3.21 \(\int \sqrt{a+b \cot ^2(x)} \tan (x) \, dx\)

Optimal. Leaf size=60 \[ \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a}}\right )-\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right ) \]

[Out]

Sqrt[a]*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a]] - Sqrt[a - b]*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]

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Rubi [A]  time = 0.096891, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3670, 446, 83, 63, 208} \[ \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a}}\right )-\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Cot[x]^2]*Tan[x],x]

[Out]

Sqrt[a]*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a]] - Sqrt[a - b]*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]

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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 83

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[(b*e - a*f)/(b*c
- a*d), Int[(e + f*x)^(p - 1)/(a + b*x), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[(e + f*x)^(p - 1)/(c + d*
x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[0, p, 1]

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

Mathematica [A]  time = 0.0266822, size = 60, normalized size = 1. \[ \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a}}\right )-\sqrt{a-b} \tanh ^{-1}\left (\frac{\sqrt{a+b \cot ^2(x)}}{\sqrt{a-b}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Cot[x]^2]*Tan[x],x]

[Out]

Sqrt[a]*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a]] - Sqrt[a - b]*ArcTanh[Sqrt[a + b*Cot[x]^2]/Sqrt[a - b]]

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Maple [C]  time = 0.289, size = 591, normalized size = 9.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*4^(1/2)/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)*(EllipticF((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(
1/2)/sin(x),((8*a^(3/2)*(a-b)^(1/2)-4*a^(1/2)*(a-b)^(1/2)*b+8*a^2-8*a*b+b^2)/b^2)^(1/2))*b+2*EllipticPi((-1+co
s(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),-1/(2*a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/
2)+2*a-b)/b)^(1/2)/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*a-2*EllipticPi((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)
-2*a+b)/b)^(1/2)/sin(x),-1/(2*a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/((2*a^(1/2
)*(a-b)^(1/2)-2*a+b)/b)^(1/2))*b-2*EllipticPi((-1+cos(x))*((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2)/sin(x),1/(2*
a^(1/2)*(a-b)^(1/2)-2*a+b)*b,(-(2*a^(1/2)*(a-b)^(1/2)+2*a-b)/b)^(1/2)/((2*a^(1/2)*(a-b)^(1/2)-2*a+b)/b)^(1/2))
*a)*sin(x)^3*(-2/b*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)+a*cos(x)-b*cos(x)-a)/(cos(x)+1))^(1/2)*(1/b
*(cos(x)*a^(1/2)*(a-b)^(1/2)-a^(1/2)*(a-b)^(1/2)-a*cos(x)+b*cos(x)+a)/(cos(x)+1))^(1/2)*((cos(x)^2*a-b*cos(x)^
2-a)/(cos(x)^2-1))^(1/2)*2^(1/2)/(-1+cos(x))/(cos(x)^2*a-b*cos(x)^2-a)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \cot \left (x\right )^{2} + a} \tan \left (x\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*cot(x)^2 + a)*tan(x), x)

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Fricas [A]  time = 1.74937, size = 949, normalized size = 15.82 \begin{align*} \left [\frac{1}{2} \, \sqrt{a} \log \left (2 \, a \tan \left (x\right )^{2} + 2 \, \sqrt{a} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}} \tan \left (x\right )^{2} + b\right ) + \frac{1}{2} \, \sqrt{a - b} \log \left (\frac{{\left (2 \, a - b\right )} \tan \left (x\right )^{2} - 2 \, \sqrt{a - b} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}} \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2} + 1}\right ), -\sqrt{-a + b} \arctan \left (-\frac{\sqrt{-a + b} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}}}{a - b}\right ) + \frac{1}{2} \, \sqrt{a} \log \left (2 \, a \tan \left (x\right )^{2} + 2 \, \sqrt{a} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}} \tan \left (x\right )^{2} + b\right ), -\sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}}}{a}\right ) + \frac{1}{2} \, \sqrt{a - b} \log \left (\frac{{\left (2 \, a - b\right )} \tan \left (x\right )^{2} - 2 \, \sqrt{a - b} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}} \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2} + 1}\right ), -\sqrt{-a} \arctan \left (\frac{\sqrt{-a} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}}}{a}\right ) - \sqrt{-a + b} \arctan \left (-\frac{\sqrt{-a + b} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}}}{a - b}\right )\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \cot ^{2}{\left (x \right )}} \tan{\left (x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*cot(x)**2)*tan(x), x)

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Giac [B]  time = 1.32114, size = 252, normalized size = 4.2 \begin{align*} \frac{1}{2} \,{\left (\frac{2 \, \sqrt{a - b} a \arctan \left (\frac{{\left (\sqrt{a - b} \sin \left (x\right ) - \sqrt{a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}\right )}^{2} - 2 \, a + b}{2 \, \sqrt{-a^{2} + a b}}\right )}{\sqrt{-a^{2} + a b}} + \sqrt{a - b} \log \left ({\left (\sqrt{a - b} \sin \left (x\right ) - \sqrt{a \sin \left (x\right )^{2} - b \sin \left (x\right )^{2} + b}\right )}^{2}\right )\right )} \mathrm{sgn}\left (\sin \left (x\right )\right ) - \frac{{\left (2 \, \sqrt{a - b} a \arctan \left (-\frac{a - b}{\sqrt{-a^{2} + a b}}\right ) + \sqrt{-a^{2} + a b} \sqrt{a - b} \log \left (b\right )\right )} \mathrm{sgn}\left (\sin \left (x\right )\right )}{2 \, \sqrt{-a^{2} + a b}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*sqrt(a - b)*a*arctan(1/2*((sqrt(a - b)*sin(x) - sqrt(a*sin(x)^2 - b*sin(x)^2 + b))^2 - 2*a + b)/sqrt(-a
^2 + a*b))/sqrt(-a^2 + a*b) + sqrt(a - b)*log((sqrt(a - b)*sin(x) - sqrt(a*sin(x)^2 - b*sin(x)^2 + b))^2))*sgn
(sin(x)) - 1/2*(2*sqrt(a - b)*a*arctan(-(a - b)/sqrt(-a^2 + a*b)) + sqrt(-a^2 + a*b)*sqrt(a - b)*log(b))*sgn(s
in(x))/sqrt(-a^2 + a*b)