Integrand size = 25, antiderivative size = 114 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(3 a+4 b) \cos (e+f x)}{3 a^2 f \sqrt {a+b \sec ^2(e+f x)}}+\frac {\cos ^3(e+f x)}{3 a f \sqrt {a+b \sec ^2(e+f x)}}-\frac {2 b (3 a+4 b) \sec (e+f x)}{3 a^3 f \sqrt {a+b \sec ^2(e+f x)}} \]
-1/3*(3*a+4*b)*cos(f*x+e)/a^2/f/(a+b*sec(f*x+e)^2)^(1/2)+1/3*cos(f*x+e)^3/ a/f/(a+b*sec(f*x+e)^2)^(1/2)-2/3*b*(3*a+4*b)*sec(f*x+e)/a^3/f/(a+b*sec(f*x +e)^2)^(1/2)
Time = 2.61 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.82 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (9 a^2+64 a b+64 b^2+8 a (a+2 b) \cos (2 (e+f x))-a^2 \cos (4 (e+f x))\right ) \sec ^3(e+f x)}{48 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]
-1/48*((a + 2*b + a*Cos[2*(e + f*x)])*(9*a^2 + 64*a*b + 64*b^2 + 8*a*(a + 2*b)*Cos[2*(e + f*x)] - a^2*Cos[4*(e + f*x)])*Sec[e + f*x]^3)/(a^3*f*(a + b*Sec[e + f*x]^2)^(3/2))
Time = 0.29 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 4622, 25, 359, 245, 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.
\(\displaystyle \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)^3}{\left (a+b \sec (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4622 |
\(\displaystyle \frac {\int -\frac {\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right )}{\left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{f}\) |
\(\Big \downarrow \) 359 |
\(\displaystyle \frac {\frac {(3 a+4 b) \int \frac {\cos ^2(e+f x)}{\left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{3 a}+\frac {\cos ^3(e+f x)}{3 a \sqrt {a+b \sec ^2(e+f x)}}}{f}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {\frac {(3 a+4 b) \left (-\frac {2 b \int \frac {1}{\left (b \sec ^2(e+f x)+a\right )^{3/2}}d\sec (e+f x)}{a}-\frac {\cos (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)}}\right )}{3 a}+\frac {\cos ^3(e+f x)}{3 a \sqrt {a+b \sec ^2(e+f x)}}}{f}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {(3 a+4 b) \left (-\frac {2 b \sec (e+f x)}{a^2 \sqrt {a+b \sec ^2(e+f x)}}-\frac {\cos (e+f x)}{a \sqrt {a+b \sec ^2(e+f x)}}\right )}{3 a}+\frac {\cos ^3(e+f x)}{3 a \sqrt {a+b \sec ^2(e+f x)}}}{f}\) |
(Cos[e + f*x]^3/(3*a*Sqrt[a + b*Sec[e + f*x]^2]) + ((3*a + 4*b)*(-(Cos[e + f*x]/(a*Sqrt[a + b*Sec[e + f*x]^2])) - (2*b*Sec[e + f*x])/(a^2*Sqrt[a + b *Sec[e + f*x]^2])))/(3*a))/f
3.2.7.3.1 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[(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 = 1.46 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.01
method | result | size |
default | \(\frac {a \left (a +b \right )^{4} \left (b +a \cos \left (f x +e \right )^{2}\right ) \left (\cos \left (f x +e \right )^{4} a^{2}-3 \cos \left (f x +e \right )^{2} a^{2}-4 \cos \left (f x +e \right )^{2} a b -6 a b -8 b^{2}\right ) \sec \left (f x +e \right )^{3}}{3 f \left (\sqrt {-a b}-a \right )^{4} \left (\sqrt {-a b}+a \right )^{4} \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}\) | \(115\) |
1/3/f*a/((-a*b)^(1/2)-a)^4/((-a*b)^(1/2)+a)^4*(a+b)^4*(b+a*cos(f*x+e)^2)*( cos(f*x+e)^4*a^2-3*cos(f*x+e)^2*a^2-4*cos(f*x+e)^2*a*b-6*a*b-8*b^2)/(a+b*s ec(f*x+e)^2)^(3/2)*sec(f*x+e)^3
Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.86 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\frac {{\left (a^{2} \cos \left (f x + e\right )^{5} - {\left (3 \, a^{2} + 4 \, a b\right )} \cos \left (f x + e\right )^{3} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a^{4} f \cos \left (f x + e\right )^{2} + a^{3} b f\right )}} \]
1/3*(a^2*cos(f*x + e)^5 - (3*a^2 + 4*a*b)*cos(f*x + e)^3 - 2*(3*a*b + 4*b^ 2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(a^4*f*cos(f* x + e)^2 + a^3*b*f)
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]
Time = 0.21 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.24 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {3 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{2}} - \frac {{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 6 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )}{a^{3}} + \frac {3 \, b}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a^{2} \cos \left (f x + e\right )} + \frac {3 \, b^{2}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a^{3} \cos \left (f x + e\right )}}{3 \, f} \]
-1/3*(3*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a^2 - ((a + b/cos(f*x + e) ^2)^(3/2)*cos(f*x + e)^3 - 6*sqrt(a + b/cos(f*x + e)^2)*b*cos(f*x + e))/a^ 3 + 3*b/(sqrt(a + b/cos(f*x + e)^2)*a^2*cos(f*x + e)) + 3*b^2/(sqrt(a + b/ cos(f*x + e)^2)*a^3*cos(f*x + e)))/f
\[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3}{{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}} \,d x \]