Integrand size = 27, antiderivative size = 124 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\left (2 a^2-3 b^2\right ) x}{2 b^3}-\frac {2 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a b^3 d}-\frac {\text {arctanh}(\cos (c+d x))}{a d}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\cos (c+d x) \sin (c+d x)}{2 b d} \] Output:
1/2*(2*a^2-3*b^2)*x/b^3-2*(a^2-b^2)^(3/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/ (a^2-b^2)^(1/2))/a/b^3/d-arctanh(cos(d*x+c))/a/d+a*cos(d*x+c)/b^2/d-1/2*co s(d*x+c)*sin(d*x+c)/b/d
Time = 0.25 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.15 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {-4 a^3 c+6 a b^2 c-4 a^3 d x+6 a b^2 d x+8 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )-4 a^2 b \cos (c+d x)+4 b^3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-4 b^3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+a b^2 \sin (2 (c+d x))}{4 a b^3 d} \] Input:
Integrate[(Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
-1/4*(-4*a^3*c + 6*a*b^2*c - 4*a^3*d*x + 6*a*b^2*d*x + 8*(a^2 - b^2)^(3/2) *ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] - 4*a^2*b*Cos[c + d*x] + 4*b^3*Log[Cos[(c + d*x)/2]] - 4*b^3*Log[Sin[(c + d*x)/2]] + a*b^2*Sin[2*( c + d*x)])/(a*b^3*d)
Time = 0.74 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.11, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.370, Rules used = {3042, 3374, 25, 3042, 3536, 3042, 3139, 1083, 217, 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.
\(\displaystyle \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^4}{\sin (c+d x) (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3374 |
\(\displaystyle -\frac {\int -\frac {\csc (c+d x) \left (2 b^2+a \sin (c+d x) b+\left (2 a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}dx}{2 b^2}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\csc (c+d x) \left (2 b^2+a \sin (c+d x) b+\left (2 a^2-3 b^2\right ) \sin ^2(c+d x)\right )}{a+b \sin (c+d x)}dx}{2 b^2}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {2 b^2+a \sin (c+d x) b+\left (2 a^2-3 b^2\right ) \sin (c+d x)^2}{\sin (c+d x) (a+b \sin (c+d x))}dx}{2 b^2}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 3536 |
\(\displaystyle \frac {-\frac {2 \left (a^2-b^2\right )^2 \int \frac {1}{a+b \sin (c+d x)}dx}{a b}+\frac {2 b^2 \int \csc (c+d x)dx}{a}+\frac {x \left (2 a^2-3 b^2\right )}{b}}{2 b^2}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {2 \left (a^2-b^2\right )^2 \int \frac {1}{a+b \sin (c+d x)}dx}{a b}+\frac {2 b^2 \int \csc (c+d x)dx}{a}+\frac {x \left (2 a^2-3 b^2\right )}{b}}{2 b^2}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 3139 |
\(\displaystyle \frac {-\frac {4 \left (a^2-b^2\right )^2 \int \frac {1}{a \tan ^2\left (\frac {1}{2} (c+d x)\right )+2 b \tan \left (\frac {1}{2} (c+d x)\right )+a}d\tan \left (\frac {1}{2} (c+d x)\right )}{a b d}+\frac {2 b^2 \int \csc (c+d x)dx}{a}+\frac {x \left (2 a^2-3 b^2\right )}{b}}{2 b^2}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 1083 |
\(\displaystyle \frac {\frac {8 \left (a^2-b^2\right )^2 \int \frac {1}{-\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )^2-4 \left (a^2-b^2\right )}d\left (2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a b d}+\frac {2 b^2 \int \csc (c+d x)dx}{a}+\frac {x \left (2 a^2-3 b^2\right )}{b}}{2 b^2}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {\frac {2 b^2 \int \csc (c+d x)dx}{a}-\frac {4 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a b d}+\frac {x \left (2 a^2-3 b^2\right )}{b}}{2 b^2}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {-\frac {4 \left (a^2-b^2\right )^{3/2} \arctan \left (\frac {2 a \tan \left (\frac {1}{2} (c+d x)\right )+2 b}{2 \sqrt {a^2-b^2}}\right )}{a b d}+\frac {x \left (2 a^2-3 b^2\right )}{b}-\frac {2 b^2 \text {arctanh}(\cos (c+d x))}{a d}}{2 b^2}+\frac {a \cos (c+d x)}{b^2 d}-\frac {\sin (c+d x) \cos (c+d x)}{2 b d}\) |
Input:
Int[(Cos[c + d*x]^3*Cot[c + d*x])/(a + b*Sin[c + d*x]),x]
Output:
(((2*a^2 - 3*b^2)*x)/b - (4*(a^2 - b^2)^(3/2)*ArcTan[(2*b + 2*a*Tan[(c + d *x)/2])/(2*Sqrt[a^2 - b^2])])/(a*b*d) - (2*b^2*ArcTanh[Cos[c + d*x]])/(a*d ))/(2*b^2) + (a*Cos[c + d*x])/(b^2*d) - (Cos[c + d*x]*Sin[c + d*x])/(2*b*d )
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2 Subst[I nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x]
Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = Fre eFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + 2*b*e*x + a *e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ [a^2 - b^2, 0]
