Integrand size = 21, antiderivative size = 148 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b \left (2 a^2-b^2\right ) \csc (c+d x)}{a^4 d}+\frac {\left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^3 d}+\frac {b \csc ^3(c+d x)}{3 a^2 d}-\frac {\csc ^4(c+d x)}{4 a d}+\frac {\left (a^2-b^2\right )^2 \log (\sin (c+d x))}{a^5 d}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^5 d} \] Output:
-b*(2*a^2-b^2)*csc(d*x+c)/a^4/d+1/2*(2*a^2-b^2)*csc(d*x+c)^2/a^3/d+1/3*b*c sc(d*x+c)^3/a^2/d-1/4*csc(d*x+c)^4/a/d+(a^2-b^2)^2*ln(sin(d*x+c))/a^5/d-(a ^2-b^2)^2*ln(a+b*sin(d*x+c))/a^5/d
Time = 0.78 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.78 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {12 a b \left (-2 a^2+b^2\right ) \csc (c+d x)+6 a^2 \left (2 a^2-b^2\right ) \csc ^2(c+d x)+4 a^3 b \csc ^3(c+d x)-3 a^4 \csc ^4(c+d x)+12 \left (a^2-b^2\right )^2 (\log (\sin (c+d x))-\log (a+b \sin (c+d x)))}{12 a^5 d} \] Input:
Integrate[Cot[c + d*x]^5/(a + b*Sin[c + d*x]),x]
Output:
(12*a*b*(-2*a^2 + b^2)*Csc[c + d*x] + 6*a^2*(2*a^2 - b^2)*Csc[c + d*x]^2 + 4*a^3*b*Csc[c + d*x]^3 - 3*a^4*Csc[c + d*x]^4 + 12*(a^2 - b^2)^2*(Log[Sin [c + d*x]] - Log[a + b*Sin[c + d*x]]))/(12*a^5*d)
Time = 0.36 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3200, 522, 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.
| \(\displaystyle \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (c+d x)^5 (a+b \sin (c+d x))}dx\) |
| \(\Big \downarrow \) 3200 |
| \(\displaystyle \frac {\int \frac {\csc ^5(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )^2}{b^5 (a+b \sin (c+d x))}d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 522 |
| \(\displaystyle \frac {\int \left (\frac {\csc ^5(c+d x)}{a b}-\frac {\csc ^4(c+d x)}{a^2}+\frac {\left (b^4-2 a^2 b^2\right ) \csc ^3(c+d x)}{a^3 b^3}+\frac {\left (2 a^2 b^2-b^4\right ) \csc ^2(c+d x)}{a^4 b^2}+\frac {\left (a^2-b^2\right )^2 \csc (c+d x)}{a^5 b}-\frac {\left (a^2-b^2\right )^2}{a^5 (a+b \sin (c+d x))}\right )d(b \sin (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b \csc ^3(c+d x)}{3 a^2}+\frac {\left (a^2-b^2\right )^2 \log (b \sin (c+d x))}{a^5}-\frac {\left (a^2-b^2\right )^2 \log (a+b \sin (c+d x))}{a^5}-\frac {b \left (2 a^2-b^2\right ) \csc (c+d x)}{a^4}+\frac {\left (2 a^2-b^2\right ) \csc ^2(c+d x)}{2 a^3}-\frac {\csc ^4(c+d x)}{4 a}}{d}\) |
Input:
Int[Cot[c + d*x]^5/(a + b*Sin[c + d*x]),x]
Output:
(-((b*(2*a^2 - b^2)*Csc[c + d*x])/a^4) + ((2*a^2 - b^2)*Csc[c + d*x]^2)/(2 *a^3) + (b*Csc[c + d*x]^3)/(3*a^2) - Csc[c + d*x]^4/(4*a) + ((a^2 - b^2)^2 *Log[b*Sin[c + d*x]])/a^5 - ((a^2 - b^2)^2*Log[a + b*Sin[c + d*x]])/a^5)/d
Int[((e_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_. ), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*(c + d*x)^n*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && IGtQ[p, 0]
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p _.), x_Symbol] :> Simp[1/f Subst[Int[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b ^2, 0] && IntegerQ[(p + 1)/2]
Time = 1.41 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.93
| method | result | size |
| derivativedivides | \(\frac {-\frac {\left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{5}}-\frac {1}{4 a \sin \left (d x +c \right )^{4}}-\frac {-2 a^{2}+b^{2}}{2 a^{3} \sin \left (d x +c \right )^{2}}+\frac {\left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{5}}-\frac {\left (2 a^{2}-b^{2}\right ) b}{a^{4} \sin \left (d x +c \right )}+\frac {b}{3 a^{2} \sin \left (d x +c \right )^{3}}}{d}\) | \(137\) |
| default | \(\frac {-\frac {\left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{5}}-\frac {1}{4 a \sin \left (d x +c \right )^{4}}-\frac {-2 a^{2}+b^{2}}{2 a^{3} \sin \left (d x +c \right )^{2}}+\frac {\left (a^{4}-2 b^{2} a^{2}+b^{4}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{5}}-\frac {\left (2 a^{2}-b^{2}\right ) b}{a^{4} \sin \left (d x +c \right )}+\frac {b}{3 a^{2} \sin \left (d x +c \right )^{3}}}{d}\) | \(137\) |
| risch | \(\frac {2 i \left (6 i a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-6 a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+3 b^{3} {\mathrm e}^{7 i \left (d x +c \right )}-6 i a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+6 i a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+14 a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-9 b^{3} {\mathrm e}^{5 i \left (d x +c \right )}+6 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}-3 i a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-14 a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+9 b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+6 a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-3 b^{3} {\mathrm e}^{i \left (d x +c \right )}\right )}{3 d \,a^{4} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a d}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{3} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{4}}{a^{5} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{a d}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{2}}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{4}}{a^{5} d}\) | \(412\) |
Input:
int(cot(d*x+c)^5/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-(a^4-2*a^2*b^2+b^4)/a^5*ln(a+b*sin(d*x+c))-1/4/a/sin(d*x+c)^4-1/2*(- 2*a^2+b^2)/a^3/sin(d*x+c)^2+(a^4-2*a^2*b^2+b^4)/a^5*ln(sin(d*x+c))-(2*a^2- b^2)/a^4*b/sin(d*x+c)+1/3/a^2*b/sin(d*x+c)^3)
Time = 0.15 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.83 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {9 \, a^{4} - 6 \, a^{2} b^{2} - 6 \, {\left (2 \, a^{4} - a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2} - 12 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + 12 \, {\left ({\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right ) - 4 \, {\left (5 \, a^{3} b - 3 \, a b^{3} - 3 \, {\left (2 \, a^{3} b - a b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{5} d \cos \left (d x + c\right )^{4} - 2 \, a^{5} d \cos \left (d x + c\right )^{2} + a^{5} d\right )}} \] Input:
integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="fricas")
Output:
1/12*(9*a^4 - 6*a^2*b^2 - 6*(2*a^4 - a^2*b^2)*cos(d*x + c)^2 - 12*((a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b ^2 + b^4)*cos(d*x + c)^2)*log(b*sin(d*x + c) + a) + 12*((a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^4 + a^4 - 2*a^2*b^2 + b^4 - 2*(a^4 - 2*a^2*b^2 + b^4)*c os(d*x + c)^2)*log(-1/2*sin(d*x + c)) - 4*(5*a^3*b - 3*a*b^3 - 3*(2*a^3*b - a*b^3)*cos(d*x + c)^2)*sin(d*x + c))/(a^5*d*cos(d*x + c)^4 - 2*a^5*d*cos (d*x + c)^2 + a^5*d)
\[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\cot ^{5}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \] Input:
integrate(cot(d*x+c)**5/(a+b*sin(d*x+c)),x)
Output:
Integral(cot(c + d*x)**5/(a + b*sin(c + d*x)), x)
Time = 0.03 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.94 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {\frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{5}} - \frac {12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{5}} - \frac {4 \, a^{2} b \sin \left (d x + c\right ) - 12 \, {\left (2 \, a^{2} b - b^{3}\right )} \sin \left (d x + c\right )^{3} - 3 \, a^{3} + 6 \, {\left (2 \, a^{3} - a b^{2}\right )} \sin \left (d x + c\right )^{2}}{a^{4} \sin \left (d x + c\right )^{4}}}{12 \, d} \] Input:
integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="maxima")
Output:
-1/12*(12*(a^4 - 2*a^2*b^2 + b^4)*log(b*sin(d*x + c) + a)/a^5 - 12*(a^4 - 2*a^2*b^2 + b^4)*log(sin(d*x + c))/a^5 - (4*a^2*b*sin(d*x + c) - 12*(2*a^2 *b - b^3)*sin(d*x + c)^3 - 3*a^3 + 6*(2*a^3 - a*b^2)*sin(d*x + c)^2)/(a^4* sin(d*x + c)^4))/d
