Integrand size = 29, antiderivative size = 170 \[ \int \cot ^2(c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {\left (a^2+2 b^2\right ) \text {arctanh}(\cos (c+d x))}{16 d}+\frac {2 a b \cot (c+d x)}{5 d}+\frac {2 a b \cot ^3(c+d x)}{15 d}+\frac {\left (a^2+2 b^2\right ) \cot (c+d x) \csc (c+d x)}{16 d}+\frac {\left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{24 d}-\frac {a b \cot (c+d x) \csc ^4(c+d x)}{15 d}-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d} \] Output:
1/16*(a^2+2*b^2)*arctanh(cos(d*x+c))/d+2/5*a*b*cot(d*x+c)/d+2/15*a*b*cot(d *x+c)^3/d+1/16*(a^2+2*b^2)*cot(d*x+c)*csc(d*x+c)/d+1/24*(a^2-2*b^2)*cot(d* x+c)*csc(d*x+c)^3/d-1/15*a*b*cot(d*x+c)*csc(d*x+c)^4/d-1/6*cot(d*x+c)*csc( d*x+c)^5*(a+b*sin(d*x+c))^2/d
Time = 1.93 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.74 \[ \int \cot ^2(c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {256 a b \cot \left (\frac {1}{2} (c+d x)\right )+30 \left (a^2+2 b^2\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )+120 a^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+240 b^2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-120 a^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-240 b^2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-30 a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )-60 b^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )+30 b^2 \sec ^4\left (\frac {1}{2} (c+d x)\right )+5 a^2 \sec ^6\left (\frac {1}{2} (c+d x)\right )-64 a b \csc ^3(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )+768 a b \csc ^5(c+d x) \sin ^6\left (\frac {1}{2} (c+d x)\right )-a \csc ^6\left (\frac {1}{2} (c+d x)\right ) (5 a+12 b \sin (c+d x))+\csc ^4\left (\frac {1}{2} (c+d x)\right ) \left (-30 b^2+4 a b \sin (c+d x)\right )-256 a b \tan \left (\frac {1}{2} (c+d x)\right )}{1920 d} \] Input:
Integrate[Cot[c + d*x]^2*Csc[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]
Output:
(256*a*b*Cot[(c + d*x)/2] + 30*(a^2 + 2*b^2)*Csc[(c + d*x)/2]^2 + 120*a^2* Log[Cos[(c + d*x)/2]] + 240*b^2*Log[Cos[(c + d*x)/2]] - 120*a^2*Log[Sin[(c + d*x)/2]] - 240*b^2*Log[Sin[(c + d*x)/2]] - 30*a^2*Sec[(c + d*x)/2]^2 - 60*b^2*Sec[(c + d*x)/2]^2 + 30*b^2*Sec[(c + d*x)/2]^4 + 5*a^2*Sec[(c + d*x )/2]^6 - 64*a*b*Csc[c + d*x]^3*Sin[(c + d*x)/2]^4 + 768*a*b*Csc[c + d*x]^5 *Sin[(c + d*x)/2]^6 - a*Csc[(c + d*x)/2]^6*(5*a + 12*b*Sin[c + d*x]) + Csc [(c + d*x)/2]^4*(-30*b^2 + 4*a*b*Sin[c + d*x]) - 256*a*b*Tan[(c + d*x)/2]) /(1920*d)
Time = 1.22 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.02, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.586, Rules used = {3042, 3368, 3042, 3527, 3042, 3510, 3042, 3500, 27, 3042, 3227, 3042, 4254, 2009, 4255, 3042, 4257}
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.
