Integrand size = 23, antiderivative size = 180 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=-\frac {5 \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{8 (a-b)^{9/2} f}-\frac {(a+2 b) \cos (e+f x)}{(a-b)^4 f}+\frac {\cos ^3(e+f x)}{3 (a-b)^3 f}-\frac {a b \sec (e+f x)}{4 (a-b)^3 f \left (a-b+b \sec ^2(e+f x)\right )^2}-\frac {b (7 a+4 b) \sec (e+f x)}{8 (a-b)^4 f \left (a-b+b \sec ^2(e+f x)\right )} \]
-(a+2*b)*cos(f*x+e)/(a-b)^4/f+1/3*cos(f*x+e)^3/(a-b)^3/f-1/4*a*b*sec(f*x+e )/(a-b)^3/f/(a-b+b*sec(f*x+e)^2)^2-1/8*b*(7*a+4*b)*sec(f*x+e)/(a-b)^4/f/(a -b+b*sec(f*x+e)^2)-5/8*(3*a+4*b)*arctan(sec(f*x+e)*b^(1/2)/(a-b)^(1/2))*b^ (1/2)/(a-b)^(9/2)/f
Time = 6.71 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.28 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {\frac {15 \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {a-b}-\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{9/2}}+\frac {15 \sqrt {b} (3 a+4 b) \arctan \left (\frac {\sqrt {a-b}+\sqrt {a} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right )}{(a-b)^{9/2}}+\frac {2 \left (3 \cos (e+f x) \left (a \left (-3+\frac {4 b^2}{(a+b+(a-b) \cos (2 (e+f x)))^2}-\frac {9 b}{a+b+(a-b) \cos (2 (e+f x))}\right )+b \left (-9-\frac {4 b}{a+b+(a-b) \cos (2 (e+f x))}\right )\right )+(a-b) \cos (3 (e+f x))\right )}{(a-b)^4}}{24 f} \]
((15*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[a - b] - Sqrt[a]*Tan[(e + f*x)/2])/S qrt[b]])/(a - b)^(9/2) + (15*Sqrt[b]*(3*a + 4*b)*ArcTan[(Sqrt[a - b] + Sqr t[a]*Tan[(e + f*x)/2])/Sqrt[b]])/(a - b)^(9/2) + (2*(3*Cos[e + f*x]*(a*(-3 + (4*b^2)/(a + b + (a - b)*Cos[2*(e + f*x)])^2 - (9*b)/(a + b + (a - b)*C os[2*(e + f*x)])) + b*(-9 - (4*b)/(a + b + (a - b)*Cos[2*(e + f*x)]))) + ( a - b)*Cos[3*(e + f*x)]))/(a - b)^4)/(24*f)
Time = 0.48 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.01, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 4147, 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.
| \(\displaystyle \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^3}{\left (a+b \tan (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4147 |
| \(\displaystyle \frac {\int -\frac {\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a-b\right )^3}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a-b\right )^3}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 361 |
| \(\displaystyle \frac {\frac {1}{4} b \int -\frac {\cos ^4(e+f x) \left (\frac {3 a \sec ^4(e+f x)}{(a-b)^3}-\frac {4 a \sec ^2(e+f x)}{(a-b)^2 b}+\frac {4}{(a-b) b}\right )}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)-\frac {a b \sec (e+f x)}{4 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {-\frac {1}{4} b \int \frac {\cos ^4(e+f x) \left (\frac {3 a \sec ^4(e+f x)}{(a-b)^3}-\frac {4 a \sec ^2(e+f x)}{(a-b)^2 b}+\frac {4}{(a-b) b}\right )}{\left (b \sec ^2(e+f x)+a-b\right )^2}d\sec (e+f x)-\frac {a b \sec (e+f x)}{4 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {-\frac {1}{4} b \left (\frac {\int \frac {\cos ^4(e+f x) \left (\frac {b^2 (7 a+4 b) \sec ^4(e+f x)}{a-b}-8 b (a+b) \sec ^2(e+f x)+8 (a-b) b\right )}{b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{2 b^2 (a-b)^3}+\frac {(7 a+4 b) \sec (e+f x)}{2 (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}\right )-\frac {a b \sec (e+f x)}{4 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {-\frac {1}{4} b \left (\frac {\int \left (8 b \cos ^4(e+f x)-\frac {8 b (a+2 b) \cos ^2(e+f x)}{a-b}+\frac {5 b^2 (3 a+4 b)}{(a-b) \left (b \sec ^2(e+f x)+a-b\right )}\right )d\sec (e+f x)}{2 b^2 (a-b)^3}+\frac {(7 a+4 b) \sec (e+f x)}{2 (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}\right )-\frac {a b \sec (e+f x)}{4 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {1}{4} b \left (\frac {\frac {5 b^{3/2} (3 a+4 b) \arctan \left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b}}\right )}{(a-b)^{3/2}}+\frac {8 b (a+2 b) \cos (e+f x)}{a-b}-\frac {8}{3} b \cos ^3(e+f x)}{2 b^2 (a-b)^3}+\frac {(7 a+4 b) \sec (e+f x)}{2 (a-b)^4 \left (a+b \sec ^2(e+f x)-b\right )}\right )-\frac {a b \sec (e+f x)}{4 (a-b)^3 \left (a+b \sec ^2(e+f x)-b\right )^2}}{f}\) |
(-1/4*(a*b*Sec[e + f*x])/((a - b)^3*(a - b + b*Sec[e + f*x]^2)^2) - (b*((( 5*b^(3/2)*(3*a + 4*b)*ArcTan[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b]])/(a - b)^ (3/2) + (8*b*(a + 2*b)*Cos[e + f*x])/(a - b) - (8*b*Cos[e + f*x]^3)/3)/(2* (a - b)^3*b^2) + ((7*a + 4*b)*Sec[e + f*x])/(2*(a - b)^4*(a - b + b*Sec[e + f*x]^2))))/4)/f
3.1.81.3.1 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[(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 = 23.30 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\frac {b \cos \left (f x +e \right )^{3}}{3}-\cos \left (f x +e \right ) a -2 b \cos \left (f x +e \right )}{\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a -b \right )}+\frac {b \left (\frac {\left (-\frac {9}{8} a^{2}+\frac {5}{8} a b +\frac {1}{2} b^{2}\right ) \cos \left (f x +e \right )^{3}+\left (-\frac {7}{8} a b -\frac {1}{2} b^{2}\right ) \cos \left (f x +e \right )}{\left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )^{2}}+\frac {5 \left (3 a +4 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{8 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{4}}}{f}\) | \(196\) |
| default | \(\frac {\frac {\frac {a \cos \left (f x +e \right )^{3}}{3}-\frac {b \cos \left (f x +e \right )^{3}}{3}-\cos \left (f x +e \right ) a -2 b \cos \left (f x +e \right )}{\left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) \left (a -b \right )}+\frac {b \left (\frac {\left (-\frac {9}{8} a^{2}+\frac {5}{8} a b +\frac {1}{2} b^{2}\right ) \cos \left (f x +e \right )^{3}+\left (-\frac {7}{8} a b -\frac {1}{2} b^{2}\right ) \cos \left (f x +e \right )}{\left (a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b \right )^{2}}+\frac {5 \left (3 a +4 b \right ) \arctan \left (\frac {\left (a -b \right ) \cos \left (f x +e \right )}{\sqrt {b \left (a -b \right )}}\right )}{8 \sqrt {b \left (a -b \right )}}\right )}{\left (a -b \right )^{4}}}{f}\) | \(196\) |
| risch | \(\frac {{\mathrm e}^{3 i \left (f x +e \right )}}{24 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}-\frac {3 \,{\mathrm e}^{i \left (f x +e \right )} a}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f \left (a -b \right )}-\frac {9 \,{\mathrm e}^{i \left (f x +e \right )} b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f \left (a -b \right )}-\frac {3 \,{\mathrm e}^{-i \left (f x +e \right )} a}{8 \left (a^{4}-4 a^{3} b +6 a^{2} b^{2}-4 a \,b^{3}+b^{4}\right ) f}-\frac {9 \,{\mathrm e}^{-i \left (f x +e \right )} b}{8 \left (a^{4}-4 a^{3} b +6 a^{2} b^{2}-4 a \,b^{3}+b^{4}\right ) f}+\frac {{\mathrm e}^{-3 i \left (f x +e \right )}}{24 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) f}+\frac {b \left (-9 a^{2} {\mathrm e}^{7 i \left (f x +e \right )}+5 a b \,{\mathrm e}^{7 i \left (f x +e \right )}+4 b^{2} {\mathrm e}^{7 i \left (f x +e \right )}-27 a^{2} {\mathrm e}^{5 i \left (f x +e \right )}-13 a b \,{\mathrm e}^{5 i \left (f x +e \right )}-4 b^{2} {\mathrm e}^{5 i \left (f x +e \right )}-27 a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-13 a b \,{\mathrm e}^{3 i \left (f x +e \right )}-4 b^{2} {\mathrm e}^{3 i \left (f x +e \right )}-9 a^{2} {\mathrm e}^{i \left (f x +e \right )}+5 a b \,{\mathrm e}^{i \left (f x +e \right )}+4 b^{2} {\mathrm e}^{i \left (f x +e \right )}\right )}{4 \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^{3}+3 a^{2} b -3 a \,b^{2}+b^{3}\right ) f \left (-a +b \right )}+\frac {15 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) a}{16 \left (a -b \right )^{5} f}+\frac {5 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{4 \left (a -b \right )^{5} f}-\frac {15 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) a}{16 \left (a -b \right )^{5} f}-\frac {5 i \sqrt {b \left (a -b \right )}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {b \left (a -b \right )}\, {\mathrm e}^{i \left (f x +e \right )}}{a -b}+1\right ) b}{4 \left (a -b \right )^{5} f}\) | \(761\) |
1/f*(1/(a^3-3*a^2*b+3*a*b^2-b^3)/(a-b)*(1/3*a*cos(f*x+e)^3-1/3*b*cos(f*x+e )^3-cos(f*x+e)*a-2*b*cos(f*x+e))+b/(a-b)^4*(((-9/8*a^2+5/8*a*b+1/2*b^2)*co s(f*x+e)^3+(-7/8*a*b-1/2*b^2)*cos(f*x+e))/(a*cos(f*x+e)^2-b*cos(f*x+e)^2+b )^2+5/8*(3*a+4*b)/(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. 369 vs. \(2 (164) = 328\).
Time = 0.40 (sec) , antiderivative size = 775, normalized size of antiderivative = 4.31 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\left [\frac {16 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 16 \, {\left (3 \, a^{3} - 2 \, a^{2} b - 5 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{5} - 50 \, {\left (3 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 15 \, {\left ({\left (3 \, a^{3} - 2 \, a^{2} b - 5 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 3 \, a b^{2} + 4 \, b^{3} + 2 \, {\left (3 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a - b}} \log \left (\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, {\left (a - b\right )} \sqrt {-\frac {b}{a - b}} \cos \left (f x + e\right ) - b}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) - 30 \, {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )}{48 \, {\left ({\left (a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{5} b - 5 \, a^{4} b^{2} + 10 \, a^{3} b^{3} - 10 \, a^{2} b^{4} + 5 \, a b^{5} - b^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{2} - 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} - 4 \, a b^{5} + b^{6}\right )} f\right )}}, \frac {8 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 8 \, {\left (3 \, a^{3} - 2 \, a^{2} b - 5 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{5} - 25 \, {\left (3 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left ({\left (3 \, a^{3} - 2 \, a^{2} b - 5 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{4} + 