Integrand size = 29, antiderivative size = 183 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a x}{2 b^2}+\frac {a \left (a^2-3 b^2\right ) x}{b^4}-\frac {2 \left (a^2-b^2\right )^{5/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^2 b^4 d}+\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {\cos (c+d x)}{b d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{b^3 d}-\frac {\cos ^3(c+d x)}{3 b d}-\frac {\cot (c+d x)}{a d}-\frac {a \cos (c+d x) \sin (c+d x)}{2 b^2 d} \] Output:
1/2*a*x/b^2+a*(a^2-3*b^2)*x/b^4-2*(a^2-b^2)^(5/2)*arctan((b+a*tan(1/2*d*x+ 1/2*c))/(a^2-b^2)^(1/2))/a^2/b^4/d+b*arctanh(cos(d*x+c))/a^2/d+cos(d*x+c)/ b/d+(a^2-3*b^2)*cos(d*x+c)/b^3/d-1/3*cos(d*x+c)^3/b/d-cot(d*x+c)/a/d-1/2*a *cos(d*x+c)*sin(d*x+c)/b^2/d
Time = 2.04 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.14 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {-12 a^5 c+30 a^3 b^2 c-12 a^5 d x+30 a^3 b^2 d x+24 \left (a^2-b^2\right )^{5/2} \arctan \left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )-3 a^2 b \left (4 a^2-9 b^2\right ) \cos (c+d x)+a^2 b^3 \cos (3 (c+d x))+6 a b^4 \cot \left (\frac {1}{2} (c+d x)\right )-12 b^5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+12 b^5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+3 a^3 b^2 \sin (2 (c+d x))-6 a b^4 \tan \left (\frac {1}{2} (c+d x)\right )}{12 a^2 b^4 d} \] Input:
Integrate[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
-1/12*(-12*a^5*c + 30*a^3*b^2*c - 12*a^5*d*x + 30*a^3*b^2*d*x + 24*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]] - 3*a^2*b*(4*a ^2 - 9*b^2)*Cos[c + d*x] + a^2*b^3*Cos[3*(c + d*x)] + 6*a*b^4*Cot[(c + d*x )/2] - 12*b^5*Log[Cos[(c + d*x)/2]] + 12*b^5*Log[Sin[(c + d*x)/2]] + 3*a^3 *b^2*Sin[2*(c + d*x)] - 6*a*b^4*Tan[(c + d*x)/2])/(a^2*b^4*d)
Time = 0.52 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {3042, 3376, 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.
\(\displaystyle \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\cos (c+d x)^6}{\sin (c+d x)^2 (a+b \sin (c+d x))}dx\) |
\(\Big \downarrow \) 3376 |
\(\displaystyle \int \left (\frac {a^3-3 a b^2}{b^4}-\frac {\left (a^2-b^2\right )^3}{a^2 b^4 (a+b \sin (c+d x))}+\frac {\left (3 b^2-a^2\right ) \sin (c+d x)}{b^3}-\frac {b \csc (c+d x)}{a^2}+\frac {a \sin ^2(c+d x)}{b^2}+\frac {\csc ^2(c+d x)}{a}-\frac {\sin ^3(c+d x)}{b}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {2 \left (a^2-b^2\right )^{5/2} \arctan \left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 b^4 d}+\frac {b \text {arctanh}(\cos (c+d x))}{a^2 d}+\frac {a x \left (a^2-3 b^2\right )}{b^4}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{b^3 d}-\frac {a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac {a x}{2 b^2}-\frac {\cot (c+d x)}{a d}-\frac {\cos ^3(c+d x)}{3 b d}+\frac {\cos (c+d x)}{b d}\) |
Input:
Int[(Cos[c + d*x]^4*Cot[c + d*x]^2)/(a + b*Sin[c + d*x]),x]
Output:
(a*x)/(2*b^2) + (a*(a^2 - 3*b^2)*x)/b^4 - (2*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*b^4*d) + (b*ArcTanh[Cos[c + d* x]])/(a^2*d) + Cos[c + d*x]/(b*d) + ((a^2 - 3*b^2)*Cos[c + d*x])/(b^3*d) - Cos[c + d*x]^3/(3*b*d) - Cot[c + d*x]/(a*d) - (a*Cos[c + d*x]*Sin[c + d*x ])/(2*b^2*d)
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(d*sin[ e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x] /; Fr eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && ( LtQ[m, -1] || (EqQ[m, -1] && GtQ[p, 0]))
Time = 7.74 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.44
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {\frac {2 \left (\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}+\left (a^{2} b -3 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (2 a^{2} b -4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+a^{2} b -\frac {7 b^{3}}{3}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+a \left (2 a^{2}-5 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}+\frac {\left (-4 a^{6}+12 a^{4} b^{2}-12 a^{2} b^{4}+4 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 a^{2} b^{4} \sqrt {a^{2}-b^{2}}}}{d}\) | \(264\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a}-\frac {1}{2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{2}}+\frac {\frac {2 \left (\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{2}+\left (a^{2} b -3 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (2 a^{2} b -4 b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-\frac {a \,b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2}+a^{2} b -\frac {7 b^{3}}{3}\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}+a \left (2 a^{2}-5 b^{2}\right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{b^{4}}+\frac {\left (-4 a^{6}+12 a^{4} b^{2}-12 a^{2} b^{4}+4 b^{6}\right ) \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{2 a^{2} b^{4} \sqrt {a^{2}-b^{2}}}}{d}\) | \(264\) |
risch | \(\frac {a^{3} x}{b^{4}}-\frac {5 a x}{2 b^{2}}-\frac {i a \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 b^{2} d}+\frac {{\mathrm e}^{i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {9 \,{\mathrm e}^{i \left (d x +c \right )}}{8 b d}+\frac {{\mathrm e}^{-i \left (d x +c \right )} a^{2}}{2 b^{3} d}-\frac {9 \,{\mathrm e}^{-i \left (d x +c \right )}}{8 b d}-\frac {2 i}{a d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}+\frac {i a \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{2} d}-\frac {\sqrt {-a^{2}+b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}+\frac {2 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2}}-\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a +\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{4}}-\frac {2 \sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,b^{2}}+\frac {\sqrt {-a^{2}+b^{2}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a -\sqrt {-a^{2}+b^{2}}}{b}\right )}{d \,a^{2}}-\frac {\cos \left (3 d x +3 c \right )}{12 b d}\) | \(519\) |
Input:
int(cos(d*x+c)^4*cot(d*x+c)^2/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/2*tan(1/2*d*x+1/2*c)/a-1/2/a/tan(1/2*d*x+1/2*c)-1/a^2*b*ln(tan(1/2* d*x+1/2*c))+2/b^4*((1/2*a*b^2*tan(1/2*d*x+1/2*c)^5+(a^2*b-3*b^3)*tan(1/2*d *x+1/2*c)^4+(2*a^2*b-4*b^3)*tan(1/2*d*x+1/2*c)^2-1/2*a*b^2*tan(1/2*d*x+1/2 *c)+a^2*b-7/3*b^3)/(1+tan(1/2*d*x+1/2*c)^2)^3+1/2*a*(2*a^2-5*b^2)*arctan(t an(1/2*d*x+1/2*c)))+1/2*(-4*a^6+12*a^4*b^2-12*a^2*b^4+4*b^6)/a^2/b^4/(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)))
Time = 0.33 (sec) , antiderivative size = 549, normalized size of antiderivative = 3.00 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\left [\frac {3 \, a^{3} b^{2} \cos \left (d x + c\right )^{3} + 3 \, b^{5} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, b^{5} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {-a^{2} + b^{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 ) \sin \left (d x + c\right ) - 3 \, {\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right ) - {\left (2 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{5} - 5 \, a^{3} b^{2}\right )} d x - 6 \, {\left (a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, a^{2} b^{4} d \sin \left (d x + c\right )}, \frac {3 \, a^{3} b^{2} \cos \left (d x + c\right )^{3} + 3 \, b^{5} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, b^{5} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 3 \, {\left (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right ) - {\left (2 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{5} - 5 \, a^{3} b^{2}\right )} d x - 6 \, {\left (a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, a^{2} b^{4} d \sin \left (d x + c\right )}\right ] \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas" )
Output:
[1/6*(3*a^3*b^2*cos(d*x + c)^3 + 3*b^5*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*b^5*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) + 3*(a^4 - 2*a^2*b ^2 + b^4)*sqrt(-a^2 + b^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))*sqr t(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2))*sin( d*x + c) - 3*(a^3*b^2 + 2*a*b^4)*cos(d*x + c) - (2*a^2*b^3*cos(d*x + c)^3 - 3*(2*a^5 - 5*a^3*b^2)*d*x - 6*(a^4*b - 2*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/(a^2*b^4*d*sin(d*x + c)), 1/6*(3*a^3*b^2*cos(d*x + c)^3 + 3*b^5*log( 1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*b^5*log(-1/2*cos(d*x + c) + 1/2)* sin(d*x + c) + 6*(a^4 - 2*a^2*b^2 + b^4)*sqrt(a^2 - b^2)*arctan(-(a*sin(d* x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*sin(d*x + c) - 3*(a^3*b^2 + 2* a*b^4)*cos(d*x + c) - (2*a^2*b^3*cos(d*x + c)^3 - 3*(2*a^5 - 5*a^3*b^2)*d* x - 6*(a^4*b - 2*a^2*b^3)*cos(d*x + c))*sin(d*x + c))/(a^2*b^4*d*sin(d*x + c))]
