Integrand size = 29, antiderivative size = 176 \[ \int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {15}{4} a b x+\frac {\left (5 a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))}{2 d}-\frac {\left (2 a^2-b^2\right ) \cos (c+d x)}{d}-\frac {\left (a^2-b^2\right ) \cos ^3(c+d x)}{3 d}+\frac {b^2 \cos ^5(c+d x)}{5 d}-\frac {2 a b \cot (c+d x)}{d}-\frac {a^2 \cot (c+d x) \csc (c+d x)}{2 d}-\frac {9 a b \cos (c+d x) \sin (c+d x)}{4 d}+\frac {a b \cos (c+d x) \sin ^3(c+d x)}{2 d} \] Output:
-15/4*a*b*x+1/2*(5*a^2-2*b^2)*arctanh(cos(d*x+c))/d-(2*a^2-b^2)*cos(d*x+c) /d-1/3*(a^2-b^2)*cos(d*x+c)^3/d+1/5*b^2*cos(d*x+c)^5/d-2*a*b*cot(d*x+c)/d- 1/2*a^2*cot(d*x+c)*csc(d*x+c)/d-9/4*a*b*cos(d*x+c)*sin(d*x+c)/d+1/2*a*b*co s(d*x+c)*sin(d*x+c)^3/d
Time = 6.81 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.42 \[ \int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {15 a b (c+d x)}{4 d}-\frac {\left (18 a^2-11 b^2\right ) \cos (c+d x)}{8 d}-\frac {\left (4 a^2-7 b^2\right ) \cos (3 (c+d x))}{48 d}+\frac {b^2 \cos (5 (c+d x))}{80 d}-\frac {a b \cot \left (\frac {1}{2} (c+d x)\right )}{d}-\frac {a^2 \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}+\frac {\left (5 a^2-2 b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {\left (-5 a^2+2 b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}+\frac {a^2 \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {a b \sin (2 (c+d x))}{d}-\frac {a b \sin (4 (c+d x))}{16 d}+\frac {a b \tan \left (\frac {1}{2} (c+d x)\right )}{d} \] Input:
Integrate[Cos[c + d*x]^3*Cot[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]
Output:
(-15*a*b*(c + d*x))/(4*d) - ((18*a^2 - 11*b^2)*Cos[c + d*x])/(8*d) - ((4*a ^2 - 7*b^2)*Cos[3*(c + d*x)])/(48*d) + (b^2*Cos[5*(c + d*x)])/(80*d) - (a* b*Cot[(c + d*x)/2])/d - (a^2*Csc[(c + d*x)/2]^2)/(8*d) + ((5*a^2 - 2*b^2)* Log[Cos[(c + d*x)/2]])/(2*d) + ((-5*a^2 + 2*b^2)*Log[Sin[(c + d*x)/2]])/(2 *d) + (a^2*Sec[(c + d*x)/2]^2)/(8*d) - (a*b*Sin[2*(c + d*x)])/d - (a*b*Sin [4*(c + d*x)])/(16*d) + (a*b*Tan[(c + d*x)/2])/d
Time = 0.71 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.11, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.414, Rules used = {3042, 3390, 3042, 3071, 252, 252, 262, 216, 4879, 360, 2341, 2009}
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.
