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

Optimal. Leaf size=63 \[ \frac{\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{a f \sqrt{a \cos ^2(e+f x)}}-\frac{\cot (e+f x)}{a f \sqrt{a \cos ^2(e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.132467, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3176, 3207, 2621, 321, 207} \[ \frac{\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{a f \sqrt{a \cos ^2(e+f x)}}-\frac{\cot (e+f x)}{a f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(a*f*Sqrt[a*Cos[e + f*x]^2]) - Cot[e + f*x]/(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 2621

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(f*a^n)^(-1), Subst
[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && Integer
Q[(n + 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 0.0756979, size = 44, normalized size = 0.7 \[ -\frac{\cot (e+f x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\sin ^2(e+f x)\right )}{a f \sqrt{a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Cot[e + f*x]*Hypergeometric2F1[-1/2, 1, 1/2, Sin[e + f*x]^2])/(a*f*Sqrt[a*Cos[e + f*x]^2]))

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Maple [A]  time = 1.17, size = 65, normalized size = 1. \begin{align*} -{\frac{\cos \left ( fx+e \right ) \left ( 2+\sin \left ( fx+e \right ) \left ( \ln \left ( -1+\sin \left ( fx+e \right ) \right ) -\ln \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) \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)^2/(a-a*sin(f*x+e)^2)^(3/2),x)

[Out]

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

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Maxima [B]  time = 1.71972, size = 297, normalized size = 4.71 \begin{align*} \frac{{\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} - 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) -{\left (\cos \left (2 \, f x + 2 \, e\right )^{2} + \sin \left (2 \, f x + 2 \, e\right )^{2} - 2 \, \cos \left (2 \, f x + 2 \, e\right ) + 1\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right ) - 4 \, \cos \left (f x + e\right ) \sin \left (2 \, f x + 2 \, e\right ) + 4 \, \cos \left (2 \, f x + 2 \, e\right ) \sin \left (f x + e\right ) - 4 \, \sin \left (f x + e\right )}{2 \,{\left (a \cos \left (2 \, f x + 2 \, e\right )^{2} + a \sin \left (2 \, f x + 2 \, e\right )^{2} - 2 \, a \cos \left (2 \, f x + 2 \, e\right ) + a\right )} \sqrt{a} f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 - 2*cos(2*f*x + 2*e) + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2 +
2*sin(f*x + e) + 1) - (cos(2*f*x + 2*e)^2 + sin(2*f*x + 2*e)^2 - 2*cos(2*f*x + 2*e) + 1)*log(cos(f*x + e)^2 +
sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - 4*cos(f*x + e)*sin(2*f*x + 2*e) + 4*cos(2*f*x + 2*e)*sin(f*x + e) - 4*s
in(f*x + e))/((a*cos(2*f*x + 2*e)^2 + a*sin(2*f*x + 2*e)^2 - 2*a*cos(2*f*x + 2*e) + a)*sqrt(a)*f)

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Fricas [A]  time = 1.74274, size = 170, normalized size = 2.7 \begin{align*} -\frac{\sqrt{a \cos \left (f x + e\right )^{2}}{\left (\log \left (-\frac{\sin \left (f x + e\right ) - 1}{\sin \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 2\right )}}{2 \, a^{2} f \cos \left (f x + e\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)^2/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

-1/2*sqrt(a*cos(f*x + e)^2)*(log(-(sin(f*x + e) - 1)/(sin(f*x + e) + 1))*sin(f*x + e) + 2)/(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 ^{2}{\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)**2/(a-a*sin(f*x+e)**2)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.30893, size = 95, normalized size = 1.51 \begin{align*} \frac{\frac{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )} + \frac{1}{a^{\frac{3}{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 )}}{2 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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