3.6.58 \(\int \frac {\cot (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [558]

Optimal. Leaf size=35 \[ -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{2 \sqrt {a} d} \]

[Out]

-1/2*arctanh((a+b*sin(d*x+c)^4)^(1/2)/a^(1/2))/d/a^(1/2)

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Rubi [A]
time = 0.05, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3308, 272, 65, 214} \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{2 \sqrt {a} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-1/2*ArcTanh[Sqrt[a + b*Sin[c + d*x]^4]/Sqrt[a]]/(Sqrt[a]*d)

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 214

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

Rule 272

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 3308

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

Rubi steps

\begin {align*} \int \frac {\cot (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx &=\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b x^2}} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac {\text {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,\sin ^4(c+d x)\right )}{4 d}\\ &=\frac {\text {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b \sin ^4(c+d x)}\right )}{2 b d}\\ &=-\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{2 \sqrt {a} d}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 35, normalized size = 1.00 \begin {gather*} -\frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \sin ^4(c+d x)}}{\sqrt {a}}\right )}{2 \sqrt {a} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-1/2*ArcTanh[Sqrt[a + b*Sin[c + d*x]^4]/Sqrt[a]]/(Sqrt[a]*d)

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Maple [F]
time = 1.81, size = 0, normalized size = 0.00 \[\int \frac {\cot \left (d x +c \right )}{\sqrt {a +b \left (\sin ^{4}\left (d x +c \right )\right )}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x)

[Out]

int(cot(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x)

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Maxima [A]
time = 0.52, size = 50, normalized size = 1.43 \begin {gather*} \frac {\log \left (\frac {\sqrt {b \sin \left (d x + c\right )^{4} + a} - \sqrt {a}}{\sqrt {b \sin \left (d x + c\right )^{4} + a} + \sqrt {a}}\right )}{4 \, \sqrt {a} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

1/4*log((sqrt(b*sin(d*x + c)^4 + a) - sqrt(a))/(sqrt(b*sin(d*x + c)^4 + a) + sqrt(a)))/(sqrt(a)*d)

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Fricas [A]
time = 0.52, size = 140, normalized size = 4.00 \begin {gather*} \left [\frac {\log \left (\frac {8 \, {\left (b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} \sqrt {a} + 2 \, a + b\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}\right )}{4 \, \sqrt {a} d}, \frac {\sqrt {-a} \arctan \left (\frac {\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b} \sqrt {-a}}{a}\right )}{2 \, a d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

[1/4*log(8*(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 - 2*sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)*sqrt
(a) + 2*a + b)/(cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1))/(sqrt(a)*d), 1/2*sqrt(-a)*arctan(sqrt(b*cos(d*x + c)^4
- 2*b*cos(d*x + c)^2 + a + b)*sqrt(-a)/a)/(a*d)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot {\left (c + d x \right )}}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sin(d*x+c)**4)**(1/2),x)

[Out]

Integral(cot(c + d*x)/sqrt(a + b*sin(c + d*x)**4), x)

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)/(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\mathrm {cot}\left (c+d\,x\right )}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)/(a + b*sin(c + d*x)^4)^(1/2),x)

[Out]

int(cot(c + d*x)/(a + b*sin(c + d*x)^4)^(1/2), x)

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