Integrand size = 26, antiderivative size = 66 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{2 \sqrt {a} f}-\frac {\sqrt {a \cos ^2(e+f x)} \csc ^2(e+f x)}{2 a f} \]
1/2*arctanh((a*cos(f*x+e)^2)^(1/2)/a^(1/2))/f/a^(1/2)-1/2*csc(f*x+e)^2*(a* cos(f*x+e)^2)^(1/2)/a/f
Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\sqrt {\cos ^2(e+f x)}\right ) \sqrt {\cos ^2(e+f x)}-\cot ^2(e+f x)}{2 f \sqrt {a \cos ^2(e+f x)}} \]
(ArcTanh[Sqrt[Cos[e + f*x]^2]]*Sqrt[Cos[e + f*x]^2] - Cot[e + f*x]^2)/(2*f *Sqrt[a*Cos[e + f*x]^2])
Time = 0.34 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {3042, 3655, 3042, 25, 3684, 8, 51, 73, 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.
\(\displaystyle \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (e+f x)^3 \sqrt {a-a \sin (e+f x)^2}}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \frac {\cot ^3(e+f x)}{\sqrt {a \cos ^2(e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\tan \left (e+f x+\frac {\pi }{2}\right )^3}{\sqrt {a \sin \left (e+f x+\frac {\pi }{2}\right )^2}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\tan \left (\frac {1}{2} (2 e+\pi )+f x\right )^3}{\sqrt {a \sin \left (\frac {1}{2} (2 e+\pi )+f x\right )^2}}dx\) |
\(\Big \downarrow \) 3684 |
\(\displaystyle -\frac {\int \frac {\cos ^2(e+f x)}{\sqrt {a \cos ^2(e+f x)} \left (1-\cos ^2(e+f x)\right )^2}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 8 |
\(\displaystyle -\frac {\int \frac {\sqrt {a \cos ^2(e+f x)}}{\left (1-\cos ^2(e+f x)\right )^2}d\cos ^2(e+f x)}{2 a f}\) |
\(\Big \downarrow \) 51 |
\(\displaystyle -\frac {\frac {\sqrt {a \cos ^2(e+f x)}}{1-\cos ^2(e+f x)}-\frac {1}{2} a \int \frac {1}{\sqrt {a \cos ^2(e+f x)} \left (1-\cos ^2(e+f x)\right )}d\cos ^2(e+f x)}{2 a f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {\frac {\sqrt {a \cos ^2(e+f x)}}{1-\cos ^2(e+f x)}-\int \frac {1}{1-\frac {\cos ^4(e+f x)}{a}}d\sqrt {a \cos ^2(e+f x)}}{2 a f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {\sqrt {a \cos ^2(e+f x)}}{1-\cos ^2(e+f x)}-\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{2 a f}\) |
-1/2*(-(Sqrt[a]*ArcTanh[Sqrt[a*Cos[e + f*x]^2]/Sqrt[a]]) + Sqrt[a*Cos[e + f*x]^2]/(1 - Cos[e + f*x]^2))/(a*f)
3.5.72.3.1 Defintions of rubi rules used
Int[(u_.)*(x_)^(m_.)*((a_.)*(x_))^(p_), x_Symbol] :> Simp[1/a^m Int[u*(a* x)^(m + p), x], x] /; FreeQ[{a, m, p}, x] && IntegerQ[m]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[A ctivateTrig[u*(a*cos[e + f*x]^2)^p], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ [a + b, 0]
Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(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)*((b*ff^(n/2)*x^(n/2))^p/(1 - ff*x)^((m + 1)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && Inte gerQ[(m - 1)/2] && IntegerQ[n/2]
Time = 0.88 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.26
method | result | size |
default | \(\frac {\cos \left (f x +e \right ) \left (2 \cos \left (f x +e \right )+\left (-\ln \left (1+\cos \left (f x +e \right )\right )+\ln \left (\cos \left (f x +e \right )-1\right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )\right )}{4 \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) f}\) | \(83\) |
risch | \(\frac {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}{\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i f x}-{\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}+\frac {\ln \left ({\mathrm e}^{i f x}+{\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}\) | \(159\) |
1/4*cos(f*x+e)*(2*cos(f*x+e)+(-ln(1+cos(f*x+e))+ln(cos(f*x+e)-1))*sin(f*x+ e)^2)/(a*cos(f*x+e)^2)^(1/2)/(cos(f*x+e)-1)/(1+cos(f*x+e))/f
Time = 0.32 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.20 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left ({\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right ) - 2 \, \cos \left (f x + e\right )\right )}}{4 \, {\left (a f \cos \left (f x + e\right )^{3} - a f \cos \left (f x + e\right )\right )}} \]
-1/4*sqrt(a*cos(f*x + e)^2)*((cos(f*x + e)^2 - 1)*log(-(cos(f*x + e) - 1)/ (cos(f*x + e) + 1)) - 2*cos(f*x + e))/(a*f*cos(f*x + e)^3 - a*f*cos(f*x + e))
\[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\int \frac {\cot ^{3}{\left (e + f x \right )}}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]
Time = 0.33 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {\frac {\log \left (\frac {2 \, \sqrt {-a \sin \left (f x + e\right )^{2} + a} \sqrt {a}}{{\left | \sin \left (f x + e\right ) \right |}} + \frac {2 \, a}{{\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} - \frac {\sqrt {-a \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right )^{2}}}{2 \, f} \]
1/2*(log(2*sqrt(-a*sin(f*x + e)^2 + a)*sqrt(a)/abs(sin(f*x + e)) + 2*a/abs (sin(f*x + e)))/sqrt(a) - sqrt(-a*sin(f*x + e)^2 + a)/(a*sin(f*x + e)^2))/ f
Exception generated. \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m operator + Error: Bad Argument Value
Timed out. \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^3}{\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2}} \,d x \]