Integrand size = 35, antiderivative size = 214 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=-\frac {(7 A+15 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \cos (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \cos ^3(c+d x) \sin (c+d x)}{2 d (a+a \cos (c+d x))^{3/2}}+\frac {(15 A+31 C) \sin (c+d x)}{5 a d \sqrt {a+a \cos (c+d x)}}+\frac {(5 A+9 C) \cos ^2(c+d x) \sin (c+d x)}{10 a d \sqrt {a+a \cos (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{10 a^2 d} \]
-1/2*(A+C)*cos(d*x+c)^3*sin(d*x+c)/d/(a+a*cos(d*x+c))^(3/2)-1/4*(7*A+15*C) *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)+1/5*(15*A+31*C)*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)+1/10*(5*A+9* C)*cos(d*x+c)^2*sin(d*x+c)/a/d/(a+a*cos(d*x+c))^(1/2)-1/10*(5*A+13*C)*sin( d*x+c)*(a+a*cos(d*x+c))^(1/2)/a^2/d
Time = 0.44 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.64 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {5 (7 A+15 C) \text {arctanh}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right )-\cos ^3\left (\frac {1}{2} (c+d x)\right ) (25 A+47 C+(20 A+39 C) \cos (c+d x)-2 C \cos (2 (c+d x))+C \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )}{5 d (a (1+\cos (c+d x)))^{3/2} \left (-1+\sin ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]
(5*(7*A + 15*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^5 - Cos[(c + d* x)/2]^3*(25*A + 47*C + (20*A + 39*C)*Cos[c + d*x] - 2*C*Cos[2*(c + d*x)] + C*Cos[3*(c + d*x)])*Sin[(c + d*x)/2])/(5*d*(a*(1 + Cos[c + d*x]))^(3/2)*( -1 + Sin[(c + d*x)/2]^2))
Time = 1.36 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.06, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.457, Rules used = {3042, 3521, 27, 3042, 3462, 27, 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.
| \(\displaystyle \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a \cos (c+d x)+a)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^{3/2}}dx\) |
| \(\Big \downarrow \) 3521 |
| \(\displaystyle \frac {\int -\frac {\cos ^2(c+d x) (2 a (A+3 C)-a (5 A+9 C) \cos (c+d x))}{2 \sqrt {\cos (c+d x) a+a}}dx}{2 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\int \frac {\cos ^2(c+d x) (2 a (A+3 C)-a (5 A+9 C) \cos (c+d x))}{\sqrt {\cos (c+d x) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a (A+3 C)-a (5 A+9 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3462 |
| \(\displaystyle -\frac {\frac {2 \int -\frac {\cos (c+d x) \left (4 a^2 (5 A+9 C)-3 a^2 (5 A+13 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {\int \frac {\cos (c+d x) \left (4 a^2 (5 A+9 C)-3 a^2 (5 A+13 C) \cos (c+d x)\right )}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {\sin \left (c+d x+\frac {\pi }{2}\right ) \left (4 a^2 (5 A+9 C)-3 a^2 (5 A+13 C) \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3447 |
| \(\displaystyle -\frac {-\frac {\int \frac {4 a^2 (5 A+9 C) \cos (c+d x)-3 a^2 (5 A+13 C) \cos ^2(c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {\int \frac {4 a^2 (5 A+9 C) \sin \left (c+d x+\frac {\pi }{2}\right )-3 a^2 (5 A+13 C) \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3502 |
| \(\displaystyle -\frac {-\frac {\frac {2 \int -\frac {3 \left (a^3 (5 A+13 C)-2 a^3 (15 A+31 C) \cos (c+d x)\right )}{2 \sqrt {\cos (c+d x) a+a}}dx}{3 a}-\frac {2 a (5 A+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {-\frac {-\frac {\int \frac {a^3 (5 A+13 C)-2 a^3 (15 A+31 C) \cos (c+d x)}{\sqrt {\cos (c+d x) a+a}}dx}{a}-\frac {2 a (5 A+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {-\frac {\int \frac {a^3 (5 A+13 C)-2 a^3 (15 A+31 C) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{a}-\frac {2 a (5 A+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle -\frac {-\frac {-\frac {5 a^3 (7 A+15 C) \int \frac {1}{\sqrt {\cos (c+d x) a+a}}dx-\frac {4 a^3 (15 A+31 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{a}-\frac {2 a (5 A+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {-\frac {-\frac {5 a^3 (7 A+15 C) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx-\frac {4 a^3 (15 A+31 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{a}-\frac {2 a (5 A+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 3128 |
| \(\displaystyle -\frac {-\frac {-\frac {-\frac {10 a^3 (7 A+15 C) \int \frac {1}{2 a-\frac {a^2 \sin ^2(c+d x)}{\cos (c+d x) a+a}}d\left (-\frac {a \sin (c+d x)}{\sqrt {\cos (c+d x) a+a}}\right )}{d}-\frac {4 a^3 (15 A+31 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{a}-\frac {2 a (5 A+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {-\frac {-\frac {\frac {5 \sqrt {2} a^{5/2} (7 A+15 C) \text {arctanh}\left (\frac {\sqrt {a} \sin (c+d x)}{\sqrt {2} \sqrt {a \cos (c+d x)+a}}\right )}{d}-\frac {4 a^3 (15 A+31 C) \sin (c+d x)}{d \sqrt {a \cos (c+d x)+a}}}{a}-\frac {2 a (5 A+13 C) \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{d}}{5 a}-\frac {2 a (5 A+9 C) \sin (c+d x) \cos ^2(c+d x)}{5 d \sqrt {a \cos (c+d x)+a}}}{4 a^2}-\frac {(A+C) \sin (c+d x) \cos ^3(c+d x)}{2 d (a \cos (c+d x)+a)^{3/2}}\) |