\[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\cos ^{4}{\left (c + d x \right )} \cot ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(cos(d*x+c)**4*cot(d*x+c)**2/(a+b*sin(d*x+c)),x)
Output:
Integral(cos(c + d*x)**4*cot(c + d*x)**2/(a + b*sin(c + d*x)), x)
Exception generated. \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^2/(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 = 0.26 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.65 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {3 \, {\left (2 \, a^{3} - 5 \, a b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {3 \, {\left (2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {12 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2} b^{4}} - \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 18 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} - 14 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \] Input:
integrate(cos(d*x+c)^4*cot(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
-1/6*(6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 3*tan(1/2*d*x + 1/2*c)/a - 3*(2*a^3 - 5*a*b^2)*(d*x + c)/b^4 - 3*(2*b*tan(1/2*d*x + 1/2*c) - a)/(a^2* tan(1/2*d*x + 1/2*c)) + 12*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(pi*floor(1 /2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a ^2 - b^2)))/(sqrt(a^2 - b^2)*a^2*b^4) - 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*tan(1/2*d*x + 1/2*c)^4 - 18*b^2*tan(1/2*d*x + 1/2*c)^4 + 12*a^2*tan( 1/2*d*x + 1/2*c)^2 - 24*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1 /2*c) + 6*a^2 - 14*b^2)/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*b^3))/d
Time = 20.64 (sec) , antiderivative size = 4892, normalized size of antiderivative = 26.73 \[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\text {Too large to display} \] Input:
int((cos(c + d*x)^4*cot(c + d*x)^2)/(a + b*sin(c + d*x)),x)
Output:
tan(c/2 + (d*x)/2)/(2*a*d) - (3*tan(c/2 + (d*x)/2)^4 + (8*tan(c/2 + (d*x)/ 2)^3*(2*a*b^2 - a^3))/b^3 + (4*tan(c/2 + (d*x)/2)^5*(3*a*b^2 - a^3))/b^3 + (tan(c/2 + (d*x)/2)^2*(2*a^2 + 3*b^2))/b^2 - (tan(c/2 + (d*x)/2)^6*(2*a^2 - b^2))/b^2 + (4*tan(c/2 + (d*x)/2)*(7*a*b^2 - 3*a^3))/(3*b^3) + 1)/(d*(2 *a*tan(c/2 + (d*x)/2) + 6*a*tan(c/2 + (d*x)/2)^3 + 6*a*tan(c/2 + (d*x)/2)^ 5 + 2*a*tan(c/2 + (d*x)/2)^7)) - (b*log(tan(c/2 + (d*x)/2)))/(a^2*d) - (at an((((-(a + b)^5*(a - b)^5)^(1/2)*((8*(10*a^2*b^12 - 14*a^14 + 136*a^4*b^1 0 - 386*a^6*b^8 + 467*a^8*b^6 - 309*a^10*b^4 + 105*a^12*b^2))/(a^2*b^8) + ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(16*b^15 - 52*a^2*b^13 + 50*a^4*b^11 + 8 6*a^6*b^9 - 155*a^8*b^7 + 76*a^10*b^5 - 12*a^12*b^3))/(a^2*b^8) + (8*tan(c /2 + (d*x)/2)*(84*a^2*b^18 - 360*a^4*b^16 + 1020*a^6*b^14 - 1264*a^8*b^12 + 661*a^10*b^10 - 140*a^12*b^8 + 8*a^14*b^6))/(a^3*b^12) + ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(32*a^2*b^14 - 64*a^4*b^12 + 50*a^6*b^10 - 14*a^8*b^8)) /(a^2*b^8) + (((8*(16*a^4*b^13 - 12*a^6*b^11))/(a^2*b^8) + (8*tan(c/2 + (d *x)/2)*(64*a^4*b^18 - 68*a^6*b^16 + 8*a^8*b^14))/(a^3*b^12))*(-(a + b)^5*( a - b)^5)^(1/2))/(a^2*b^4) + (8*tan(c/2 + (d*x)/2)*(64*a^2*b^19 - 148*a^4* b^17 + 152*a^6*b^15 - 80*a^8*b^13 + 16*a^10*b^11))/(a^3*b^12)))/(a^2*b^4)) )/(a^2*b^4) + (8*tan(c/2 + (d*x)/2)*(4*b^19 - 24*a^2*b^17 + 460*a^4*b^15 - 1300*a^6*b^13 + 1769*a^8*b^11 - 1408*a^10*b^9 + 652*a^12*b^7 - 160*a^14*b ^5 + 16*a^16*b^3))/(a^3*b^12))*1i)/(a^2*b^4) + ((-(a + b)^5*(a - b)^5)^...
\[ \int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\cos \left (d x +c \right )^{4} \cot \left (d x +c \right )^{2}}{\sin \left (d x +c \right ) b +a}d x \] Input:
int(cos(d*x+c)^4*cot(d*x+c)^2/(a+b*sin(d*x+c)),x)
Output:
int(cos(d*x+c)^4*cot(d*x+c)^2/(a+b*sin(d*x+c)),x)