Integrand size = 23, antiderivative size = 140 \[ \int \frac {\cos ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{\sqrt {a} d}+\frac {28 \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}+\frac {2 \cos ^2(c+d x) \sin (c+d x)}{5 d \sqrt {a+a \cos (c+d x)}}-\frac {2 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 a d} \]
-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)+28/15*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)+2/5*cos(d*x+c)^2*sin(d*x +c)/d/(a+a*cos(d*x+c))^(1/2)-2/15*sin(d*x+c)*(a+a*cos(d*x+c))^(1/2)/a/d
Time = 0.17 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.72 \[ \int \frac {\cos ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {\left (-15 \sqrt {2} \text {arctanh}\left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right )+\sqrt {1-\cos (c+d x)} (29-2 \cos (c+d x)+3 \cos (2 (c+d x)))\right ) \sin (c+d x)}{15 d \sqrt {1-\cos (c+d x)} \sqrt {a (1+\cos (c+d x))}} \]
((-15*Sqrt[2]*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2]] + Sqrt[1 - Cos[c + d*x]]*( 29 - 2*Cos[c + d*x] + 3*Cos[2*(c + d*x)]))*Sin[c + d*x])/(15*d*Sqrt[1 - Co s[c + d*x]]*Sqrt[a*(1 + Cos[c + d*x])])
Time = 0.79 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.10, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {3042, 3257, 25, 3042, 3447, 3042, 3502, 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.
| \(\displaystyle \int \frac {\cos ^3(c+d x)}{\sqrt {a \cos (c+d x)+a}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^3}{\sqrt {a \sin \left (c+d x+\frac {\pi }{2}\right )+a}}dx\) |
| \(\Big \downarrow \) 3257 |
| \(\displaystyle \frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}-\frac {\int -\frac {\cos (c+d x) (4 a-a \cos (c+d x))}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\cos (c+d x) (4 a-a \cos (c+d x))}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (4 a-a \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle \frac {\int \frac {4 a \cos (c+d x)-a \cos ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {4 a \sin \left (c+d x+\frac {\pi }{2}\right )-a \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle \frac {\frac {2 \int -\frac {a^2-14 a^2 \cos (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {-\frac {\int \frac {a^2-14 a^2 \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {\int \frac {a^2-14 a^2 \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{3 a}-\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {-\frac {15 a^2 \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {28 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {-\frac {15 a^2 \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {28 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {-\frac {-\frac {30 a^2 \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {28 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {-\frac {\frac {15 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {28 a^2 \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{3 a}-\frac {2 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{3 d}}{5 a}+\frac {2 \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}\) |
(2*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]]) + ((-2*Sqrt [a + a*Cos[c + d*x]]*Sin[c + d*x])/(3*d) - ((15*Sqrt[2]*a^(3/2)*ArcTanh[(S qrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d - (28*a^2*Sin[ c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(3*a))/(5*a)
3.2.23.3.1 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[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)/Sqrt[(a_) + (b_.)*sin[(e_. ) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*d*Cos[e + f*x]*((c + d*Sin[e + f*x]) ^(n - 1)/(f*(2*n - 1)*Sqrt[a + b*Sin[e + f*x]])), x] - Simp[1/(b*(2*n - 1)) Int[((c + d*Sin[e + f*x])^(n - 2)/Sqrt[a + b*Sin[e + f*x]])*Simp[a*c*d - b*(2*d^2*(n - 1) + c^2*(2*n - 1)) + d*(a*d - b*c*(4*n - 3))*Sin[e + f*x], 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] && GtQ[n, 1] && IntegerQ[2*n]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Co s[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*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
Time = 1.29 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.31
| method | result | size |
| default | \(-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-24 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+20 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+15 \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a -30 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\right )}{15 a^{\frac {3}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(183\) |
-1/15*cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*(a*si n(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^4+20*(a*sin(1/2*d*x+1 /2*c)^2)^(1/2)*a^(1/2)*sin(1/2*d*x+1/2*c)^2+15*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-30*a^(1/2)*(a*sin(1/2*d*x+1/2*c )^2)^(1/2))/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 = 0.26 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\frac {4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (3 \, \cos \left (d x + c\right )^{2} - \cos \left (d x + c\right ) + 13\right )} \sin \left (d x + c\right ) + \frac {15 \, \sqrt {2} {\left (a \cos \left (d x + c\right ) + a\right )} \log \left (-\frac {\cos \left (d x + c\right )^{2} + \frac {2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt {a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt {a}}}{30 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]
1/30*(4*sqrt(a*cos(d*x + c) + a)*(3*cos(d*x + c)^2 - cos(d*x + c) + 13)*si n(d*x + c) + 15*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)/(c os(d*x + c)^2 + 2*cos(d*x + c) + 1))/sqrt(a))/(a*d*cos(d*x + c) + a*d)
Timed out. \[ \int \frac {\cos ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 908518 vs. \(2 (119) = 238\).