| \(\displaystyle \int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\cos (c+d x)^6 (a+b \sin (c+d x))^2}{\sin (c+d x)^3}dx\) |
| \(\Big \downarrow \) 3390 |
| \(\displaystyle \int \cos ^3(c+d x) \cot ^3(c+d x) \left (a^2+b^2 \sin ^2(c+d x)\right )dx+2 a b \int \cos ^4(c+d x) \cot ^2(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cos (c+d x)^3 \cot (c+d x)^3 \left (a^2+b^2 \sin (c+d x)^2\right )dx+2 a b \int \sin \left (c+d x+\frac {\pi }{2}\right )^4 \tan \left (c+d x+\frac {\pi }{2}\right )^2dx\) |
| \(\Big \downarrow \) 3071 |
| \(\displaystyle \int \cos (c+d x)^3 \cot (c+d x)^3 \left (a^2+b^2 \sin (c+d x)^2\right )dx-\frac {2 a b \int \frac {\cot ^6(c+d x)}{\left (\cot ^2(c+d x)+1\right )^3}d\cot (c+d x)}{d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \int \cos (c+d x)^3 \cot (c+d x)^3 \left (a^2+b^2 \sin (c+d x)^2\right )dx-\frac {2 a b \left (\frac {5}{4} \int \frac {\cot ^4(c+d x)}{\left (\cot ^2(c+d x)+1\right )^2}d\cot (c+d x)-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle \int \cos (c+d x)^3 \cot (c+d x)^3 \left (a^2+b^2 \sin (c+d x)^2\right )dx-\frac {2 a b \left (\frac {5}{4} \left (\frac {3}{2} \int \frac {\cot ^2(c+d x)}{\cot ^2(c+d x)+1}d\cot (c+d x)-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \int \cos (c+d x)^3 \cot (c+d x)^3 \left (a^2+b^2 \sin (c+d x)^2\right )dx-\frac {2 a b \left (\frac {5}{4} \left (\frac {3}{2} \left (\cot (c+d x)-\int \frac {1}{\cot ^2(c+d x)+1}d\cot (c+d x)\right )-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \int \cos (c+d x)^3 \cot (c+d x)^3 \left (a^2+b^2 \sin (c+d x)^2\right )dx-\frac {2 a b \left (\frac {5}{4} \left (\frac {3}{2} (\cot (c+d x)-\arctan (\cot (c+d x)))-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 4879 |
| \(\displaystyle -\frac {\int \frac {\cos ^6(c+d x) \left (a^2+b^2-b^2 \cos ^2(c+d x)\right )}{\left (1-\cos ^2(c+d x)\right )^2}d\cos (c+d x)}{d}-\frac {2 a b \left (\frac {5}{4} \left (\frac {3}{2} (\cot (c+d x)-\arctan (\cot (c+d x)))-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 360 |
| \(\displaystyle -\frac {\frac {a^2 \cos (c+d x)}{2 \left (1-\cos ^2(c+d x)\right )}-\frac {1}{2} \int \frac {-2 b^2 \cos ^6(c+d x)+2 a^2 \cos ^4(c+d x)+2 a^2 \cos ^2(c+d x)+a^2}{1-\cos ^2(c+d x)}d\cos (c+d x)}{d}-\frac {2 a b \left (\frac {5}{4} \left (\frac {3}{2} (\cot (c+d x)-\arctan (\cot (c+d x)))-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2341 |
| \(\displaystyle -\frac {\frac {a^2 \cos (c+d x)}{2 \left (1-\cos ^2(c+d x)\right )}-\frac {1}{2} \int \left (2 b^2 \cos ^4(c+d x)-2 \left (a^2-b^2\right ) \cos ^2(c+d x)-2 \left (2 a^2-b^2\right )+\frac {5 a^2-2 b^2}{1-\cos ^2(c+d x)}\right )d\cos (c+d x)}{d}-\frac {2 a b \left (\frac {5}{4} \left (\frac {3}{2} (\cot (c+d x)-\arctan (\cot (c+d x)))-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}\right )}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {1}{2} \left (-\left (5 a^2-2 b^2\right ) \text {arctanh}(\cos (c+d x))+\frac {2}{3} \left (a^2-b^2\right ) \cos ^3(c+d x)+2 \left (2 a^2-b^2\right ) \cos (c+d x)-\frac {2}{5} b^2 \cos ^5(c+d x)\right )+\frac {a^2 \cos (c+d x)}{2 \left (1-\cos ^2(c+d x)\right )}}{d}-\frac {2 a b \left (\frac {5}{4} \left (\frac {3}{2} (\cot (c+d x)-\arctan (\cot (c+d x)))-\frac {\cot ^3(c+d x)}{2 \left (\cot ^2(c+d x)+1\right )}\right )-\frac {\cot ^5(c+d x)}{4 \left (\cot ^2(c+d x)+1\right )^2}\right )}{d}\) |
Input:
Int[Cos[c + d*x]^3*Cot[c + d*x]^3*(a + b*Sin[c + d*x])^2,x]
Output:
-(((a^2*Cos[c + d*x])/(2*(1 - Cos[c + d*x]^2)) + (-((5*a^2 - 2*b^2)*ArcTan h[Cos[c + d*x]]) + 2*(2*a^2 - b^2)*Cos[c + d*x] + (2*(a^2 - b^2)*Cos[c + d *x]^3)/3 - (2*b^2*Cos[c + d*x]^5)/5)/2)/d) - (2*a*b*(-1/4*Cot[c + d*x]^5/( 1 + Cot[c + d*x]^2)^2 + (5*((3*(-ArcTan[Cot[c + d*x]] + Cot[c + d*x]))/2 - Cot[c + d*x]^3/(2*(1 + Cot[c + d*x]^2))))/4))/d
