Integrand size = 24, antiderivative size = 78 \[ \int \frac {\csc ^3(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=-\frac {5 \text {arctanh}(\cos (c+d x))}{2 a^2 d}+\frac {5 \sec (c+d x)}{2 a^2 d}+\frac {5 \sec ^3(c+d x)}{6 a^2 d}-\frac {\csc ^2(c+d x) \sec ^3(c+d x)}{2 a^2 d} \]
-5/2*arctanh(cos(d*x+c))/a^2/d+5/2*sec(d*x+c)/a^2/d+5/6*sec(d*x+c)^3/a^2/d -1/2*csc(d*x+c)^2*sec(d*x+c)^3/a^2/d
Leaf count is larger than twice the leaf count of optimal. \(208\) vs. \(2(78)=156\).
Time = 0.38 (sec) , antiderivative size = 208, normalized size of antiderivative = 2.67 \[ \int \frac {\csc ^3(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {2 \csc ^8(c+d x) \left (22-40 \cos (2 (c+d x))+13 \cos (3 (c+d x))-30 \cos (4 (c+d x))+13 \cos (5 (c+d x))+15 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+15 \cos (5 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-15 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-15 \cos (5 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (-26-30 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+30 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{3 a^2 d \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )^3} \]
(2*Csc[c + d*x]^8*(22 - 40*Cos[2*(c + d*x)] + 13*Cos[3*(c + d*x)] - 30*Cos [4*(c + d*x)] + 13*Cos[5*(c + d*x)] + 15*Cos[3*(c + d*x)]*Log[Cos[(c + d*x )/2]] + 15*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 15*Cos[3*(c + d*x)]*Lo g[Sin[(c + d*x)/2]] - 15*Cos[5*(c + d*x)]*Log[Sin[(c + d*x)/2]] + Cos[c + d*x]*(-26 - 30*Log[Cos[(c + d*x)/2]] + 30*Log[Sin[(c + d*x)/2]])))/(3*a^2* d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2)^3)
Time = 0.31 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.85, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {3042, 3654, 3042, 3102, 252, 254, 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.
\(\displaystyle \int \frac {\csc ^3(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x)^3 \left (a-a \sin (c+d x)^2\right )^2}dx\) |
\(\Big \downarrow \) 3654 |
\(\displaystyle \frac {\int \csc ^3(c+d x) \sec ^4(c+d x)dx}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc (c+d x)^3 \sec (c+d x)^4dx}{a^2}\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int \frac {\sec ^6(c+d x)}{\left (1-\sec ^2(c+d x)\right )^2}d\sec (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {\sec ^5(c+d x)}{2 \left (1-\sec ^2(c+d x)\right )}-\frac {5}{2} \int \frac {\sec ^4(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {\frac {\sec ^5(c+d x)}{2 \left (1-\sec ^2(c+d x)\right )}-\frac {5}{2} \int \left (-\sec ^2(c+d x)+\frac {1}{1-\sec ^2(c+d x)}-1\right )d\sec (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {\sec ^5(c+d x)}{2 \left (1-\sec ^2(c+d x)\right )}-\frac {5}{2} \left (\text {arctanh}(\sec (c+d x))-\frac {1}{3} \sec ^3(c+d x)-\sec (c+d x)\right )}{a^2 d}\) |
(Sec[c + d*x]^5/(2*(1 - Sec[c + d*x]^2)) - (5*(ArcTanh[Sec[c + d*x]] - Sec [c + d*x] - Sec[c + d*x]^3/3))/2)/(a^2*d)
3.1.54.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[1/(f*a^n) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 )/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[ a^p Int[ActivateTrig[u*cos[e + f*x]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]
Time = 0.89 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {\frac {1}{4+4 \cos \left (d x +c \right )}-\frac {5 \ln \left (1+\cos \left (d x +c \right )\right )}{4}+\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {2}{\cos \left (d x +c \right )}+\frac {1}{4 \cos \left (d x +c \right )-4}+\frac {5 \ln \left (\cos \left (d x +c \right )-1\right )}{4}}{d \,a^{2}}\) | \(75\) |
default | \(\frac {\frac {1}{4+4 \cos \left (d x +c \right )}-\frac {5 \ln \left (1+\cos \left (d x +c \right )\right )}{4}+\frac {1}{3 \cos \left (d x +c \right )^{3}}+\frac {2}{\cos \left (d x +c \right )}+\frac {1}{4 \cos \left (d x +c \right )-4}+\frac {5 \ln \left (\cos \left (d x +c \right )-1\right )}{4}}{d \,a^{2}}\) | \(75\) |
parallelrisch | \(\frac {60 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-165 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 \left (\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+225 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-130}{24 a^{2} d \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}\) | \(127\) |
