3.485 \(\int \frac{\cot ^4(e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=38 \[ -\frac{\cot (e+f x) \csc ^2(e+f x)}{3 a f \sqrt{a \cos ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.126279, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3176, 3207, 2606, 30} \[ -\frac{\cot (e+f x) \csc ^2(e+f x)}{3 a f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4/(a - a*Sin[e + f*x]^2)^(3/2),x]

[Out]

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

Rule 2606

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0343927, size = 29, normalized size = 0.76 \[ -\frac{\cot ^3(e+f x)}{3 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^4/(a - a*Sin[e + f*x]^2)^(3/2),x]

[Out]

-Cot[e + f*x]^3/(3*f*(a*Cos[e + f*x]^2)^(3/2))

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Maple [A]  time = 0.528, size = 35, normalized size = 0.9 \begin{align*} -{\frac{\cos \left ( fx+e \right ) }{3\,a \left ( \sin \left ( fx+e \right ) \right ) ^{3}f}{\frac{1}{\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4/(a-a*sin(f*x+e)^2)^(3/2),x)

[Out]

-1/3/a*cos(f*x+e)/sin(f*x+e)^3/(a*cos(f*x+e)^2)^(1/2)/f

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Maxima [B]  time = 1.66848, size = 516, normalized size = 13.58 \begin{align*} \frac{8 \,{\left (\cos \left (3 \, f x + 3 \, e\right ) \sin \left (6 \, f x + 6 \, e\right ) - 3 \, \cos \left (3 \, f x + 3 \, e\right ) \sin \left (4 \, f x + 4 \, e\right ) -{\left (3 \, \cos \left (2 \, f x + 2 \, e\right ) - 1\right )} \sin \left (3 \, f x + 3 \, e\right ) - \cos \left (6 \, f x + 6 \, e\right ) \sin \left (3 \, f x + 3 \, e\right ) + 3 \, \cos \left (4 \, f x + 4 \, e\right ) \sin \left (3 \, f x + 3 \, e\right ) + 3 \, \cos \left (3 \, f x + 3 \, e\right ) \sin \left (2 \, f x + 2 \, e\right )\right )} \sqrt{a}}{3 \,{\left (a^{2} \cos \left (6 \, f x + 6 \, e\right )^{2} + 9 \, a^{2} \cos \left (4 \, f x + 4 \, e\right )^{2} + 9 \, a^{2} \cos \left (2 \, f x + 2 \, e\right )^{2} + a^{2} \sin \left (6 \, f x + 6 \, e\right )^{2} + 9 \, a^{2} \sin \left (4 \, f x + 4 \, e\right )^{2} - 18 \, a^{2} \sin \left (4 \, f x + 4 \, e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 9 \, a^{2} \sin \left (2 \, f x + 2 \, e\right )^{2} - 6 \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) + a^{2} - 2 \,{\left (3 \, a^{2} \cos \left (4 \, f x + 4 \, e\right ) - 3 \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) + a^{2}\right )} \cos \left (6 \, f x + 6 \, e\right ) - 6 \,{\left (3 \, a^{2} \cos \left (2 \, f x + 2 \, e\right ) - a^{2}\right )} \cos \left (4 \, f x + 4 \, e\right ) - 6 \,{\left (a^{2} \sin \left (4 \, f x + 4 \, e\right ) - a^{2} \sin \left (2 \, f x + 2 \, e\right )\right )} \sin \left (6 \, f x + 6 \, e\right )\right )} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

8/3*(cos(3*f*x + 3*e)*sin(6*f*x + 6*e) - 3*cos(3*f*x + 3*e)*sin(4*f*x + 4*e) - (3*cos(2*f*x + 2*e) - 1)*sin(3*
f*x + 3*e) - cos(6*f*x + 6*e)*sin(3*f*x + 3*e) + 3*cos(4*f*x + 4*e)*sin(3*f*x + 3*e) + 3*cos(3*f*x + 3*e)*sin(
2*f*x + 2*e))*sqrt(a)/((a^2*cos(6*f*x + 6*e)^2 + 9*a^2*cos(4*f*x + 4*e)^2 + 9*a^2*cos(2*f*x + 2*e)^2 + a^2*sin
(6*f*x + 6*e)^2 + 9*a^2*sin(4*f*x + 4*e)^2 - 18*a^2*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 9*a^2*sin(2*f*x + 2*e)
^2 - 6*a^2*cos(2*f*x + 2*e) + a^2 - 2*(3*a^2*cos(4*f*x + 4*e) - 3*a^2*cos(2*f*x + 2*e) + a^2)*cos(6*f*x + 6*e)
- 6*(3*a^2*cos(2*f*x + 2*e) - a^2)*cos(4*f*x + 4*e) - 6*(a^2*sin(4*f*x + 4*e) - a^2*sin(2*f*x + 2*e))*sin(6*f
*x + 6*e))*f)

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Fricas [A]  time = 1.6556, size = 117, normalized size = 3.08 \begin{align*} \frac{\sqrt{a \cos \left (f x + e\right )^{2}}}{3 \,{\left (a^{2} f \cos \left (f x + e\right )^{3} - a^{2} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

1/3*sqrt(a*cos(f*x + e)^2)/((a^2*f*cos(f*x + e)^3 - a^2*f*cos(f*x + e))*sin(f*x + e))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{4}{\left (e + f x \right )}}{\left (- a \left (\sin{\left (e + f x \right )} - 1\right ) \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**4/(a-a*sin(f*x+e)**2)**(3/2),x)

[Out]

Integral(cot(e + f*x)**4/(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))**(3/2), x)

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Giac [B]  time = 1.37892, size = 153, normalized size = 4.03 \begin{align*} \frac{\frac{3 \, \sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + \sqrt{a}}{a^{2} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right ) \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3}} + \frac{a^{\frac{9}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 3 \, a^{\frac{9}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{6} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

1/24*((3*sqrt(a)*tan(1/2*f*x + 1/2*e)^2 + sqrt(a))/(a^2*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)*tan(1/2*f*x + 1/2*e)^3
) + (a^(9/2)*tan(1/2*f*x + 1/2*e)^3 + 3*a^(9/2)*tan(1/2*f*x + 1/2*e))/(a^6*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)))/f