\(\int \cot ^3(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx\) [461]
Optimal result
Integrand size = 26, antiderivative size = 87 \[
\int \cot ^3(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\frac {3 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}-\frac {3 \sqrt {a \cos ^2(e+f x)}}{2 f}-\frac {\left (a \cos ^2(e+f x)\right )^{3/2} \csc ^2(e+f x)}{2 a f}
\]
[Out]
-1/2*(a*cos(f*x+e)^2)^(3/2)*csc(f*x+e)^2/a/f+3/2*arctanh((a*cos(f*x+e)^2)^(1/2)/a^(1/2))*a^(1/2)/f-3/2*(a*cos(
f*x+e)^2)^(1/2)/f
Rubi [A] (verified)
Time = 0.14 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.00, number of
steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {3255, 3284, 16, 43, 52, 65,
212} \[
\int \cot ^3(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\frac {3 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )}{2 f}-\frac {3 \sqrt {a \cos ^2(e+f x)}}{2 f}-\frac {\csc ^2(e+f x) \left (a \cos ^2(e+f x)\right )^{3/2}}{2 a f}
\]
[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 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 52
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 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 \sqrt {a \cos ^2(e+f x)} \cot ^3(e+f x) \, dx \\ & = -\frac {\text {Subst}\left (\int \frac {x \sqrt {a x}}{(1-x)^2} \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {\text {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 \text {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) \text {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 \text {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} \text {arctanh}\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] (verified)
Time = 1.52 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.11
\[
\int \cot ^3(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\frac {2 \sqrt {a} \text {arctanh}\left (\frac {\sqrt {a \cos ^2(e+f x)}}{\sqrt {a}}\right )+\sqrt {a \cos ^2(e+f x)} \left (\frac {\text {arctanh}\left (\sqrt {\cos ^2(e+f x)}\right )}{\sqrt {\cos ^2(e+f x)}}+(-2+\cos (2 (e+f x))) \csc ^2(e+f x)\right )}{2 f}
\]
[In]
Integrate[Cot[e + f*x]^3*Sqrt[a - a*Sin[e + f*x]^2],x]
[Out]
(2*Sqrt[a]*ArcTanh[Sqrt[a*Cos[e + f*x]^2]/Sqrt[a]] + Sqrt[a*Cos[e + f*x]^2]*(ArcTanh[Sqrt[Cos[e + f*x]^2]]/Sqr
t[Cos[e + f*x]^2] + (-2 + Cos[2*(e + f*x)])*Csc[e + f*x]^2))/(2*f)
Maple [A] (verified)
Time = 0.96 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.17
| | |
| method | result | size |
| | |
| default |
\(\frac {a \cos \left (f x +e \right ) \left (4 \cos \left (f x +e \right ) \left (\sin ^{2}\left (f x +e \right )\right )+2 \cos \left (f x +e \right )+\left (-3 \ln \left (1+\cos \left (f x +e \right )\right )+3 \ln \left (\cos \left (f x +e \right )-1\right )\right ) \left (\sin ^{2}\left (f x +e \right )\right )\right )}{4 \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (\cos \left (f x +e \right )-1\right ) \left (1+\cos \left (f x +e \right )\right ) f}\) |
\(102\) |
| risch |
\(-\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, \left (3 \ln \left ({\mathrm e}^{i f x}-{\mathrm e}^{-i e}\right ) {\mathrm e}^{5 i \left (f x +e \right )}-3 \ln \left ({\mathrm e}^{i f x}+{\mathrm e}^{-i e}\right ) {\mathrm e}^{5 i \left (f x +e \right )}+{\mathrm e}^{6 i \left (f x +e \right )}-6 \ln \left ({\mathrm e}^{i f x}-{\mathrm e}^{-i e}\right ) {\mathrm e}^{3 i \left (f x +e \right )}+6 \ln \left ({\mathrm e}^{i f x}+{\mathrm e}^{-i e}\right ) {\mathrm e}^{3 i \left (f x +e \right )}-3 \,{\mathrm e}^{4 i \left (f x +e \right )}+3 \ln \left ({\mathrm e}^{i f x}-{\mathrm e}^{-i e}\right ) {\mathrm e}^{i \left (f x +e \right )}-3 \ln \left ({\mathrm e}^{i f x}+{\mathrm e}^{-i e}\right ) {\mathrm e}^{i \left (f x +e \right )}-3 \,{\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}\) |
\(241\) |
| | |
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[In]
int(cot(f*x+e)^3*(a-a*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/4*a*cos(f*x+e)*(4*cos(f*x+e)*sin(f*x+e)^2+2*cos(f*x+e)+(-3*ln(1+cos(f*x+e))+3*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 = 0.31 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.01
\[
\int \cot ^3(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=-\frac {\sqrt {a \cos \left (f x + e\right )^{2}} {\left (4 \, \cos \left (f x + e\right )^{3} + 3 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} \log \left (-\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right ) - 6 \, \cos \left (f x + e\right )\right )}}{4 \, {\left (f \cos \left (f x + e\right )^{3} - f \cos \left (f x + e\right )\right )}}
\]
[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))
Sympy [F]
\[
\int \cot ^3(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\int \sqrt {- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )} \cot ^{3}{\left (e + f x \right )}\, dx
\]