Time = 17.66 (sec) , antiderivative size = 908518, normalized size of antiderivative = 6489.41 \[ \int \frac {\cos ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\text {Too large to display} \]
-1/1680*(84*(sqrt(2)*cos(3/2*d*x + 3/2*c)^2*sin(d*x + c) + 2*sqrt(2)*cos(3 /2*d*x + 3/2*c)*cos(1/2*d*x + 1/2*c)*sin(d*x + c) + sqrt(2)*sin(3/2*d*x + 3/2*c)^2*sin(d*x + c) + 2*sqrt(2)*sin(3/2*d*x + 3/2*c)*sin(d*x + c)*sin(1/ 2*d*x + 1/2*c) + (sqrt(2)*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*sin(1/2*d*x + 1 /2*c)^2)*sin(d*x + c))*cos(7/2*d*x + 7/2*c)^3 - 84*((sqrt(2)*cos(d*x + c) + sqrt(2))*cos(3/2*d*x + 3/2*c)^2 + (sqrt(2)*cos(d*x + c) + sqrt(2))*sin(3 /2*d*x + 3/2*c)^2 + sqrt(2)*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*sin(1/2*d*x + 1/2*c)^2 + 2*(sqrt(2)*cos(d*x + c)*cos(1/2*d*x + 1/2*c) + sqrt(2)*cos(1/2 *d*x + 1/2*c))*cos(3/2*d*x + 3/2*c) + (sqrt(2)*cos(1/2*d*x + 1/2*c)^2 + sq rt(2)*sin(1/2*d*x + 1/2*c)^2)*cos(d*x + c) + 2*(sqrt(2)*cos(d*x + c)*sin(1 /2*d*x + 1/2*c) + sqrt(2)*sin(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c))*sin( 7/2*d*x + 7/2*c)^3 - 24*((sqrt(2)*cos(d*x + c)^2 + sqrt(2)*sin(d*x + c)^2 + 2*sqrt(2)*cos(d*x + c) + sqrt(2))*cos(3/2*d*x + 3/2*c)^2 + (sqrt(2)*cos( 1/2*d*x + 1/2*c)^2 + sqrt(2)*sin(1/2*d*x + 1/2*c)^2)*cos(d*x + c)^2 + (sqr t(2)*cos(d*x + c)^2 + sqrt(2)*sin(d*x + c)^2 + 2*sqrt(2)*cos(d*x + c) + sq rt(2))*sin(3/2*d*x + 3/2*c)^2 + (sqrt(2)*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)* sin(1/2*d*x + 1/2*c)^2)*sin(d*x + c)^2 + sqrt(2)*cos(1/2*d*x + 1/2*c)^2 + sqrt(2)*sin(1/2*d*x + 1/2*c)^2 + 2*(sqrt(2)*cos(d*x + c)^2*cos(1/2*d*x + 1 /2*c) + sqrt(2)*cos(1/2*d*x + 1/2*c)*sin(d*x + c)^2 + 2*sqrt(2)*cos(d*x + c)*cos(1/2*d*x + 1/2*c) + sqrt(2)*cos(1/2*d*x + 1/2*c))*cos(3/2*d*x + 3...
Time = 0.41 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.99 \[ \int \frac {\cos ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=-\frac {\frac {15 \, \sqrt {2} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {15 \, \sqrt {2} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{\sqrt {a} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sqrt {2} {\left (12 \, a^{\frac {9}{2}} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, a^{\frac {9}{2}} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, a^{\frac {9}{2}} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{5} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{30 \, d} \]
-1/30*(15*sqrt(2)*log(sin(1/2*d*x + 1/2*c) + 1)/(sqrt(a)*sgn(cos(1/2*d*x + 1/2*c))) - 15*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)*(12*a^(9/2)*sin(1/2*d*x + 1/2*c)^5 - 10*a^(9/2 )*sin(1/2*d*x + 1/2*c)^3 + 15*a^(9/2)*sin(1/2*d*x + 1/2*c))/(a^5*sgn(cos(1 /2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\cos ^3(c+d x)}{\sqrt {a+a \cos (c+d x)}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^3}{\sqrt {a+a\,\cos \left (c+d\,x\right )}} \,d x \]