Integrand size = 26, antiderivative size = 96 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {\cot (e+f x)}{f \sqrt {a \cos ^2(e+f x)}}+\frac {2 \cot (e+f x) \csc ^2(e+f x)}{3 f \sqrt {a \cos ^2(e+f x)}}-\frac {\cot (e+f x) \csc ^4(e+f x)}{5 f \sqrt {a \cos ^2(e+f x)}} \]
-cot(f*x+e)/f/(a*cos(f*x+e)^2)^(1/2)+2/3*cot(f*x+e)*csc(f*x+e)^2/f/(a*cos( f*x+e)^2)^(1/2)-1/5*cot(f*x+e)*csc(f*x+e)^4/f/(a*cos(f*x+e)^2)^(1/2)
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.51 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {\cot (e+f x) \left (15-10 \csc ^2(e+f x)+3 \csc ^4(e+f x)\right )}{15 f \sqrt {a \cos ^2(e+f x)}} \]
Time = 0.42 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.58, 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, 3686, 3042, 25, 3086, 210, 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.
\(\displaystyle \int \frac {\cot ^6(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\tan (e+f x)^6 \sqrt {a-a \sin (e+f x)^2}}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \frac {\cot ^6(e+f x)}{\sqrt {a \cos ^2(e+f x)}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan \left (e+f x+\frac {\pi }{2}\right )^6}{\sqrt {a \sin \left (e+f x+\frac {\pi }{2}\right )^2}}dx\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \frac {\cos (e+f x) \int \cot ^5(e+f x) \csc (e+f x)dx}{\sqrt {a \cos ^2(e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\cos (e+f x) \int -\sec \left (e+f x-\frac {\pi }{2}\right ) \tan \left (e+f x-\frac {\pi }{2}\right )^5dx}{\sqrt {a \cos ^2(e+f x)}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\cos (e+f x) \int \sec \left (\frac {1}{2} (2 e-\pi )+f x\right ) \tan \left (\frac {1}{2} (2 e-\pi )+f x\right )^5dx}{\sqrt {a \cos ^2(e+f x)}}\) |
\(\Big \downarrow \) 3086 |
\(\displaystyle -\frac {\cos (e+f x) \int \left (\csc ^2(e+f x)-1\right )^2d\csc (e+f x)}{f \sqrt {a \cos ^2(e+f x)}}\) |
\(\Big \downarrow \) 210 |
\(\displaystyle -\frac {\cos (e+f x) \int \left (\csc ^4(e+f x)-2 \csc ^2(e+f x)+1\right )d\csc (e+f x)}{f \sqrt {a \cos ^2(e+f x)}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\cos (e+f x) \left (\frac {1}{5} \csc ^5(e+f x)-\frac {2}{3} \csc ^3(e+f x)+\csc (e+f x)\right )}{f \sqrt {a \cos ^2(e+f x)}}\) |
-((Cos[e + f*x]*(Csc[e + f*x] - (2*Csc[e + f*x]^3)/3 + Csc[e + f*x]^5/5))/ (f*Sqrt[a*Cos[e + f*x]^2]))
3.5.77.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^2 )^p, x], x] /; FreeQ[{a, b}, x] && IGtQ[p, 0]
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^( n_.), x_Symbol] :> Simp[a/f Subst[Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2 ), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2 ] && !(IntegerQ[m/2] && LtQ[0, m, n + 1])
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[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Simp[(b*ff^n)^IntPart[p]*((b*Sin[e + f*x]^ n)^FracPart[p]/(Sin[e + f*x]/ff)^(n*FracPart[p])) Int[ActivateTrig[u]*(Si n[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] || MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) / ; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig]])
Time = 0.67 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.77
method | result | size |
default | \(-\frac {\cos \left (f x +e \right ) \left (15 \left (\cos ^{4}\left (f x +e \right )\right )-20 \left (\cos ^{2}\left (f x +e \right )\right )+8\right )}{15 \left (\cos \left (f x +e \right )-1\right )^{2} \left (1+\cos \left (f x +e \right )\right )^{2} \sin \left (f x +e \right ) \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f}\) | \(74\) |
risch | \(-\frac {2 i \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \left (15 \,{\mathrm e}^{8 i \left (f x +e \right )}-20 \,{\mathrm e}^{6 i \left (f x +e \right )}+58 \,{\mathrm e}^{4 i \left (f x +e \right )}-20 \,{\mathrm e}^{2 i \left (f x +e \right )}+15\right )}{15 \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5}}\) | \(103\) |
-1/15*cos(f*x+e)*(15*cos(f*x+e)^4-20*cos(f*x+e)^2+8)/(cos(f*x+e)-1)^2/(1+c os(f*x+e))^2/sin(f*x+e)/(a*cos(f*x+e)^2)^(1/2)/f
Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.82 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {{\left (15 \, \cos \left (f x + e\right )^{4} - 20 \, \cos \left (f x + e\right )^{2} + 8\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{15 \, {\left (a f \cos \left (f x + e\right )^{5} - 2 \, a f \cos \left (f x + e\right )^{3} + a f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \]
-1/15*(15*cos(f*x + e)^4 - 20*cos(f*x + e)^2 + 8)*sqrt(a*cos(f*x + e)^2)/( (a*f*cos(f*x + e)^5 - 2*a*f*cos(f*x + e)^3 + a*f*cos(f*x + e))*sin(f*x + e ))
\[ \int \frac {\cot ^6(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\int \frac {\cot ^{6}{\left (e + f x \right )}}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 1236 vs. \(2 (86) = 172\).
Time = 0.38 (sec) , antiderivative size = 1236, normalized size of antiderivative = 12.88 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\text {Too large to display} \]
2/15*((15*sin(9*f*x + 9*e) - 20*sin(7*f*x + 7*e) + 58*sin(5*f*x + 5*e) - 2 0*sin(3*f*x + 3*e) + 15*sin(f*x + e))*cos(10*f*x + 10*e) + 75*(sin(8*f*x + 8*e) - 2*sin(6*f*x + 6*e) + 2*sin(4*f*x + 4*e) - sin(2*f*x + 2*e))*cos(9* f*x + 9*e) + 5*(20*sin(7*f*x + 7*e) - 58*sin(5*f*x + 5*e) + 20*sin(3*f*x + 3*e) - 15*sin(f*x + e))*cos(8*f*x + 8*e) + 100*(2*sin(6*f*x + 6*e) - 2*si n(4*f*x + 4*e) + sin(2*f*x + 2*e))*cos(7*f*x + 7*e) + 10*(58*sin(5*f*x + 5 *e) - 20*sin(3*f*x + 3*e) + 15*sin(f*x + e))*cos(6*f*x + 6*e) + 290*(2*sin (4*f*x + 4*e) - sin(2*f*x + 2*e))*cos(5*f*x + 5*e) + 50*(4*sin(3*f*x + 3*e ) - 3*sin(f*x + e))*cos(4*f*x + 4*e) - (15*cos(9*f*x + 9*e) - 20*cos(7*f*x + 7*e) + 58*cos(5*f*x + 5*e) - 20*cos(3*f*x + 3*e) + 15*cos(f*x + e))*sin (10*f*x + 10*e) - 15*(5*cos(8*f*x + 8*e) - 10*cos(6*f*x + 6*e) + 10*cos(4* f*x + 4*e) - 5*cos(2*f*x + 2*e) + 1)*sin(9*f*x + 9*e) - 5*(20*cos(7*f*x + 7*e) - 58*cos(5*f*x + 5*e) + 20*cos(3*f*x + 3*e) - 15*cos(f*x + e))*sin(8* f*x + 8*e) - 20*(10*cos(6*f*x + 6*e) - 10*cos(4*f*x + 4*e) + 5*cos(2*f*x + 2*e) - 1)*sin(7*f*x + 7*e) - 10*(58*cos(5*f*x + 5*e) - 20*cos(3*f*x + 3*e ) + 15*cos(f*x + e))*sin(6*f*x + 6*e) - 58*(10*cos(4*f*x + 4*e) - 5*cos(2* f*x + 2*e) + 1)*sin(5*f*x + 5*e) - 50*(4*cos(3*f*x + 3*e) - 3*cos(f*x + e) )*sin(4*f*x + 4*e) - 20*(5*cos(2*f*x + 2*e) - 1)*sin(3*f*x + 3*e) + 100*co s(3*f*x + 3*e)*sin(2*f*x + 2*e) - 75*cos(f*x + e)*sin(2*f*x + 2*e) + 75*co s(2*f*x + 2*e)*sin(f*x + e) - 15*sin(f*x + e))*sqrt(a)/((a*cos(10*f*x +...
Exception generated. \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\text {Exception raised: TypeError} \]
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
Time = 20.96 (sec) , antiderivative size = 491, normalized size of antiderivative = 5.11 \[ \int \frac {\cot ^6(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,4{}\mathrm {i}}{a\,f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,32{}\mathrm {i}}{3\,a\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^2\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,352{}\mathrm {i}}{15\,a\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^3\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,128{}\mathrm {i}}{5\,a\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^4\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}-\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,64{}\mathrm {i}}{5\,a\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^5\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )} \]
- (exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f* x*1i)*1i)/2)^2)^(1/2)*4i)/(a*f*(exp(e*2i + f*x*2i) - 1)*(exp(e*1i + f*x*1i ) + exp(e*3i + f*x*3i))) - (exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1 i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2)*32i)/(3*a*f*(exp(e*2i + f*x *2i) - 1)^2*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (exp(e*3i + f*x*3 i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/ 2)*352i)/(15*a*f*(exp(e*2i + f*x*2i) - 1)^3*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (exp(e*3i + f*x*3i)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - ( exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2)*128i)/(5*a*f*(exp(e*2i + f*x*2i) - 1)^4 *(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i))) - (exp(e*3i + f*x*3i)*(a - a*( (exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2)*64i)/(5* a*f*(exp(e*2i + f*x*2i) - 1)^5*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i)))