3.461 \(\int \cot ^3(e+f x) \sqrt{a-a \sin ^2(e+f x)} \, dx\)

Optimal. Leaf size=87 \[ -\frac{3 \sqrt{a \cos ^2(e+f x)}}{2 f}-\frac{\csc ^2(e+f x) \left (a \cos ^2(e+f x)\right )^{3/2}}{2 a f}+\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \cos ^2(e+f x)}}{\sqrt{a}}\right )}{2 f} \]

[Out]

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

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Rubi [A]  time = 0.118243, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {3176, 3205, 16, 47, 50, 63, 206} \[ -\frac{3 \sqrt{a \cos ^2(e+f x)}}{2 f}-\frac{\csc ^2(e+f x) \left (a \cos ^2(e+f x)\right )^{3/2}}{2 a f}+\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \cos ^2(e+f x)}}{\sqrt{a}}\right )}{2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sqrt[a]*ArcTanh[Sqrt[a*Cos[e + f*x]^2]/Sqrt[a]])/(2*f) - (3*Sqrt[a*Cos[e + f*x]^2])/(2*f) - ((a*Cos[e + f*x
]^2)^(3/2)*Csc[e + f*x]^2)/(2*a*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 47

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

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 \cot ^3(e+f x) \sqrt{a-a \sin ^2(e+f x)} \, dx &=\int \sqrt{a \cos ^2(e+f x)} \cot ^3(e+f x) \, dx\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x \sqrt{a x}}{(1-x)^2} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(a x)^{3/2}}{(1-x)^2} \, dx,x,\cos ^2(e+f x)\right )}{2 a f}\\ &=-\frac{\left (a \cos ^2(e+f x)\right )^{3/2} \csc ^2(e+f x)}{2 a f}+\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt{a x}}{1-x} \, dx,x,\cos ^2(e+f x)\right )}{4 f}\\ &=-\frac{3 \sqrt{a \cos ^2(e+f x)}}{2 f}-\frac{\left (a \cos ^2(e+f x)\right )^{3/2} \csc ^2(e+f x)}{2 a f}+\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a x}} \, dx,x,\cos ^2(e+f x)\right )}{4 f}\\ &=-\frac{3 \sqrt{a \cos ^2(e+f x)}}{2 f}-\frac{\left (a \cos ^2(e+f x)\right )^{3/2} \csc ^2(e+f x)}{2 a f}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^2}{a}} \, dx,x,\sqrt{a \cos ^2(e+f x)}\right )}{2 f}\\ &=\frac{3 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \cos ^2(e+f x)}}{\sqrt{a}}\right )}{2 f}-\frac{3 \sqrt{a \cos ^2(e+f x)}}{2 f}-\frac{\left (a \cos ^2(e+f x)\right )^{3/2} \csc ^2(e+f x)}{2 a f}\\ \end{align*}

Mathematica [A]  time = 0.432015, size = 88, normalized size = 1.01 \[ -\frac{\sec (e+f x) \sqrt{a \cos ^2(e+f x)} \left (8 \cos (e+f x)+\csc ^2\left (\frac{1}{2} (e+f x)\right )-\sec ^2\left (\frac{1}{2} (e+f x)\right )+12 \log \left (\sin \left (\frac{1}{2} (e+f x)\right )\right )-12 \log \left (\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{8 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a*Cos[e + f*x]^2]*(8*Cos[e + f*x] + Csc[(e + f*x)/2]^2 - 12*Log[Cos[(e + f*x)/2]] + 12*Log[Sin[(e + f*x
)/2]] - Sec[(e + f*x)/2]^2)*Sec[e + f*x])/(8*f)

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Maple [A]  time = 1.508, size = 83, normalized size = 1. \begin{align*} -{\frac{1}{f}\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{3}{2\,f}\sqrt{a}\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 ) }-{\frac{1}{2\,f \left ( \sin \left ( fx+e \right ) \right ) ^{2}}\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)^3*(a-a*sin(f*x+e)^2)^(1/2),x)

[Out]

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

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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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6959, size = 230, normalized size = 2.64 \begin{align*} -\frac{\sqrt{a \cos \left (f x + e\right )^{2}}{\left (4 \, \cos \left (f x + e\right )^{3} + 3 \,{\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 ) - 6 \, \cos \left (f x + e\right )\right )}}{4 \,{\left (f \cos \left (f x + e\right )^{3} - 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)^(1/2),x, algorithm="fricas")

[Out]

-1/4*sqrt(a*cos(f*x + e)^2)*(4*cos(f*x + e)^3 + 3*(cos(f*x + e)^2 - 1)*log(-(cos(f*x + e) - 1)/(cos(f*x + e) +
1)) - 6*cos(f*x + e))/(f*cos(f*x + e)^3 - f*cos(f*x + e))

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

[Out]

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

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Giac [B]  time = 1.24256, size = 231, normalized size = 2.66 \begin{align*} -\frac{{\left (\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} - 6 \, \log \left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 1\right ) + \frac{3 \, \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 )^{4} - 14 \, \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} - \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 )^{4} + \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2}}\right )} \sqrt{a}}{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)^(1/2),x, algorithm="giac")

[Out]

-1/8*(sgn(tan(1/2*f*x + 1/2*e)^4 - 1)*tan(1/2*f*x + 1/2*e)^2 - 6*log(tan(1/2*f*x + 1/2*e)^2)*sgn(tan(1/2*f*x +
1/2*e)^4 - 1) + (3*sgn(tan(1/2*f*x + 1/2*e)^4 - 1)*tan(1/2*f*x + 1/2*e)^4 - 14*sgn(tan(1/2*f*x + 1/2*e)^4 - 1
)*tan(1/2*f*x + 1/2*e)^2 - sgn(tan(1/2*f*x + 1/2*e)^4 - 1))/(tan(1/2*f*x + 1/2*e)^4 + tan(1/2*f*x + 1/2*e)^2))
*sqrt(a)/f