[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)
Maxima [A] (verification not implemented)
none
Time = 0.31 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.14
\[
\int \cot ^3(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\frac {3 \, \sqrt {a} \log \left (\frac {2 \, \sqrt {-a \sin \left (f x + e\right )^{2} + a} \sqrt {a}}{{\left | \sin \left (f x + e\right ) \right |}} + \frac {2 \, a}{{\left | \sin \left (f x + e\right ) \right |}}\right ) - 3 \, \sqrt {-a \sin \left (f x + e\right )^{2} + a} - \frac {{\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}{a \sin \left (f x + e\right )^{2}}}{2 \, f}
\]
[In]
integrate(cot(f*x+e)^3*(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
1/2*(3*sqrt(a)*log(2*sqrt(-a*sin(f*x + e)^2 + a)*sqrt(a)/abs(sin(f*x + e)) + 2*a/abs(sin(f*x + e))) - 3*sqrt(-
a*sin(f*x + e)^2 + a) - (-a*sin(f*x + e)^2 + a)^(3/2)/(a*sin(f*x + e)^2))/f
Giac [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 1996 vs. \(2 (71) = 142\).
Time = 1.19 (sec) , antiderivative size = 1996, normalized size of antiderivative = 22.94
\[
\int \cot ^3(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\text {Too large to display}
\]
[In]
integrate(cot(f*x+e)^3*(a-a*sin(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
1/8*(12*sqrt(a)*log(abs(tan(1/2*f*x)*tan(1/2*e) - 1))*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1
/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/
2*f*x)*tan(1/2*e) + 1) - 12*sqrt(a)*log(abs(tan(1/2*f*x) + tan(1/2*e)))*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*ta
n(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1
/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1) + 16*(2*sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*ta
n(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan
(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2*f*x)*tan(1/2*e) + sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3
*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*
tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2*e)^2 - sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2
*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*
f*x)*tan(1/2*e) + 1))/((tan(1/2*f*x)^2 + 1)*(tan(1/2*e)^2 + 1)) + (2*sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 -
4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 -
tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2*f*x)^3*tan(1/2*e)^7 + sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2
*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*
e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2*f*x)^2*tan(1/2*e)^8 + 2*sqrt(a)*sgn(tan(1/2*f*x)^
4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)
*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2*f*x)^3*tan(1/2*e)^5 - 4*sqrt(a)*sgn(tan(
1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan
(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2*f*x)^2*tan(1/2*e)^6 - 2*sqrt(a)
*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e
) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2*f*x)*tan(1/2*e)^7 + 2*
sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*f*x)^3*ta
n(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2*f*x)^3*tan(1/2*
e)^3 + 6*sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2*
f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2*f*x)*t
an(1/2*e)^5 + 2*sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*t
an(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2
*e)^6 + 2*sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*tan(1/2
*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/2*f*x)^
3*tan(1/2*e) + 4*sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*x)^4 - 4*
tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) + 1)*tan(1/
2*f*x)^2*tan(1/2*e)^2 + 6*sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f*
x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) +
1)*tan(1/2*f*x)*tan(1/2*e)^3 - sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1
/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*
e) + 1)*tan(1/2*f*x)^2 - 2*sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(1/2*f
*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2*e) +
1)*tan(1/2*f*x)*tan(1/2*e) - 2*sqrt(a)*sgn(tan(1/2*f*x)^4*tan(1/2*e)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e)^3 - tan(
1/2*f*x)^4 - 4*tan(1/2*f*x)^3*tan(1/2*e) - 4*tan(1/2*f*x)*tan(1/2*e)^3 - tan(1/2*e)^4 - 4*tan(1/2*f*x)*tan(1/2
*e) + 1)*tan(1/2*e)^2)/((tan(1/2*f*x)^2*tan(1/2*e) + tan(1/2*f*x)*tan(1/2*e)^2 - tan(1/2*f*x) - tan(1/2*e))^2*
tan(1/2*e)^2))/f
Mupad [F(-1)]
Timed out. \[
\int \cot ^3(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx=\int {\mathrm {cot}\left (e+f\,x\right )}^3\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x
\]
[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)