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

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

[Out]

1/2*arctanh((a*cos(f*x+e)^2)^(1/2)/a^(1/2))/f/a^(1/2)-1/2*csc(f*x+e)^2*(a*cos(f*x+e)^2)^(1/2)/a/f

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Rubi [A]  time = 0.11, antiderivative size = 66, normalized size of antiderivative = 1.00, 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, 47, 63, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{2 \sqrt {a} f}-\frac {\csc ^2(e+f x) \sqrt {a \cos ^2(e+f x)}}{2 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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])

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
]

Rubi steps

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

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Mathematica [A]  time = 0.17, size = 80, normalized size = 1.21 \[ \frac {\cos (e+f x) \left (-\csc ^2\left (\frac {1}{2} (e+f x)\right )+\sec ^2\left (\frac {1}{2} (e+f x)\right )-4 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{8 f \sqrt {a \cos ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[e + f*x]*(-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*Sqrt[a*Cos[e + f*x]^2])

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fricas [A]  time = 0.45, size = 79, normalized size = 1.20 \[ -\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left ({\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right ) - 2 \, \cos \left (f x + e\right )\right )}}{4 \, {\left (a f \cos \left (f x + e\right )^{3} - a f \cos \left (f x + e\right )\right )}} \]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

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]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)2/f/sqrt(a)*(-1/16*tan((f*x+exp(1))/2)^2/sign(tan((f*x+exp(1))/2)^4-1)+1/16*(-2*tan(
(f*x+exp(1))/2)^2+1)/tan((f*x+exp(1))/2)^2/sign(tan((f*x+exp(1))/2)^4-1)+1/8*ln(tan((f*x+exp(1))/2)^2)/sign(ta
n((f*x+exp(1))/2)^4-1))

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maple [A]  time = 1.72, size = 69, normalized size = 1.05 \[ \frac {\ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}+2 a}{\sin \left (f x +e \right )}\right )}{2 \sqrt {a}\, f}-\frac {\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}}{2 f a \sin \left (f x +e \right )^{2}} \]

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]

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

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maxima [A]  time = 0.41, size = 81, normalized size = 1.23 \[ \frac {\frac {\log \left (\frac {2 \, \sqrt {-a \sin \left (f x + e\right )^{2} + a} \sqrt {a}}{{\left | \sin \left (f x + e\right ) \right |}} + \frac {2 \, a}{{\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} - \frac {\sqrt {-a \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right )^{2}}}{2 \, f} \]

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]

1/2*(log(2*sqrt(-a*sin(f*x + e)^2 + a)*sqrt(a)/abs(sin(f*x + e)) + 2*a/abs(sin(f*x + e)))/sqrt(a) - sqrt(-a*si
n(f*x + e)^2 + a)/(a*sin(f*x + e)^2))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {{\mathrm {cot}\left (e+f\,x\right )}^3}{\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{3}{\left (e + f x \right )}}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]

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

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