Integrand size = 25, antiderivative size = 206 \[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=-\frac {\cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 b f}-\frac {2 (a-b) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {a (2 a-b) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \]
-1/3*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/b/f-2/3*(a-b)*Elliptic E(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(a+b*sin(f*x+e) ^2)^(1/2)/b^2/f/(1+b*sin(f*x+e)^2/a)^(1/2)+1/3*a*(2*a-b)*EllipticF(sin(f*x +e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(1/ 2)/b^2/f/(a+b*sin(f*x+e)^2)^(1/2)
Time = 0.73 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.79 \[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\frac {-4 \sqrt {2} a (a-b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+2 \sqrt {2} a (2 a-b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+b (-2 a-b+b \cos (2 (e+f x))) \sin (2 (e+f x))}{6 \sqrt {2} b^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]
(-4*Sqrt[2]*a*(a - b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + 2*Sqrt[2]*a*(2*a - b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/ a]*EllipticF[e + f*x, -(b/a)] + b*(-2*a - b + b*Cos[2*(e + f*x)])*Sin[2*(e + f*x)])/(6*Sqrt[2]*b^2*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])
Time = 0.40 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.95, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3042, 3667, 381, 399, 323, 321, 330, 327}
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.
| \(\displaystyle \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (e+f x)^4}{\sqrt {a+b \sin (e+f x)^2}}dx\) |
| \(\Big \downarrow \) 3667 |
| \(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \frac {\sin ^4(e+f x)}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{f}\) |
| \(\Big \downarrow \) 381 |
| \(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\int \frac {a-2 (a-b) \sin ^2(e+f x)}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{3 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {a+b \sin ^2(e+f x)}}{3 b}\right )}{f}\) |
| \(\Big \downarrow \) 399 |
| \(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {a (2 a-b) \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{b}-\frac {2 (a-b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}}{3 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {a+b \sin ^2(e+f x)}}{3 b}\right )}{f}\) |
| \(\Big \downarrow \) 323 |
| \(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {a (2 a-b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}d\sin (e+f x)}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a-b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}}{3 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {a+b \sin ^2(e+f x)}}{3 b}\right )}{f}\) |
| \(\Big \downarrow \) 321 |
| \(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {a (2 a-b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a-b) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}}{3 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {a+b \sin ^2(e+f x)}}{3 b}\right )}{f}\) |
| \(\Big \downarrow \) 330 |
| \(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {a (2 a-b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a-b) \sqrt {a+b \sin ^2(e+f x)} \int \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}}{3 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {a+b \sin ^2(e+f x)}}{3 b}\right )}{f}\) |
| \(\Big \downarrow \) 327 |
| \(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {a (2 a-b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 (a-b) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}}{3 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {a+b \sin ^2(e+f x)}}{3 b}\right )}{f}\) |
(Sqrt[Cos[e + f*x]^2]*Sec[e + f*x]*(-1/3*(Sin[e + f*x]*Sqrt[1 - Sin[e + f* x]^2]*Sqrt[a + b*Sin[e + f*x]^2])/b + ((-2*(a - b)*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2])/(b*Sqrt[1 + (b*Sin[e + f*x]^2 )/a]) + (a*(2*a - b)*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sqrt[1 + (b*S in[e + f*x]^2)/a])/(b*Sqrt[a + b*Sin[e + f*x]^2]))/(3*b)))/f
3.2.48.3.1 Defintions of rubi rules used
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S imp[(1/(Sqrt[a]*Sqrt[c]*Rt[-d/c, 2]))*EllipticF[ArcSin[Rt[-d/c, 2]*x], b*(c /(a*d))], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0] && !(NegQ[b/a] && SimplerSqrtQ[-b/a, -d/c])
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> S imp[Sqrt[1 + (d/c)*x^2]/Sqrt[c + d*x^2] Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + ( d/c)*x^2]), x], x] /; FreeQ[{a, b, c, d}, x] && !GtQ[c, 0]
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ (Sqrt[a]/(Sqrt[c]*Rt[-d/c, 2]))*EllipticE[ArcSin[Rt[-d/c, 2]*x], b*(c/(a*d) )], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[a, 0]
Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[ Sqrt[a + b*x^2]/Sqrt[1 + (b/a)*x^2] Int[Sqrt[1 + (b/a)*x^2]/Sqrt[c + d*x^ 2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && !GtQ[a, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_ ), x_Symbol] :> Simp[e^3*(e*x)^(m - 3)*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(b*d*(m + 2*(p + q) + 1))), x] - Simp[e^4/(b*d*(m + 2*(p + q) + 1)) Int[(e*x)^(m - 4)*(a + b*x^2)^p*(c + d*x^2)^q*Simp[a*c*(m - 3) + (a*d*(m + 2*q - 1) + b*c*(m + 2*p - 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, p, q }, x] && NeQ[b*c - a*d, 0] && GtQ[m, 3] && IntBinomialQ[a, b, c, d, e, m, 2 , p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_) ^2]), x_Symbol] :> Simp[f/b Int[Sqrt[a + b*x^2]/Sqrt[c + d*x^2], x], x] + Simp[(b*e - a*f)/b Int[1/(Sqrt[a + b*x^2]*Sqrt[c + d*x^2]), x], x] /; Fr eeQ[{a, b, c, d, e, f}, x] && !((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c])))))
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff^(m + 1 )*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])) Subst[Int[x^m*((a + b*ff^2*x^2) ^p/Sqrt[1 - ff^2*x^2]), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && !IntegerQ[p]
Time = 1.51 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.30
| method | result | size |
| default | \(\frac {b^{2} \left (\sin ^{5}\left (f x +e \right )\right )+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}-a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b +a b \left (\sin ^{3}\left (f x +e \right )\right )-b^{2} \left (\sin ^{3}\left (f x +e \right )\right )-a b \sin \left (f x +e \right )}{3 b^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(268\) |
1/3*(b^2*sin(f*x+e)^5+2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)* EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2-a*(cos(f*x+e)^2)^(1/2)*((a+b*sin( f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-2*(cos(f*x+e)^2) ^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a ^2+2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e ),(-1/a*b)^(1/2))*a*b+a*b*sin(f*x+e)^3-b^2*sin(f*x+e)^3-a*b*sin(f*x+e))/b^ 2/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f
\[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
integral(-(cos(f*x + e)^4 - 2*cos(f*x + e)^2 + 1)*sqrt(-b*cos(f*x + e)^2 + a + b)/(b*cos(f*x + e)^2 - a - b), x)
Timed out. \[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
\[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int { \frac {\sin \left (f x + e\right )^{4}}{\sqrt {b \sin \left (f x + e\right )^{2} + a}} \,d x } \]
Timed out. \[ \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^4}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \]