Integrand size = 23, antiderivative size = 105 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {7 \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {\sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {2 \sin (c+d x)}{a d \sqrt {a+a \cos (c+d x)}} \]
1/2*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-7/4*arctanh(1/2*sin(d*x+c)*a^(1/2) *2^(1/2)/(a+a*cos(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+2*sin(d*x+c)/a/d/(a+a*c os(d*x+c))^(1/2)
Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.95 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {\left (-7 \sqrt {2} \text {arctanh}\left (\sqrt {\sin ^2\left (\frac {1}{2} (c+d x)\right )}\right ) (1+\cos (c+d x))+2 \sqrt {1-\cos (c+d x)} (5+4 \cos (c+d x))\right ) \sin (c+d x)}{4 d \sqrt {1-\cos (c+d x)} (a (1+\cos (c+d x)))^{3/2}} \]
((-7*Sqrt[2]*ArcTanh[Sqrt[Sin[(c + d*x)/2]^2]]*(1 + Cos[c + d*x]) + 2*Sqrt [1 - Cos[c + d*x]]*(5 + 4*Cos[c + d*x]))*Sin[c + d*x])/(4*d*Sqrt[1 - Cos[c + d*x]]*(a*(1 + Cos[c + d*x]))^(3/2))
Time = 0.47 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {3042, 3237, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x)}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3237 |
| \(\displaystyle \frac {\int -\frac {3 a-4 a \cos (c+d x)}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}+\frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {\int \frac {3 a-4 a \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {\int \frac {3 a-4 a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {7 a \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {8 a \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {7 a \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {8 a \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle \frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {-\frac {14 a \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {8 a \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sin (c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}-\frac {\frac {7 \sqrt {2} \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {8 a \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{4 a^2}\) |
Sin[c + d*x]/(2*d*(a + a*Cos[c + d*x])^(3/2)) - ((7*Sqrt[2]*Sqrt[a]*ArcTan h[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/d - (8*a*Sin [c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/(4*a^2)
3.2.33.3.1 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[b*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/(a*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[(a + b*Sin[e + f*x])^(m + 1)*(a*m - b*(2* m + 1)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 1.56 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.65
| method | result | size |
| default | \(\frac {\sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-7 \sqrt {2}\, \ln \left (\frac {4 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+4 a}{\cos \left (\frac {d x}{2}+\frac {c}{2}\right )}\right ) a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {2}\, \sqrt {a}+\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a^{\frac {5}{2}} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\) | \(173\) |
1/4/cos(1/2*d*x+1/2*c)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-7*2^(1/2)*ln(2*(2* a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a)/cos(1/2*d*x+1/2*c))*a*cos(1/2* d*x+1/2*c)^2+8*cos(1/2*d*x+1/2*c)^2*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*2^(1/2) *a^(1/2)+2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2))/a^(5/2)/sin(1/2*d *x+1/2*c)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d
Time = 0.25 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.56 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {7 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (4 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
1/8*(7*sqrt(2)*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*sqrt(a)*log(-(a*cos(d *x + c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(a)*sin(d*x + c) - 2*a* cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*sqrt(a*cos( d*x + c) + a)*(4*cos(d*x + c) + 5)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2 *a^2*d*cos(d*x + c) + a^2*d)
\[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\left (a \left (\cos {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Time = 0.46 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.38 \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {\frac {7 \, \sqrt {2} \log \left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {7 \, \sqrt {2} \log \left (-\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {16 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {2 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (\sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{\frac {3}{2}} \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{8 \, d} \]
-1/8*(7*sqrt(2)*log(sin(1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos(1/2*d*x + 1 /2*c))) - 7*sqrt(2)*log(-sin(1/2*d*x + 1/2*c) + 1)/(a^(3/2)*sgn(cos(1/2*d* x + 1/2*c))) - 16*sqrt(2)*sin(1/2*d*x + 1/2*c)/(a^(3/2)*sgn(cos(1/2*d*x + 1/2*c))) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c)/((sin(1/2*d*x + 1/2*c)^2 - 1)*a^ (3/2)*sgn(cos(1/2*d*x + 1/2*c))))/d
Timed out. \[ \int \frac {\cos ^2(c+d x)}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]