Integrand size = 31, antiderivative size = 94 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{a^{3/2} d}-\frac {\cos (c+d x)}{a d \sqrt {a+a \sin (c+d x)}}-\frac {\cot (c+d x) \sqrt {a+a \sin (c+d x)}}{a^2 d} \] Output:
3*arctanh(a^(1/2)*cos(d*x+c)/(a+a*sin(d*x+c))^(1/2))/a^(3/2)/d-cos(d*x+c)/ a/d/(a+a*sin(d*x+c))^(1/2)-cot(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(220\) vs. \(2(94)=188\).
Time = 1.76 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.34 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^3 \left (2-8 \cos \left (\frac {1}{2} (c+d x)\right )-\cot \left (\frac {1}{4} (c+d x)\right )+6 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-6 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {2 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )-\sin \left (\frac {1}{4} (c+d x)\right )}-\frac {2 \sin \left (\frac {1}{4} (c+d x)\right )}{\cos \left (\frac {1}{4} (c+d x)\right )+\sin \left (\frac {1}{4} (c+d x)\right )}+8 \sin \left (\frac {1}{2} (c+d x)\right )-\tan \left (\frac {1}{4} (c+d x)\right )\right )}{4 d (a (1+\sin (c+d x)))^{3/2}} \] Input:
Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^2)/(a + a*Sin[c + d*x])^(3/2),x]
Output:
((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(2 - 8*Cos[(c + d*x)/2] - Cot[(c + d*x)/4] + 6*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 6*Log[1 - Cos [(c + d*x)/2] + Sin[(c + d*x)/2]] + (2*Sin[(c + d*x)/4])/(Cos[(c + d*x)/4] - Sin[(c + d*x)/4]) - (2*Sin[(c + d*x)/4])/(Cos[(c + d*x)/4] + Sin[(c + d *x)/4]) + 8*Sin[(c + d*x)/2] - Tan[(c + d*x)/4]))/(4*d*(a*(1 + Sin[c + d*x ]))^(3/2))
Time = 1.04 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.50, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.452, Rules used = {3042, 3359, 3042, 3252, 219, 3523, 27, 3042, 3242, 27, 2011, 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.
\(\displaystyle \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a \sin (c+d x)+a)^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x)^2 (a \sin (c+d x)+a)^{3/2}}dx\) |
\(\Big \downarrow \) 3359 |
\(\displaystyle \frac {\int \csc ^2(c+d x) \sqrt {\sin (c+d x) a+a} \left (\sin ^2(c+d x)+1\right )dx}{a^2}-\frac {2 \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin (c+d x) a+a} \left (\sin (c+d x)^2+1\right )}{\sin (c+d x)^2}dx}{a^2}-\frac {2 \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx}{a^2}\) |
\(\Big \downarrow \) 3252 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin (c+d x) a+a} \left (\sin (c+d x)^2+1\right )}{\sin (c+d x)^2}dx}{a^2}+\frac {4 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{a d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\int \frac {\sqrt {\sin (c+d x) a+a} \left (\sin (c+d x)^2+1\right )}{\sin (c+d x)^2}dx}{a^2}+\frac {4 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\) |
\(\Big \downarrow \) 3523 |
\(\displaystyle \frac {\frac {\int \frac {1}{2} \csc (c+d x) (\sin (c+d x) a+a)^{3/2}dx}{a}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}}{a^2}+\frac {4 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \csc (c+d x) (\sin (c+d x) a+a)^{3/2}dx}{2 a}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}}{a^2}+\frac {4 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {(\sin (c+d x) a+a)^{3/2}}{\sin (c+d x)}dx}{2 a}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}}{a^2}+\frac {4 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\) |
\(\Big \downarrow \) 3242 |
\(\displaystyle \frac {\frac {2 \int \frac {\csc (c+d x) \left (\sin (c+d x) a^2+a^2\right )}{2 \sqrt {\sin (c+d x) a+a}}dx-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}}{a^2}+\frac {4 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {\csc (c+d x) \left (\sin (c+d x) a^2+a^2\right )}{\sqrt {\sin (c+d x) a+a}}dx-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}}{a^2}+\frac {4 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle \frac {\frac {a \int \csc (c+d x) \sqrt {\sin (c+d x) a+a}dx-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}}{a^2}+\frac {4 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {a \int \frac {\sqrt {\sin (c+d x) a+a}}{\sin (c+d x)}dx-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}}{a^2}+\frac {4 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\) |
\(\Big \downarrow \) 3252 |
\(\displaystyle \frac {\frac {-\frac {2 a^2 \int \frac {1}{a-\frac {a^2 \cos ^2(c+d x)}{\sin (c+d x) a+a}}d\frac {a \cos (c+d x)}{\sqrt {\sin (c+d x) a+a}}}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}}{a^2}+\frac {4 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {4 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{a^{3/2} d}+\frac {\frac {-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{d}-\frac {2 a^2 \cos (c+d x)}{d \sqrt {a \sin (c+d x)+a}}}{2 a}-\frac {\cot (c+d x) \sqrt {a \sin (c+d x)+a}}{d}}{a^2}\) |
Input:
Int[(Cos[c + d*x]^2*Cot[c + d*x]^2)/(a + a*Sin[c + d*x])^(3/2),x]
Output:
