Integrand size = 25, antiderivative size = 79 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a+b-b \cos ^2(e+f x)}}\right )}{b^{3/2} f}+\frac {a \cos (e+f x)}{b (a+b) f \sqrt {a+b-b \cos ^2(e+f x)}} \]
-arctan(cos(f*x+e)*b^(1/2)/(a+b-b*cos(f*x+e)^2)^(1/2))/b^(3/2)/f+a*cos(f*x +e)/b/(a+b)/f/(a+b-b*cos(f*x+e)^2)^(1/2)
Time = 0.47 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.22 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {\frac {\sqrt {2} a b \cos (e+f x)}{(a+b) \sqrt {2 a+b-b \cos (2 (e+f x))}}+\sqrt {-b} \log \left (\sqrt {2} \sqrt {-b} \cos (e+f x)+\sqrt {2 a+b-b \cos (2 (e+f x))}\right )}{b^2 f} \]
((Sqrt[2]*a*b*Cos[e + f*x])/((a + b)*Sqrt[2*a + b - b*Cos[2*(e + f*x)]]) + Sqrt[-b]*Log[Sqrt[2]*Sqrt[-b]*Cos[e + f*x] + Sqrt[2*a + b - b*Cos[2*(e + f*x)]]])/(b^2*f)
Time = 0.27 (sec) , antiderivative size = 78, 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, 298, 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.
\(\displaystyle \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin (e+f x)^3}{\left (a+b \sin (e+f x)^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 3665 |
\(\displaystyle -\frac {\int \frac {1-\cos ^2(e+f x)}{\left (-b \cos ^2(e+f x)+a+b\right )^{3/2}}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle -\frac {\frac {\int \frac {1}{\sqrt {-b \cos ^2(e+f x)+a+b}}d\cos (e+f x)}{b}-\frac {a \cos (e+f x)}{b (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {\frac {\int \frac {1}{\frac {b \cos ^2(e+f x)}{-b \cos ^2(e+f x)+a+b}+1}d\frac {\cos (e+f x)}{\sqrt {-b \cos ^2(e+f x)+a+b}}}{b}-\frac {a \cos (e+f x)}{b (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {\frac {\arctan \left (\frac {\sqrt {b} \cos (e+f x)}{\sqrt {a-b \cos ^2(e+f x)+b}}\right )}{b^{3/2}}-\frac {a \cos (e+f x)}{b (a+b) \sqrt {a-b \cos ^2(e+f x)+b}}}{f}\) |
-((ArcTan[(Sqrt[b]*Cos[e + f*x])/Sqrt[a + b - b*Cos[e + f*x]^2]]/b^(3/2) - (a*Cos[e + f*x])/(b*(a + b)*Sqrt[a + b - b*Cos[e + f*x]^2]))/f)
3.2.53.3.1 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[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 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. \(155\) vs. \(2(71)=142\).
Time = 1.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.97
method | result | size |
default | \(\frac {\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (\frac {\arctan \left (\frac {\sqrt {b}\, \left (\sin ^{2}\left (f x +e \right )-\frac {-a +b}{2 b}\right )}{\sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{2 b^{\frac {3}{2}}}+\frac {a \left (\cos ^{2}\left (f x +e \right )\right )}{b \left (a +b \right ) \sqrt {-\left (-b \left (\sin ^{2}\left (f x +e \right )\right )-a \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{\cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(156\) |
(-(-b*sin(f*x+e)^2-a)*cos(f*x+e)^2)^(1/2)*(1/2/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))+1/b*a*c os(f*x+e)^2/(a+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. 227 vs. \(2 (71) = 142\).
Time = 0.40 (sec) , antiderivative size = 564, normalized size of antiderivative = 7.14 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\left [-\frac {8 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a b \cos \left (f x + e\right ) + {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 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 ({\left (a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f\right )}}, -\frac {4 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} a b \cos \left (f x + e\right ) - {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - 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 ({\left (a b^{3} + b^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f\right )}}\right ] \]
[-1/8*(8*sqrt(-b*cos(f*x + e)^2 + a + b)*a*b*cos(f*x + e) + ((a*b + b^2)*c os(f*x + e)^2 - a^2 - 2*a*b - b^2)*sqrt(-b)*log(128*b^4*cos(f*x + e)^8 - 2 56*(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)))/((a*b^3 + b^4)*f*cos(f*x + e)^2 - (a^2*b^2 + 2*a*b^3 + b^4)*f), -1/4* (4*sqrt(-b*cos(f*x + e)^2 + a + b)*a*b*cos(f*x + e) - ((a*b + b^2)*cos(f*x + e)^2 - 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))))/((a*b^3 + b^4)*f*cos(f*x + e)^2 - (a^ 2*b^2 + 2*a*b^3 + b^4)*f)]
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Timed out} \]
Time = 0.30 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.03 \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\frac {\arcsin \left (\frac {b \cos \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{b^{\frac {3}{2}}} + \frac {\cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (a + b\right )}} - \frac {\cos \left (f x + e\right )}{\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} b}}{f} \]
-(arcsin(b*cos(f*x + e)/sqrt((a + b)*b))/b^(3/2) + cos(f*x + e)/(sqrt(-b*c os(f*x + e)^2 + a + b)*(a + b)) - cos(f*x + e)/(sqrt(-b*cos(f*x + e)^2 + a + b)*b))/f
\[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int { \frac {\sin \left (f x + e\right )^{3}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sin ^3(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^3}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]