Integrand size = 21, antiderivative size = 114 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {2 b \csc (c+d x)}{a^3 d}-\frac {\csc ^2(c+d x)}{2 a^2 d}-\frac {\left (a^2-3 b^2\right ) \log (\sin (c+d x))}{a^4 d}+\frac {\left (a^2-3 b^2\right ) \log (a+b \sin (c+d x))}{a^4 d}-\frac {a^2-b^2}{a^3 d (a+b \sin (c+d x))} \]
2*b*csc(d*x+c)/a^3/d-1/2*csc(d*x+c)^2/a^2/d-(a^2-3*b^2)*ln(sin(d*x+c))/a^4 /d+(a^2-3*b^2)*ln(a+b*sin(d*x+c))/a^4/d+(-a^2+b^2)/a^3/d/(a+b*sin(d*x+c))
Time = 0.42 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.84 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {-4 a b \csc (c+d x)+a^2 \csc ^2(c+d x)+2 \left (a^2-3 b^2\right ) \log (\sin (c+d x))-2 \left (a^2-3 b^2\right ) \log (a+b \sin (c+d x))+\frac {2 a (a-b) (a+b)}{a+b \sin (c+d x)}}{2 a^4 d} \]
-1/2*(-4*a*b*Csc[c + d*x] + a^2*Csc[c + d*x]^2 + 2*(a^2 - 3*b^2)*Log[Sin[c + d*x]] - 2*(a^2 - 3*b^2)*Log[a + b*Sin[c + d*x]] + (2*a*(a - b)*(a + b)) /(a + b*Sin[c + d*x]))/(a^4*d)
Time = 0.30 (sec) , antiderivative size = 105, 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.
\(\displaystyle \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (c+d x)^3 (a+b \sin (c+d x))^2}dx\) |
\(\Big \downarrow \) 3200 |
\(\displaystyle \frac {\int \frac {\csc ^3(c+d x) \left (b^2-b^2 \sin ^2(c+d x)\right )}{b^3 (a+b \sin (c+d x))^2}d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 522 |
\(\displaystyle \frac {\int \left (\frac {\csc ^3(c+d x)}{a^2 b}-\frac {2 \csc ^2(c+d x)}{a^3}+\frac {\left (3 b^2-a^2\right ) \csc (c+d x)}{a^4 b}+\frac {a^2-3 b^2}{a^4 (a+b \sin (c+d x))}+\frac {a^2-b^2}{a^3 (a+b \sin (c+d x))^2}\right )d(b \sin (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 b \csc (c+d x)}{a^3}-\frac {\csc ^2(c+d x)}{2 a^2}-\frac {\left (a^2-3 b^2\right ) \log (b \sin (c+d x))}{a^4}+\frac {\left (a^2-3 b^2\right ) \log (a+b \sin (c+d x))}{a^4}-\frac {a^2-b^2}{a^3 (a+b \sin (c+d x))}}{d}\) |
((2*b*Csc[c + d*x])/a^3 - Csc[c + d*x]^2/(2*a^2) - ((a^2 - 3*b^2)*Log[b*Si n[c + d*x]])/a^4 + ((a^2 - 3*b^2)*Log[a + b*Sin[c + d*x]])/a^4 - (a^2 - b^ 2)/(a^3*(a + b*Sin[c + d*x])))/d
3.2.85.3.1 Defintions of rubi rules used
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 = 2.03 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.92
method | result | size |
derivativedivides | \(\frac {-\frac {1}{2 a^{2} \sin \left (d x +c \right )^{2}}+\frac {\left (-a^{2}+3 b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{4}}+\frac {2 b}{a^{3} \sin \left (d x +c \right )}+\frac {\left (a^{2}-3 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{4}}-\frac {a^{2}-b^{2}}{a^{3} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(105\) |
default | \(\frac {-\frac {1}{2 a^{2} \sin \left (d x +c \right )^{2}}+\frac {\left (-a^{2}+3 b^{2}\right ) \ln \left (\sin \left (d x +c \right )\right )}{a^{4}}+\frac {2 b}{a^{3} \sin \left (d x +c \right )}+\frac {\left (a^{2}-3 b^{2}\right ) \ln \left (a +b \sin \left (d x +c \right )\right )}{a^{4}}-\frac {a^{2}-b^{2}}{a^{3} \left (a +b \sin \left (d x +c \right )\right )}}{d}\) | \(105\) |
risch | \(-\frac {2 i \left (-3 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-3 b^{2} {\mathrm e}^{5 i \left (d x +c \right )}+3 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-4 a^{2} {\mathrm e}^{3 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )} b^{2}+a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+a^{2} {\mathrm e}^{i \left (d x +c \right )}\right )}{\left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}-b +2 i a \,{\mathrm e}^{i \left (d x +c \right )}\right ) d \,a^{3}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )}{a^{2} d}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right ) b^{2}}{a^{4} d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{a^{2} d}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right ) b^{2}}{a^{4} d}\) | \(282\) |
1/d*(-1/2/a^2/sin(d*x+c)^2+(-a^2+3*b^2)/a^4*ln(sin(d*x+c))+2*b/a^3/sin(d*x +c)+(a^2-3*b^2)/a^4*ln(a+b*sin(d*x+c))-(a^2-b^2)/a^3/(a+b*sin(d*x+c)))
Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (112) = 224\).
Time = 0.30 (sec) , antiderivative size = 259, normalized size of antiderivative = 2.27 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {3 \, a^{2} b \sin \left (d x + c\right ) - 3 \, a^{3} + 6 \, a b^{2} + 2 \, {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (a^{3} - 3 \, a b^{2} - {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{2} b - 3 \, b^{3} - {\left (a^{2} b - 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) - 2 \, {\left (a^{3} - 3 \, a b^{2} - {\left (a^{3} - 3 \, a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (a^{2} b - 3 \, b^{3} - {\left (a^{2} b - 3 \, b^{3}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \sin \left (d x + c\right )\right )}{2 \, {\left (a^{5} d \cos \left (d x + c\right )^{2} - a^{5} d + {\left (a^{4} b d \cos \left (d x + c\right )^{2} - a^{4} b d\right )} \sin \left (d x + c\right )\right )}} \]
-1/2*(3*a^2*b*sin(d*x + c) - 3*a^3 + 6*a*b^2 + 2*(a^3 - 3*a*b^2)*cos(d*x + c)^2 + 2*(a^3 - 3*a*b^2 - (a^3 - 3*a*b^2)*cos(d*x + c)^2 + (a^2*b - 3*b^3 - (a^2*b - 3*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(b*sin(d*x + c) + a) - 2*(a^3 - 3*a*b^2 - (a^3 - 3*a*b^2)*cos(d*x + c)^2 + (a^2*b - 3*b^3 - (a^2 *b - 3*b^3)*cos(d*x + c)^2)*sin(d*x + c))*log(-1/2*sin(d*x + c)))/(a^5*d*c os(d*x + c)^2 - a^5*d + (a^4*b*d*cos(d*x + c)^2 - a^4*b*d)*sin(d*x + c))
\[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\int \frac {\cot ^{3}{\left (c + d x \right )}}{\left (a + b \sin {\left (c + d x \right )}\right )^{2}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.02 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {\frac {3 \, a b \sin \left (d x + c\right ) - 2 \, {\left (a^{2} - 3 \, b^{2}\right )} \sin \left (d x + c\right )^{2} - a^{2}}{a^{3} b \sin \left (d x + c\right )^{3} + a^{4} \sin \left (d x + c\right )^{2}} + \frac {2 \, {\left (a^{2} - 3 \, b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4}} - \frac {2 \, {\left (a^{2} - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right )\right )}{a^{4}}}{2 \, d} \]
1/2*((3*a*b*sin(d*x + c) - 2*(a^2 - 3*b^2)*sin(d*x + c)^2 - a^2)/(a^3*b*si n(d*x + c)^3 + a^4*sin(d*x + c)^2) + 2*(a^2 - 3*b^2)*log(b*sin(d*x + c) + a)/a^4 - 2*(a^2 - 3*b^2)*log(sin(d*x + c))/a^4)/d
Time = 0.35 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.45 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=-\frac {\frac {2 \, {\left (a^{2} - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) \right |}\right )}{a^{4}} - \frac {2 \, {\left (a^{2} b - 3 \, b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b} + \frac {2 \, {\left (a^{2} b \sin \left (d x + c\right ) - 3 \, b^{3} \sin \left (d x + c\right ) + 2 \, a^{3} - 4 \, a b^{2}\right )}}{{\left (b \sin \left (d x + c\right ) + a\right )} a^{4}} - \frac {3 \, a^{2} \sin \left (d x + c\right )^{2} - 9 \, b^{2} \sin \left (d x + c\right )^{2} + 4 \, a b \sin \left (d x + c\right ) - a^{2}}{a^{4} \sin \left (d x + c\right )^{2}}}{2 \, d} \]
-1/2*(2*(a^2 - 3*b^2)*log(abs(sin(d*x + c)))/a^4 - 2*(a^2*b - 3*b^3)*log(a bs(b*sin(d*x + c) + a))/(a^4*b) + 2*(a^2*b*sin(d*x + c) - 3*b^3*sin(d*x + c) + 2*a^3 - 4*a*b^2)/((b*sin(d*x + c) + a)*a^4) - (3*a^2*sin(d*x + c)^2 - 9*b^2*sin(d*x + c)^2 + 4*a*b*sin(d*x + c) - a^2)/(a^4*sin(d*x + c)^2))/d
Time = 6.25 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.06 \[ \int \frac {\cot ^3(c+d x)}{(a+b \sin (c+d x))^2} \, dx=\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^3\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2}{2}-8\,b^2\right )+\frac {a^2}{2}-\frac {4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,a^2\,b-2\,b^3\right )}{a}-3\,a\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+4\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+8\,b\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{8\,a^2\,d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a^2-3\,b^2\right )}{a^4\,d}+\frac {\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )\,\left (a^2-3\,b^2\right )}{a^4\,d} \]
(b*tan(c/2 + (d*x)/2))/(a^3*d) - (tan(c/2 + (d*x)/2)^2*(a^2/2 - 8*b^2) + a ^2/2 - (4*tan(c/2 + (d*x)/2)^3*(3*a^2*b - 2*b^3))/a - 3*a*b*tan(c/2 + (d*x )/2))/(d*(4*a^4*tan(c/2 + (d*x)/2)^2 + 4*a^4*tan(c/2 + (d*x)/2)^4 + 8*a^3* b*tan(c/2 + (d*x)/2)^3)) - tan(c/2 + (d*x)/2)^2/(8*a^2*d) - (log(tan(c/2 + (d*x)/2))*(a^2 - 3*b^2))/(a^4*d) + (log(a + 2*b*tan(c/2 + (d*x)/2) + a*ta n(c/2 + (d*x)/2)^2)*(a^2 - 3*b^2))/(a^4*d)