Integrand size = 26, antiderivative size = 85 \[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}-\frac {a \sqrt {\cos (c+d x)} \sin (c+d x)}{d \sqrt {a-a \cos (c+d x)}} \]
arctanh(sin(d*x+c)*a^(1/2)/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^(1/2))*a^(1/2 )/d-a*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(a-a*cos(d*x+c))^(1/2)
Result contains complex when optimal does not.
Time = 0.53 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.79 \[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=-\frac {i \left (\left (1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}}-e^{i (c+d x)} \text {arcsinh}\left (e^{i (c+d x)}\right )-e^{i (c+d x)} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}{d \left (-1+e^{i (c+d x)}\right ) \sqrt {1+e^{2 i (c+d x)}}} \]
((-I)*((1 + E^(I*(c + d*x)))*Sqrt[1 + E^((2*I)*(c + d*x))] - E^(I*(c + d*x ))*ArcSinh[E^(I*(c + d*x))] - E^(I*(c + d*x))*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]])*Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])/(d*(-1 + E^(I*(c + d*x)))*Sqrt[1 + E^((2*I)*(c + d*x))])
Time = 0.39 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3042, 3249, 3042, 3254, 220}
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.
\(\displaystyle \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 3249 |
\(\displaystyle -\frac {1}{2} \int \frac {\sqrt {a-a \cos (c+d x)}}{\sqrt {\cos (c+d x)}}dx-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{2} \int \frac {\sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}\) |
\(\Big \downarrow \) 3254 |
\(\displaystyle -\frac {a \int \frac {1}{\frac {a^2 \sin (c+d x) \tan (c+d x)}{a-a \cos (c+d x)}-a}d\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}}{d}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}\) |
\(\Big \downarrow \) 220 |
\(\displaystyle \frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{d}-\frac {a \sin (c+d x) \sqrt {\cos (c+d x)}}{d \sqrt {a-a \cos (c+d x)}}\) |
(Sqrt[a]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos [c + d*x]])])/d - (a*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(d*Sqrt[a - a*Cos[c + d*x]])
3.3.64.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(- 1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[-2*b*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^n/(f*(2*n + 1)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[2*n*((b*c + a*d)/(b*( 2*n + 1))) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && IntegerQ[2*n]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b + d*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]))], x ] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Time = 12.95 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.41
method | result | size |
default | \(-\frac {\csc \left (d x +c \right ) \left (\cos \left (d x +c \right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}-\operatorname {arctanh}\left (\sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right ) \sqrt {-a \left (\cos \left (d x +c \right )-1\right )}}{d \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\) | \(120\) |
-1/d*csc(d*x+c)*(cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)+(cos(d*x+c)/ (1+cos(d*x+c)))^(1/2)-arctanh((cos(d*x+c)/(1+cos(d*x+c)))^(1/2)))*cos(d*x+ c)^(1/2)*(-a*(cos(d*x+c)-1))^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
Time = 0.28 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.67 \[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=\frac {\sqrt {a} \log \left (-\frac {4 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (2 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \sqrt {\cos \left (d x + c\right )} + {\left (8 \, a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 4 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{4 \, d \sin \left (d x + c\right )} \]
1/4*(sqrt(a)*log(-(4*sqrt(-a*cos(d*x + c) + a)*(2*cos(d*x + c)^2 + 3*cos(d *x + c) + 1)*sqrt(a)*sqrt(cos(d*x + c)) + (8*a*cos(d*x + c)^2 + 8*a*cos(d* x + c) + a)*sin(d*x + c))/sin(d*x + c))*sin(d*x + c) - 4*sqrt(-a*cos(d*x + c) + a)*(cos(d*x + c) + 1)*sqrt(cos(d*x + c)))/(d*sin(d*x + c))
\[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=\int \sqrt {- a \left (\cos {\left (c + d x \right )} - 1\right )} \sqrt {\cos {\left (c + d x \right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 795 vs. \(2 (73) = 146\).
Time = 0.41 (sec) , antiderivative size = 795, normalized size of antiderivative = 9.35 \[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=\text {Too large to display} \]
-1/4*(2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1) ^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - (cos(d*x + c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c ) + 1)))*sqrt(-a) + sqrt(-a)*(arctan2(-(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2 *c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), c os(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(sin(2*d* x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^ 2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))) + 1) - arctan2(-(cos(2*d*x + 2*c)^2 + sin(2* d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - cos(d*x + c)*sin(1/2*arctan2(s in(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin (2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2 *d*x + 2*c), cos(2*d*x + 2*c) + 1))) - 1) + arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d *x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^ 2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2* d*x + 2*c) + 1)) + 1) - arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^...
\[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=\int { \sqrt {-a \cos \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )} \,d x } \]
Timed out. \[ \int \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)} \, dx=\int \sqrt {\cos \left (c+d\,x\right )}\,\sqrt {a-a\,\cos \left (c+d\,x\right )} \,d x \]