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

Optimal. Leaf size=66 \[ -\frac{\csc ^2(e+f x) \sqrt{a \cos ^2(e+f x)}}{2 a^2 f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cos ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{3/2} f} \]

[Out]

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

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Rubi [A]  time = 0.126603, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {3176, 3205, 16, 51, 63, 206} \[ -\frac{\csc ^2(e+f x) \sqrt{a \cos ^2(e+f x)}}{2 a^2 f}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a \cos ^2(e+f x)}}{\sqrt{a}}\right )}{2 a^{3/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3205

Int[((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFact
ors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(b*ff^(n/2)*x^(n/2))^p)/(1 - ff*x
)^((m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{b, e, f, p}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n/2
]

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.172236, size = 82, normalized size = 1.24 \[ -\frac{\cos ^3(e+f x) \left (\csc ^2\left (\frac{1}{2} (e+f x)\right )-\sec ^2\left (\frac{1}{2} (e+f x)\right )-4 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )+4 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{8 f \left (a \cos ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[e + f*x]^3*(Csc[(e + f*x)/2]^2 + 4*Log[Cos[(e + f*x)/2]] - 4*Log[Sin[(e + f*x)/2]] - Sec[(e + f*x)/2]^2)
)/(8*f*(a*Cos[e + f*x]^2)^(3/2))

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Maple [A]  time = 1.306, size = 69, normalized size = 1.1 \begin{align*} -{\frac{1}{2\,{a}^{2}f \left ( \sin \left ( fx+e \right ) \right ) ^{2}}\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}-{\frac{1}{2\,f}\ln \left ({\frac{1}{\sin \left ( fx+e \right ) } \left ( 2\,a+2\,\sqrt{a}\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6094, size = 212, normalized size = 3.21 \begin{align*} -\frac{\sqrt{a \cos \left (f x + e\right )^{2}}{\left ({\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac{\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1}\right ) - 2 \, \cos \left (f x + e\right )\right )}}{4 \,{\left (a^{2} f \cos \left (f x + e\right )^{3} - a^{2} f \cos \left (f x + e\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(a*cos(f*x + e)^2)*((cos(f*x + e)^2 - 1)*log(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1)) - 2*cos(f*x + e)
)/(a^2*f*cos(f*x + e)^3 - a^2*f*cos(f*x + e))

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

[Out]

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

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Giac [B]  time = 1.35796, size = 174, normalized size = 2.64 \begin{align*} -\frac{\frac{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )} + \frac{2 \, \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}\right )}{a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right )} - \frac{2 \, \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 )^{2}}}{8 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(tan(1/2*f*x + 1/2*e)^2/(a^(3/2)*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)) + 2*log(tan(1/2*f*x + 1/2*e)^2)/(a^(3/2
)*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)) - (2*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)^2))/f