| \(\displaystyle \int \cot ^2(c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^2 (a+b \sin (c+d x))^2}{\sin (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3368 |
| \(\displaystyle \int \left (1-\sin ^2(c+d x)\right ) \csc ^7(c+d x) (a+b \sin (c+d x))^2dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (1-\sin (c+d x)^2\right ) (a+b \sin (c+d x))^2}{\sin (c+d x)^7}dx\) |
| \(\Big \downarrow \) 3527 |
| \(\displaystyle \frac {1}{6} \int \csc ^6(c+d x) (a+b \sin (c+d x)) \left (-3 b \sin ^2(c+d x)-a \sin (c+d x)+2 b\right )dx-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \int \frac {(a+b \sin (c+d x)) \left (-3 b \sin (c+d x)^2-a \sin (c+d x)+2 b\right )}{\sin (c+d x)^6}dx-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3510 |
| \(\displaystyle \frac {1}{6} \left (-\frac {1}{5} \int \csc ^5(c+d x) \left (15 b^2 \sin ^2(c+d x)+12 a b \sin (c+d x)+5 \left (a^2-2 b^2\right )\right )dx-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (-\frac {1}{5} \int \frac {15 b^2 \sin (c+d x)^2+12 a b \sin (c+d x)+5 \left (a^2-2 b^2\right )}{\sin (c+d x)^5}dx-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {5 \left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {1}{4} \int 3 \csc ^4(c+d x) \left (16 a b+5 \left (a^2+2 b^2\right ) \sin (c+d x)\right )dx\right )-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {5 \left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3}{4} \int \csc ^4(c+d x) \left (16 a b+5 \left (a^2+2 b^2\right ) \sin (c+d x)\right )dx\right )-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {5 \left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3}{4} \int \frac {16 a b+5 \left (a^2+2 b^2\right ) \sin (c+d x)}{\sin (c+d x)^4}dx\right )-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3227 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {5 \left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3}{4} \left (5 \left (a^2+2 b^2\right ) \int \csc ^3(c+d x)dx+16 a b \int \csc ^4(c+d x)dx\right )\right )-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {5 \left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3}{4} \left (5 \left (a^2+2 b^2\right ) \int \csc (c+d x)^3dx+16 a b \int \csc (c+d x)^4dx\right )\right )-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {5 \left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3}{4} \left (5 \left (a^2+2 b^2\right ) \int \csc (c+d x)^3dx-\frac {16 a b \int \left (\cot ^2(c+d x)+1\right )d\cot (c+d x)}{d}\right )\right )-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {5 \left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3}{4} \left (5 \left (a^2+2 b^2\right ) \int \csc (c+d x)^3dx-\frac {16 a b \left (\frac {1}{3} \cot ^3(c+d x)+\cot (c+d x)\right )}{d}\right )\right )-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {5 \left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3}{4} \left (5 \left (a^2+2 b^2\right ) \left (\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {16 a b \left (\frac {1}{3} \cot ^3(c+d x)+\cot (c+d x)\right )}{d}\right )\right )-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {5 \left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3}{4} \left (5 \left (a^2+2 b^2\right ) \left (\frac {1}{2} \int \csc (c+d x)dx-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {16 a b \left (\frac {1}{3} \cot ^3(c+d x)+\cot (c+d x)\right )}{d}\right )\right )-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {1}{6} \left (\frac {1}{5} \left (\frac {5 \left (a^2-2 b^2\right ) \cot (c+d x) \csc ^3(c+d x)}{4 d}-\frac {3}{4} \left (5 \left (a^2+2 b^2\right ) \left (-\frac {\text {arctanh}(\cos (c+d x))}{2 d}-\frac {\cot (c+d x) \csc (c+d x)}{2 d}\right )-\frac {16 a b \left (\frac {1}{3} \cot ^3(c+d x)+\cot (c+d x)\right )}{d}\right )\right )-\frac {2 a b \cot (c+d x) \csc ^4(c+d x)}{5 d}\right )-\frac {\cot (c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2}{6 d}\) |
Input:
Int[Cot[c + d*x]^2*Csc[c + d*x]^5*(a + b*Sin[c + d*x])^2,x]
Output:
((-2*a*b*Cot[c + d*x]*Csc[c + d*x]^4)/(5*d) + ((5*(a^2 - 2*b^2)*Cot[c + d* x]*Csc[c + d*x]^3)/(4*d) - (3*((-16*a*b*(Cot[c + d*x] + Cot[c + d*x]^3/3)) /d + 5*(a^2 + 2*b^2)*(-1/2*ArcTanh[Cos[c + d*x]]/d - (Cot[c + d*x]*Csc[c + d*x])/(2*d))))/4)/5)/6 - (Cot[c + d*x]*Csc[c + d*x]^5*(a + b*Sin[c + d*x] )^2)/(6*d)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x _)]), x_Symbol] :> Simp[c Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b Int [(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
Int[cos[(e_.) + (f_.)*(x_)]^2*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*(1 - Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, m, n }, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || IntegersQ[2*m, 2*n])