Int[cos[(e_.) + (f_.)*(x_)]^4*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[a*(n + 3)*Cos[e + f* x]*(d*Sin[e + f*x])^(n + 1)*((a + b*Sin[e + f*x])^(m + 1)/(b^2*d*f*(m + n + 3)*(m + n + 4))), x] + (-Simp[Cos[e + f*x]*(d*Sin[e + f*x])^(n + 2)*((a + b*Sin[e + f*x])^(m + 1)/(b*d^2*f*(m + n + 4))), x] - Simp[1/(b^2*(m + n + 3 )*(m + n + 4)) Int[(d*Sin[e + f*x])^n*(a + b*Sin[e + f*x])^m*Simp[a^2*(n + 1)*(n + 3) - b^2*(m + n + 3)*(m + n + 4) + a*b*m*Sin[e + f*x] - (a^2*(n + 2)*(n + 3) - b^2*(m + n + 3)*(m + n + 5))*Sin[e + f*x]^2, x], x], x]) /; F reeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 - b^2, 0] && (IGtQ[m, 0] || Integ ersQ[2*m, 2*n]) && !m < -1 && !LtQ[n, -1] && NeQ[m + n + 3, 0] && NeQ[m + n + 4, 0]
Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 2)/(((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.) *(x_)])), x_Symbol] :> Simp[C*(x/(b*d)), x] + (Simp[(A*b^2 - a*b*B + a^2*C) /(b*(b*c - a*d)) Int[1/(a + b*Sin[e + f*x]), x], x] - Simp[(c^2*C - B*c*d + A*d^2)/(d*(b*c - a*d)) Int[1/(c + d*Sin[e + f*x]), x], x]) /; FreeQ[{a , b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 2.61 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.45
method | result | size |
derivativedivides | \(\frac {\frac {\left (-2 a^{4}+4 a^{2} b^{2}-2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \,b^{3} \sqrt {a^{2}-b^{2}}}+\frac {\frac {2 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a b -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{2}+a b \right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (2 a^{2}-3 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\) | \(180\) |
default | \(\frac {\frac {\left (-2 a^{4}+4 a^{2} b^{2}-2 b^{4}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a \,b^{3} \sqrt {a^{2}-b^{2}}}+\frac {\frac {2 \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} b^{2}}{2}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a b -\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}}{2}+a b \right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\left (2 a^{2}-3 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{3}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}}{d}\) | \(180\) |
risch | \(\frac {x \,a^{2}}{b^{3}}-\frac {3 x}{2 b}+\frac {a \,{\mathrm e}^{i \left (d x +c \right )}}{2 b^{2} d}+\frac {a \,{\mathrm e}^{-i \left (d x +c \right )}}{2 b^{2} d}+\frac {i \sqrt {a^{2}-b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (-a +\sqrt {a^{2}-b^{2}}\right )}{b}\right )}{d \,b^{3}}-\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \left (-a +\sqrt {a^{2}-b^{2}}\right )}{b}\right )}{d b a}-\frac {i \sqrt {a^{2}-b^{2}}\, a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (a +\sqrt {a^{2}-b^{2}}\right )}{b}\right )}{d \,b^{3}}+\frac {i \sqrt {a^{2}-b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \left (a +\sqrt {a^{2}-b^{2}}\right )}{b}\right )}{d b a}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a d}-\frac {\sin \left (2 d x +2 c \right )}{4 b d}\) | \(320\) |
Input:
int(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*((-2*a^4+4*a^2*b^2-2*b^4)/a/b^3/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/ 2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+2/b^3*((1/2*tan(1/2*d*x+1/2*c)^3*b^2+ta n(1/2*d*x+1/2*c)^2*a*b-1/2*tan(1/2*d*x+1/2*c)*b^2+a*b)/(1+tan(1/2*d*x+1/2* c)^2)^2+1/2*(2*a^2-3*b^2)*arctan(tan(1/2*d*x+1/2*c)))+1/a*ln(tan(1/2*d*x+1 /2*c)))
Time = 0.19 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.82 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\left [-\frac {a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} b \cos \left (d x + c\right ) + b^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - b^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x - {\left (-a^{2} + b^{2}\right )}^{\frac {3}{2}} \log \left (-\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} - 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right )}{2 \, a b^{3} d}, -\frac {a b^{2} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 2 \, a^{2} b \cos \left (d x + c\right ) + b^{3} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - b^{3} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left (2 \, a^{3} - 3 \, a b^{2}\right )} d x - 2 \, {\left (a^{2} - b^{2}\right )}^{\frac {3}{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right )}{2 \, a b^{3} d}\right ] \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
