3.486 \(\int \frac{\cot ^6(e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=77 \[ \frac{\cot (e+f x) \csc ^2(e+f x)}{3 a f \sqrt{a \cos ^2(e+f x)}}-\frac{\cot (e+f x) \csc ^4(e+f x)}{5 a f \sqrt{a \cos ^2(e+f x)}} \]

[Out]

(Cot[e + f*x]*Csc[e + f*x]^2)/(3*a*f*Sqrt[a*Cos[e + f*x]^2]) - (Cot[e + f*x]*Csc[e + f*x]^4)/(5*a*f*Sqrt[a*Cos
[e + f*x]^2])

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Rubi [A]  time = 0.144077, antiderivative size = 77, normalized size of antiderivative = 1., 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, 2606, 14} \[ \frac{\cot (e+f x) \csc ^2(e+f x)}{3 a f \sqrt{a \cos ^2(e+f x)}}-\frac{\cot (e+f x) \csc ^4(e+f x)}{5 a f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^6/(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]) - (Cot[e + f*x]*Csc[e + f*x]^4)/(5*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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.0932803, size = 41, normalized size = 0.53 \[ -\frac{\cot ^3(e+f x) \left (3 \csc ^2(e+f x)-5\right )}{15 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^6/(a - a*Sin[e + f*x]^2)^(3/2),x]

[Out]

-(Cot[e + f*x]^3*(-5 + 3*Csc[e + f*x]^2))/(15*f*(a*Cos[e + f*x]^2)^(3/2))

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Maple [A]  time = 0.855, size = 67, normalized size = 0.9 \begin{align*} -{\frac{\cos \left ( fx+e \right ) \left ( 5\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}-2 \right ) }{15\, \left ( -1+\cos \left ( fx+e \right ) \right ) ^{2} \left ( \cos \left ( fx+e \right ) +1 \right ) ^{2}a\sin \left ( fx+e \right ) 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)^6/(a-a*sin(f*x+e)^2)^(3/2),x)

[Out]

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

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Maxima [B]  time = 1.74951, size = 1435, normalized size = 18.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

8/15*((5*sin(7*f*x + 7*e) + 2*sin(5*f*x + 5*e) + 5*sin(3*f*x + 3*e))*cos(10*f*x + 10*e) - 5*(5*sin(7*f*x + 7*e
) + 2*sin(5*f*x + 5*e) + 5*sin(3*f*x + 3*e))*cos(8*f*x + 8*e) - 25*(2*sin(6*f*x + 6*e) - 2*sin(4*f*x + 4*e) +
sin(2*f*x + 2*e))*cos(7*f*x + 7*e) + 10*(2*sin(5*f*x + 5*e) + 5*sin(3*f*x + 3*e))*cos(6*f*x + 6*e) + 10*(2*sin
(4*f*x + 4*e) - sin(2*f*x + 2*e))*cos(5*f*x + 5*e) - (5*cos(7*f*x + 7*e) + 2*cos(5*f*x + 5*e) + 5*cos(3*f*x +
3*e))*sin(10*f*x + 10*e) + 5*(5*cos(7*f*x + 7*e) + 2*cos(5*f*x + 5*e) + 5*cos(3*f*x + 3*e))*sin(8*f*x + 8*e) +
5*(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*(2*cos(5*f*x + 5
*e) + 5*cos(3*f*x + 3*e))*sin(6*f*x + 6*e) - 2*(10*cos(4*f*x + 4*e) - 5*cos(2*f*x + 2*e) + 1)*sin(5*f*x + 5*e)
+ 50*cos(3*f*x + 3*e)*sin(4*f*x + 4*e) + 5*(5*cos(2*f*x + 2*e) - 1)*sin(3*f*x + 3*e) - 50*cos(4*f*x + 4*e)*si
n(3*f*x + 3*e) - 25*cos(3*f*x + 3*e)*sin(2*f*x + 2*e))*sqrt(a)/((a^2*cos(10*f*x + 10*e)^2 + 25*a^2*cos(8*f*x +
8*e)^2 + 100*a^2*cos(6*f*x + 6*e)^2 + 100*a^2*cos(4*f*x + 4*e)^2 + 25*a^2*cos(2*f*x + 2*e)^2 + a^2*sin(10*f*x
+ 10*e)^2 + 25*a^2*sin(8*f*x + 8*e)^2 + 100*a^2*sin(6*f*x + 6*e)^2 + 100*a^2*sin(4*f*x + 4*e)^2 - 100*a^2*sin
(4*f*x + 4*e)*sin(2*f*x + 2*e) + 25*a^2*sin(2*f*x + 2*e)^2 - 10*a^2*cos(2*f*x + 2*e) + a^2 - 2*(5*a^2*cos(8*f*
x + 8*e) - 10*a^2*cos(6*f*x + 6*e) + 10*a^2*cos(4*f*x + 4*e) - 5*a^2*cos(2*f*x + 2*e) + a^2)*cos(10*f*x + 10*e
) - 10*(10*a^2*cos(6*f*x + 6*e) - 10*a^2*cos(4*f*x + 4*e) + 5*a^2*cos(2*f*x + 2*e) - a^2)*cos(8*f*x + 8*e) - 2
0*(10*a^2*cos(4*f*x + 4*e) - 5*a^2*cos(2*f*x + 2*e) + a^2)*cos(6*f*x + 6*e) - 20*(5*a^2*cos(2*f*x + 2*e) - a^2
)*cos(4*f*x + 4*e) - 10*(a^2*sin(8*f*x + 8*e) - 2*a^2*sin(6*f*x + 6*e) + 2*a^2*sin(4*f*x + 4*e) - a^2*sin(2*f*
x + 2*e))*sin(10*f*x + 10*e) - 50*(2*a^2*sin(6*f*x + 6*e) - 2*a^2*sin(4*f*x + 4*e) + a^2*sin(2*f*x + 2*e))*sin
(8*f*x + 8*e) - 100*(2*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 = 1.68613, size = 185, normalized size = 2.4 \begin{align*} -\frac{\sqrt{a \cos \left (f x + e\right )^{2}}{\left (5 \, \cos \left (f x + e\right )^{2} - 2\right )}}{15 \,{\left (a^{2} f \cos \left (f x + e\right )^{5} - 2 \, 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)^6/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/15*sqrt(a*cos(f*x + e)^2)*(5*cos(f*x + e)^2 - 2)/((a^2*f*cos(f*x + e)^5 - 2*a^2*f*cos(f*x + e)^3 + a^2*f*co
s(f*x + e))*sin(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)**6/(a-a*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 1.42406, size = 204, normalized size = 2.65 \begin{align*} -\frac{\frac{30 \, \sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 5 \, \sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 3 \, \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 )^{5}} - \frac{3 \, a^{\frac{17}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 5 \, a^{\frac{17}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 30 \, a^{\frac{17}{2}} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{10} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )}}{480 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

-1/480*((30*sqrt(a)*tan(1/2*f*x + 1/2*e)^4 + 5*sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - 3*sqrt(a))/(a^2*sgn(tan(1/2*f*
x + 1/2*e)^4 - 1)*tan(1/2*f*x + 1/2*e)^5) - (3*a^(17/2)*tan(1/2*f*x + 1/2*e)^5 - 5*a^(17/2)*tan(1/2*f*x + 1/2*
e)^3 - 30*a^(17/2)*tan(1/2*f*x + 1/2*e))/(a^10*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)))/f