Integrand size = 25, antiderivative size = 146 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b) \cos (e+f x)}{a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\cos ^3(e+f x)}{3 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b (a+2 b) \sec (e+f x)}{3 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b (a+2 b) \sec (e+f x)}{3 a^4 f \sqrt {a+b \sec ^2(e+f x)}} \]
-(a+2*b)*cos(f*x+e)/a^2/f/(a+b*sec(f*x+e)^2)^(3/2)+1/3*cos(f*x+e)^3/a/f/(a +b*sec(f*x+e)^2)^(3/2)-4/3*b*(a+2*b)*sec(f*x+e)/a^3/f/(a+b*sec(f*x+e)^2)^( 3/2)-8/3*b*(a+2*b)*sec(f*x+e)/a^4/f/(a+b*sec(f*x+e)^2)^(1/2)
Time = 2.19 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.88 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (26 a^3+264 a^2 b+640 a b^2+512 b^3+3 a \left (11 a^2+96 a b+128 b^2\right ) \cos (2 (e+f x))+6 a^2 (a+4 b) \cos (4 (e+f x))-a^3 \cos (6 (e+f x))\right ) \sec ^5(e+f x)}{192 a^4 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]
-1/192*((a + 2*b + a*Cos[2*(e + f*x)])*(26*a^3 + 264*a^2*b + 640*a*b^2 + 5 12*b^3 + 3*a*(11*a^2 + 96*a*b + 128*b^2)*Cos[2*(e + f*x)] + 6*a^2*(a + 4*b )*Cos[4*(e + f*x)] - a^3*Cos[6*(e + f*x)])*Sec[e + f*x]^5)/(a^4*f*(a + b*S ec[e + f*x]^2)^(5/2))
Time = 0.30 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4622, 25, 359, 245, 209, 208}
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.
| \(\displaystyle \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^3}{\left (a+b \sec (e+f x)^2\right )^{5/2}}dx\) |
| \(\Big \downarrow \) 4622 |
| \(\displaystyle \frac {\int -\frac {\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\int \frac {\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{f}\) |
| \(\Big \downarrow \) 359 |
| \(\displaystyle \frac {\frac {(a+2 b) \int \frac {\cos ^2(e+f x)}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{a}+\frac {\cos ^3(e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 245 |
| \(\displaystyle \frac {\frac {(a+2 b) \left (-\frac {4 b \int \frac {1}{\left (b \sec ^2(e+f x)+a\right )^{5/2}}d\sec (e+f x)}{a}-\frac {\cos (e+f x)}{a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}+\frac {\cos ^3(e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 209 |
| \(\displaystyle \frac {\frac {(a+2 b) \left (-\frac {4 b \left (\frac {2 \int \frac {1}{\left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{3 a}+\frac {\sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cos (e+f x)}{a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}+\frac {\cos ^3(e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\) |
| \(\Big \downarrow \) 208 |
| \(\displaystyle \frac {\frac {(a+2 b) \left (-\frac {4 b \left (\frac {2 \sec (e+f x)}{3 a^2 \sqrt {a+b \sec ^2(e+f x)}}+\frac {\sec (e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}-\frac {\cos (e+f x)}{a \left (a+b \sec ^2(e+f x)\right )^{3/2}}\right )}{a}+\frac {\cos ^3(e+f x)}{3 a \left (a+b \sec ^2(e+f x)\right )^{3/2}}}{f}\) |
(Cos[e + f*x]^3/(3*a*(a + b*Sec[e + f*x]^2)^(3/2)) + ((a + 2*b)*(-(Cos[e + f*x]/(a*(a + b*Sec[e + f*x]^2)^(3/2))) - (4*b*(Sec[e + f*x]/(3*a*(a + b*S ec[e + f*x]^2)^(3/2)) + (2*Sec[e + f*x])/(3*a^2*Sqrt[a + b*Sec[e + f*x]^2] )))/a))/a)/f
3.2.20.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(a*e^2*(m + 1)) Int[(e*x)^(m + 2)* (a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !ILtQ[p, -1]
Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si mp[1/(f*ff^m) Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p /x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
Time = 1.53 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.01
| method | result | size |
| default | \(-\frac {a \left (b +a \cos \left (f x +e \right )^{2}\right ) \left (a +b \right )^{5} \left (a^{3} \cos \left (f x +e \right )^{6}-3 \cos \left (f x +e \right )^{4} a^{3}-6 \cos \left (f x +e \right )^{4} a^{2} b -12 \cos \left (f x +e \right )^{2} a^{2} b -24 \cos \left (f x +e \right )^{2} a \,b^{2}-8 a \,b^{2}-16 b^{3}\right ) \sec \left (f x +e \right )^{5}}{3 f \left (\sqrt {-a b}+a \right )^{5} \left (\sqrt {-a b}-a \right )^{5} \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}\) | \(147\) |
-1/3/f*a/((-a*b)^(1/2)+a)^5/((-a*b)^(1/2)-a)^5*(b+a*cos(f*x+e)^2)*(a+b)^5* (a^3*cos(f*x+e)^6-3*cos(f*x+e)^4*a^3-6*cos(f*x+e)^4*a^2*b-12*cos(f*x+e)^2* a^2*b-24*cos(f*x+e)^2*a*b^2-8*a*b^2-16*b^3)/(a+b*sec(f*x+e)^2)^(5/2)*sec(f *x+e)^5
Time = 0.38 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.95 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\frac {{\left (a^{3} \cos \left (f x + e\right )^{7} - 3 \, {\left (a^{3} + 2 \, a^{2} b\right )} \cos \left (f x + e\right )^{5} - 12 \, {\left (a^{2} b + 2 \, a b^{2}\right )} \cos \left (f x + e\right )^{3} - 8 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a^{6} f \cos \left (f x + e\right )^{4} + 2 \, a^{5} b f \cos \left (f x + e\right )^{2} + a^{4} b^{2} f\right )}} \]
1/3*(a^3*cos(f*x + e)^7 - 3*(a^3 + 2*a^2*b)*cos(f*x + e)^5 - 12*(a^2*b + 2 *a*b^2)*cos(f*x + e)^3 - 8*(a*b^2 + 2*b^3)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(a^6*f*cos(f*x + e)^4 + 2*a^5*b*f*cos(f*x + e)^ 2 + a^4*b^2*f)
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.34 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {3 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3}} - \frac {{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{4}} + \frac {6 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{3} \cos \left (f x + e\right )^{3}} + \frac {9 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{4} \cos \left (f x + e\right )^{3}}}{3 \, f} \]
-1/3*(3*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a^3 - ((a + b/cos(f*x + e) ^2)^(3/2)*cos(f*x + e)^3 - 9*sqrt(a + b/cos(f*x + e)^2)*b*cos(f*x + e))/a^ 4 + (6*(a + b/cos(f*x + e)^2)*b*cos(f*x + e)^2 - b^2)/((a + b/cos(f*x + e) ^2)^(3/2)*a^3*cos(f*x + e)^3) + (9*(a + b/cos(f*x + e)^2)*b^2*cos(f*x + e) ^2 - b^3)/((a + b/cos(f*x + e)^2)^(3/2)*a^4*cos(f*x + e)^3))/f
\[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{5/2}} \,d x \]