Integrand size = 24, antiderivative size = 50 \[ \int \cot (e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=-\frac {\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{f}+\frac {\sqrt {a \cos ^2(e+f x)}}{f} \]
Time = 0.08 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.94 \[ \int \cot (e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\frac {-\sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a \cos ^2(e+f x)}}{f} \]
Time = 0.32 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.04, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 3655, 3042, 25, 3684, 60, 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.
\(\displaystyle \int \cot (e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a-a \sin (e+f x)^2}}{\tan (e+f x)}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \cot (e+f x) \sqrt {a \cos ^2(e+f x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan \left (e+f x+\frac {\pi }{2}\right ) \left (-\sqrt {a \sin \left (e+f x+\frac {\pi }{2}\right )^2}\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sqrt {a \sin \left (\frac {1}{2} (2 e+\pi )+f x\right )^2} \tan \left (\frac {1}{2} (2 e+\pi )+f x\right )dx\) |
\(\Big \downarrow \) 3684 |
\(\displaystyle -\frac {\int \frac {\sqrt {a \cos ^2(e+f x)}}{1-\cos ^2(e+f x)}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle -\frac {a \int \frac {1}{\sqrt {a \cos ^2(e+f x)} \left (1-\cos ^2(e+f x)\right )}d\cos ^2(e+f x)-2 \sqrt {a \cos ^2(e+f x)}}{2 f}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {2 \int \frac {1}{1-\frac {\cos ^4(e+f x)}{a}}d\sqrt {a \cos ^2(e+f x)}-2 \sqrt {a \cos ^2(e+f x)}}{2 f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )-2 \sqrt {a \cos ^2(e+f x)}}{2 f}\) |
3.5.60.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
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 = 0.83 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.08
method | result | size |
default | \(\frac {a \cos \left (f x +e \right ) \left (2 \cos \left (f x +e \right )+\ln \left (\cos \left (f x +e \right )-1\right )-\ln \left (1+\cos \left (f x +e \right )\right )\right )}{2 \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f}\) | \(54\) |
risch | \(\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {\ln \left ({\mathrm e}^{i f x}+{\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\ln \left ({\mathrm e}^{i f x}-{\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\) | \(232\) |
Time = 0.30 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.14 \[ \int \cot (e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (2 \, \cos \left (f x + e\right ) - \log \left (-\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1}\right )\right )}}{2 \, f \cos \left (f x + e\right )} \]
1/2*sqrt(a*cos(f*x + e)^2)*(2*cos(f*x + e) - log(-(cos(f*x + e) + 1)/(cos( f*x + e) - 1)))/(f*cos(f*x + e))
\[ \int \cot (e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \cot {\left (e + f x \right )}\, dx \]
Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.40 \[ \int \cot (e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=-\frac {\sqrt {a} \log \left (\frac {2 \, \sqrt {-a \sin \left (f x + e\right )^{2} + a} \sqrt {a}}{{\left | \sin \left (f x + e\right ) \right |}} + \frac {2 \, a}{{\left | \sin \left (f x + e\right ) \right |}}\right ) - \sqrt {-a \sin \left (f x + e\right )^{2} + a}}{f} \]
-(sqrt(a)*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*sin(f*x + e)^2 + a))/f
Leaf count of result is larger than twice the leaf count of optimal. 524 vs. \(2 (42) = 84\).
Time = 0.56 (sec) , antiderivative size = 524, normalized size of antiderivative = 10.48 \[ \int \cot (e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\text {Too large to display} \]
-(sqrt(a)*log(abs(tan(1/2*f*x)*tan(1/2*e) - 1))*sgn(tan(1/2*f*x)^4*tan(1/2 *e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3* tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*t an(1/2*e) + 1) - sqrt(a)*log(abs(tan(1/2*f*x) + tan(1/2*e)))*sgn(tan(1/2*f *x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*ta n(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*t an(1/2*f*x)*tan(1/2*e) + 1) + 2*(2*sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1 /2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/ 2*e) + 1)*tan(1/2*f*x)*tan(1/2*e) + sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^ 4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan( 1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1 /2*e) + 1)*tan(1/2*e)^2 - sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan( 1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4 *tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1) )/((tan(1/2*f*x)^2 + 1)*(tan(1/2*e)^2 + 1)))/f
Timed out. \[ \int \cot (e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\int \mathrm {cot}\left (e+f\,x\right )\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x \]