Integrand size = 25, antiderivative size = 220 \[ \int \cos ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {2 (a+3 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{15 b f}-\frac {\cos (e+f x) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b f}-\frac {\left (2 a^2+7 a b-3 b^2\right ) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{15 b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {2 a (a+b) (a+3 b) \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{15 b^2 f \sqrt {a+b \sin ^2(e+f x)}} \]
-1/5*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2)/b/f+2/15*(a+3*b)*cos(f *x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/b/f-1/15*(2*a^2+7*a*b-3*b^2)*(co s(f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticE(sin(f*x+e),(-b/a)^(1/2))*(a+b*sin(f *x+e)^2)^(1/2)/b^2/f/(1+b*sin(f*x+e)^2/a)^(1/2)+2/15*a*(a+b)*(a+3*b)*(cos( f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticF(sin(f*x+e),(-b/a)^(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 = 1.79 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.90 \[ \int \cos ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\frac {-16 a \left (2 a^2+7 a b-3 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+32 a \left (a^2+4 a b+3 b^2\right ) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )-\sqrt {2} b \left (8 a^2-32 a b-15 b^2-4 (4 a-3 b) b \cos (2 (e+f x))+3 b^2 \cos (4 (e+f x))\right ) \sin (2 (e+f x))}{240 b^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]
(-16*a*(2*a^2 + 7*a*b - 3*b^2)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*Elli pticE[e + f*x, -(b/a)] + 32*a*(a^2 + 4*a*b + 3*b^2)*Sqrt[(2*a + b - b*Cos[ 2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] - Sqrt[2]*b*(8*a^2 - 32*a*b - 15*b^2 - 4*(4*a - 3*b)*b*Cos[2*(e + f*x)] + 3*b^2*Cos[4*(e + f*x)])*Sin[2* (e + f*x)])/(240*b^2*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])
Time = 0.45 (sec) , antiderivative size = 260, normalized size of antiderivative = 1.18, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3671, 318, 403, 25, 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.
\(\displaystyle \int \cos ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \cos (e+f x)^4 \sqrt {a+b \sin (e+f x)^2}dx\) |
\(\Big \downarrow \) 3671 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \int \left (1-\sin ^2(e+f x)\right )^{3/2} \sqrt {b \sin ^2(e+f x)+a}d\sin (e+f x)}{f}\) |
\(\Big \downarrow \) 318 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\int \frac {\sqrt {b \sin ^2(e+f x)+a} \left (-2 (a+3 b) \sin ^2(e+f x)+a+5 b\right )}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{5 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b}\right )}{f}\) |
\(\Big \downarrow \) 403 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {2}{3} (a+3 b) \sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \sqrt {a+b \sin ^2(e+f x)}-\frac {1}{3} \int -\frac {a (a+9 b)-\left (2 a^2+7 b a-3 b^2\right ) \sin ^2(e+f x)}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{5 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b}\right )}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {1}{3} \int \frac {a (a+9 b)-\left (2 a^2+7 b a-3 b^2\right ) \sin ^2(e+f x)}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)+\frac {2}{3} (a+3 b) \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b}\right )}{f}\) |
\(\Big \downarrow \) 399 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {1}{3} \left (\frac {2 a (a+b) (a+3 b) \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {b \sin ^2(e+f x)+a}}d\sin (e+f x)}{b}-\frac {\left (2 a^2+7 a b-3 b^2\right ) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}\right )+\frac {2}{3} (a+3 b) \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b}\right )}{f}\) |
\(\Big \downarrow \) 323 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {1}{3} \left (\frac {2 a (a+b) (a+3 b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \int \frac {1}{\sqrt {1-\sin ^2(e+f x)} \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}d\sin (e+f x)}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (2 a^2+7 a b-3 b^2\right ) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}\right )+\frac {2}{3} (a+3 b) \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b}\right )}{f}\) |