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] : > Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Simp[1/(2*b^(m/2 + 1)*(p + 1)) Int[(a + b*x^2)^(p + 1)*Expan dToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)] - (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[m/2, 0] & & (IntegerQ[p] || EqQ[m + 2*p + 1, 0])
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq* (a + b*x^2)^p, x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[b*(ff/f) Subst[I nt[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[2*a*(b/d) Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e + f* x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 - b^2, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa ctors[Cos[v], x]}, -d/Coefficient[v, x, 1] Subst[Int[SubstFor[1, Cos[v]/d , u/Sin[v], x], x], x, Cos[v]/d]], x] /; !FalseQ[v] && FunctionOfQ[Nonfree Factors[Cos[v], x], u/Sin[v], x]]
Time = 7.15 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.06
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{7}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{5}}{2}-\frac {5 \cos \left (d x +c \right )^{3}}{6}-\frac {5 \cos \left (d x +c \right )}{2}-\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{7}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+b^{2} \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\) | \(187\) |
| default | \(\frac {a^{2} \left (-\frac {\cos \left (d x +c \right )^{7}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )^{5}}{2}-\frac {5 \cos \left (d x +c \right )^{3}}{6}-\frac {5 \cos \left (d x +c \right )}{2}-\frac {5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )+2 a b \left (-\frac {\cos \left (d x +c \right )^{7}}{\sin \left (d x +c \right )}-\left (\cos \left (d x +c \right )^{5}+\frac {5 \cos \left (d x +c \right )^{3}}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )-\frac {15 d x}{8}-\frac {15 c}{8}\right )+b^{2} \left (\frac {\cos \left (d x +c \right )^{5}}{5}+\frac {\cos \left (d x +c \right )^{3}}{3}+\cos \left (d x +c \right )+\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )\right )}{d}\) | \(187\) |
| risch | \(-\frac {15 a b x}{4}-\frac {{\mathrm e}^{3 i \left (d x +c \right )} a^{2}}{24 d}+\frac {7 \,{\mathrm e}^{3 i \left (d x +c \right )} b^{2}}{96 d}-\frac {i a b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}-\frac {9 a^{2} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {11 \,{\mathrm e}^{i \left (d x +c \right )} b^{2}}{16 d}-\frac {9 a^{2} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {11 \,{\mathrm e}^{-i \left (d x +c \right )} b^{2}}{16 d}-\frac {i a \left (i a \,{\mathrm e}^{3 i \left (d x +c \right )}+i a \,{\mathrm e}^{i \left (d x +c \right )}+4 b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 b \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {{\mathrm e}^{-3 i \left (d x +c \right )} a^{2}}{24 d}+\frac {7 \,{\mathrm e}^{-3 i \left (d x +c \right )} b^{2}}{96 d}+\frac {i a b \,{\mathrm e}^{2 i \left (d x +c \right )}}{2 d}-\frac {5 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b^{2}}{d}+\frac {5 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b^{2}}{d}+\frac {\cos \left (5 d x +5 c \right ) b^{2}}{80 d}-\frac {a b \sin \left (4 d x +4 c \right )}{16 d}\) | \(351\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)^3*(a+b*sin(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
1/d*(a^2*(-1/2/sin(d*x+c)^2*cos(d*x+c)^7-1/2*cos(d*x+c)^5-5/6*cos(d*x+c)^3 -5/2*cos(d*x+c)-5/2*ln(csc(d*x+c)-cot(d*x+c)))+2*a*b*(-1/sin(d*x+c)*cos(d* x+c)^7-(cos(d*x+c)^5+5/4*cos(d*x+c)^3+15/8*cos(d*x+c))*sin(d*x+c)-15/8*d*x -15/8*c)+b^2*(1/5*cos(d*x+c)^5+1/3*cos(d*x+c)^3+cos(d*x+c)+ln(csc(d*x+c)-c ot(d*x+c))))
