Integrand size = 25, antiderivative size = 125 \[ \int \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {(a-3 b) (a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{8 b^{3/2} f}+\frac {(a-3 b) \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{8 b f}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}{4 b f} \]
1/8*(a-3*b)*(a+b)*arctan(cos(f*x+e)*b^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/b^ (3/2)/f-1/4*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(3/2)/b/f+1/8*(a-3*b)*cos(f*x+ e)*(a+b-b*cos(f*x+e)^2)^(1/2)/b/f
Time = 0.48 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.95 \[ \int \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {\frac {\cos (e+f x) \sqrt {2 a+b-b \cos (2 (e+f x))} (-a-4 b+b \cos (2 (e+f x)))}{\sqrt {2} b}+\frac {(a+b) (-a+3 b) \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{(-b)^{3/2}}}{8 f} \]
((Cos[e + f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*(-a - 4*b + b*Cos[2*(e + f*x)]))/(Sqrt[2]*b) + ((a + b)*(-a + 3*b)*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f* x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/(-b)^(3/2))/(8*f)
Time = 0.28 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3665, 299, 211, 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.
| \(\displaystyle \int \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x)^3 \sqrt {a+b \sin (e+f x)^2}dx\) |
| \(\Big \downarrow \) 3665 |
| \(\displaystyle -\frac {\int \left (1-\cos ^2(e+f x)\right ) \sqrt {-b \cos ^2(e+f x)+a+b}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle -\frac {\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \int \sqrt {-b \cos ^2(e+f x)+a+b}d\cos (e+f x)}{4 b}}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle -\frac {\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \left (\frac {1}{2} (a+b) \int \frac {1}{\sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}\right )}{4 b}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle -\frac {\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \left (\frac {1}{2} (a+b) \int \frac {1}{\frac {b \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}+1}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}\right )}{4 b}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {\frac {\cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{4 b}-\frac {(a-3 b) \left (\frac {(a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}\right )}{4 b}}{f}\) |
-(((Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^(3/2))/(4*b) - ((a - 3*b)*(((a + b)*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(2*Sq rt[b]) + (Cos[e + f*x]*Sqrt[a + b - b*Cos[e + f*x]^2])/2))/(4*b))/f)
3.2.22.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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. \(310\) vs. \(2(109)=218\).
Time = 1.13 (sec) , antiderivative size = 311, normalized size of antiderivative = 2.49
| method | result | size |
| default | \(-\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (-4 \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{\frac {5}{2}} \left (\cos ^{2}\left (f x +e \right )\right )+10 \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{\frac {5}{2}}+2 a \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{\frac {3}{2}}+\arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right ) a^{2} b -2 \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right ) a \,b^{2}-3 b^{3} \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )\right )}{16 b^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(311\) |
-1/16*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(-4*(-b*cos(f*x+e)^4+(a+b)*c os(f*x+e)^2)^(1/2)*b^(5/2)*cos(f*x+e)^2+10*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+ e)^2)^(1/2)*b^(5/2)+2*a*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^(3/2) +arctan(1/2*(-2*b*cos(f*x+e)^2+a+b)/b^(1/2)/(-b*cos(f*x+e)^4+(a+b)*cos(f*x +e)^2)^(1/2))*a^2*b-2*arctan(1/2*(-2*b*cos(f*x+e)^2+a+b)/b^(1/2)/(-b*cos(f *x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2))*a*b^2-3*b^3*arctan(1/2*(-2*b*cos(f*x+e) ^2+a+b)/b^(1/2)/(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)))/b^(5/2)/cos(f *x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
Time = 0.52 (sec) , antiderivative size = 501, normalized size of antiderivative = 4.01 \[ \int \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\left [\frac {{\left (a^{2} - 2 \, a b - 3 \, b^{2}\right )} \sqrt {-b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{6} + 160 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 32 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}\right ) + 8 \, {\left (2 \, b^{2} \cos \left (f x + e\right )^{3} - {\left (a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{64 \, b^{2} f}, -\frac {{\left (a^{2} - 2 \, a b - 3 \, b^{2}\right )} \sqrt {b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{5} - 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )}}\right ) - 4 \, {\left (2 \, b^{2} \cos \left (f x + e\right )^{3} - {\left (a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{32 \, b^{2} f}\right ] \]
[1/64*((a^2 - 2*a*b - 3*b^2)*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)) + 8*(2*b^2*cos(f*x + e)^3 - (a*b + 5*b^2)*cos(f*x + e))*sqrt(-b*cos(f*x + e) ^2 + a + b))/(b^2*f), -1/32*((a^2 - 2*a*b - 3*b^2)*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*(2*b^2*cos(f *x + e)^3 - (a*b + 5*b^2)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/( b^2*f)]
Timed out. \[ \int \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\text {Timed out} \]
Time = 0.34 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.41 \[ \int \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {\frac {{\left (a + b\right )} a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} + \frac {{\left (a + b\right )} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - \frac {4 \, a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} - 4 \, \sqrt {b} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) - 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right ) - \frac {2 \, {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \cos \left (f x + e\right )}{b} + \frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \cos \left (f x + e\right )}{b}}{8 \, f} \]
1/8*((a + b)*a*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/b^(3/2) + (a + b)*ar csin(b*cos(f*x + e)/sqrt((a + b)*b))/sqrt(b) - 4*a*arcsin(b*cos(f*x + e)/s qrt((a + b)*b))/sqrt(b) - 4*sqrt(b)*arcsin(b*cos(f*x + e)/sqrt((a + b)*b)) - 4*sqrt(-b*cos(f*x + e)^2 + a + b)*cos(f*x + e) - 2*(-b*cos(f*x + e)^2 + a + b)^(3/2)*cos(f*x + e)/b + sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)*cos (f*x + e)/b)/f
Leaf count of result is larger than twice the leaf count of optimal. 2404 vs. \(2 (109) = 218\).
Time = 0.50 (sec) , antiderivative size = 2404, normalized size of antiderivative = 19.23 \[ \int \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\text {Too large to display} \]
-1/4*((a^2 - 2*a*b - 3*b^2)*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f *x + 1/2*e)^2 + a) + sqrt(a))/sqrt(b))/b^(3/2) - 2*((sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4 *b*tan(1/2*f*x + 1/2*e)^2 + a))^7*a^2 - 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2 *f*x + 1/2*e)^2 + a))^7*a*b - 3*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*t an(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2 *e)^2 + a))^7*b^2 + 7*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) )^6*a^(5/2) + 18*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/ 2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^6*a ^(3/2)*b - 21*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e )^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^6*sqrt (a)*b^2 + 21*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e) ^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*a^3 + 122*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a *tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*a^2*b + 121*( sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1 /2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*a*b^2 - 44*(sqrt...
Timed out. \[ \int \sin ^3(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int {\sin \left (e+f\,x\right )}^3\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \]