3.467 \(\int \cot ^6(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx\)

Optimal. Leaf size=124 \[ -\frac {\tan (e+f x) \sqrt {a \cos ^2(e+f x)}}{f}-\frac {\csc ^5(e+f x) \sec (e+f x) \sqrt {a \cos ^2(e+f x)}}{5 f}+\frac {\csc ^3(e+f x) \sec (e+f x) \sqrt {a \cos ^2(e+f x)}}{f}-\frac {3 \csc (e+f x) \sec (e+f x) \sqrt {a \cos ^2(e+f x)}}{f} \]

[Out]

-3*csc(f*x+e)*sec(f*x+e)*(a*cos(f*x+e)^2)^(1/2)/f+csc(f*x+e)^3*sec(f*x+e)*(a*cos(f*x+e)^2)^(1/2)/f-1/5*csc(f*x
+e)^5*sec(f*x+e)*(a*cos(f*x+e)^2)^(1/2)/f-(a*cos(f*x+e)^2)^(1/2)*tan(f*x+e)/f

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Rubi [A]  time = 0.12, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3176, 3207, 2590, 270} \[ -\frac {\tan (e+f x) \sqrt {a \cos ^2(e+f x)}}{f}-\frac {\csc ^5(e+f x) \sec (e+f x) \sqrt {a \cos ^2(e+f x)}}{5 f}+\frac {\csc ^3(e+f x) \sec (e+f x) \sqrt {a \cos ^2(e+f x)}}{f}-\frac {3 \csc (e+f x) \sec (e+f x) \sqrt {a \cos ^2(e+f x)}}{f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^6*Sqrt[a - a*Sin[e + f*x]^2],x]

[Out]

(-3*Sqrt[a*Cos[e + f*x]^2]*Csc[e + f*x]*Sec[e + f*x])/f + (Sqrt[a*Cos[e + f*x]^2]*Csc[e + f*x]^3*Sec[e + f*x])
/f - (Sqrt[a*Cos[e + f*x]^2]*Csc[e + f*x]^5*Sec[e + f*x])/(5*f) - (Sqrt[a*Cos[e + f*x]^2]*Tan[e + f*x])/f

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[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]

Rule 3176

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]

Rule 3207

Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[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
]])

Rubi steps

\begin {align*} \int \cot ^6(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx &=\int \sqrt {a \cos ^2(e+f x)} \cot ^6(e+f x) \, dx\\ &=\left (\sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \int \cos (e+f x) \cot ^6(e+f x) \, dx\\ &=-\frac {\left (\sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^6} \, dx,x,-\sin (e+f x)\right )}{f}\\ &=-\frac {\left (\sqrt {a \cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \left (-1+\frac {1}{x^6}-\frac {3}{x^4}+\frac {3}{x^2}\right ) \, dx,x,-\sin (e+f x)\right )}{f}\\ &=-\frac {3 \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)}{f}-\frac {\sqrt {a \cos ^2(e+f x)} \csc ^5(e+f x) \sec (e+f x)}{5 f}-\frac {\sqrt {a \cos ^2(e+f x)} \tan (e+f x)}{f}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 67, normalized size = 0.54 \[ \frac {(235 \cos (2 (e+f x))-90 \cos (4 (e+f x))+5 \cos (6 (e+f x))-182) \csc ^5(e+f x) \sec (e+f x) \sqrt {a \cos ^2(e+f x)}}{160 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^6*Sqrt[a - a*Sin[e + f*x]^2],x]

[Out]

(Sqrt[a*Cos[e + f*x]^2]*(-182 + 235*Cos[2*(e + f*x)] - 90*Cos[4*(e + f*x)] + 5*Cos[6*(e + f*x)])*Csc[e + f*x]^
5*Sec[e + f*x])/(160*f)

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fricas [A]  time = 0.46, size = 86, normalized size = 0.69 \[ \frac {{\left (5 \, \cos \left (f x + e\right )^{6} - 30 \, \cos \left (f x + e\right )^{4} + 40 \, \cos \left (f x + e\right )^{2} - 16\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{5 \, {\left (f \cos \left (f x + e\right )^{5} - 2 \, f \cos \left (f x + e\right )^{3} + f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6*(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

1/5*(5*cos(f*x + e)^6 - 30*cos(f*x + e)^4 + 40*cos(f*x + e)^2 - 16)*sqrt(a*cos(f*x + e)^2)/((f*cos(f*x + e)^5
- 2*f*cos(f*x + e)^3 + f*cos(f*x + e))*sin(f*x + e))

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6*(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)2/f*sqrt(a)*(sign(tan((f*x+exp(1))/2)^4-1)/(tan((f*x+exp(1))/2)+1/tan((f*x+exp(1))/2
))+1/1073741824*(-67108864*(tan((f*x+exp(1))/2)+1/tan((f*x+exp(1))/2))^3*sign(tan((f*x+exp(1))/2)^4-1)+1677721
6/5*(tan((f*x+exp(1))/2)+1/tan((f*x+exp(1))/2))^5*sign(tan((f*x+exp(1))/2)^4-1)[phone]*(tan((f*x+exp(1))/2)
+1/tan((f*x+exp(1))/2))*sign(tan((f*x+exp(1))/2)^4-1)))

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maple [A]  time = 1.10, size = 65, normalized size = 0.52 \[ -\frac {\cos \left (f x +e \right ) a \left (5 \left (\sin ^{6}\left (f x +e \right )\right )+15 \left (\sin ^{4}\left (f x +e \right )\right )-5 \left (\sin ^{2}\left (f x +e \right )\right )+1\right )}{5 \sin \left (f x +e \right )^{5} \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^6*(a-a*sin(f*x+e)^2)^(1/2),x)

[Out]

-1/5*cos(f*x+e)*a*(5*sin(f*x+e)^6+15*sin(f*x+e)^4-5*sin(f*x+e)^2+1)/sin(f*x+e)^5/(a*cos(f*x+e)^2)^(1/2)/f

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maxima [A]  time = 0.42, size = 68, normalized size = 0.55 \[ -\frac {16 \, \sqrt {a} \tan \left (f x + e\right )^{6} + 8 \, \sqrt {a} \tan \left (f x + e\right )^{4} - 2 \, \sqrt {a} \tan \left (f x + e\right )^{2} + \sqrt {a}}{5 \, \sqrt {\tan \left (f x + e\right )^{2} + 1} f \tan \left (f x + e\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6*(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/5*(16*sqrt(a)*tan(f*x + e)^6 + 8*sqrt(a)*tan(f*x + e)^4 - 2*sqrt(a)*tan(f*x + e)^2 + sqrt(a))/(sqrt(tan(f*x
+ e)^2 + 1)*f*tan(f*x + e)^5)

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mupad [B]  time = 26.64, size = 555, normalized size = 4.48 \[ -\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}\,12{}\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}}{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}\,144{}\mathrm {i}}{5\,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\,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\,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 )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(e + f*x)^6*(a - a*sin(e + f*x)^2)^(1/2),x)

[Out]

- ((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)*12i)/(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 - (exp(e*1i + f*x*1i)*1i)/2)^2)^(1/2)*16i)/(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)*144i)/(5*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*f*(e
xp(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*f*(exp(e*2i + f*x*2i) - 1)^5*(exp(e*1i + f*x*1i)
+ exp(e*3i + f*x*3i)))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \cot ^{6}{\left (e + f x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**6*(a-a*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(sqrt(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))*cot(e + f*x)**6, x)

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