3.398 \(\int \frac{\cot (x)}{\sqrt{a+b \tan ^4(x)}} \, dx\)

Optimal. Leaf size=70 \[ \frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{a+b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^4(x)}}{\sqrt{a}}\right )}{2 \sqrt{a}} \]

[Out]

ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(2*Sqrt[a + b]) - ArcTanh[Sqrt[a + b*Tan[x]^4]/Sq
rt[a]]/(2*Sqrt[a])

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Rubi [A]  time = 0.158019, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.533, Rules used = {3670, 1252, 961, 725, 206, 266, 63, 208} \[ \frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{a+b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^4(x)}}{\sqrt{a}}\right )}{2 \sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[x]/Sqrt[a + b*Tan[x]^4],x]

[Out]

ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(2*Sqrt[a + b]) - ArcTanh[Sqrt[a + b*Tan[x]^4]/Sq
rt[a]]/(2*Sqrt[a])

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 961

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIn
tegrand[(d + e*x)^m*(f + g*x)^n*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] &&
NeQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0])) &&  !(IGtQ[m, 0] || IGtQ[n, 0])

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 (x)}{\sqrt{a+b \tan ^4(x)}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{x \left (1+x^2\right ) \sqrt{a+b x^4}} \, dx,x,\tan (x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x (1+x) \sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{1}{(-1-x) \sqrt{a+b x^2}}+\frac{1}{x \sqrt{a+b x^2}}\right ) \, dx,x,\tan ^2(x)\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{(-1-x) \sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )+\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x^2}} \, dx,x,\tan ^2(x)\right )\\ &=\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\tan ^4(x)\right )-\frac{1}{2} \operatorname{Subst}\left (\int \frac{1}{a+b-x^2} \, dx,x,\frac{-a+b \tan ^2(x)}{\sqrt{a+b \tan ^4(x)}}\right )\\ &=\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{a+b}}+\frac{\operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan ^4(x)}\right )}{2 b}\\ &=\frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{a+b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^4(x)}}{\sqrt{a}}\right )}{2 \sqrt{a}}\\ \end{align*}

Mathematica [A]  time = 0.0484304, size = 70, normalized size = 1. \[ \frac{\tanh ^{-1}\left (\frac{a-b \tan ^2(x)}{\sqrt{a+b} \sqrt{a+b \tan ^4(x)}}\right )}{2 \sqrt{a+b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^4(x)}}{\sqrt{a}}\right )}{2 \sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]/Sqrt[a + b*Tan[x]^4],x]

[Out]

ArcTanh[(a - b*Tan[x]^2)/(Sqrt[a + b]*Sqrt[a + b*Tan[x]^4])]/(2*Sqrt[a + b]) - ArcTanh[Sqrt[a + b*Tan[x]^4]/Sq
rt[a]]/(2*Sqrt[a])

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Maple [F]  time = 0.148, size = 0, normalized size = 0. \begin{align*} \int{\cot \left ( x \right ){\frac{1}{\sqrt{a+b \left ( \tan \left ( x \right ) \right ) ^{4}}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)/(a+b*tan(x)^4)^(1/2),x)

[Out]

int(cot(x)/(a+b*tan(x)^4)^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (x\right )}{\sqrt{b \tan \left (x\right )^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cot(x)/sqrt(b*tan(x)^4 + a), x)

