Integrand size = 23, antiderivative size = 78 \[ \int \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {(a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{2 \sqrt {b} f}-\frac {\cos (e+f x) \sqrt {a+b-b \cos ^2(e+f x)}}{2 f} \]
-1/2*(a+b)*arctan(cos(f*x+e)*b^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/f/b^(1/2) -1/2*cos(f*x+e)*(a+b-b*cos(f*x+e)^2)^(1/2)/f
Time = 0.22 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.19 \[ \int \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {\sqrt {2} \cos (e+f x) \sqrt {2 a+b-b \cos (2 (e+f x))}+\frac {2 (a+b) \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{\sqrt {-b}}}{4 f} \]
-1/4*(Sqrt[2]*Cos[e + f*x]*Sqrt[2*a + b - b*Cos[2*(e + f*x)]] + (2*(a + b) *Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/ Sqrt[-b])/f
Time = 0.24 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 3665, 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.
\(\displaystyle \int \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sin (e+f x) \sqrt {a+b \sin (e+f x)^2}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\frac {\int \sqrt {-b \cos ^2(e+f x)+a+b}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle -\frac {\frac {1}{2} (a+b) \int \frac {1}{\sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {\frac {1}{2} (a+b) \int \frac {1}{\frac {b \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}+1}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {(a+b) \arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{2 \sqrt {b}}+\frac {1}{2} \cos (e+f x) \sqrt {a-b \cos ^2(e+f x)+b}}{f}\) |
-((((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)/f)
3.2.23.3.1 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. \(183\) vs. \(2(66)=132\).
Time = 0.78 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.36
method | result | size |
default | \(-\frac {\sqrt {\left (\cos ^{2}\left (f x +e \right )\right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )}\, \left (-b \arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )+2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}-\arctan \left (\frac {-2 b \left (\cos ^{2}\left (f x +e \right )\right )+a +b}{2 \sqrt {b}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right ) a \right )}{4 \sqrt {b}\, \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(184\) |
-1/4*(cos(f*x+e)^2*(a+b*sin(f*x+e)^2))^(1/2)*(-b*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))+2*b^(1/2)*(- b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)-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^(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. 161 vs. \(2 (66) = 132\).
Time = 0.37 (sec) , antiderivative size = 433, normalized size of antiderivative = 5.55 \[ \int \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\left [-\frac {8 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right ) + {\left (a + b\right )} \sqrt {-b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{6} + 160 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} - 32 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{7} - 24 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{5} + 10 \, {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}\right )}{16 \, b f}, \frac {{\left (a + b\right )} \sqrt {b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{5} - 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )}}\right ) - 4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b \cos \left (f x + e\right )}{8 \, b f}\right ] \]
[-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*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*sqrt(- b*cos(f*x + e)^2 + a + b)*b*cos(f*x + e))/(b*f)]
\[ \int \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \sqrt {a + b \sin ^{2}{\left (e + f x \right )}} \sin {\left (e + f x \right )}\, dx \]
Time = 0.34 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.90 \[ \int \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=-\frac {\frac {a \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {b}} + \sqrt {b} \arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right ) + \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )}{2 \, f} \]
-1/2*(a*arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/sqrt(b) + sqrt(b)*arcsin(b* cos(f*x + e)/sqrt((a + b)*b)) + sqrt(-b*cos(f*x + e)^2 + a + b)*cos(f*x + e))/f
Leaf count of result is larger than twice the leaf count of optimal. 723 vs. \(2 (66) = 132\).
Time = 0.41 (sec) , antiderivative size = 723, normalized size of antiderivative = 9.27 \[ \int \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\text {Too large to display} \]
((a + b)*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))/sqrt(b) + 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))^3*a - (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))^3 *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))^2*a^(3/2) + 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))^2*sqrt(a)*b + 3*(sq rt(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))*a^2 + 9*(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))*a*b + 4*(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*ta n(1/2*f*x + 1/2*e)^2 + a))*b^2 + a^(5/2) + 3*a^(3/2)*b + 4*sqrt(a)*b^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))^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))*sqrt(a) + a + 4*b)^2)/f
Timed out. \[ \int \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int \sin \left (e+f\,x\right )\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \]