Integrand size = 31, antiderivative size = 148 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{d}+\frac {61 a \cos (c+d x)}{15 d \sqrt {a+a \sin (c+d x)}}+\frac {4 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{15 d}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{d}-\frac {2 \cos (c+d x) (a+a \sin (c+d x))^{3/2}}{5 a d} \]
-2/5*cos(d*x+c)*(a+a*sin(d*x+c))^(3/2)/a/d-arctanh(cos(d*x+c)*a^(1/2)/(a+a *sin(d*x+c))^(1/2))*a^(1/2)/d+61/15*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+ 4/15*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/d-cot(d*x+c)*(a+a*sin(d*x+c))^(1/2) /d
Time = 1.17 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.74 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\csc ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-155 \cos \left (\frac {1}{2} (c+d x)\right )+87 \cos \left (\frac {3}{2} (c+d x)\right )+5 \cos \left (\frac {5}{2} (c+d x)\right )+3 \cos \left (\frac {7}{2} (c+d x)\right )+155 \sin \left (\frac {1}{2} (c+d x)\right )-30 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+30 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \sin (c+d x)+87 \sin \left (\frac {3}{2} (c+d x)\right )-5 \sin \left (\frac {5}{2} (c+d x)\right )+3 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{30 d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )-\sec \left (\frac {1}{4} (c+d x)\right )\right ) \left (\csc \left (\frac {1}{4} (c+d x)\right )+\sec \left (\frac {1}{4} (c+d x)\right )\right )} \]
(Csc[(c + d*x)/2]^4*Sqrt[a*(1 + Sin[c + d*x])]*(-155*Cos[(c + d*x)/2] + 87 *Cos[(3*(c + d*x))/2] + 5*Cos[(5*(c + d*x))/2] + 3*Cos[(7*(c + d*x))/2] + 155*Sin[(c + d*x)/2] - 30*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin [c + d*x] + 30*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[c + d*x] + 87*Sin[(3*(c + d*x))/2] - 5*Sin[(5*(c + d*x))/2] + 3*Sin[(7*(c + d*x))/2] ))/(30*d*(1 + Cot[(c + d*x)/2])*(Csc[(c + d*x)/4] - Sec[(c + d*x)/4])*(Csc [(c + d*x)/4] + Sec[(c + d*x)/4]))
Time = 1.33 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.32, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.516, Rules used = {3042, 3360, 3042, 3238, 27, 3042, 3230, 3042, 3125, 3523, 27, 3042, 3460, 3042, 3252, 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.
| \(\displaystyle \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a \sin (c+d x)+a} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^4 \sqrt {a \sin (c+d x)+a}}{\sin (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3360 |
| \(\displaystyle \int \sin ^2(c+d x) \sqrt {\sin (c+d x) a+a}dx+\int \csc ^2(c+d x) \sqrt {\sin (c+d x) a+a} \left (1-2 \sin ^2(c+d x)\right )dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x)^2 \sqrt {\sin (c+d x) a+a}dx+\int \frac {\sqrt {\sin (c+d x) a+a} \left (1-2 \sin (c+d x)^2\right )}{\sin (c+d x)^2}dx\) |
| \(\Big \downarrow \) 3238 |
| \(\displaystyle \int \frac {\sqrt {\sin (c+d x) a+a} \left (1-2 \sin (c+d x)^2\right )}{\sin (c+d x)^2}dx+\frac {2 \int \frac {1}{2} (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \int \frac {\sqrt {\sin (c+d x) a+a} \left (1-2 \sin (c+d x)^2\right )}{\sin (c+d x)^2}dx+\frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int (3 a-2 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{5 a}+\int \frac {\sqrt {\sin (c+d x) a+a} \left (1-2 \sin (c+d x)^2\right )}{\sin (c+d x)^2}dx-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 3230 |
| \(\displaystyle \int \frac {\sqrt {\sin (c+d x) a+a} \left (1-2 \sin (c+d x)^2\right )}{\sin (c+d x)^2}dx+\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {\sin (c+d x) a+a} \left (1-2 \sin (c+d x)^2\right )}{\sin (c+d x)^2}dx+\frac {\frac {7}{3} a \int \sqrt {\sin (c+d x) a+a}dx+\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 3125 |
| \(\displaystyle \int \frac {\sqrt {\sin (c+d x) a+a} \left (1-2 \sin (c+d x)^2\right )}{\sin (c+d x)^2}dx+\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}\) |
| \(\Big \downarrow \) 3523 |
| \(\displaystyle \frac {\int \frac {1}{2} \csc (c+d x) (a-5 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{a}+\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \csc (c+d x) (a-5 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}dx}{2 a}+\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {(a-5 a \sin (c+d x)) \sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx}{2 a}+\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 3460 |
| \(\displaystyle \frac {a \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx+\frac {10 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{2 a}+\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {a \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx+\frac {10 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{2 a}+\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 3252 |
| \(\displaystyle \frac {\frac {10 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {2 a^2 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}}{2 a}+\frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {4 a \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{3 d}-\frac {14 a^2 \cos (c+d x)}{3 d \sqrt {a \sin (c+d x)+a}}}{5 a}+\frac {\frac {10 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}}{2 a}-\frac {2 \cos (c+d x) (a \sin (c+d x)+a)^{3/2}}{5 a d}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}\) |
-((Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d) - (2*Cos[c + d*x]*(a + a*Sin[ c + d*x])^(3/2))/(5*a*d) + ((-2*a^(3/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqr t[a + a*Sin[c + d*x]]])/d + (10*a^2*Cos[c + d*x])/(d*Sqrt[a + a*Sin[c + d* x]]))/(2*a) + ((-14*a^2*Cos[c + d*x])/(3*d*Sqrt[a + a*Sin[c + d*x]]) + (4* a*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/(3*d))/(5*a)
3.5.46.3.1 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[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[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)]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + ( f_.)*(x_)]), x_Symbol] :> Simp[-2*(b/f) Subst[Int[1/(b*c + a*d - d*x^2), x], x, b*(Cos[e + f*x]/Sqrt[a + b*Sin[e + f*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]
Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[1/d^4 Int[(d*Sin[e + f*x])^(n + 4)*(a + b*Sin[e + f*x])^m, x], x] + Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - 2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !IGtQ[m, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + ( f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp [-2*b*B*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(2*n + 3)*Sqrt[a + b*Sin[e + f*x]])), x] + Simp[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(b *d*(2*n + 3)) Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[n, -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[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Simp[1/(b*d*(n + 1)*(c^2 - d^2)) Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a *d*m + b*c*(n + 1)) + c*C*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(m + n + 2) + C* (c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])
Time = 0.12 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.09
| method | result | size |
| default | \(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (\sin \left (d x +c \right ) \left (6 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} a^{\frac {3}{2}}-20 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} a^{\frac {5}{2}}-30 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {7}{2}}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) a^{4}\right )+15 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {7}{2}}\right )}{15 a^{\frac {7}{2}} \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(162\) |
-1/15*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)/a^(7/2)*(sin(d*x+c)*(6*(a-a *sin(d*x+c))^(5/2)*a^(3/2)-20*(a-a*sin(d*x+c))^(3/2)*a^(5/2)-30*(a-a*sin(d *x+c))^(1/2)*a^(7/2)+15*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))*a^4)+15*(a -a*sin(d*x+c))^(1/2)*a^(7/2))/sin(d*x+c)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2) /d
Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (128) = 256\).
Time = 0.29 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.16 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {15 \, {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) - 4 \, {\left (6 \, \cos \left (d x + c\right )^{4} + 8 \, \cos \left (d x + c\right )^{3} + 40 \, \cos \left (d x + c\right )^{2} + {\left (6 \, \cos \left (d x + c\right )^{3} - 2 \, \cos \left (d x + c\right )^{2} + 38 \, \cos \left (d x + c\right ) + 61\right )} \sin \left (d x + c\right ) - 23 \, \cos \left (d x + c\right ) - 61\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{60 \, {\left (d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right ) + d\right )} \sin \left (d x + c\right ) - d\right )}} \]
1/60*(15*(cos(d*x + c)^2 - (cos(d*x + c) + 1)*sin(d*x + c) - 1)*sqrt(a)*lo g((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 + (cos(d*x + c) + 3)*sin(d*x + c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqrt(a ) - 9*a*cos(d*x + c) + (a*cos(d*x + c)^2 + 8*a*cos(d*x + c) - a)*sin(d*x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - cos(d*x + c) - 1)) - 4*(6*cos(d*x + c)^4 + 8*cos(d*x + c)^3 + 40*cos (d*x + c)^2 + (6*cos(d*x + c)^3 - 2*cos(d*x + c)^2 + 38*cos(d*x + c) + 61) *sin(d*x + c) - 23*cos(d*x + c) - 61)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^2 - (d*cos(d*x + c) + d)*sin(d*x + c) - d)
Timed out. \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\text {Timed out} \]
\[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int { \sqrt {a \sin \left (d x + c\right ) + a} \cos \left (d x + c\right )^{4} \csc \left (d x + c\right )^{2} \,d x } \]
Time = 0.33 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.41 \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\frac {\sqrt {2} {\left (96 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 160 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 120 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {60 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1}\right )} \sqrt {a}}{60 \, d} \]
1/60*sqrt(2)*(96*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^5 - 160*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d* x + 1/2*c)^3 - 15*sqrt(2)*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1 /2*c))/abs(2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c)))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 120*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c) - 60*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)/(2*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1))*sqrt(a)/d
Timed out. \[ \int \cos ^2(c+d x) \cot ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^4\,\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\sin \left (c+d\,x\right )}^2} \,d x \]