risch | \(\frac {15 \,{\mathrm e}^{9 i \left (d x +c \right )}+20 \,{\mathrm e}^{7 i \left (d x +c \right )}-22 \,{\mathrm e}^{5 i \left (d x +c \right )}+20 \,{\mathrm e}^{3 i \left (d x +c \right )}+15 \,{\mathrm e}^{i \left (d x +c \right )}}{3 d \,a^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d \,a^{2}}+\frac {5 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d \,a^{2}}\) | \(132\) |
norman | \(\frac {\frac {1}{8 a d}+\frac {\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}+\frac {75 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}-\frac {65 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d a}-\frac {55 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d a}}{a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {5 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} d}\) | \(135\) |
1/d/a^2*(1/4/(1+cos(d*x+c))-5/4*ln(1+cos(d*x+c))+1/3/cos(d*x+c)^3+2/cos(d* x+c)+1/4/(cos(d*x+c)-1)+5/4*ln(cos(d*x+c)-1))
Time = 0.25 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.51 \[ \int \frac {\csc ^3(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {30 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} - 15 \, {\left (\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 15 \, {\left (\cos \left (d x + c\right )^{5} - \cos \left (d x + c\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{5} - a^{2} d \cos \left (d x + c\right )^{3}\right )}} \]
1/12*(30*cos(d*x + c)^4 - 20*cos(d*x + c)^2 - 15*(cos(d*x + c)^5 - cos(d*x + c)^3)*log(1/2*cos(d*x + c) + 1/2) + 15*(cos(d*x + c)^5 - cos(d*x + c)^3 )*log(-1/2*cos(d*x + c) + 1/2) - 4)/(a^2*d*cos(d*x + c)^5 - a^2*d*cos(d*x + c)^3)
\[ \int \frac {\csc ^3(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\int \frac {\csc ^{3}{\left (c + d x \right )}}{\sin ^{4}{\left (c + d x \right )} - 2 \sin ^{2}{\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.23 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.10 \[ \int \frac {\csc ^3(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {\frac {2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} - 2\right )}}{a^{2} \cos \left (d x + c\right )^{5} - a^{2} \cos \left (d x + c\right )^{3}} - \frac {15 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{2}} + \frac {15 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a^{2}}}{12 \, d} \]
1/12*(2*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 - 2)/(a^2*cos(d*x + c)^5 - a^2*cos(d*x + c)^3) - 15*log(cos(d*x + c) + 1)/a^2 + 15*log(cos(d*x + c) - 1)/a^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (70) = 140\).
Time = 0.33 (sec) , antiderivative size = 175, normalized size of antiderivative = 2.24 \[ \int \frac {\csc ^3(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=-\frac {\frac {3 \, {\left (\frac {10 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - 1\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}{a^{2} {\left (\cos \left (d x + c\right ) - 1\right )}} - \frac {30 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a^{2}} + \frac {3 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{a^{2} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {16 \, {\left (\frac {12 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {9 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 7\right )}}{a^{2} {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{3}}}{24 \, d} \]
-1/24*(3*(10*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)*(cos(d*x + c) + 1) /(a^2*(cos(d*x + c) - 1)) - 30*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a^2 + 3*(cos(d*x + c) - 1)/(a^2*(cos(d*x + c) + 1)) - 16*(12*(cos(d *x + c) - 1)/(cos(d*x + c) + 1) + 9*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1 )^2 + 7)/(a^2*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^3))/d
Time = 12.94 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.90 \[ \int \frac {\csc ^3(c+d x)}{\left (a-a \sin ^2(c+d x)\right )^2} \, dx=\frac {-\frac {5\,{\cos \left (c+d\,x\right )}^4}{2}+\frac {5\,{\cos \left (c+d\,x\right )}^2}{3}+\frac {1}{3}}{d\,\left (a^2\,{\cos \left (c+d\,x\right )}^3-a^2\,{\cos \left (c+d\,x\right )}^5\right )}-\frac {5\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{2\,a^2\,d} \]