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Fricas [A]  time = 3.1643, size = 1226, normalized size = 17.51 \begin{align*} \left [\frac{\sqrt{a + b} a \log \left (\frac{{\left (a b + 2 \, b^{2}\right )} \tan \left (x\right )^{4} - 2 \, a b \tan \left (x\right )^{2} - 2 \, \sqrt{b \tan \left (x\right )^{4} + a}{\left (b \tan \left (x\right )^{2} - a\right )} \sqrt{a + b} + 2 \, a^{2} + a b}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right ) +{\left (a + b\right )} \sqrt{a} \log \left (-\frac{b \tan \left (x\right )^{4} - 2 \, \sqrt{b \tan \left (x\right )^{4} + a} \sqrt{a} + 2 \, a}{\tan \left (x\right )^{4}}\right )}{4 \,{\left (a^{2} + a b\right )}}, \frac{2 \, \sqrt{-a}{\left (a + b\right )} \arctan \left (\frac{\sqrt{b \tan \left (x\right )^{4} + a} \sqrt{-a}}{a}\right ) + \sqrt{a + b} a \log \left (\frac{{\left (a b + 2 \, b^{2}\right )} \tan \left (x\right )^{4} - 2 \, a b \tan \left (x\right )^{2} - 2 \, \sqrt{b \tan \left (x\right )^{4} + a}{\left (b \tan \left (x\right )^{2} - a\right )} \sqrt{a + b} + 2 \, a^{2} + a b}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right )}{4 \,{\left (a^{2} + a b\right )}}, \frac{2 \, a \sqrt{-a - b} \arctan \left (\frac{\sqrt{b \tan \left (x\right )^{4} + a}{\left (b \tan \left (x\right )^{2} - a\right )} \sqrt{-a - b}}{{\left (a b + b^{2}\right )} \tan \left (x\right )^{4} + a^{2} + a b}\right ) +{\left (a + b\right )} \sqrt{a} \log \left (-\frac{b \tan \left (x\right )^{4} - 2 \, \sqrt{b \tan \left (x\right )^{4} + a} \sqrt{a} + 2 \, a}{\tan \left (x\right )^{4}}\right )}{4 \,{\left (a^{2} + a b\right )}}, \frac{a \sqrt{-a - b} \arctan \left (\frac{\sqrt{b \tan \left (x\right )^{4} + a}{\left (b \tan \left (x\right )^{2} - a\right )} \sqrt{-a - b}}{{\left (a b + b^{2}\right )} \tan \left (x\right )^{4} + a^{2} + a b}\right ) + \sqrt{-a}{\left (a + b\right )} \arctan \left (\frac{\sqrt{b \tan \left (x\right )^{4} + a} \sqrt{-a}}{a}\right )}{2 \,{\left (a^{2} + a b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(sqrt(a + b)*a*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqr
t(a + b) + 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)) + (a + b)*sqrt(a)*log(-(b*tan(x)^4 - 2*sqrt(b*tan(x)^4 +
a)*sqrt(a) + 2*a)/tan(x)^4))/(a^2 + a*b), 1/4*(2*sqrt(-a)*(a + b)*arctan(sqrt(b*tan(x)^4 + a)*sqrt(-a)/a) + sq
rt(a + b)*a*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 - 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(a + b)
+ 2*a^2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1)))/(a^2 + a*b), 1/4*(2*a*sqrt(-a - b)*arctan(sqrt(b*tan(x)^4 + a)*(
b*tan(x)^2 - a)*sqrt(-a - b)/((a*b + b^2)*tan(x)^4 + a^2 + a*b)) + (a + b)*sqrt(a)*log(-(b*tan(x)^4 - 2*sqrt(b
*tan(x)^4 + a)*sqrt(a) + 2*a)/tan(x)^4))/(a^2 + a*b), 1/2*(a*sqrt(-a - b)*arctan(sqrt(b*tan(x)^4 + a)*(b*tan(x
)^2 - a)*sqrt(-a - b)/((a*b + b^2)*tan(x)^4 + a^2 + a*b)) + sqrt(-a)*(a + b)*arctan(sqrt(b*tan(x)^4 + a)*sqrt(
-a)/a))/(a^2 + a*b)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot{\left (x \right )}}{\sqrt{a + b \tan ^{4}{\left (x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(x)/(a+b*tan(x)**4)**(1/2),x)

[Out]

Integral(cot(x)/sqrt(a + b*tan(x)**4), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (x\right )}{\sqrt{b \tan \left (x\right )^{4} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(cot(x)/sqrt(b*tan(x)^4 + a), x)