\(\Big \downarrow \) 321 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {1}{3} \left (\frac {2 a (a+b) (a+3 b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (2 a^2+7 a b-3 b^2\right ) \int \frac {\sqrt {b \sin ^2(e+f x)+a}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b}\right )+\frac {2}{3} (a+3 b) \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b}\right )}{f}\) |
\(\Big \downarrow \) 330 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {1}{3} \left (\frac {2 a (a+b) (a+3 b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (2 a^2+7 a b-3 b^2\right ) \sqrt {a+b \sin ^2(e+f x)} \int \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}{\sqrt {1-\sin ^2(e+f x)}}d\sin (e+f x)}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}\right )+\frac {2}{3} (a+3 b) \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b}\right )}{f}\) |
\(\Big \downarrow \) 327 |
\(\displaystyle \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \left (\frac {\frac {1}{3} \left (\frac {2 a (a+b) (a+3 b) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{b \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (2 a^2+7 a b-3 b^2\right ) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{b \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}\right )+\frac {2}{3} (a+3 b) \sqrt {1-\sin ^2(e+f x)} \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{5 b}-\frac {\sin (e+f x) \sqrt {1-\sin ^2(e+f x)} \left (a+b \sin ^2(e+f x)\right )^{3/2}}{5 b}\right )}{f}\) |
(Sqrt[Cos[e + f*x]^2]*Sec[e + f*x]*(-1/5*(Sin[e + f*x]*Sqrt[1 - Sin[e + f* x]^2]*(a + b*Sin[e + f*x]^2)^(3/2))/b + ((2*(a + 3*b)*Sin[e + f*x]*Sqrt[1 - Sin[e + f*x]^2]*Sqrt[a + b*Sin[e + f*x]^2])/3 + (-(((2*a^2 + 7*a*b - 3*b ^2)*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])) + (2*a*(a + b)*(a + 3*b)*EllipticF[ArcSi n[Sin[e + f*x]], -(b/a)]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])/(b*Sqrt[a + b*Sin [e + f*x]^2]))/3)/(5*b)))/f
3.4.29.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
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_) + (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[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[ff*(Sqrt[ Cos[e + f*x]^2]/(f*Cos[e + f*x])) Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && !IntegerQ[p]
Time = 3.78 (sec) , antiderivative size = 432, normalized size of antiderivative = 1.96
method | result | size |
default | \(\frac {-3 b^{3} \left (\cos ^{6}\left (f x +e \right )\right ) \sin \left (f x +e \right )+4 \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right ) a \,b^{2}+\left (-a^{2} b +2 a \,b^{2}+3 b^{3}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}+8 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b +6 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{3}-7 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2} b +3 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a \,b^{2}}{15 b^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\) | \(432\) |
1/15*(-3*b^3*cos(f*x+e)^6*sin(f*x+e)+4*cos(f*x+e)^4*sin(f*x+e)*a*b^2+(-a^2 *b+2*a*b^2+3*b^3)*cos(f*x+e)^2*sin(f*x+e)+2*(cos(f*x+e)^2)^(1/2)*(-b/a*cos (f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^3+8*(cos(f *x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/ a*b)^(1/2))*a^2*b+6*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2) *EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2-2*(cos(f*x+e)^2)^(1/2)*(-b/a*c os(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^3-7*(cos (f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(- 1/a*b)^(1/2))*a^2*b+3*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/ 2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2)/b^2/cos(f*x+e)/(a+b*sin(f*x +e)^2)^(1/2)/f
\[ \int \cos ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \cos ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\text {Timed out} \]
\[ \int \cos ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{4} \,d x } \]
\[ \int \cos ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int { \sqrt {b \sin \left (f x + e\right )^{2} + a} \cos \left (f x + e\right )^{4} \,d x } \]
Timed out. \[ \int \cos ^4(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx=\int {\cos \left (e+f\,x\right )}^4\,\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a} \,d x \]