Integrand size = 23, antiderivative size = 114 \[ \int \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {3 (a+b)^2 \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{8 \sqrt {b} f}-\frac {3 (a+b) \cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{8 f}-\frac {\cos (e+f x) \left (a+b-b \cos ^2(e+f x)\right )^{3/2}}{4 f} \]
-1/4*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(3/2)/f-3/8*(a+b)^2*arctan(cos(f*x+e) *b^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/f/b^(1/2)-3/8*(a+b)*cos(f*x+e)*(a+b-b *cos(f*x+e)^2)^(1/2)/f
Time = 0.31 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.99 \[ \int \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {\frac {\cos (e+f x) \sqrt {2 a+b-b \cos (2 (e+f x))} (5 a+4 b-b \cos (2 (e+f x)))}{\sqrt {2}}+\frac {3 (a+b)^2 \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{\sqrt {-b}}}{8 f} \]
-1/8*((Cos[e + f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*(5*a + 4*b - b*Cos[ 2*(e + f*x)]))/Sqrt[2] + (3*(a + b)^2*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/Sqrt[-b])/f
Time = 0.26 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 3665, 211, 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.
| \(\displaystyle \int \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (e+f x) \left (a+b \sin (e+f x)^2\right )^{3/2}dx\) |
| \(\Big \downarrow \) 3665 |
| \(\displaystyle -\frac {\int \left (-b \cos ^2(e+f x)+a+b\right )^{3/2}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle -\frac {\frac {3}{4} (a+b) \int \sqrt {-b \cos ^2(e+f x)+a+b}d\cos (e+f x)+\frac {1}{4} \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{f}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle -\frac {\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{\sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}\right )+\frac {1}{4} \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{f}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle -\frac {\frac {3}{4} (a+b) \left (\frac {1}{2} (a+b) \int \frac {1}{\frac {b \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}+1}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}\right )+\frac {1}{4} \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle -\frac {\frac {3}{4} (a+b) \left (\frac {(a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}\right )+\frac {1}{4} \cos (e+f x) \left (a-b \cos ^2(e+f x)+b\right )^{3/2}}{f}\) |
-(((Cos[e + f*x]*(a + b - b*Cos[e + f*x]^2)^(3/2))/4 + (3*(a + b)*(((a + b )*ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]])/(2*Sqrt[b ]) + (Cos[e + f*x]*Sqrt[a + b - b*Cos[e + f*x]^2])/2))/4)/f)
3.2.33.3.1 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[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. \(308\) vs. \(2(98)=196\).
Time = 1.12 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.71
| method | result | size |
| default | \(-\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (-4 b^{\frac {3}{2}} \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (\cos ^{2}\left (f x +e \right )\right )+10 b^{\frac {3}{2}} \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}+10 a \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {b}-3 \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right ) a^{2}-6 b \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right ) a -3 b^{2} \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )\right )}{16 \sqrt {b}\, \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(309\) |
-1/16*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(-4*b^(3/2)*(-b*cos(f*x+e)^4 +(a+b)*cos(f*x+e)^2)^(1/2)*cos(f*x+e)^2+10*b^(3/2)*(-b*cos(f*x+e)^4+(a+b)* cos(f*x+e)^2)^(1/2)+10*a*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^(1/2 )-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))*a^2-6*b*arctan(1/2*(-2*b*cos(f*x+e)^2+a+b)/b^(1/2)/(-b*co s(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2))*a-3*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)))/b^(1/2)/cos(f* x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
Time = 0.54 (sec) , antiderivative size = 495, normalized size of antiderivative = 4.34 \[ \int \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\left [-\frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{6} + 160 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 32 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}\right ) - 8 \, {\left (2 \, b^{2} \cos \left (f x + e\right )^{3} - 5 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{64 \, b f}, \frac {3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{5} - 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )}}\right ) + 4 \, {\left (2 \, b^{2} \cos \left (f x + e\right )^{3} - 5 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{32 \, b f}\right ] \]
[-1/64*(3*(a^2 + 2*a*b + 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)*co s(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 - 5*(a*b + b^2)*cos(f*x + e))*sqrt(-b*cos(f*x + e )^2 + a + b))/(b*f), 1/32*(3*(a^2 + 2*a*b + 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 - 5*(a*b + b^2)*cos(f*x + e))*sqrt(-b*cos(f*x + e)^2 + a + b))/(b* f)]
Timed out. \[ \int \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Timed out} \]
Time = 0.31 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.93 \[ \int \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=-\frac {\frac {3 \, {\left (a + b\right )} a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + 3 \, {\left (a + b\right )} \sqrt {b} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + 2 \, {\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{\frac {3}{2}} \cos \left (f x + e\right ) + 3 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )} \cos \left (f x + e\right )}{8 \, f} \]
-1/8*(3*(a + b)*a*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/sqrt(b) + 3*(a + b)*sqrt(b)*arcsin(b*cos(f*x + e)/sqrt((a + b)*b)) + 2*(-b*cos(f*x + e)^2 + a + b)^(3/2)*cos(f*x + e) + 3*sqrt(-b*cos(f*x + e)^2 + a + b)*(a + b)*cos (f*x + e))/f
Leaf count of result is larger than twice the leaf count of optimal. 2403 vs. \(2 (98) = 196\).
Time = 0.63 (sec) , antiderivative size = 2403, normalized size of antiderivative = 21.08 \[ \int \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\text {Too large to display} \]
1/4*(3*(a^2 + 2*a*b + b^2)*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - s qrt(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))/sqrt(b) + 2*(5*(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 - 6*(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* 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*b^2 + 35*(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) + 22*(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*sq rt(a)*b^2 + 105*(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 + 246*(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 + 10 5*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*ta n(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*(s...
Timed out. \[ \int \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx=\int \sin \left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]