Integrand size = 26, antiderivative size = 95 \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \]
-arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^ (1/2))*2^(1/2)/d/a^(1/2)+2*sin(d*x+c)/d/cos(d*x+c)^(1/2)/(a-a*cos(d*x+c))^ (1/2)
Result contains complex when optimal does not.
Time = 0.53 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.65 \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\frac {2 \left (-\frac {e^{-\frac {1}{2} i (c+d x)} \left (1+e^{2 i (c+d x)}\right ) \text {arctanh}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{\sqrt {2}}+2 \sqrt {1+e^{2 i (c+d x)}} \cos \left (\frac {1}{2} (c+d x)\right )\right ) \sin \left (\frac {1}{2} (c+d x)\right )}{d \sqrt {1+e^{2 i (c+d x)}} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}} \]
(2*(-(((1 + E^((2*I)*(c + d*x)))*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sq rt[1 + E^((2*I)*(c + d*x))])])/(Sqrt[2]*E^((I/2)*(c + d*x)))) + 2*Sqrt[1 + E^((2*I)*(c + d*x))]*Cos[(c + d*x)/2])*Sin[(c + d*x)/2])/(d*Sqrt[1 + E^(( 2*I)*(c + d*x))]*Sqrt[Cos[c + d*x]]*Sqrt[a - a*Cos[c + d*x]])
Time = 0.39 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3042, 3258, 27, 3042, 3261, 221}
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.
| \(\displaystyle \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx\) |
| \(\Big \downarrow \) 3258 |
| \(\displaystyle \frac {\int \frac {a}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}dx}{a}+\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}dx+\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \sqrt {a-a \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {2 a \int \frac {1}{2 a^2-\frac {a^3 \sin (c+d x) \tan (c+d x)}{a-a \cos (c+d x)}}d\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}}{d}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 \sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {\cos (c+d x)} \sqrt {a-a \cos (c+d x)}}\right )}{\sqrt {a} d}\) |
-((Sqrt[2]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt [a - a*Cos[c + d*x]])])/(Sqrt[a]*d)) + (2*Sin[c + d*x])/(d*Sqrt[Cos[c + d* x]]*Sqrt[a - a*Cos[c + d*x]])
3.3.79.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. ) + (f_.)*(x_)]], x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(2* b*(n + 1)*(c^2 - d^2)) Int[(c + d*Sin[e + f*x])^(n + 1)*(Simp[a*d - 2*b*c *(n + 1) + b*d*(2*n + 3)*Sin[e + f*x], x]/Sqrt[a + b*Sin[e + f*x]]), 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] && LtQ[n, -1] && IntegerQ[2*n]
Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[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 = 5.45 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.89
| method | result | size |
| default | \(\frac {\left (-\operatorname {arctanh}\left (\frac {\sqrt {2}}{2 \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}+\sqrt {2}\right ) \sin \left (d x +c \right ) \sqrt {2}}{d \sqrt {-a \left (\cos \left (d x +c \right )-1\right )}\, \sqrt {\cos \left (d x +c \right )}}\) | \(85\) |
1/d*(-arctanh(1/2*2^(1/2)/(cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*(cos(d*x+c)/( 1+cos(d*x+c)))^(1/2)+2^(1/2))*sin(d*x+c)/(-a*(cos(d*x+c)-1))^(1/2)/cos(d*x +c)^(1/2)*2^(1/2)
Time = 0.29 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.60 \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\frac {\sqrt {2} \sqrt {a} \cos \left (d x + c\right ) \log \left (-\frac {\frac {2 \, \sqrt {2} \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {a}} - {\left (3 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 4 \, \sqrt {-a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) + 1\right )} \sqrt {\cos \left (d x + c\right )}}{2 \, a d \cos \left (d x + c\right ) \sin \left (d x + c\right )} \]
1/2*(sqrt(2)*sqrt(a)*cos(d*x + c)*log(-(2*sqrt(2)*sqrt(-a*cos(d*x + c) + a )*(cos(d*x + c) + 1)*sqrt(cos(d*x + c))/sqrt(a) - (3*cos(d*x + c) + 1)*sin (d*x + c))/((cos(d*x + c) - 1)*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)))/(a*d*cos(d*x + c)*si n(d*x + c))
\[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\int \frac {1}{\sqrt {- a \left (\cos {\left (c + d x \right )} - 1\right )} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]
Result contains complex when optimal does not.
Time = 0.40 (sec) , antiderivative size = 351, normalized size of antiderivative = 3.69 \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\frac {2 \, \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) \sin \left (d x + c\right ) - 2 \, {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) - \sqrt {2} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \arctan \left (\frac {2 \, \sqrt {2} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )}{\sqrt {a} {\left | e^{\left (i \, d x + i \, c\right )} - 1 \right |}}, \frac {2 \, {\left (\sqrt {2} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sqrt {a} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) - \sqrt {-a} {\left | e^{\left (i \, d x + i \, c\right )} - 1 \right |} + 2 \, \sqrt {a}\right )}}{a {\left | e^{\left (i \, d x + i \, c\right )} - 1 \right |}}\right )}{{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sqrt {-a} d} \]
(2*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))*sin(d*x + c) - 2*(cos(d*x + c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - sqrt(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)*arctan2(2*sqrt(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)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))/(sqrt(a)*abs(e^(I*d*x + I*c) - 1)), 2*(sqrt(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)*sqrt(a)*cos(1/ 2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - sqrt(-a)*abs(e^(I*d*x + I*c) - 1) + 2*sqrt(a))/(a*abs(e^(I*d*x + I*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)*sqrt(-a)*d)
Leaf count of result is larger than twice the leaf count of optimal. 279 vs. \(2 (80) = 160\).
Time = 0.70 (sec) , antiderivative size = 279, normalized size of antiderivative = 2.94 \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=-\frac {\frac {4 \, {\left (\frac {\sqrt {2} \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2}}{\sqrt {a} \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2}}{\sqrt {a} \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{\sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1}} - \frac {\sqrt {2} \log \left (\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} - \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {\sqrt {2} \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 3 \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {\sqrt {2} \log \left ({\left | -\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + \sqrt {\tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{4} - 6 \, \tan \left (\frac {1}{4} \, d x + \frac {1}{4} \, c\right )^{2} + 1} + 1 \right |}\right )}{\sqrt {a} \mathrm {sgn}\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{2 \, d} \]
-1/2*(4*(sqrt(2)*tan(1/4*d*x + 1/4*c)^2/(sqrt(a)*sgn(sin(1/2*d*x + 1/2*c)) ) - sqrt(2)/(sqrt(a)*sgn(sin(1/2*d*x + 1/2*c))))/sqrt(tan(1/4*d*x + 1/4*c) ^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) - sqrt(2)*log(tan(1/4*d*x + 1/4*c)^2 - sqrt(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 1)/(sqrt(a)* sgn(sin(1/2*d*x + 1/2*c))) + sqrt(2)*log(abs(-tan(1/4*d*x + 1/4*c)^2 + sqr t(tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 3))/(sqrt(a)*sg n(sin(1/2*d*x + 1/2*c))) + sqrt(2)*log(abs(-tan(1/4*d*x + 1/4*c)^2 + sqrt( tan(1/4*d*x + 1/4*c)^4 - 6*tan(1/4*d*x + 1/4*c)^2 + 1) + 1))/(sqrt(a)*sgn( sin(1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) \sqrt {a-a \cos (c+d x)}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {a-a\,\cos \left (c+d\,x\right )}} \,d x \]