Integrand size = 26, antiderivative size = 91 \[ \int \cot ^4(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\frac {2 \sqrt {a \cos ^2(e+f x)} \csc (e+f x) \sec (e+f x)}{f}-\frac {\sqrt {a \cos ^2(e+f x)} \csc ^3(e+f x) \sec (e+f x)}{3 f}+\frac {\sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{f} \] Output:
2*(a*cos(f*x+e)^2)^(1/2)*csc(f*x+e)*sec(f*x+e)/f-1/3*(a*cos(f*x+e)^2)^(1/2 )*csc(f*x+e)^3*sec(f*x+e)/f+(a*cos(f*x+e)^2)^(1/2)*tan(f*x+e)/f
Time = 0.09 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.52 \[ \int \cot ^4(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=-\frac {\sqrt {a \cos ^2(e+f x)} \left (-3-6 \csc ^2(e+f x)+\csc ^4(e+f x)\right ) \tan (e+f x)}{3 f} \] Input:
Integrate[Cot[e + f*x]^4*Sqrt[a - a*Sin[e + f*x]^2],x]
Output:
-1/3*(Sqrt[a*Cos[e + f*x]^2]*(-3 - 6*Csc[e + f*x]^2 + Csc[e + f*x]^4)*Tan[ e + f*x])/f
Time = 0.41 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.59, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3042, 3655, 3042, 3686, 3042, 3070, 244, 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.
\(\displaystyle \int \cot ^4(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a-a \sin (e+f x)^2}}{\tan (e+f x)^4}dx\) |
\(\Big \downarrow \) 3655 |
\(\displaystyle \int \cot ^4(e+f x) \sqrt {a \cos ^2(e+f x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan \left (e+f x+\frac {\pi }{2}\right )^4 \sqrt {a \sin \left (e+f x+\frac {\pi }{2}\right )^2}dx\) |
\(\Big \downarrow \) 3686 |
\(\displaystyle \sec (e+f x) \sqrt {a \cos ^2(e+f x)} \int \cos (e+f x) \cot ^4(e+f x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \sec (e+f x) \sqrt {a \cos ^2(e+f x)} \int \sin \left (e+f x+\frac {\pi }{2}\right ) \tan \left (e+f x+\frac {\pi }{2}\right )^4dx\) |
\(\Big \downarrow \) 3070 |
\(\displaystyle -\frac {\sec (e+f x) \sqrt {a \cos ^2(e+f x)} \int \csc ^4(e+f x) \left (1-\sin ^2(e+f x)\right )^2d(-\sin (e+f x))}{f}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle -\frac {\sec (e+f x) \sqrt {a \cos ^2(e+f x)} \int \left (\csc ^4(e+f x)-2 \csc ^2(e+f x)+1\right )d(-\sin (e+f x))}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\sec (e+f x) \sqrt {a \cos ^2(e+f x)} \left (-\sin (e+f x)+\frac {1}{3} \csc ^3(e+f x)-2 \csc (e+f x)\right )}{f}\) |
Input:
Int[Cot[e + f*x]^4*Sqrt[a - a*Sin[e + f*x]^2],x]
Output:
-((Sqrt[a*Cos[e + f*x]^2]*Sec[e + f*x]*(-2*Csc[e + f*x] + Csc[e + f*x]^3/3 - Sin[e + f*x]))/f)
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Simp[-f^(-1) Subst[Int[(1 - x^2)^((m + n - 1)/2)/x^n, x], x, Cos[e + f *x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]
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 = 0.64 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82
method | result | size |
default | \(-\frac {\cos \left (f x +e \right ) a \left (3 \sin \left (f x +e \right )^{4}+6 \sin \left (f x +e \right )^{2}-1\right )}{3 \left (\cos \left (f x +e \right )-1\right ) \left (\cos \left (f x +e \right )+1\right ) \sin \left (f x +e \right ) \sqrt {a \cos \left (f x +e \right )^{2}}\, f}\) | \(75\) |
risch | \(-\frac {i \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left (3 \,{\mathrm e}^{8 i \left (f x +e \right )}-36 \,{\mathrm e}^{6 i \left (f x +e \right )}+50 \,{\mathrm e}^{4 i \left (f x +e \right )}-36 \,{\mathrm e}^{2 i \left (f x +e \right )}+3\right )}{6 \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{3} f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\) | \(105\) |
Input:
int(cot(f*x+e)^4*(a-a*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/3*cos(f*x+e)*a*(3*sin(f*x+e)^4+6*sin(f*x+e)^2-1)/(cos(f*x+e)-1)/(cos(f* x+e)+1)/sin(f*x+e)/(a*cos(f*x+e)^2)^(1/2)/f
Time = 0.09 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.73 \[ \int \cot ^4(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=-\frac {{\left (3 \, \cos \left (f x + e\right )^{4} - 12 \, \cos \left (f x + e\right )^{2} + 8\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{3 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \] Input:
integrate(cot(f*x+e)^4*(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")
Output:
-1/3*(3*cos(f*x + e)^4 - 12*cos(f*x + e)^2 + 8)*sqrt(a*cos(f*x + e)^2)/((f *cos(f*x + e)^3 - f*cos(f*x + e))*sin(f*x + e))
\[ \int \cot ^4(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \cot ^{4}{\left (e + f x \right )}\, dx \] Input:
integrate(cot(f*x+e)**4*(a-a*sin(f*x+e)**2)**(1/2),x)
Output:
Integral(sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))*cot(e + f*x)**4, x )
Time = 0.13 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.63 \[ \int \cot ^4(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\frac {8 \, \sqrt {a} \tan \left (f x + e\right )^{4} + 4 \, \sqrt {a} \tan \left (f x + e\right )^{2} - \sqrt {a}}{3 \, \sqrt {\tan \left (f x + e\right )^{2} + 1} f \tan \left (f x + e\right )^{3}} \] Input:
integrate(cot(f*x+e)^4*(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
Output:
1/3*(8*sqrt(a)*tan(f*x + e)^4 + 4*sqrt(a)*tan(f*x + e)^2 - sqrt(a))/(sqrt( tan(f*x + e)^2 + 1)*f*tan(f*x + e)^3)
Exception generated. \[ \int \cot ^4(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cot(f*x+e)^4*(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Time = 39.46 (sec) , antiderivative size = 364, normalized size of antiderivative = 4.00 \[ \int \cot ^4(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\frac {\left (\frac {1{}\mathrm {i}}{f}-\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}}{f}\right )\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}+\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,8{}\mathrm {i}}{f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,16{}\mathrm {i}}{3\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^2\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )}+\frac {{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\sqrt {a-a\,{\left (\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}-\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2}\right )}^2}\,16{}\mathrm {i}}{3\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}-1\right )}^3\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\right )} \] Input:
int(cot(e + f*x)^4*(a - a*sin(e + f*x)^2)^(1/2),x)
Output:
((1i/f - (exp(e*2i + f*x*2i)*1i)/f)*(a - a*((exp(- e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2))/(exp(e*2i + f*x*2i) + 1) + (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)*8i)/(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*1i)*1i)/2 - (ex p(e*1i + f*x*1i)*1i)/2)^2)^(1/2)*16i)/(3*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*3i)*(a - a*((exp( - e*1i - f*x*1i)*1i)/2 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2)*16i)/(3*f*(ex p(e*2i + f*x*2i) - 1)^3*(exp(e*1i + f*x*1i) + exp(e*3i + f*x*3i)))
\[ \int \cot ^4(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\sqrt {a}\, \left (\int \sqrt {-\sin \left (f x +e \right )^{2}+1}\, \cot \left (f x +e \right )^{4}d x \right ) \] Input:
int(cot(f*x+e)^4*(a-a*sin(f*x+e)^2)^(1/2),x)
Output:
sqrt(a)*int(sqrt( - sin(e + f*x)**2 + 1)*cot(e + f*x)**4,x)