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f _.)*(x_)]^2), x_Symbol] :> Simp[(-(b*c - a*d))*(A*b^2 - a*b*B + a^2*C)*Cos[ e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b^2*f*(m + 1)*(a^2 - b^2))), x] - S imp[1/(b^2*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*( m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1)) ))*Sin[e + f*x] - b*C*d*(m + 1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; F reeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[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/(d*(n + 1)*(c^2 - d^ 2)) Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A* d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n + 2) - b *c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*( A*d^2*(m + n + 2) + C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.56 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.17
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{6 \sin \left (d x +c \right )^{6}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{16 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{16}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{16}\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{3}}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \cos \left (d x +c \right )^{3}}{15 \sin \left (d x +c \right )^{3}}\right )+b^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\) | \(199\) |
| default | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{6 \sin \left (d x +c \right )^{6}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{16 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{16}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{16}\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{3}}{5 \sin \left (d x +c \right )^{5}}-\frac {2 \cos \left (d x +c \right )^{3}}{15 \sin \left (d x +c \right )^{3}}\right )+b^{2} \left (-\frac {\cos \left (d x +c \right )^{3}}{4 \sin \left (d x +c \right )^{4}}-\frac {\cos \left (d x +c \right )^{3}}{8 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{8}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{8}\right )}{d}\) | \(199\) |
| risch | \(-\frac {15 a^{2} {\mathrm e}^{11 i \left (d x +c \right )}+30 b^{2} {\mathrm e}^{11 i \left (d x +c \right )}-640 i a b \,{\mathrm e}^{6 i \left (d x +c \right )}-85 a^{2} {\mathrm e}^{9 i \left (d x +c \right )}+150 b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+64 i a b -570 a^{2} {\mathrm e}^{7 i \left (d x +c \right )}-180 b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-570 a^{2} {\mathrm e}^{5 i \left (d x +c \right )}-180 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-384 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-85 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+150 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}+960 i a b \,{\mathrm e}^{8 i \left (d x +c \right )}+15 a^{2} {\mathrm e}^{i \left (d x +c \right )}+30 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{120 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{6}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{16 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{8 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{16 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{8 d}\) | \(316\) |
Input:
int(cot(d*x+c)^2*csc(d*x+c)^5*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(-1/6/sin(d*x+c)^6*cos(d*x+c)^3-1/8/sin(d*x+c)^4*cos(d*x+c)^3-1/1 6/sin(d*x+c)^2*cos(d*x+c)^3-1/16*cos(d*x+c)-1/16*ln(csc(d*x+c)-cot(d*x+c)) )+2*a*b*(-1/5/sin(d*x+c)^5*cos(d*x+c)^3-2/15/sin(d*x+c)^3*cos(d*x+c)^3)+b^ 2*(-1/4/sin(d*x+c)^4*cos(d*x+c)^3-1/8/sin(d*x+c)^2*cos(d*x+c)^3-1/8*cos(d* x+c)-1/8*ln(csc(d*x+c)-cot(d*x+c))))
Time = 0.09 (sec) , antiderivative size = 283, normalized size of antiderivative = 1.66 \[ \int \cot ^2(c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {30 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{5} - 80 \, a^{2} \cos \left (d x + c\right )^{3} - 30 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right ) - 15 \, {\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left ({\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{6} - 3 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - a^{2} - 2 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 64 \, {\left (2 \, a b \cos \left (d x + c\right )^{5} - 5 \, a b \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )}{480 \, {\left (d \cos \left (d x + c\right )^{6} - 3 \, d \cos \left (d x + c\right )^{4} + 3 \, d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="frica s")
Output:
-1/480*(30*(a^2 + 2*b^2)*cos(d*x + c)^5 - 80*a^2*cos(d*x + c)^3 - 30*(a^2 + 2*b^2)*cos(d*x + c) - 15*((a^2 + 2*b^2)*cos(d*x + c)^6 - 3*(a^2 + 2*b^2) *cos(d*x + c)^4 + 3*(a^2 + 2*b^2)*cos(d*x + c)^2 - a^2 - 2*b^2)*log(1/2*co s(d*x + c) + 1/2) + 15*((a^2 + 2*b^2)*cos(d*x + c)^6 - 3*(a^2 + 2*b^2)*cos (d*x + c)^4 + 3*(a^2 + 2*b^2)*cos(d*x + c)^2 - a^2 - 2*b^2)*log(-1/2*cos(d *x + c) + 1/2) + 64*(2*a*b*cos(d*x + c)^5 - 5*a*b*cos(d*x + c)^3)*sin(d*x + c))/(d*cos(d*x + c)^6 - 3*d*cos(d*x + c)^4 + 3*d*cos(d*x + c)^2 - d)
Timed out. \[ \int \cot ^2(c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\text {Timed out} \] Input:
integrate(cot(d*x+c)**2*csc(d*x+c)**5*(a+b*sin(d*x+c))**2,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.09 \[ \int \cot ^2(c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {5 \, a^{2} {\left (\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{5} - 8 \, \cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{6} - 3 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2} - 1} - 3 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + 30 \, b^{2} {\left (\frac {2 \, {\left (\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1} - \log \left (\cos \left (d x + c\right ) + 1\right ) + \log \left (\cos \left (d x + c\right ) - 1\right )\right )} + \frac {64 \, {\left (5 \, \tan \left (d x + c\right )^{2} + 3\right )} a b}{\tan \left (d x + c\right )^{5}}}{480 \, d} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-1/480*(5*a^2*(2*(3*cos(d*x + c)^5 - 8*cos(d*x + c)^3 - 3*cos(d*x + c))/(c os(d*x + c)^6 - 3*cos(d*x + c)^4 + 3*cos(d*x + c)^2 - 1) - 3*log(cos(d*x + c) + 1) + 3*log(cos(d*x + c) - 1)) + 30*b^2*(2*(cos(d*x + c)^3 + cos(d*x + c))/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1) - log(cos(d*x + c) + 1) + lo g(cos(d*x + c) - 1)) + 64*(5*tan(d*x + c)^2 + 3)*a*b/tan(d*x + c)^5)/d
Time = 0.17 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.62 \[ \int \cot ^2(c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {5 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 24 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 30 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 40 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, {\left (a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {294 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 588 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 240 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 40 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 30 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 5 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6}}}{1920 \, d} \] Input:
integrate(cot(d*x+c)^2*csc(d*x+c)^5*(a+b*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/1920*(5*a^2*tan(1/2*d*x + 1/2*c)^6 + 24*a*b*tan(1/2*d*x + 1/2*c)^5 + 15* a^2*tan(1/2*d*x + 1/2*c)^4 + 30*b^2*tan(1/2*d*x + 1/2*c)^4 + 40*a*b*tan(1/ 2*d*x + 1/2*c)^3 - 15*a^2*tan(1/2*d*x + 1/2*c)^2 - 240*a*b*tan(1/2*d*x + 1 /2*c) - 120*(a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + (294*a^2*tan(1/ 2*d*x + 1/2*c)^6 + 588*b^2*tan(1/2*d*x + 1/2*c)^6 + 240*a*b*tan(1/2*d*x + 1/2*c)^5 + 15*a^2*tan(1/2*d*x + 1/2*c)^4 - 40*a*b*tan(1/2*d*x + 1/2*c)^3 - 15*a^2*tan(1/2*d*x + 1/2*c)^2 - 30*b^2*tan(1/2*d*x + 1/2*c)^2 - 24*a*b*ta n(1/2*d*x + 1/2*c) - 5*a^2)/tan(1/2*d*x + 1/2*c)^6)/d
Time = 34.08 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.44 \[ \int \cot ^2(c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{384\,d}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{128\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {a^2}{16}+\frac {b^2}{8}\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {a^2}{6}-\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{2}+b^2\right )+\frac {4\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}-8\,a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\frac {4\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{5}\right )}{64\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^2}{128}+\frac {b^2}{64}\right )}{d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{48\,d}+\frac {a\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{80\,d}-\frac {a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{8\,d} \] Input:
int((cot(c + d*x)^2*(a + b*sin(c + d*x))^2)/sin(c + d*x)^5,x)
Output:
(a^2*tan(c/2 + (d*x)/2)^6)/(384*d) - (a^2*tan(c/2 + (d*x)/2)^2)/(128*d) - (log(tan(c/2 + (d*x)/2))*(a^2/16 + b^2/8))/d - (cot(c/2 + (d*x)/2)^6*(a^2/ 6 - (a^2*tan(c/2 + (d*x)/2)^4)/2 + tan(c/2 + (d*x)/2)^2*(a^2/2 + b^2) + (4 *a*b*tan(c/2 + (d*x)/2)^3)/3 - 8*a*b*tan(c/2 + (d*x)/2)^5 + (4*a*b*tan(c/2 + (d*x)/2))/5))/(64*d) + (tan(c/2 + (d*x)/2)^4*(a^2/128 + b^2/64))/d + (a *b*tan(c/2 + (d*x)/2)^3)/(48*d) + (a*b*tan(c/2 + (d*x)/2)^5)/(80*d) - (a*b *tan(c/2 + (d*x)/2))/(8*d)
Time = 0.17 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.17 \[ \int \cot ^2(c+d x) \csc ^5(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {64 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a b +15 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{2}+30 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{2}+32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b +10 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2}-60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-96 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -40 \cos \left (d x +c \right ) a^{2}-15 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{6} a^{2}-30 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{6} b^{2}}{240 \sin \left (d x +c \right )^{6} d} \] Input:
int(cot(d*x+c)^2*csc(d*x+c)^5*(a+b*sin(d*x+c))^2,x)
Output:
(64*cos(c + d*x)*sin(c + d*x)**5*a*b + 15*cos(c + d*x)*sin(c + d*x)**4*a** 2 + 30*cos(c + d*x)*sin(c + d*x)**4*b**2 + 32*cos(c + d*x)*sin(c + d*x)**3 *a*b + 10*cos(c + d*x)*sin(c + d*x)**2*a**2 - 60*cos(c + d*x)*sin(c + d*x) **2*b**2 - 96*cos(c + d*x)*sin(c + d*x)*a*b - 40*cos(c + d*x)*a**2 - 15*lo g(tan((c + d*x)/2))*sin(c + d*x)**6*a**2 - 30*log(tan((c + d*x)/2))*sin(c + d*x)**6*b**2)/(240*sin(c + d*x)**6*d)