Integrand size = 23, antiderivative size = 89 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {(2 a+b) \csc ^2(c+d x)}{2 a^2 d}-\frac {\csc ^4(c+d x)}{4 a d}+\frac {(a+b)^2 \log (\sin (c+d x))}{a^3 d}-\frac {(a+b)^2 \log \left (a+b \sin ^2(c+d x)\right )}{2 a^3 d} \] Output:
1/2*(2*a+b)*csc(d*x+c)^2/a^2/d-1/4*csc(d*x+c)^4/a/d+(a+b)^2*ln(sin(d*x+c)) /a^3/d-1/2*(a+b)^2*ln(a+b*sin(d*x+c)^2)/a^3/d
Time = 0.59 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.81 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {2 a (2 a+b) \csc ^2(c+d x)-a^2 \csc ^4(c+d x)+2 (a+b)^2 \left (2 \log (\sin (c+d x))-\log \left (a+b \sin ^2(c+d x)\right )\right )}{4 a^3 d} \] Input:
Integrate[Cot[c + d*x]^5/(a + b*Sin[c + d*x]^2),x]
Output:
(2*a*(2*a + b)*Csc[c + d*x]^2 - a^2*Csc[c + d*x]^4 + 2*(a + b)^2*(2*Log[Si n[c + d*x]] - Log[a + b*Sin[c + d*x]^2]))/(4*a^3*d)
Time = 0.29 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.91, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 3673, 99, 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.
| \(\displaystyle \int \frac {\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (c+d x)^5 \left (a+b \sin (c+d x)^2\right )}dx\) |
| \(\Big \downarrow \) 3673 |
| \(\displaystyle \frac {\int \frac {\csc ^6(c+d x) \left (1-\sin ^2(c+d x)\right )^2}{b \sin ^2(c+d x)+a}d\sin ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle \frac {\int \left (\frac {\csc ^6(c+d x)}{a}+\frac {(-2 a-b) \csc ^4(c+d x)}{a^2}+\frac {(a+b)^2 \csc ^2(c+d x)}{a^3}-\frac {b (a+b)^2}{a^3 \left (b \sin ^2(c+d x)+a\right )}\right )d\sin ^2(c+d x)}{2 d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {(a+b)^2 \log \left (\sin ^2(c+d x)\right )}{a^3}-\frac {(a+b)^2 \log \left (a+b \sin ^2(c+d x)\right )}{a^3}+\frac {(2 a+b) \csc ^2(c+d x)}{a^2}-\frac {\csc ^4(c+d x)}{2 a}}{2 d}\) |
Input:
Int[Cot[c + d*x]^5/(a + b*Sin[c + d*x]^2),x]
Output:
(((2*a + b)*Csc[c + d*x]^2)/a^2 - Csc[c + d*x]^4/(2*a) + ((a + b)^2*Log[Si n[c + d*x]^2])/a^3 - ((a + b)^2*Log[a + b*Sin[c + d*x]^2])/a^3)/(2*d)
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^ (m_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x]^2, x]}, Simp[ff^((m + 1)/2)/(2*f) Subst[Int[x^((m - 1)/2)*((a + b*ff*x)^p/(1 - ff*x)^((m + 1 )/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && Integ erQ[(m - 1)/2]
Time = 3.63 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.81
| method | result | size |
| derivativedivides | \(\frac {-\frac {1}{16 a \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {-7 a -4 b}{16 a^{2} \left (\cos \left (d x +c \right )+1\right )}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (\cos \left (d x +c \right )+1\right )}{2 a^{3}}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (a +b -b \cos \left (d x +c \right )^{2}\right )}{2 a^{3}}-\frac {1}{16 a \left (\cos \left (d x +c \right )-1\right )^{2}}-\frac {7 a +4 b}{16 a^{2} \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (\cos \left (d x +c \right )-1\right )}{2 a^{3}}}{d}\) | \(161\) |
| default | \(\frac {-\frac {1}{16 a \left (\cos \left (d x +c \right )+1\right )^{2}}-\frac {-7 a -4 b}{16 a^{2} \left (\cos \left (d x +c \right )+1\right )}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (\cos \left (d x +c \right )+1\right )}{2 a^{3}}-\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (a +b -b \cos \left (d x +c \right )^{2}\right )}{2 a^{3}}-\frac {1}{16 a \left (\cos \left (d x +c \right )-1\right )^{2}}-\frac {7 a +4 b}{16 a^{2} \left (\cos \left (d x +c \right )-1\right )}+\frac {\left (a^{2}+2 a b +b^{2}\right ) \ln \left (\cos \left (d x +c \right )-1\right )}{2 a^{3}}}{d}\) | \(161\) |