Time = 0.11 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.39 \[ \int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {12 \, b^{2} \cos \left (d x + c\right )^{7} - 225 \, a b d x \cos \left (d x + c\right )^{2} - 4 \, {\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{5} + 225 \, a b d x - 20 \, {\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{3} + 30 \, {\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right ) + 15 \, {\left ({\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 5 \, a^{2} + 2 \, b^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left ({\left (5 \, a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 5 \, a^{2} + 2 \, b^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 15 \, {\left (2 \, a b \cos \left (d x + c\right )^{5} + 5 \, a b \cos \left (d x + c\right )^{3} - 15 \, a b \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{60 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="frica s")
Output:
1/60*(12*b^2*cos(d*x + c)^7 - 225*a*b*d*x*cos(d*x + c)^2 - 4*(5*a^2 - 2*b^ 2)*cos(d*x + c)^5 + 225*a*b*d*x - 20*(5*a^2 - 2*b^2)*cos(d*x + c)^3 + 30*( 5*a^2 - 2*b^2)*cos(d*x + c) + 15*((5*a^2 - 2*b^2)*cos(d*x + c)^2 - 5*a^2 + 2*b^2)*log(1/2*cos(d*x + c) + 1/2) - 15*((5*a^2 - 2*b^2)*cos(d*x + c)^2 - 5*a^2 + 2*b^2)*log(-1/2*cos(d*x + c) + 1/2) - 15*(2*a*b*cos(d*x + c)^5 + 5*a*b*cos(d*x + c)^3 - 15*a*b*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^ 2 - d)
\[ \int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \cos ^{3}{\left (c + d x \right )} \cot ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)**3*(a+b*sin(d*x+c))**2,x)
Output:
Integral((a + b*sin(c + d*x))**2*cos(c + d*x)**3*cot(c + d*x)**3, x)
Time = 0.12 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.08 \[ \int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=-\frac {5 \, {\left (4 \, \cos \left (d x + c\right )^{3} - \frac {6 \, \cos \left (d x + c\right )}{\cos \left (d x + c\right )^{2} - 1} + 24 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} a^{2} + 15 \, {\left (15 \, d x + 15 \, c + \frac {15 \, \tan \left (d x + c\right )^{4} + 25 \, \tan \left (d x + c\right )^{2} + 8}{\tan \left (d x + c\right )^{5} + 2 \, \tan \left (d x + c\right )^{3} + \tan \left (d x + c\right )}\right )} a b - 2 \, {\left (6 \, \cos \left (d x + c\right )^{5} + 10 \, \cos \left (d x + c\right )^{3} + 30 \, \cos \left (d x + c\right ) - 15 \, \log \left (\cos \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\cos \left (d x + c\right ) - 1\right )\right )} b^{2}}{60 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="maxim a")
Output:
-1/60*(5*(4*cos(d*x + c)^3 - 6*cos(d*x + c)/(cos(d*x + c)^2 - 1) + 24*cos( d*x + c) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1))*a^2 + 15*( 15*d*x + 15*c + (15*tan(d*x + c)^4 + 25*tan(d*x + c)^2 + 8)/(tan(d*x + c)^ 5 + 2*tan(d*x + c)^3 + tan(d*x + c)))*a*b - 2*(6*cos(d*x + c)^5 + 10*cos(d *x + c)^3 + 30*cos(d*x + c) - 15*log(cos(d*x + c) + 1) + 15*log(cos(d*x + c) - 1))*b^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 346 vs. \(2 (162) = 324\).
Time = 0.23 (sec) , antiderivative size = 346, normalized size of antiderivative = 1.97 \[ \int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 450 \, {\left (d x + c\right )} a b + 120 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 60 \, {\left (5 \, a^{2} - 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right ) + \frac {15 \, {\left (30 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}} + \frac {4 \, {\left (135 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 180 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 180 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{8} + 150 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 600 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 360 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} - 800 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 560 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 150 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 520 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 280 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 135 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 140 \, a^{2} + 92 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{120 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)^3*(a+b*sin(d*x+c))^2,x, algorithm="giac" )