(4*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(a^(3/2)*d) + (-((Cot[c + d*x]*Sqrt[a + a*Sin[c + d*x]])/d) + ((-2*a^(3/2)*ArcTanh[(Sqr t[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/d - (2*a^2*Cos[c + d*x])/(d* Sqrt[a + a*Sin[c + d*x]]))/(2*a))/a^2
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[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*x])
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f*x ])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n))), x] + Simp[1/(d*(m + n)) Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c* (m - 2) + b^2*d*(n + 1) + a^2*d*(m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*n - 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 1] && !LtQ[ n, -1] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[ c, 0]))
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[-2/(a*b*d) Int[(d* Sin[e + f*x])^(n + 1)*(a + b*Sin[e + f*x])^(m + 2), x], x] + Simp[1/a^2 I nt[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^(m + 2)*(1 + Sin[e + f*x]^2), x] , x] /; FreeQ[{a, b, d, e, f, n}, x] && EqQ[a^2 - b^2, 0] && 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[(-(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 = 0.41 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.35
method | result | size |
default | \(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (2 \sqrt {a -a \sin \left (d x +c \right )}\, \sin \left (d x +c \right ) \sqrt {a}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}}{\sqrt {a}}\right ) \sin \left (d x +c \right ) a +\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {a}\right )}{a^{\frac {5}{2}} \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\) | \(127\) |
Input:
int(cos(d*x+c)^2*cot(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x,method=_RETURNVERBO SE)
Output:
-1/a^(5/2)*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(2*(a-a*sin(d*x+c))^(1 /2)*sin(d*x+c)*a^(1/2)-3*arctanh((a-a*sin(d*x+c))^(1/2)/a^(1/2))*sin(d*x+c )*a+(a-a*sin(d*x+c))^(1/2)*a^(1/2))/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. 291 vs. \(2 (84) = 168\).
Time = 0.10 (sec) , antiderivative size = 291, normalized size of antiderivative = 3.10 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {3 \, {\left (\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) - 1\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} + 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (2 \, \cos \left (d x + c\right )^{2} + {\left (2 \, \cos \left (d x + c\right ) + 1\right )} \sin \left (d x + c\right ) + \cos \left (d x + c\right ) - 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} - a^{2} d - {\left (a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )} \sin \left (d x + c\right )\right )}} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="f ricas")
Output:
1/4*(3*(cos(d*x + c)^2 - (cos(d*x + c) + 1)*sin(d*x + c) - 1)*sqrt(a)*log( (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*(2*cos(d*x + c)^2 + (2*cos(d*x + c) + 1)*sin(d* x + c) + cos(d*x + c) - 1)*sqrt(a*sin(d*x + c) + a))/(a^2*d*cos(d*x + c)^2 - a^2*d - (a^2*d*cos(d*x + c) + a^2*d)*sin(d*x + c))
\[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(cos(d*x+c)**2*cot(d*x+c)**2/(a+a*sin(d*x+c))**(3/2),x)
Output:
Integral(cos(c + d*x)**2*cot(c + d*x)**2/(a*(sin(c + d*x) + 1))**(3/2), x)
Timed out. \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="m axima")
Output:
Timed out
Time = 0.20 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.76 \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} \sqrt {a} {\left (\frac {3 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} + \frac {8 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}\right )}}{4 \, d} \] Input:
integrate(cos(d*x+c)^2*cot(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x, algorithm="g iac")
Output:
1/4*sqrt(2)*sqrt(a)*(3*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)))/(a^2*sgn(co s(-1/4*pi + 1/2*d*x + 1/2*c))) + 8*sin(-1/4*pi + 1/2*d*x + 1/2*c)/(a^2*sgn (cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 4*sin(-1/4*pi + 1/2*d*x + 1/2*c)/((2*s in(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c ))))/d
Timed out. \[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,{\mathrm {cot}\left (c+d\,x\right )}^2}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{3/2}} \,d x \] Input:
int((cos(c + d*x)^2*cot(c + d*x)^2)/(a + a*sin(c + d*x))^(3/2),x)
Output:
int((cos(c + d*x)^2*cot(c + d*x)^2)/(a + a*sin(c + d*x))^(3/2), x)
\[ \int \frac {\cos ^2(c+d x) \cot ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx=\frac {\sqrt {a}\, \left (\int \frac {\sqrt {\sin \left (d x +c \right )+1}\, \cos \left (d x +c \right )^{2} \cot \left (d x +c \right )^{2}}{\sin \left (d x +c \right )^{2}+2 \sin \left (d x +c \right )+1}d x \right )}{a^{2}} \] Input:
int(cos(d*x+c)^2*cot(d*x+c)^2/(a+a*sin(d*x+c))^(3/2),x)
Output:
(sqrt(a)*int((sqrt(sin(c + d*x) + 1)*cos(c + d*x)**2*cot(c + d*x)**2)/(sin (c + d*x)**2 + 2*sin(c + d*x) + 1),x))/a**2