Integrand size = 21, antiderivative size = 73 \[ \int \frac {\cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \sin (c+d x)}{d \sqrt {a+a \cos (c+d x)}} \]
-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)+2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)
Time = 0.03 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.73 \[ \int \frac {\cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (\text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{d \sqrt {a (1+\cos (c+d x))}} \]
(-2*Cos[(c + d*x)/2]*(ArcTanh[Sin[(c + d*x)/2]] - 2*Sin[(c + d*x)/2]))/(d* Sqrt[a*(1 + Cos[c + d*x])])
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {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.
\(\displaystyle \int \frac {\cos (c+d x)}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {2 \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}+\frac {2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{\sqrt {a} d}\) |
-((Sqrt[2]*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x] ])])/(Sqrt[a]*d)) + (2*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])
3.2.25.3.1 Defintions of rubi rules used
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)]
Time = 1.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.64
method | result | size |
default | \(\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (2 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \right )}{a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(120\) |
cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*a^(1/2)*(a*si n(1/2*d*x+1/2*c)^2)^(1/2)-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 = 0.31 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.67 \[ \int \frac {\cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\frac {\sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}} + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{2 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
1/2*(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) + 4*sqrt(a*cos(d*x + c) + a)*sin(d*x + c ))/(a*d*cos(d*x + c) + a*d)
\[ \int \frac {\cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {\cos {\left (c + d x \right )}}{\sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 18948 vs. \(2 (62) = 124\).
Time = 0.58 (sec) , antiderivative size = 18948, normalized size of antiderivative = 259.56 \[ \int \frac {\cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \]
-1/12*(12*sqrt(2)*cos(3/2*d*x + 3/2*c)^3*sin(d*x + c) - 12*(sqrt(2)*cos(d* x + c) + sqrt(2))*sin(3/2*d*x + 3/2*c)^3 - 8*sqrt(2)*sin(1/2*d*x + 1/2*c)^ 3 + ((3*sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*si n(1/2*d*x + 1/2*c) + 1) - 3*sqrt(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) - 8*sqrt(2)*sin(1/2*d*x + 1/2* c))*cos(d*x + c)^2 + (3*sqrt(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) - 3*sqrt(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) - 8*sqrt(2)*si n(1/2*d*x + 1/2*c))*sin(d*x + c)^2 + 24*sqrt(2)*cos(1/2*d*x + 1/2*c)*sin(d *x + c) + 2*(3*sqrt(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) - 3*sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + si n(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 8*sqrt(2)*sin(1/2*d*x + 1/2*c))*cos(d*x + c) + 3*sqrt(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) - 3*sqrt(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) - 8*sqrt(2 )*sin(1/2*d*x + 1/2*c))*cos(3/2*d*x + 3/2*c)^2 - (8*sqrt(2)*sin(1/2*d*x + 1/2*c)^3 - 3*(sqrt(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) - sqrt(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))*cos(1/2*d*x + 1/2*c)^2 - 3*(sqrt(2)*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(...
Time = 0.34 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.37 \[ \int \frac {\cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\frac {\sqrt {2} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{2 \, d} \]
-1/2*(sqrt(2)*log(sin(1/2*d*x + 1/2*c) + 1)/(sqrt(a)*sgn(cos(1/2*d*x + 1/2 *c))) - sqrt(2)*log(-sin(1/2*d*x + 1/2*c) + 1)/(sqrt(a)*sgn(cos(1/2*d*x + 1/2*c))) - 4*sqrt(2)*sin(1/2*d*x + 1/2*c)/(sqrt(a)*sgn(cos(1/2*d*x + 1/2*c ))))/d
Time = 0.13 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.82 \[ \int \frac {\cos (c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {2\,\left (2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )-\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |1\right )\right )\,\sqrt {\frac {a+a\,\cos \left (c+d\,x\right )}{2\,a}}}{d\,\sqrt {a+a\,\cos \left (c+d\,x\right )}} \]