3 \, a b^{2} + 4 \, b^{3} + 2 \, {\left (3 \, a^{2} b + a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a - b}} \arctan \left (-\frac {{\left (a - b\right )} \sqrt {\frac {b}{a - b}} \cos \left (f x + e\right )}{b}\right ) - 15 \, {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )}{24 \, {\left ({\left (a^{6} - 6 \, a^{5} b + 15 \, a^{4} b^{2} - 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} - 6 \, a b^{5} + b^{6}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{5} b - 5 \, a^{4} b^{2} + 10 \, a^{3} b^{3} - 10 \, a^{2} b^{4} + 5 \, a b^{5} - b^{6}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{4} b^{2} - 4 \, a^{3} b^{3} + 6 \, a^{2} b^{4} - 4 \, a b^{5} + b^{6}\right )} f\right )}}\right ] \]
[1/48*(16*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 16*(3*a^3 - 2*a ^2*b - 5*a*b^2 + 4*b^3)*cos(f*x + e)^5 - 50*(3*a^2*b + a*b^2 - 4*b^3)*cos( f*x + e)^3 + 15*((3*a^3 - 2*a^2*b - 5*a*b^2 + 4*b^3)*cos(f*x + e)^4 + 3*a* b^2 + 4*b^3 + 2*(3*a^2*b + a*b^2 - 4*b^3)*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*(3*a*b^2 + 4*b^3)*cos(f*x + e))/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*f*cos(f *x + e)^4 + 2*(a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6 )*f*cos(f*x + e)^2 + (a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a*b^5 + b^6)*f), 1/24*(8*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 8*(3*a^3 - 2*a^2 *b - 5*a*b^2 + 4*b^3)*cos(f*x + e)^5 - 25*(3*a^2*b + a*b^2 - 4*b^3)*cos(f* x + e)^3 - 15*((3*a^3 - 2*a^2*b - 5*a*b^2 + 4*b^3)*cos(f*x + e)^4 + 3*a*b^ 2 + 4*b^3 + 2*(3*a^2*b + a*b^2 - 4*b^3)*cos(f*x + e)^2)*sqrt(b/(a - b))*ar ctan(-(a - b)*sqrt(b/(a - b))*cos(f*x + e)/b) - 15*(3*a*b^2 + 4*b^3)*cos(f *x + e))/((a^6 - 6*a^5*b + 15*a^4*b^2 - 20*a^3*b^3 + 15*a^2*b^4 - 6*a*b^5 + b^6)*f*cos(f*x + e)^4 + 2*(a^5*b - 5*a^4*b^2 + 10*a^3*b^3 - 10*a^2*b^4 + 5*a*b^5 - b^6)*f*cos(f*x + e)^2 + (a^4*b^2 - 4*a^3*b^3 + 6*a^2*b^4 - 4*a* b^5 + b^6)*f)]
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Exception raised: ValueError} \]
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. 541 vs. \(2 (164) = 328\).
Time = 1.00 (sec) , antiderivative size = 541, normalized size of antiderivative = 3.01 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\frac {a^{6} f^{17} \cos \left (f x + e\right )^{3} - 6 \, a^{5} b f^{17} \cos \left (f x + e\right )^{3} + 15 \, a^{4} b^{2} f^{17} \cos \left (f x + e\right )^{3} - 20 \, a^{3} b^{3} f^{17} \cos \left (f x + e\right )^{3} + 15 \, a^{2} b^{4} f^{17} \cos \left (f x + e\right )^{3} - 6 \, a b^{5} f^{17} \cos \left (f x + e\right )^{3} + b^{6} f^{17} \cos \left (f x + e\right )^{3} - 3 \, a^{6} f^{17} \cos \left (f x + e\right ) + 9 \, a^{5} b f^{17} \cos \left (f x + e\right ) - 30 \, a^{3} b^{3} f^{17} \cos \left (f x + e\right ) + 45 \, a^{2} b^{4} f^{17} \cos \left (f x + e\right ) - 27 \, a b^{5} f^{17} \cos \left (f x + e\right ) + 6 \, b^{6} f^{17} \cos \left (f x + e\right )}{3 \, {\left (a^{9} f^{18} - 9 \, a^{8} b f^{18} + 36 \, a^{7} b^{2} f^{18} - 84 \, a^{6} b^{3} f^{18} + 126 \, a^{5} b^{4} f^{18} - 126 \, a^{4} b^{5} f^{18} + 84 \, a^{3} b^{6} f^{18} - 36 \, a^{2} b^{7} f^{18} + 9 \, a b^{8} f^{18} - b^{9} f^{18}\right )}} + \frac {5 \, {\left (3 \, a b + 4 \, b^{2}\right )} \arctan \left (\frac {a \cos \left (f x + e\right ) - b \cos \left (f x + e\right )}{\sqrt {a b - b^{2}}}\right )}{8 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \sqrt {a b - b^{2}} f} - \frac {\frac {9 \, a^{2} b \cos \left (f x + e\right )^{3}}{f} - \frac {5 \, a b^{2} \cos \left (f x + e\right )^{3}}{f} - \frac {4 \, b^{3} \cos \left (f x + e\right )^{3}}{f} + \frac {7 \, a b^{2} \cos \left (f x + e\right )}{f} + \frac {4 \, b^{3} \cos \left (f x + e\right )}{f}}{8 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} {\left (a \cos \left (f x + e\right )^{2} - b \cos \left (f x + e\right )^{2} + b\right )}^{2}} \]
1/3*(a^6*f^17*cos(f*x + e)^3 - 6*a^5*b*f^17*cos(f*x + e)^3 + 15*a^4*b^2*f^ 17*cos(f*x + e)^3 - 20*a^3*b^3*f^17*cos(f*x + e)^3 + 15*a^2*b^4*f^17*cos(f *x + e)^3 - 6*a*b^5*f^17*cos(f*x + e)^3 + b^6*f^17*cos(f*x + e)^3 - 3*a^6* f^17*cos(f*x + e) + 9*a^5*b*f^17*cos(f*x + e) - 30*a^3*b^3*f^17*cos(f*x + e) + 45*a^2*b^4*f^17*cos(f*x + e) - 27*a*b^5*f^17*cos(f*x + e) + 6*b^6*f^1 7*cos(f*x + e))/(a^9*f^18 - 9*a^8*b*f^18 + 36*a^7*b^2*f^18 - 84*a^6*b^3*f^ 18 + 126*a^5*b^4*f^18 - 126*a^4*b^5*f^18 + 84*a^3*b^6*f^18 - 36*a^2*b^7*f^ 18 + 9*a*b^8*f^18 - b^9*f^18) + 5/8*(3*a*b + 4*b^2)*arctan((a*cos(f*x + e) - b*cos(f*x + e))/sqrt(a*b - b^2))/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*sqrt(a*b - b^2)*f) - 1/8*(9*a^2*b*cos(f*x + e)^3/f - 5*a*b^2*cos(f* x + e)^3/f - 4*b^3*cos(f*x + e)^3/f + 7*a*b^2*cos(f*x + e)/f + 4*b^3*cos(f *x + e)/f)/((a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*(a*cos(f*x + e)^2 - b*cos(f*x + e)^2 + b)^2)
Time = 14.78 (sec) , antiderivative size = 1154, normalized size of antiderivative = 6.41 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \tan ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
- ((6*a*b^2 + 83*a^2*b + 16*a^3)/(12*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^ 3)) + (tan(e/2 + (f*x)/2)^2*(299*a*b^2 - 8*a^3 + 24*b^3))/(6*(a - b)*(3*a* b^2 - 3*a^2*b + a^3 - b^3)) + (5*a*tan(e/2 + (f*x)/2)^12*(3*a*b + 4*b^2))/ (4*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (tan(e/2 + (f*x)/2)^10*(28*a *b^3 - 32*a^3*b + 8*a^4 + 8*b^4 + 93*a^2*b^2))/(2*a*(a - b)*(3*a*b^2 - 3*a ^2*b + a^3 - b^3)) + (tan(e/2 + (f*x)/2)^6*(546*a*b^3 - 144*a^3*b + 56*a^4 + 36*b^4 + 31*a^2*b^2))/(3*a*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + ( tan(e/2 + (f*x)/2)^4*(1208*a*b^3 + 71*a^3*b - 96*a^4 + 48*b^4 + 344*a^2*b^ 2))/(12*a*(a - b)*(3*a*b^2 - 3*a^2*b + a^3 - b^3)) + (tan(e/2 + (f*x)/2)^8 *(1704*a*b^3 + 569*a^3*b - 176*a^4 + 144*b^4 - 666*a^2*b^2))/(12*a*(a - b) *(3*a*b^2 - 3*a^2*b + a^3 - b^3)))/(f*(tan(e/2 + (f*x)/2)^2*(8*a*b - a^2) + tan(e/2 + (f*x)/2)^12*(8*a*b - a^2) + a^2*tan(e/2 + (f*x)/2)^14 + tan(e/ 2 + (f*x)/2)^4*(8*a*b - 3*a^2 + 16*b^2) + tan(e/2 + (f*x)/2)^10*(8*a*b - 3 *a^2 + 16*b^2) + tan(e/2 + (f*x)/2)^6*(3*a^2 - 16*a*b + 48*b^2) + tan(e/2 + (f*x)/2)^8*(3*a^2 - 16*a*b + 48*b^2) + a^2)) - (5*b^(1/2)*atan((2*(tan(e /2 + (f*x)/2)^2*((5*b^(1/2)*(3*a + 4*b)*(240*a^11*b + 320*a^2*b^10 - 2320* a^3*b^9 + 7040*a^4*b^8 - 11200*a^5*b^7 + 8960*a^6*b^6 - 1120*a^7*b^5 - 448 0*a^8*b^4 + 4160*a^9*b^3 - 1600*a^10*b^2))/(16*a*(a - b)^(17/2)) - (25*b^( 1/2)*(a - 2*b)*(3*a + 4*b)^2*(1792*a^14*b - 128*a^15 + 256*a^2*b^13 - 3200 *a^3*b^12 + 18432*a^4*b^11 - 64768*a^5*b^10 + 154880*a^6*b^9 - 266112*a...