Integrand size = 24, antiderivative size = 58 \[ \int \frac {\csc ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {3 \text {arctanh}(\cos (c+d x))}{2 a d}+\frac {3 \sec (c+d x)}{2 a d}-\frac {\csc ^2(c+d x) \sec (c+d x)}{2 a d} \]
Leaf count is larger than twice the leaf count of optimal. \(146\) vs. \(2(58)=116\).
Time = 0.26 (sec) , antiderivative size = 146, normalized size of antiderivative = 2.52 \[ \int \frac {\csc ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\csc ^4(c+d x) \left (2-6 \cos (2 (c+d x))+2 \cos (3 (c+d x))+3 \cos (3 (c+d x)) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )-3 \cos (3 (c+d x)) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )+\cos (c+d x) \left (-2-3 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+3 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{2 a d \left (\csc ^2\left (\frac {1}{2} (c+d x)\right )-\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]
(Csc[c + d*x]^4*(2 - 6*Cos[2*(c + d*x)] + 2*Cos[3*(c + d*x)] + 3*Cos[3*(c + d*x)]*Log[Cos[(c + d*x)/2]] - 3*Cos[3*(c + d*x)]*Log[Sin[(c + d*x)/2]] + Cos[c + d*x]*(-2 - 3*Log[Cos[(c + d*x)/2]] + 3*Log[Sin[(c + d*x)/2]])))/( 2*a*d*(Csc[(c + d*x)/2]^2 - Sec[(c + d*x)/2]^2))
Time = 0.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.93, 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, 262, 219}
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.
\(\displaystyle \int \frac {\csc ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin (c+d x)^3 \left (a-a \sin (c+d x)^2\right )}dx\) |
\(\Big \downarrow \) 3654 |
\(\displaystyle \frac {\int \csc ^3(c+d x) \sec ^2(c+d x)dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc (c+d x)^3 \sec (c+d x)^2dx}{a}\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int \frac {\sec ^4(c+d x)}{\left (1-\sec ^2(c+d x)\right )^2}d\sec (c+d x)}{a d}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {\sec ^3(c+d x)}{2 \left (1-\sec ^2(c+d x)\right )}-\frac {3}{2} \int \frac {\sec ^2(c+d x)}{1-\sec ^2(c+d x)}d\sec (c+d x)}{a d}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {\sec ^3(c+d x)}{2 \left (1-\sec ^2(c+d x)\right )}-\frac {3}{2} \left (\int \frac {1}{1-\sec ^2(c+d x)}d\sec (c+d x)-\sec (c+d x)\right )}{a d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\sec ^3(c+d x)}{2 \left (1-\sec ^2(c+d x)\right )}-\frac {3}{2} (\text {arctanh}(\sec (c+d x))-\sec (c+d x))}{a d}\) |
((-3*(ArcTanh[Sec[c + d*x]] - Sec[c + d*x]))/2 + Sec[c + d*x]^3/(2*(1 - Se c[c + d*x]^2)))/(a*d)
3.1.40.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
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 = 0.64 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {\frac {1}{4+4 \cos \left (d x +c \right )}-\frac {3 \ln \left (1+\cos \left (d x +c \right )\right )}{4}+\frac {1}{\cos \left (d x +c \right )}+\frac {1}{4 \cos \left (d x +c \right )-4}+\frac {3 \ln \left (\cos \left (d x +c \right )-1\right )}{4}}{d a}\) | \(63\) |
default | \(\frac {\frac {1}{4+4 \cos \left (d x +c \right )}-\frac {3 \ln \left (1+\cos \left (d x +c \right )\right )}{4}+\frac {1}{\cos \left (d x +c \right )}+\frac {1}{4 \cos \left (d x +c \right )-4}+\frac {3 \ln \left (\cos \left (d x +c \right )-1\right )}{4}}{d a}\) | \(63\) |
parallelrisch | \(\frac {\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\cot ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-18}{8 d a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\) | \(83\) |
norman | \(\frac {\frac {1}{8 a d}+\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d a}-\frac {9 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d a}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a d}\) | \(94\) |
risch | \(\frac {3 \,{\mathrm e}^{5 i \left (d x +c \right )}-2 \,{\mathrm e}^{3 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{2 d a}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{2 d a}\) | \(109\) |
1/d/a*(1/4/(1+cos(d*x+c))-3/4*ln(1+cos(d*x+c))+1/cos(d*x+c)+1/4/(cos(d*x+c )-1)+3/4*ln(cos(d*x+c)-1))
Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.69 \[ \int \frac {\csc ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {6 \, \cos \left (d x + c\right )^{2} - 3 \, {\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 3 \, {\left (\cos \left (d x + c\right )^{3} - \cos \left (d x + c\right )\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4}{4 \, {\left (a d \cos \left (d x + c\right )^{3} - a d \cos \left (d x + c\right )\right )}} \]
1/4*(6*cos(d*x + c)^2 - 3*(cos(d*x + c)^3 - cos(d*x + c))*log(1/2*cos(d*x + c) + 1/2) + 3*(cos(d*x + c)^3 - cos(d*x + c))*log(-1/2*cos(d*x + c) + 1/ 2) - 4)/(a*d*cos(d*x + c)^3 - a*d*cos(d*x + c))
\[ \int \frac {\csc ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=- \frac {\int \frac {\csc ^{3}{\left (c + d x \right )}}{\sin ^{2}{\left (c + d x \right )} - 1}\, dx}{a} \]
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.21 \[ \int \frac {\csc ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 2\right )}}{a \cos \left (d x + c\right )^{3} - a \cos \left (d x + c\right )} - \frac {3 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a} + \frac {3 \, \log \left (\cos \left (d x + c\right ) - 1\right )}{a}}{4 \, d} \]
1/4*(2*(3*cos(d*x + c)^2 - 2)/(a*cos(d*x + c)^3 - a*cos(d*x + c)) - 3*log( cos(d*x + c) + 1)/a + 3*log(cos(d*x + c) - 1)/a)/d
Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (52) = 104\).
Time = 0.33 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.57 \[ \int \frac {\csc ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {6 \, \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{a} + \frac {\frac {14 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {3 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + \frac {{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )}} - \frac {\cos \left (d x + c\right ) - 1}{a {\left (\cos \left (d x + c\right ) + 1\right )}}}{8 \, d} \]
1/8*(6*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1))/a + (14*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 1)/(a*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + (cos(d*x + c) - 1)^2/(co s(d*x + c) + 1)^2)) - (cos(d*x + c) - 1)/(a*(cos(d*x + c) + 1)))/d
Time = 14.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.95 \[ \int \frac {\csc ^3(c+d x)}{a-a \sin ^2(c+d x)} \, dx=-\frac {\frac {3\,{\cos \left (c+d\,x\right )}^2}{2}-1}{d\,\left (a\,\cos \left (c+d\,x\right )-a\,{\cos \left (c+d\,x\right )}^3\right )}-\frac {3\,\mathrm {atanh}\left (\cos \left (c+d\,x\right )\right )}{2\,a\,d} \]