Time = 0.13 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.03 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{5} d} - \frac {{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{5} b d} + \frac {4 \, a^{3} b \sin \left (d x + c\right ) - 3 \, a^{4} - 12 \, {\left (2 \, a^{3} b - a b^{3}\right )} \sin \left (d x + c\right )^{3} + 6 \, {\left (2 \, a^{4} - a^{2} b^{2}\right )} \sin \left (d x + c\right )^{2}}{12 \, a^{5} d \sin \left (d x + c\right )^{4}} \] Input:
integrate(cot(d*x+c)^5/(a+b*sin(d*x+c)),x, algorithm="giac")
Output:
(a^4 - 2*a^2*b^2 + b^4)*log(abs(sin(d*x + c)))/(a^5*d) - (a^4*b - 2*a^2*b^ 3 + b^5)*log(abs(b*sin(d*x + c) + a))/(a^5*b*d) + 1/12*(4*a^3*b*sin(d*x + c) - 3*a^4 - 12*(2*a^3*b - a*b^3)*sin(d*x + c)^3 + 6*(2*a^4 - a^2*b^2)*sin (d*x + c)^2)/(a^5*d*sin(d*x + c)^4)
Time = 17.96 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.90 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3}{16\,a}-\frac {b^2}{8\,a^3}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{64\,a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {b}{8\,a^2}+\frac {2\,b\,\left (\frac {3}{8\,a}-\frac {b^2}{4\,a^3}\right )}{a}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a\,b^2-3\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (14\,a^2\,b-8\,b^3\right )+\frac {a^3}{4}-\frac {2\,a^2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{3}}{16\,a^4\,d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}-\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{a^5\,d}+\frac {b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{24\,a^2\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )}{a^5\,d} \] Input:
int(cot(c + d*x)^5/(a + b*sin(c + d*x)),x)
Output:
(tan(c/2 + (d*x)/2)^2*(3/(16*a) - b^2/(8*a^3)))/d - tan(c/2 + (d*x)/2)^4/( 64*a*d) - (tan(c/2 + (d*x)/2)*(b/(8*a^2) + (2*b*(3/(8*a) - b^2/(4*a^3)))/a ))/d - (tan(c/2 + (d*x)/2)^2*(2*a*b^2 - 3*a^3) + tan(c/2 + (d*x)/2)^3*(14* a^2*b - 8*b^3) + a^3/4 - (2*a^2*b*tan(c/2 + (d*x)/2))/3)/(16*a^4*d*tan(c/2 + (d*x)/2)^4) - (log(a + 2*b*tan(c/2 + (d*x)/2) + a*tan(c/2 + (d*x)/2)^2) *(a^4 + b^4 - 2*a^2*b^2))/(a^5*d) + (b*tan(c/2 + (d*x)/2)^3)/(24*a^2*d) + (log(tan(c/2 + (d*x)/2))*(a^4 + b^4 - 2*a^2*b^2))/(a^5*d)
Time = 0.18 (sec) , antiderivative size = 318, normalized size of antiderivative = 2.15 \[ \int \frac {\cot ^5(c+d x)}{a+b \sin (c+d x)} \, dx=\frac {-96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{4} a^{4}+192 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{4} a^{2} b^{2}-96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} a +2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b +a \right ) \sin \left (d x +c \right )^{4} b^{4}+96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a^{4}-192 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} a^{2} b^{2}+96 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{4} b^{4}-39 \sin \left (d x +c \right )^{4} a^{4}+24 \sin \left (d x +c \right )^{4} a^{2} b^{2}-192 \sin \left (d x +c \right )^{3} a^{3} b +96 \sin \left (d x +c \right )^{3} a \,b^{3}+96 \sin \left (d x +c \right )^{2} a^{4}-48 \sin \left (d x +c \right )^{2} a^{2} b^{2}+32 \sin \left (d x +c \right ) a^{3} b -24 a^{4}}{96 \sin \left (d x +c \right )^{4} a^{5} d} \] Input:
int(cot(d*x+c)^5/(a+b*sin(d*x+c)),x)
Output:
( - 96*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)* *4*a**4 + 192*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**4*a**2*b**2 - 96*log(tan((c + d*x)/2)**2*a + 2*tan((c + d*x)/2)*b + a)*sin(c + d*x)**4*b**4 + 96*log(tan((c + d*x)/2))*sin(c + d*x)**4*a**4 - 192*log(tan((c + d*x)/2))*sin(c + d*x)**4*a**2*b**2 + 96*log(tan((c + d* x)/2))*sin(c + d*x)**4*b**4 - 39*sin(c + d*x)**4*a**4 + 24*sin(c + d*x)**4 *a**2*b**2 - 192*sin(c + d*x)**3*a**3*b + 96*sin(c + d*x)**3*a*b**3 + 96*s in(c + d*x)**2*a**4 - 48*sin(c + d*x)**2*a**2*b**2 + 32*sin(c + d*x)*a**3* b - 24*a**4)/(96*sin(c + d*x)**4*a**5*d)