[-1/2*(a*b^2*cos(d*x + c)*sin(d*x + c) - 2*a^2*b*cos(d*x + c) + b^3*log(1/ 2*cos(d*x + c) + 1/2) - b^3*log(-1/2*cos(d*x + c) + 1/2) - (2*a^3 - 3*a*b^ 2)*d*x - (-a^2 + b^2)^(3/2)*log(-((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin (d*x + c) - a^2 - b^2 - 2*(a*cos(d*x + c)*sin(d*x + c) + b*cos(d*x + c))*s qrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)))/( a*b^3*d), -1/2*(a*b^2*cos(d*x + c)*sin(d*x + c) - 2*a^2*b*cos(d*x + c) + b ^3*log(1/2*cos(d*x + c) + 1/2) - b^3*log(-1/2*cos(d*x + c) + 1/2) - (2*a^3 - 3*a*b^2)*d*x - 2*(a^2 - b^2)^(3/2)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a ^2 - b^2)*cos(d*x + c))))/(a*b^3*d)]
Timed out. \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Timed out} \] Input:
integrate(cos(d*x+c)**3*cot(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
Timed out
Exception generated. \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f or more de
Time = 0.15 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.48 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {2 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a} + \frac {{\left (2 \, a^{2} - 3 \, b^{2}\right )} {\left (d x + c\right )}}{b^{3}} - \frac {4 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a b^{3}} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} b^{2}}}{2 \, d} \] Input:
integrate(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
1/2*(2*log(abs(tan(1/2*d*x + 1/2*c)))/a + (2*a^2 - 3*b^2)*(d*x + c)/b^3 - 4*(a^4 - 2*a^2*b^2 + b^4)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arcta n((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a*b^3) + 2*(b*tan(1/2*d*x + 1/2*c)^3 + 2*a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c) + 2*a)/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*b^2))/d
Time = 42.72 (sec) , antiderivative size = 1320, normalized size of antiderivative = 10.65 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
int((cos(c + d*x)^3*cot(c + d*x))/(a + b*sin(c + d*x)),x)
Output:
log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))/(a*d) - sin(2*c + 2*d*x)/(4*b*d ) - (3*atan((2*a^3*cos(c/2 + (d*x)/2) + 2*b^3*sin(c/2 + (d*x)/2) - 3*a*b^2 *cos(c/2 + (d*x)/2))/(2*b^3*cos(c/2 + (d*x)/2) - 2*a^3*sin(c/2 + (d*x)/2) + 3*a*b^2*sin(c/2 + (d*x)/2))))/(b*d) + (a*cos(c + d*x))/(b^2*d) + (2*a^2* atan((2*a^3*cos(c/2 + (d*x)/2) + 2*b^3*sin(c/2 + (d*x)/2) - 3*a*b^2*cos(c/ 2 + (d*x)/2))/(2*b^3*cos(c/2 + (d*x)/2) - 2*a^3*sin(c/2 + (d*x)/2) + 3*a*b ^2*sin(c/2 + (d*x)/2))))/(b^3*d) + (atan((b^6*sin(c/2 + (d*x)/2)*(b^6 - a^ 6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2)*64i - a^12*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*16i - a^6*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3 *a^2*b^4 + 3*a^4*b^2)^(3/2)*16i - a^3*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2)*42i + a^3*b^9*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*66i - a^5*b^7*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*176i + a^7*b^5*cos(c/2 + (d*x)/2)*(b^6 - a^ 6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*178i - a^9*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*81i - a^2*b^4*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2)*116i + a^4*b^2*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(3/2)*72i + a^2*b^10*sin(c/2 + (d*x)/2)*(b ^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*148i - a^4*b^8*sin(c/2 + (d*x)/2)* (b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*460i + a^6*b^6*sin(c/2 + (d*x)/2 )*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*577i - a^8*b^4*sin(c/2 + (d...
Time = 3.60 (sec) , antiderivative size = 169, normalized size of antiderivative = 1.36 \[ \int \frac {\cos ^3(c+d x) \cot (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) a^{2}+4 \sqrt {a^{2}-b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a +b}{\sqrt {a^{2}-b^{2}}}\right ) b^{2}-\cos \left (d x +c \right ) \sin \left (d x +c \right ) a \,b^{2}+2 \cos \left (d x +c \right ) a^{2} b +2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{3}+2 a^{3} c +2 a^{3} d x -3 a \,b^{2} c -3 a \,b^{2} d x}{2 a \,b^{3} d} \] Input:
int(cos(d*x+c)^3*cot(d*x+c)/(a+b*sin(d*x+c)),x)
Output:
( - 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2))*a **2 + 4*sqrt(a**2 - b**2)*atan((tan((c + d*x)/2)*a + b)/sqrt(a**2 - b**2)) *b**2 - cos(c + d*x)*sin(c + d*x)*a*b**2 + 2*cos(c + d*x)*a**2*b + 2*log(t an((c + d*x)/2))*b**3 + 2*a**3*c + 2*a**3*d*x - 3*a*b**2*c - 3*a*b**2*d*x) /(2*a*b**3*d)