*b^2)*cos(f*x+e)^3+(-7/8*a^2*b-a*b^2)*co s(f*x+e))/(a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)^2+1/8*(15*a^2+40*a*b+8*b^2)/(b *(a-b))^(1/2)*arctan((a-b)*cos(f*x+e)/(b*(a-b))^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 490 vs. \(2 (244) = 488\).
Time = 0.40 (sec) , antiderivative size = 1018, normalized size of antiderivative = 3.86 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
[-1/240*(48*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^9 - 1 6*(10*a^4 - 31*a^3*b + 33*a^2*b^2 - 13*a*b^3 + b^4)*cos(f*x + e)^7 + 16*(1 5*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b^4)*cos(f*x + e)^5 + 50*(15* a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)^3 + 15*((15*a^4 + 10*a ^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b^4)*cos(f*x + e)^4 + 15*a^2*b^2 + 40*a*b ^3 + 8*b^4 + 2*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)^2)* sqrt(-b/(a - b))*log(-((a - b)*cos(f*x + e)^2 - 2*(a - b)*sqrt(-b/(a - b)) *cos(f*x + e) - b)/((a - b)*cos(f*x + e)^2 + b)) + 30*(15*a^2*b^2 + 40*a*b ^3 + 8*b^4)*cos(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 /120*(24*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^9 - 8*(1 0*a^4 - 31*a^3*b + 33*a^2*b^2 - 13*a*b^3 + b^4)*cos(f*x + e)^7 + 8*(15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b^4)*cos(f*x + e)^5 + 25*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)^3 + 15*((15*a^4 + 10*a^3*b - 57*a^2*b^2 + 24*a*b^3 + 8*b^4)*cos(f*x + e)^4 + 15*a^2*b^2 + 40*a*b^3 + 8*b^4 + 2*(15*a^3*b + 25*a^2*b^2 - 32*a*b^3 - 8*b^4)*cos(f*x + e)^2)*sqrt( b/(a - b))*arctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) + 15*(15*a^2*b^ 2 + 40*a*b^3 + 8*b^4)*cos(f*x + e))/((a^7 - 7*a^6*b + 21*a^5*b^2 - 35*a...
Timed out. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(b-a>0)', see `assume?` for more details)Is
Leaf count of result is larger than twice the leaf count of optimal. 834 vs. \(2 (244) = 488\).
Time = 1.04 (sec) , antiderivative size = 834, normalized size of antiderivative = 3.16 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
-1/120*(15*(15*a^2*b + 40*a*b^2 + 8*b^3)*arctan(-(a*cos(f*x + e) - b*cos(f *x + e) - b)/(sqrt(a*b - b^2)*cos(f*x + e) + sqrt(a*b - b^2)))/((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 - b^2)) + 30*(9*a ^3*b + 6*a^2*b^2 + 27*a^3*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 32*a^2 *b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 40*a*b^3*(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 - 5 4*a^2*b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 24*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 48*b^4*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 9*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 16*a^2*b^2*( cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 8*a*b^3*(cos(f*x + e) - 1)^3/(c os(f*x + e) + 1)^3)/((a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*(a + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)^2) - 16*(8*a^2 + 59*a*b + 23*b^2 - 40*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 250*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 70*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 + 320*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 140*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 - 90*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 + 45*b^2*(cos(f*x + e) - 1)^4/(cos(f...
Time = 15.63 (sec) , antiderivative size = 1536, normalized size of antiderivative = 5.82 \[ \int \frac {\sin ^5(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
(b^(1/2)*atan(((a - b)^11*(2*tan(e/2 + (f*x)/2)^2*((b^(1/2)*(40*a*b + 15*a ^2 + 8*b^2)*(640*a^3*b^12 - 128*a^2*b^13 - 240*a^14*b + 400*a^4*b^11 - 110 40*a^5*b^10 + 39120*a^6*b^9 - 73344*a^7*b^8 + 84000*a^8*b^7 - 58560*a^9*b^ 6 + 20640*a^10*b^5 + 1280*a^11*b^4 - 4528*a^12*b^3 + 1760*a^13*b^2))/(16*a *(a - b)^(21/2)) - (b^(1/2)*(a - 2*b)*(40*a*b + 15*a^2 + 8*b^2)^2*(128*a^1 8 - 2176*a^17*b + 256*a^2*b^16 - 3968*a^3*b^15 + 28800*a^4*b^14 - 129920*a ^5*b^13 + 407680*a^6*b^12 - 943488*a^7*b^11 + 1665664*a^8*b^10 - 2288000*a ^9*b^9 + 2471040*a^10*b^8 - 2104960*a^11*b^7 + 1409408*a^12*b^6 - 733824*a ^13*b^5 + 291200*a^14*b^4 - 85120*a^15*b^3 + 17280*a^16*b^2))/(512*a*(a - b)^(33/2))) - (b^(1/2)*(a - 2*b)*(40*a*b + 15*a^2 + 8*b^2)^2*(1920*a^17*b - 128*a^18 + 128*a^3*b^15 - 1920*a^4*b^14 + 13440*a^5*b^13 - 58240*a^6*b^1 2 + 174720*a^7*b^11 - 384384*a^8*b^10 + 640640*a^9*b^9 - 823680*a^10*b^8 + 823680*a^11*b^7 - 640640*a^12*b^6 + 384384*a^13*b^5 - 174720*a^14*b^4 + 5 8240*a^15*b^3 - 13440*a^16*b^2))/(256*a*(a - b)^(33/2))))/(225*a^16*b + 64 *a^2*b^15 - 1680*a^4*b^13 + 3920*a^5*b^12 + 7665*a^6*b^11 - 50778*a^7*b^10 + 104685*a^8*b^9 - 111960*a^9*b^8 + 57330*a^10*b^7 + 2660*a^11*b^6 - 2028 6*a^12*b^5 + 9240*a^13*b^4 - 35*a^14*b^3 - 1050*a^15*b^2))*(40*a*b + 15*a^ 2 + 8*b^2))/(8*f*(a - b)^(11/2)) - ((607*a^3*b + 64*a^4 + 274*a^2*b^2)/(60 *(a - b)*(a^4 - 4*a^3*b - 4*a*b^3 + b^4 + 6*a^2*b^2)) + (tan(e/2 + (f*x)/2 )^14*(128*a*b^3 + 15*a^3*b + 24*b^4 + 85*a^2*b^2))/(2*(a - b)*(a^4 - 4*...