\(\int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx\) [472]

Optimal result

Integrand size = 26, antiderivative size = 66 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{2 \sqrt {a} f}-\frac {\sqrt {a \cos ^2(e+f x)} \csc ^2(e+f x)}{2 a f} \]

[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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3255, 3284, 16, 43, 65, 212} \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{2 \sqrt {a} f}-\frac {\csc ^2(e+f x) \sqrt {a \cos ^2(e+f x)}}{2 a f} \]

[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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.20 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.88 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {\text {arctanh}\left (\sqrt {\cos ^2(e+f x)}\right ) \sqrt {\cos ^2(e+f x)}-\cot ^2(e+f x)}{2 f \sqrt {a \cos ^2(e+f x)}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.88 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.26

method	result	size
default	\(\frac {\cos \left (f x +e \right ) \left (2 \cos \left (f x +e \right )+\left (-\ln \left (1+\cos \left (f x +e \right )\right )+\ln \left (\cos \left (f x +e \right )-1\right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )\right )}{4 \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) f}\)	\(83\)
risch	\(\frac {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}{\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i f x}-{\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}+\frac {\ln \left ({\mathrm e}^{i f x}+{\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}\)	\(159\)

[In]

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

[Out]

1/4*cos(f*x+e)*(2*cos(f*x+e)+(-ln(1+cos(f*x+e))+ln(cos(f*x+e)-1))*sin(f*x+e)^2)/(a*cos(f*x+e)^2)^(1/2)/(cos(f*
x+e)-1)/(1+cos(f*x+e))/f

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.20 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left ({\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right ) - 2 \, \cos \left (f x + e\right )\right )}}{4 \, {\left (a f \cos \left (f x + e\right )^{3} - a f \cos \left (f x + e\right )\right )}} \]

[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))

Sympy [F]

\[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\int \frac {\cot ^{3}{\left (e + f x \right )}}{\sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )}}\, dx \]

[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)

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.23 \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\frac {\frac {\log \left (\frac {2 \, \sqrt {-a \sin \left (f x + e\right )^{2} + a} \sqrt {a}}{{\left | \sin \left (f x + e\right ) \right |}} + \frac {2 \, a}{{\left | \sin \left (f x + e\right ) \right |}}\right )}{\sqrt {a}} - \frac {\sqrt {-a \sin \left (f x + e\right )^{2} + a}}{a \sin \left (f x + e\right )^{2}}}{2 \, f} \]

[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

Giac [F(-2)]

Exception generated. \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\text {Exception raised: TypeError} \]

[In]

integrate(cot(f*x+e)^3/(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:index.cc index_m operator + Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {\cot ^3(e+f x)}{\sqrt {a-a \sin ^2(e+f x)}} \, dx=\int \frac {{\mathrm {cot}\left (e+f\,x\right )}^3}{\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2}} \,d x \]

[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)