Integrand size = 23, antiderivative size = 104 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}-\frac {4 \sin (c+d x)}{3 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{3 a d} \]
arctanh(1/2*sin(d*x+c)*a^(1/2)*2^(1/2)/(a+a*cos(d*x+c))^(1/2))*2^(1/2)/d/a ^(1/2)-4/3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/3*sin(d*x+c)*(a+a*cos(d*x +c))^(1/2)/a/d
Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.81 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\left (\sqrt {2} \text {arctanh}\left (\frac {\sqrt {1-\cos (c+d x)}}{\sqrt {2}}\right )-\frac {2}{3} (1-\cos (c+d x))^{3/2}\right ) \sin (c+d x)}{d \sqrt {1-\cos (c+d x)} \sqrt {a (1+\cos (c+d x))}} \]
((Sqrt[2]*ArcTanh[Sqrt[1 - Cos[c + d*x]]/Sqrt[2]] - (2*(1 - Cos[c + d*x])^ (3/2))/3)*Sin[c + d*x])/(d*Sqrt[1 - Cos[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x] )])
Time = 0.47 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.08, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3238, 27, 3042, 3230, 3042, 3128, 219}
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.
\(\displaystyle \int \frac {\cos ^2(c+d x)}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 3238 |
\(\displaystyle \frac {2 \int \frac {a-2 a \cos (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {a-2 a \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{3 a}+\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {a-2 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}+\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {3 a \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {4 a \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}+\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}+\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {-\frac {6 a \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {4 a \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}+\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {3 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {4 a \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}+\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 a d}\) |
(2*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*a*d) + ((3*Sqrt[2]*Sqrt[a]*Ar cTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d - (4*a *Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(3*a)
3.2.24.3.1 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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( f*(m + 1))), x] + Simp[(a*d*m + b*c*(m + 1))/(b*(m + 1)) Int[(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && !LtQ[m, -2^(-1)]
Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-Cos[e + f*x])*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2 ))), x] + Simp[1/(b*(m + 2)) Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) - a*Si n[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] && ! LtQ[m, -2^(-1)]
Time = 1.41 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.27
method | result | size |
default | \(\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-4 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{3 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(132\) |
1/3*cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*(a*sin(1 /2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^2+3*ln(4*(a^(1/2)*(a*sin (1/2*d*x+1/2*c)^2)^(1/2)+a)/cos(1/2*d*x+1/2*c))*a)/a^(3/2)/sin(1/2*d*x+1/2 *c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.26 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right ) + \frac {3 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} - \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
1/6*(4*sqrt(a*cos(d*x + c) + a)*(cos(d*x + c) - 1)*sin(d*x + c) + 3*sqrt(2 )*(a*cos(d*x + c) + a)*log(-(cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*cos(d*x + c ) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos( d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 19437 vs. \(2 (87) = 174\).
Time = 0.66 (sec) , antiderivative size = 19437, normalized size of antiderivative = 186.89 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \]
1/60*(20*(cos(d*x + c) + 1)*sin(5/2*d*x + 5/2*c)^3 + 8*(cos(d*x + c)^2 + s in(d*x + c)^2 + 2*cos(d*x + c) + 1)*sin(3/2*d*x + 3/2*c)^3 - 20*cos(5/2*d* x + 5/2*c)^3*sin(d*x + c) + 2*(15*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d* x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c)^2 + 15* (log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2 *c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2 *d*x + 1/2*c) + 1))*sin(d*x + c)^2 + 30*(log(cos(1/2*d*x + 1/2*c)^2 + sin( 1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c )^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c) + 4*(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*cos(d*x + c) + 1)*sin(3/2*d*x + 3/ 2*c) - 20*cos(3/2*d*x + 3/2*c)*sin(d*x + c) + 15*log(cos(1/2*d*x + 1/2*c)^ 2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 15*log(cos(1/2* d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos (5/2*d*x + 5/2*c)^2 + 30*((log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2* c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2* d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c)^2 + (log(cos(1/ 2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2* c) + 1))*sin(d*x + c)^2 + 2*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x +...
Time = 0.36 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\frac {8 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {3 \, \sqrt {2} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {3 \, \sqrt {2} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{6 \, d} \]
-1/6*(8*sqrt(2)*sin(1/2*d*x + 1/2*c)^3/(sqrt(a)*sgn(cos(1/2*d*x + 1/2*c))) - 3*sqrt(2)*log(sin(1/2*d*x + 1/2*c) + 1)/(sqrt(a)*sgn(cos(1/2*d*x + 1/2* c))) + 3*sqrt(2)*log(-sin(1/2*d*x + 1/2*c) + 1)/(sqrt(a)*sgn(cos(1/2*d*x + 1/2*c))))/d
Time = 14.23 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.93 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2\,\sin \left (c+d\,x\right )\,\sqrt {a+a\,\cos \left (c+d\,x\right )}}{3\,a\,d}-\frac {2\,\left (4\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )-3\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )\right )\,\sqrt {\frac {a+a\,\cos \left (c+d\,x\right )}{2\,a}}}{3\,a^2\,d\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \]