Integrand size = 23, antiderivative size = 175 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}+\frac {4 a E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{3 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{3 b^2 d \sqrt {a+b \sin (c+d x)}} \]
2/3*cos(d*x+c)*(a+b*sin(d*x+c))^(1/2)/b/d-4/3*a*(sin(1/2*c+1/4*Pi+1/2*d*x) ^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^ (1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/b^2/d/((a+b*sin(d*x+c))/(a+b ))^(1/2)+4/3*(a^2-b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*P i+1/2*d*x)*EllipticF(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(( a+b*sin(d*x+c))/(a+b))^(1/2)/b^2/d/(a+b*sin(d*x+c))^(1/2)
Time = 0.47 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\frac {2 b \cos (c+d x) (a+b \sin (c+d x))-4 a (a+b) E\left (\frac {1}{4} (-2 c+\pi -2 d x)|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}+4 \left (a^2-b^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 c+\pi -2 d x),\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{3 b^2 d \sqrt {a+b \sin (c+d x)}} \]
(2*b*Cos[c + d*x]*(a + b*Sin[c + d*x]) - 4*a*(a + b)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)] + 4*(a^2 - b ^2)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d* x])/(a + b)])/(3*b^2*d*Sqrt[a + b*Sin[c + d*x]])
Time = 0.77 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.02, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 3174, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}
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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^2}{\sqrt {a+b \sin (c+d x)}}dx\) |
| \(\Big \downarrow \) 3174 |
| \(\displaystyle \frac {2 \int \frac {b+a \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}}dx}{3 b}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \int \frac {b+a \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}}dx}{3 b}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}\) |
| \(\Big \downarrow \) 3231 |
| \(\displaystyle \frac {2 \left (\frac {a \int \sqrt {a+b \sin (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )}{3 b}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {a \int \sqrt {a+b \sin (c+d x)}dx}{b}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )}{3 b}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}\) |
| \(\Big \downarrow \) 3134 |
| \(\displaystyle \frac {2 \left (\frac {a \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )}{3 b}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {a \sqrt {a+b \sin (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}dx}{b \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )}{3 b}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}\) |
| \(\Big \downarrow \) 3132 |
| \(\displaystyle \frac {2 \left (\frac {2 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}}dx}{b}\right )}{3 b}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}\) |
| \(\Big \downarrow \) 3142 |
| \(\displaystyle \frac {2 \left (\frac {2 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{b \sqrt {a+b \sin (c+d x)}}\right )}{3 b}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 \left (\frac {2 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}}dx}{b \sqrt {a+b \sin (c+d x)}}\right )}{3 b}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}\) |
| \(\Big \downarrow \) 3140 |
| \(\displaystyle \frac {2 \left (\frac {2 a \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \sin (c+d x)}}\right )}{3 b}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)}}{3 b d}\) |
(2*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]])/(3*b*d) + (2*((2*a*EllipticE[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(b*d*Sqrt[(a + b *Sin[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*EllipticF[(c - Pi/2 + d*x)/2, (2 *b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(b*d*Sqrt[a + b*Sin[c + d *x]])))/(3*b)
3.6.13.3.1 Defintions of rubi rules used
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)] Int[Sqrt[a/(a + b) + ( b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 , 0] && !GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ {a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]] Int[1/Sqrt[a/(a + b) + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && !GtQ[a + b, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x _)])^(m_), x_Symbol] :> Simp[g*(g*Cos[e + f*x])^(p - 1)*((a + b*Sin[e + f*x ])^(m + 1)/(b*f*(m + p))), x] + Simp[g^2*((p - 1)/(b*(m + p))) Int[(g*Cos [e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; F reeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b Int[1/Sqrt[a + b*Sin[e + f*x ]], x], x] + Simp[d/b Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b , c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(461\) vs. \(2(225)=450\).
Time = 1.02 (sec) , antiderivative size = 462, normalized size of antiderivative = 2.64
| method | result | size |
| default | \(\frac {\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, F\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{2} b}{3}-\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, F\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) b^{3}}{3}-\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, E\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a^{3}}{3}+\frac {4 \sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}\, \sqrt {-\frac {\left (\sin \left (d x +c \right )-1\right ) b}{a +b}}\, \sqrt {-\frac {\left (1+\sin \left (d x +c \right )\right ) b}{a -b}}\, E\left (\sqrt {\frac {a +b \sin \left (d x +c \right )}{a -b}}, \sqrt {\frac {a -b}{a +b}}\right ) a \,b^{2}}{3}-\frac {2 \left (\sin ^{3}\left (d x +c \right )\right ) b^{3}}{3}-\frac {2 \left (\sin ^{2}\left (d x +c \right )\right ) a \,b^{2}}{3}+\frac {2 \sin \left (d x +c \right ) b^{3}}{3}+\frac {2 a \,b^{2}}{3}}{b^{3} \cos \left (d x +c \right ) \sqrt {a +b \sin \left (d x +c \right )}\, d}\) | \(462\) |
2/3*(2*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1 +sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b )/(a+b))^(1/2))*a^2*b-2*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/ (a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/( a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^3-2*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(s in(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+ b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3+2*((a+b*sin(d*x+c))/(a -b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2) *EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^2-sin(d *x+c)^3*b^3-sin(d*x+c)^2*a*b^2+sin(d*x+c)*b^3+a*b^2)/b^3/cos(d*x+c)/(a+b*s in(d*x+c))^(1/2)/d
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.12 (sec) , antiderivative size = 408, normalized size of antiderivative = 2.33 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=-\frac {2 \, {\left (3 i \, \sqrt {2} a \sqrt {i \, b} b {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right )\right ) - 3 i \, \sqrt {2} a \sqrt {-i \, b} b {\rm weierstrassZeta}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right )\right ) - 3 \, \sqrt {b \sin \left (d x + c\right ) + a} b^{2} \cos \left (d x + c\right ) + \sqrt {2} {\left (2 \, a^{2} - 3 \, b^{2}\right )} \sqrt {i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 i \, a^{3} - 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) - 2 i \, a}{3 \, b}\right ) + \sqrt {2} {\left (2 \, a^{2} - 3 \, b^{2}\right )} \sqrt {-i \, b} {\rm weierstrassPInverse}\left (-\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (-8 i \, a^{3} + 9 i \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 i \, a}{3 \, b}\right )\right )}}{9 \, b^{3} d} \]
-2/9*(3*I*sqrt(2)*a*sqrt(I*b)*b*weierstrassZeta(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, weierstrassPInverse(-4/3*(4*a^2 - 3*b^2)/ b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d* x + c) - 2*I*a)/b)) - 3*I*sqrt(2)*a*sqrt(-I*b)*b*weierstrassZeta(-4/3*(4*a ^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, weierstrassPInverse(-4/ 3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b)) - 3*sqrt(b*sin(d*x + c) + a)*b^2*co s(d*x + c) + sqrt(2)*(2*a^2 - 3*b^2)*sqrt(I*b)*weierstrassPInverse(-4/3*(4 *a^2 - 3*b^2)/b^2, -8/27*(8*I*a^3 - 9*I*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) - 2*I*a)/b) + sqrt(2)*(2*a^2 - 3*b^2)*sqrt(-I*b)*weie rstrassPInverse(-4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(-8*I*a^3 + 9*I*a*b^2)/b^3 , 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*I*a)/b))/(b^3*d)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt {a + b \sin {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right ) + a}} \,d x } \]
\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right ) + a}} \,d x } \]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{\sqrt {a+b\,\sin \left (c+d\,x\right )}} \,d x \]