| risch | \(-\frac {2 \left (2 \,{\mathrm e}^{6 i \left (d x +c \right )} a +{\mathrm e}^{6 i \left (d x +c \right )} b -2 \,{\mathrm e}^{4 i \left (d x +c \right )} a -2 b \,{\mathrm e}^{4 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{2 i \left (d x +c \right )}\right )}{d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a d}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b}{a^{2} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{4 i \left (d x +c \right )}-\frac {2 \left (2 a +b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{b}+1\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{4 i \left (d x +c \right )}-\frac {2 \left (2 a +b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{b}+1\right ) b}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{4 i \left (d x +c \right )}-\frac {2 \left (2 a +b \right ) {\mathrm e}^{2 i \left (d x +c \right )}}{b}+1\right ) b^{2}}{2 d \,a^{3}}\) | \(277\) |
Input:
int(cot(d*x+c)^5/(a+b*sin(d*x+c)^2),x,method=_RETURNVERBOSE)
Output:
1/d*(-1/16/a/(cos(d*x+c)+1)^2-1/16*(-7*a-4*b)/a^2/(cos(d*x+c)+1)+1/2*(a^2+ 2*a*b+b^2)/a^3*ln(cos(d*x+c)+1)-1/2*(a^2+2*a*b+b^2)/a^3*ln(a+b-b*cos(d*x+c )^2)-1/16/a/(cos(d*x+c)-1)^2-1/16*(7*a+4*b)/a^2/(cos(d*x+c)-1)+1/2*(a^2+2* a*b+b^2)/a^3*ln(cos(d*x+c)-1))
Leaf count of result is larger than twice the leaf count of optimal. 198 vs. \(2 (83) = 166\).
Time = 0.14 (sec) , antiderivative size = 198, normalized size of antiderivative = 2.22 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {2 \, {\left (2 \, a^{2} + a b\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 2 \, a b + 2 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \log \left (-b \cos \left (d x + c\right )^{2} + a + b\right ) - 4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \log \left (\frac {1}{2} \, \sin \left (d x + c\right )\right )}{4 \, {\left (a^{3} d \cos \left (d x + c\right )^{4} - 2 \, a^{3} d \cos \left (d x + c\right )^{2} + a^{3} d\right )}} \] Input:
integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="fricas")
Output:
-1/4*(2*(2*a^2 + a*b)*cos(d*x + c)^2 - 3*a^2 - 2*a*b + 2*((a^2 + 2*a*b + b ^2)*cos(d*x + c)^4 - 2*(a^2 + 2*a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*log(-b*cos(d*x + c)^2 + a + b) - 4*((a^2 + 2*a*b + b^2)*cos(d*x + c)^ 4 - 2*(a^2 + 2*a*b + b^2)*cos(d*x + c)^2 + a^2 + 2*a*b + b^2)*log(1/2*sin( d*x + c)))/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^2 + a^3*d)
\[ \int \frac {\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\int \frac {\cot ^{5}{\left (c + d x \right )}}{a + b \sin ^{2}{\left (c + d x \right )}}\, dx \] Input:
integrate(cot(d*x+c)**5/(a+b*sin(d*x+c)**2),x)
Output:
Integral(cot(c + d*x)**5/(a + b*sin(c + d*x)**2), x)
Time = 0.03 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.03 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=-\frac {\frac {2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (b \sin \left (d x + c\right )^{2} + a\right )}{a^{3}} - \frac {2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\sin \left (d x + c\right )^{2}\right )}{a^{3}} - \frac {2 \, {\left (2 \, a + b\right )} \sin \left (d x + c\right )^{2} - a}{a^{2} \sin \left (d x + c\right )^{4}}}{4 \, d} \] Input:
integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="maxima")
Output:
-1/4*(2*(a^2 + 2*a*b + b^2)*log(b*sin(d*x + c)^2 + a)/a^3 - 2*(a^2 + 2*a*b + b^2)*log(sin(d*x + c)^2)/a^3 - (2*(2*a + b)*sin(d*x + c)^2 - a)/(a^2*si n(d*x + c)^4))/d
Time = 0.67 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.21 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{3} d} - \frac {{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right )^{2} + a \right |}\right )}{2 \, a^{3} b d} + \frac {2 \, {\left (2 \, a^{2} + a b\right )} \sin \left (d x + c\right )^{2} - a^{2}}{4 \, a^{3} d \sin \left (d x + c\right )^{4}} \] Input:
integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)^2),x, algorithm="giac")
Output:
(a^2 + 2*a*b + b^2)*log(abs(sin(d*x + c)))/(a^3*d) - 1/2*(a^2*b + 2*a*b^2 + b^3)*log(abs(b*sin(d*x + c)^2 + a))/(a^3*b*d) + 1/4*(2*(2*a^2 + a*b)*sin (d*x + c)^2 - a^2)/(a^3*d*sin(d*x + c)^4)
Time = 34.82 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.16 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (a^2+2\,a\,b+b^2\right )}{a^3\,d}-\frac {\ln \left (a+a\,{\mathrm {tan}\left (c+d\,x\right )}^2+b\,{\mathrm {tan}\left (c+d\,x\right )}^2\right )\,\left (a^2+2\,a\,b+b^2\right )}{2\,a^3\,d}-\frac {\frac {1}{4\,a}-\frac {{\mathrm {tan}\left (c+d\,x\right )}^2\,\left (a+b\right )}{2\,a^2}}{d\,{\mathrm {tan}\left (c+d\,x\right )}^4} \] Input:
int(cot(c + d*x)^5/(a + b*sin(c + d*x)^2),x)
Output:
(log(tan(c + d*x))*(2*a*b + a^2 + b^2))/(a^3*d) - (log(a + a*tan(c + d*x)^ 2 + b*tan(c + d*x)^2)*(2*a*b + a^2 + b^2))/(2*a^3*d) - (1/(4*a) - (tan(c + d*x)^2*(a + b))/(2*a^2))/(d*tan(c + d*x)^4)
Time = 0.21 (sec) , antiderivative size = 521, normalized size of antiderivative = 5.85 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin ^2(c+d x)} \, dx=\frac {-16 \,\mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a^{2}-32 \,\mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a b -16 \,\mathrm {log}\left (-\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} b^{2}-16 \,\mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a^{2}-32 \,\mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a b -16 \,\mathrm {log}\left (\sqrt {2 \sqrt {b}\, \sqrt {a +b}-a -2 b}+\sqrt {a}\, \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} b^{2}-16 \,\mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +a +2 b \right ) \sin \left (d x +c \right )^{4} a^{2}-32 \,\mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +a +2 b \right ) \sin \left (d x +c \right )^{4} a b -16 \,\mathrm {log}\left (2 \sqrt {b}\, \sqrt {a +b}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +a +2 b \right ) \sin \left (d x +c \right )^{4} b^{2}+32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a^{2}+64 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a b +32 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} b^{2}-13 \sin \left (d x +c \right )^{4} a^{2}-8 \sin \left (d x +c \right )^{4} a b +32 \sin \left (d x +c \right )^{2} a^{2}+16 \sin \left (d x +c \right )^{2} a b -8 a^{2}}{32 \sin \left (d x +c \right )^{4} a^{3} d} \] Input:
int(cot(d*x+c)^5/(a+b*sin(d*x+c)^2),x)
Output:
( - 16*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x )/2))*sin(c + d*x)**4*a**2 - 32*log( - sqrt(2*sqrt(b)*sqrt(a + b) - a - 2* b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**4*a*b - 16*log( - sqrt(2*sqrt (b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**4*b** 2 - 16*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2 ))*sin(c + d*x)**4*a**2 - 32*log(sqrt(2*sqrt(b)*sqrt(a + b) - a - 2*b) + s qrt(a)*tan((c + d*x)/2))*sin(c + d*x)**4*a*b - 16*log(sqrt(2*sqrt(b)*sqrt( a + b) - a - 2*b) + sqrt(a)*tan((c + d*x)/2))*sin(c + d*x)**4*b**2 - 16*lo g(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*sin(c + d*x)**4 *a**2 - 32*log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*si n(c + d*x)**4*a*b - 16*log(2*sqrt(b)*sqrt(a + b) + tan((c + d*x)/2)**2*a + a + 2*b)*sin(c + d*x)**4*b**2 + 32*log(tan((c + d*x)/2))*sin(c + d*x)**4* a**2 + 64*log(tan((c + d*x)/2))*sin(c + d*x)**4*a*b + 32*log(tan((c + d*x) /2))*sin(c + d*x)**4*b**2 - 13*sin(c + d*x)**4*a**2 - 8*sin(c + d*x)**4*a* b + 32*sin(c + d*x)**2*a**2 + 16*sin(c + d*x)**2*a*b - 8*a**2)/(32*sin(c + d*x)**4*a**3*d)