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

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]

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

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Rubi [A]  time = 0.0711073, antiderivative size = 35, normalized size of antiderivative = 1., 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 = {3229, 266, 63, 208} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^4(c+d x)}}{\sqrt{a}}\right )}{2 \sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3229

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.0254163, size = 35, normalized size = 1. \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^4(c+d x)}}{\sqrt{a}}\right )}{2 \sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.717, size = 0, normalized size = 0. \begin{align*} \int{\cot \left ( dx+c \right ){\frac{1}{\sqrt{a+b \left ( \sin \left ( dx+c \right ) \right ) ^{4}}}}}\, dx \end{align*}

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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.47433, size = 369, normalized size = 10.54 \begin{align*} \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{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot{\left (c + d x \right )}}{\sqrt{a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \end{align*}

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 [A]  time = 1.14464, size = 42, normalized size = 1.2 \begin{align*} \frac{\arctan \left (\frac{\sqrt{b \sin \left (d x + c\right )^{4} + a}}{\sqrt{-a}}\right )}{2 \, \sqrt{-a} d} \end{align*}

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]

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