3.5.81 \(\int \frac {\cot (e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\) [481]

Optimal. Leaf size=53 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f}+\frac {1}{a f \sqrt {a \cos ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {3255, 3284, 53, 65, 212} \begin {gather*} \frac {1}{a f \sqrt {a \cos ^2(e+f x)}}-\frac {\tanh ^{-1}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{a^{3/2} f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTanh[Sqrt[a*Cos[e + f*x]^2]/Sqrt[a]]/(a^(3/2)*f)) + 1/(a*f*Sqrt[a*Cos[e + f*x]^2])

Rule 53

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 65

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 212

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

Rule 3255

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 3284

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

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Mathematica [A]
time = 0.05, size = 55, normalized size = 1.04 \begin {gather*} \frac {1+\cos (e+f x) \left (-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+\log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )}{a f \sqrt {a \cos ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 16.43, size = 75, normalized size = 1.42

method	result	size
default	\(\frac {-\ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}+2 a}{\sin \left (f x +e \right )}\right ) a^{2} \left (\cos ^{2}\left (f x +e \right )\right )+\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, a^{\frac {3}{2}}}{a^{\frac {7}{2}} \cos \left (f x +e \right )^{2} f}\)	\(75\)
risch	\(\frac {2}{a \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, f}+\frac {2 \ln \left ({\mathrm e}^{i f x}-{\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, a}-\frac {2 \ln \left ({\mathrm e}^{i f x}+{\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, a}\)	\(144\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a^(7/2)/cos(f*x+e)^2*(-ln(2/sin(f*x+e)*(a^(1/2)*(a*cos(f*x+e)^2)^(1/2)+a))*a^2*cos(f*x+e)^2+(a*cos(f*x+e)^2)
^(1/2)*a^(3/2))/f

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Maxima [A]
time = 0.31, size = 77, normalized size = 1.45 \begin {gather*} -\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 )}{a^{\frac {3}{2}}} - \frac {1}{\sqrt {-a \sin \left (f x + e\right )^{2} + a} a}}{f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(log(2*sqrt(-a*sin(f*x + e)^2 + a)*sqrt(a)/abs(sin(f*x + e)) + 2*a/abs(sin(f*x + e)))/a^(3/2) - 1/(sqrt(-a*si
n(f*x + e)^2 + a)*a))/f

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Fricas [A]
time = 0.41, size = 58, normalized size = 1.09 \begin {gather*} -\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (\cos \left (f x + e\right ) \log \left (-\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1}\right ) - 2\right )}}{2 \, a^{2} f \cos \left (f x + e\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot {\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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 59, normalized size = 1.11 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {-a \sin \left (f x + e\right )^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a f} + \frac {1}{\sqrt {-a \sin \left (f x + e\right )^{2} + a} a f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

arctan(sqrt(-a*sin(f*x + e)^2 + a)/sqrt(-a))/(sqrt(-a)*a*f) + 1/(sqrt(-a*sin(f*x + e)^2 + a)*a*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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