Output:
1/120*(15*a^2*tan(1/2*d*x + 1/2*c)^2 - 450*(d*x + c)*a*b + 120*a*b*tan(1/2 *d*x + 1/2*c) - 60*(5*a^2 - 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c))) + 15*(30 *a^2*tan(1/2*d*x + 1/2*c)^2 - 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 8*a*b*tan(1/ 2*d*x + 1/2*c) - a^2)/tan(1/2*d*x + 1/2*c)^2 + 4*(135*a*b*tan(1/2*d*x + 1/ 2*c)^9 - 180*a^2*tan(1/2*d*x + 1/2*c)^8 + 180*b^2*tan(1/2*d*x + 1/2*c)^8 + 150*a*b*tan(1/2*d*x + 1/2*c)^7 - 600*a^2*tan(1/2*d*x + 1/2*c)^6 + 360*b^2 *tan(1/2*d*x + 1/2*c)^6 - 800*a^2*tan(1/2*d*x + 1/2*c)^4 + 560*b^2*tan(1/2 *d*x + 1/2*c)^4 - 150*a*b*tan(1/2*d*x + 1/2*c)^3 - 520*a^2*tan(1/2*d*x + 1 /2*c)^2 + 280*b^2*tan(1/2*d*x + 1/2*c)^2 - 135*a*b*tan(1/2*d*x + 1/2*c) - 140*a^2 + 92*b^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d
Time = 23.49 (sec) , antiderivative size = 484, normalized size of antiderivative = 2.75 \[ \int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx =\text {Too large to display} \] Input:
int(cos(c + d*x)^3*cot(c + d*x)^3*(a + b*sin(c + d*x))^2,x)
Output:
(a^2*tan(c/2 + (d*x)/2)^2)/(8*d) - (log(tan(c/2 + (d*x)/2))*((5*a^2)/2 - b ^2))/d - (tan(c/2 + (d*x)/2)^10*((49*a^2)/2 - 24*b^2) + tan(c/2 + (d*x)/2) ^8*((165*a^2)/2 - 48*b^2) + tan(c/2 + (d*x)/2)^2*((127*a^2)/6 - (184*b^2)/ 15) + tan(c/2 + (d*x)/2)^4*((223*a^2)/3 - (112*b^2)/3) + tan(c/2 + (d*x)/2 )^6*((335*a^2)/3 - (224*b^2)/3) + a^2/2 + 38*a*b*tan(c/2 + (d*x)/2)^3 + 60 *a*b*tan(c/2 + (d*x)/2)^5 + 40*a*b*tan(c/2 + (d*x)/2)^7 - 14*a*b*tan(c/2 + (d*x)/2)^11 + 4*a*b*tan(c/2 + (d*x)/2))/(d*(4*tan(c/2 + (d*x)/2)^2 + 20*t an(c/2 + (d*x)/2)^4 + 40*tan(c/2 + (d*x)/2)^6 + 40*tan(c/2 + (d*x)/2)^8 + 20*tan(c/2 + (d*x)/2)^10 + 4*tan(c/2 + (d*x)/2)^12)) + (15*a*b*atan((225*a ^2*b^2)/(4*(15*a*b^3 - (75*a^3*b)/2 + (225*a^2*b^2*tan(c/2 + (d*x)/2))/4)) - (15*a*b^3*tan(c/2 + (d*x)/2))/(15*a*b^3 - (75*a^3*b)/2 + (225*a^2*b^2*t an(c/2 + (d*x)/2))/4) + (75*a^3*b*tan(c/2 + (d*x)/2))/(2*(15*a*b^3 - (75*a ^3*b)/2 + (225*a^2*b^2*tan(c/2 + (d*x)/2))/4))))/(2*d) + (a*b*tan(c/2 + (d *x)/2))/d
Time = 0.18 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.47 \[ \int \cos ^3(c+d x) \cot ^3(c+d x) (a+b \sin (c+d x))^2 \, dx=\frac {24 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{6} b^{2}+60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{5} a b +40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} a^{2}-88 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} b^{2}-270 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a b -280 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a^{2}+184 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} b^{2}-240 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a b -60 \cos \left (d x +c \right ) a^{2}-300 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a^{2}+120 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b^{2}+355 \sin \left (d x +c \right )^{2} a^{2}-450 \sin \left (d x +c \right )^{2} a b d x -184 \sin \left (d x +c \right )^{2} b^{2}}{120 \sin \left (d x +c \right )^{2} d} \] Input:
int(cos(d*x+c)^3*cot(d*x+c)^3*(a+b*sin(d*x+c))^2,x)
Output:
(24*cos(c + d*x)*sin(c + d*x)**6*b**2 + 60*cos(c + d*x)*sin(c + d*x)**5*a* b + 40*cos(c + d*x)*sin(c + d*x)**4*a**2 - 88*cos(c + d*x)*sin(c + d*x)**4 *b**2 - 270*cos(c + d*x)*sin(c + d*x)**3*a*b - 280*cos(c + d*x)*sin(c + d* x)**2*a**2 + 184*cos(c + d*x)*sin(c + d*x)**2*b**2 - 240*cos(c + d*x)*sin( c + d*x)*a*b - 60*cos(c + d*x)*a**2 - 300*log(tan((c + d*x)/2))*sin(c + d* x)**2*a**2 + 120*log(tan((c + d*x)/2))*sin(c + d*x)**2*b**2 + 355*sin(c + d*x)**2*a**2 - 450*sin(c + d*x)**2*a*b*d*x - 184*sin(c + d*x)**2*b**2)/(12 0*sin(c + d*x)**2*d)