Integrand size = 25, antiderivative size = 83 \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {(a-b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 b^{3/2} f}-\frac {\cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{2 b f} \]
1/2*(a-b)*arctan(cos(f*x+e)*b^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/b^(3/2)/f- 1/2*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/b/f
Time = 0.37 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.27 \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\cos (e+f x) \sqrt {2 a+b-b \cos (2 (e+f x))}}{2 \sqrt {2} b f}+\frac {(a-b) \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{2 \sqrt {-b} b f} \]
-1/2*(Cos[e + f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])/(Sqrt[2]*b*f) + ((a - b)*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x) ]]])/(2*Sqrt[-b]*b*f)
Time = 0.27 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3665, 299, 224, 216}
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.
\(\displaystyle \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)^3}{\sqrt {a+b \sin (e+f x)^2}}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\frac {\int \frac {1-\cos ^2(e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle -\frac {\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 b}-\frac {(a-b) \int \frac {1}{\sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{2 b}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 b}-\frac {(a-b) \int \frac {1}{\frac {b \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}+1}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}}{2 b}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {\cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{2 b}-\frac {(a-b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 b^{3/2}}}{f}\) |
-((-1/2*((a - b)*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x] ^2]])/b^(3/2) + (Cos[e + f*x]*Sqrt[a + b - b*Cos[e + f*x]^2])/(2*b))/f)
3.2.44.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Leaf count of result is larger than twice the leaf count of optimal. \(147\) vs. \(2(71)=142\).
Time = 0.91 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.78
method | result | size |
default | \(\frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (-\frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{2 b}+\frac {\left (-a +b \right ) \arctan \left (\frac {\sqrt {b}\, \left (\sin ^{2}\left (f x +e \right )-\frac {-a +b}{2 b}\right )}{\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{4 b^{\frac {3}{2}}}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(148\) |
(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)*(-1/2/b*(-(-b*sin(f*x+e)^2-a)*co s(f*x+e)^2)^(1/2)+1/4*(-a+b)/b^(3/2)*arctan(b^(1/2)*(sin(f*x+e)^2-1/2*(-a+ b)/b)/(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)))/cos(f*x+e)/(a+b*sin(f*x+ e)^2)^(1/2)/f
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (71) = 142\).
Time = 0.36 (sec) , antiderivative size = 438, normalized size of antiderivative = 5.28 \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\left [-\frac {8 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right ) - {\left (a - b\right )} \sqrt {-b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{6} + 160 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 32 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}\right )}{16 \, b^{2} f}, -\frac {{\left (a - b\right )} \sqrt {b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{5} - 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )}}\right ) + 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right )}{8 \, b^{2} f}\right ] \]
[-1/16*(8*sqrt(-b*cos(f*x + e)^2 + a + b)*b*cos(f*x + e) - (a - b)*sqrt(-b )*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + b^4)*cos(f*x + e)^6 + 160*(a^2 *b^2 + 2*a*b^3 + b^4)*cos(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 - 32*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*cos(f*x + e)^2 + 8*(16*b^3 *cos(f*x + e)^7 - 24*(a*b^2 + b^3)*cos(f*x + e)^5 + 10*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^3 - (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*cos(f*x + e))*sqrt(- b*cos(f*x + e)^2 + a + b)*sqrt(-b)))/(b^2*f), -1/8*((a - b)*sqrt(b)*arctan (1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + b^2)*cos(f*x + e)^2 + a^2 + 2*a*b + b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)/(2*b^3*cos(f*x + e)^5 - 3*(a* b^2 + b^3)*cos(f*x + e)^3 + (a^2*b + 2*a*b^2 + b^3)*cos(f*x + e))) + 4*sqr t(-b*cos(f*x + e)^2 + a + b)*b*cos(f*x + e))/(b^2*f)]
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\text {Timed out} \]
Time = 0.40 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {\frac {a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} - \frac {\arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )}{b}}{2 \, f} \]
1/2*(a*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/b^(3/2) - arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/sqrt(b) - sqrt(-b*cos(f*x + e)^2 + a + b)*cos(f*x + e )/b)/f
\[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]