3.5.61 \(\int \cot ^3(e+f x) \sqrt {a-a \sin ^2(e+f x)} \, dx\) [461]

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

[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

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Rubi [A]
time = 0.08, antiderivative size = 87, normalized size of antiderivative = 1.00, 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 = {3255, 3284, 16, 43, 52, 65, 212} \begin {gather*} -\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} \end {gather*}

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 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*} \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 {\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} \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*}

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Mathematica [A]
time = 0.30, size = 88, normalized size = 1.01 \begin {gather*} -\frac {\sqrt {a \cos ^2(e+f x)} \left (8 \cos (e+f x)+\csc ^2\left (\frac {1}{2} (e+f x)\right )-12 \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )+12 \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )-\sec ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sec (e+f x)}{8 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(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])/f

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Maple [A]
time = 11.07, size = 78, normalized size = 0.90

method	result	size
default	\(\frac {-\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}-\frac {\sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}}{2 \sin \left (f x +e \right )^{2}}+\frac {3 \sqrt {a}\, \ln \left (\frac {2 \sqrt {a}\, \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}+2 a}{\sin \left (f x +e \right )}\right )}{2}}{f}\)	\(78\)
risch	\(-\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {\sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{\left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2} f}-\frac {3 \ln \left ({\mathrm e}^{i f x}-{\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}+\frac {3 \ln \left ({\mathrm e}^{i f x}+{\mathrm e}^{-i e}\right ) \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, {\mathrm e}^{i \left (f x +e \right )}}{2 f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}\)	\(285\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.55, size = 105, normalized size = 1.21 \begin {gather*} \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} \end {gather*}

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*(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

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Fricas [A]
time = 0.39, size = 88, normalized size = 1.01 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 {gather*}

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] Leaf count of result is larger than twice the leaf count of optimal. 171 vs. \(2 (75) = 150\).
time = 0.57, size = 171, normalized size = 1.97 \begin {gather*} -\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 {gather*}

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {cot}\left (e+f\,x\right )}^3\,\sqrt {a-a\,{\sin \left (e+f\,x\right )}^2} \,d x \end {gather*}

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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