-1/2*((A + C)*Cos[c + d*x]^3*Sin[c + d*x])/(d*(a + a*Cos[c + d*x])^(3/2)) - ((-2*a*(5*A + 9*C)*Cos[c + d*x]^2*Sin[c + d*x])/(5*d*Sqrt[a + a*Cos[c + d*x]]) - ((-2*a*(5*A + 13*C)*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/d - (( 5*Sqrt[2]*a^(5/2)*(7*A + 15*C)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqr t[a + a*Cos[c + d*x]])])/d - (4*a^3*(15*A + 31*C)*Sin[c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]]))/a)/(5*a))/(4*a^2)
3.2.12.3.1 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[((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_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^n/(f*(m + n + 1))), x] + Simp[1/(b*(m + n + 1)) Int[(a + b*Sin[e + f*x])^m*(c + d*S in[e + f*x])^(n - 1)*Simp[A*b*c*(m + n + 1) + B*(a*c*m + b*d*n) + (A*b*d*(m + n + 1) + B*(a*d*m + b*c*n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[n, 0] && (IntegerQ[n] || EqQ[m + 1/2, 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]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[a*(A + C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(f*(b*c - a*d)*(2*m + 1))), x] + Simp[1/(b*(b*c - a*d)*(2*m + 1)) I nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[A*(a*c*(m + 1) - b*d*(2*m + n + 2)) - C*(a*c*m + b*d*(n + 1)) + (a*A*d*(m + n + 2) + C*(b* c*(2*m + 1) - a*d*(m - n - 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)]
Leaf count of result is larger than twice the leaf count of optimal. \(412\) vs. \(2(187)=374\).
Time = 6.95 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.93
| method | result | size |
| default | \(\frac {\sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-32 C \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 C \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+35 A \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 ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -40 A \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+75 C \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 ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -80 C \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-35 A \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 +45 A \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}-75 C \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 +85 C \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\right )}{20 \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}\) | \(413\) |
| parts | \(\frac {A \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}+\frac {C \sqrt {2}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-32 \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {a}\, \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+32 \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 )+75 \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 ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -80 \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 )-75 \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 +85 \sqrt {a}\, \sqrt {a \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\right )}{20 \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}\) | \(440\) |
1/20/cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-32*C*a^(1 /2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^6+32*C*a^(1/2)*(a*si n(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^4+35*A*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))*sin(1/2*d*x+1/2*c)^2*a-40*A* a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+75*C*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))*sin(1/2*d*x+1/2 *c)^2*a-80*C*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2-3 5*A*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+ 45*A*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^(1/2)-75*C*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+85*C*(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 = 0.27 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.02 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\frac {5 \, \sqrt {2} {\left ({\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right ) + 7 \, A + 15 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a \cos \left (d x + c\right ) + a} \sqrt {a} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left (4 \, C \cos \left (d x + c\right )^{3} - 4 \, C \cos \left (d x + c\right )^{2} + 4 \, {\left (5 \, A + 9 \, C\right )} \cos \left (d x + c\right ) + 25 \, A + 49 \, C\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
1/40*(5*sqrt(2)*((7*A + 15*C)*cos(d*x + c)^2 + 2*(7*A + 15*C)*cos(d*x + c) + 7*A + 15*C)*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*(4*C*cos(d*x + c)^3 - 4*C*cos(d*x + c)^2 + 4*(5*A + 9*C)*cos(d*x + c) + 25*A + 49*C)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)) /(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Unable to divide, perhaps due to ro unding error%%%{%%{[%%%{%%{[2475880078570760549798248448,0]:[1,0,-2]%%},[1 0]%%%},0]
Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+a \cos (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]