3.579 \(\int \cot (x) \sqrt {a+b \sin ^n(x)} \, dx\)

Optimal. Leaf size=47 \[ \frac {2 \sqrt {a+b \sin ^n(x)}}{n}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^n(x)}}{\sqrt {a}}\right )}{n} \]

[Out]

-2*arctanh((a+b*sin(x)^n)^(1/2)/a^(1/2))*a^(1/2)/n+2*(a+b*sin(x)^n)^(1/2)/n

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Rubi [A]  time = 0.08, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3230, 266, 50, 63, 208} \[ \frac {2 \sqrt {a+b \sin ^n(x)}}{n}-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^n(x)}}{\sqrt {a}}\right )}{n} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]*Sqrt[a + b*Sin[x]^n],x]

[Out]

(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sin[x]^n]/Sqrt[a]])/n + (2*Sqrt[a + b*Sin[x]^n])/n

Rule 50

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 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 208

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3230

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + b*(c*ff*x)^n)^p)/(1 - ff^2*x^2)^(
(m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 0]

Rubi steps

\begin {align*} \int \cot (x) \sqrt {a+b \sin ^n(x)} \, dx &=\operatorname {Subst}\left (\int \frac {\sqrt {a+b x^n}}{x} \, dx,x,\sin (x)\right )\\ &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x}}{x} \, dx,x,\sin ^n(x)\right )}{n}\\ &=\frac {2 \sqrt {a+b \sin ^n(x)}}{n}+\frac {a \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^n(x)\right )}{n}\\ &=\frac {2 \sqrt {a+b \sin ^n(x)}}{n}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^n(x)}\right )}{b n}\\ &=-\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^n(x)}}{\sqrt {a}}\right )}{n}+\frac {2 \sqrt {a+b \sin ^n(x)}}{n}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 45, normalized size = 0.96 \[ \frac {2 \sqrt {a+b \sin ^n(x)}-2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^n(x)}}{\sqrt {a}}\right )}{n} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]*Sqrt[a + b*Sin[x]^n],x]

[Out]

(-2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sin[x]^n]/Sqrt[a]] + 2*Sqrt[a + b*Sin[x]^n])/n

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fricas [A]  time = 0.46, size = 97, normalized size = 2.06 \[ \left [\frac {\sqrt {a} \log \left (\frac {b \sin \relax (x)^{n} - 2 \, \sqrt {b \sin \relax (x)^{n} + a} \sqrt {a} + 2 \, a}{\sin \relax (x)^{n}}\right ) + 2 \, \sqrt {b \sin \relax (x)^{n} + a}}{n}, \frac {2 \, {\left (\sqrt {-a} \arctan \left (\frac {\sqrt {b \sin \relax (x)^{n} + a} \sqrt {-a}}{a}\right ) + \sqrt {b \sin \relax (x)^{n} + a}\right )}}{n}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(a+b*sin(x)^n)^(1/2),x, algorithm="fricas")

[Out]

[(sqrt(a)*log((b*sin(x)^n - 2*sqrt(b*sin(x)^n + a)*sqrt(a) + 2*a)/sin(x)^n) + 2*sqrt(b*sin(x)^n + a))/n, 2*(sq
rt(-a)*arctan(sqrt(b*sin(x)^n + a)*sqrt(-a)/a) + sqrt(b*sin(x)^n + a))/n]

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giac [A]  time = 0.15, size = 46, normalized size = 0.98 \[ \frac {2 \, {\left (\frac {a b \arctan \left (\frac {\sqrt {b \sin \relax (x)^{n} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \sqrt {b \sin \relax (x)^{n} + a} b\right )}}{b n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(a+b*sin(x)^n)^(1/2),x, algorithm="giac")

[Out]

2*(a*b*arctan(sqrt(b*sin(x)^n + a)/sqrt(-a))/sqrt(-a) + sqrt(b*sin(x)^n + a)*b)/(b*n)

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maple [A]  time = 0.11, size = 38, normalized size = 0.81 \[ \frac {2 \sqrt {a +b \left (\sin ^{n}\relax (x )\right )}-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {a +b \left (\sin ^{n}\relax (x )\right )}}{\sqrt {a}}\right )}{n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)*(a+b*sin(x)^n)^(1/2),x)

[Out]

1/n*(2*(a+b*sin(x)^n)^(1/2)-2*a^(1/2)*arctanh((a+b*sin(x)^n)^(1/2)/a^(1/2)))

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maxima [A]  time = 0.89, size = 57, normalized size = 1.21 \[ \frac {\sqrt {a} \log \left (\frac {\sqrt {b \sin \relax (x)^{n} + a} - \sqrt {a}}{\sqrt {b \sin \relax (x)^{n} + a} + \sqrt {a}}\right )}{n} + \frac {2 \, \sqrt {b \sin \relax (x)^{n} + a}}{n} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(a+b*sin(x)^n)^(1/2),x, algorithm="maxima")

[Out]

sqrt(a)*log((sqrt(b*sin(x)^n + a) - sqrt(a))/(sqrt(b*sin(x)^n + a) + sqrt(a)))/n + 2*sqrt(b*sin(x)^n + a)/n

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \mathrm {cot}\relax (x)\,\sqrt {a+b\,{\sin \relax (x)}^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)*(a + b*sin(x)^n)^(1/2),x)

[Out]

int(cot(x)*(a + b*sin(x)^n)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \sin ^{n}{\relax (x )}} \cot {\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)*(a+b*sin(x)**n)**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(x)**n)*cot(x), x)

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