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

Optimal. Leaf size=54 \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \cot (x)}{\sqrt{a+b \cot ^2(x)}}\right )}{\sqrt{a-b}}+\frac{\tan (x) \sqrt{a+b \cot ^2(x)}}{a} \]

[Out]

ArcTan[(Sqrt[a - b]*Cot[x])/Sqrt[a + b*Cot[x]^2]]/Sqrt[a - b] + (Sqrt[a + b*Cot[x]^2]*Tan[x])/a

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Rubi [A]  time = 0.0912968, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {3670, 480, 12, 377, 203} \[ \frac{\tan ^{-1}\left (\frac{\sqrt{a-b} \cot (x)}{\sqrt{a+b \cot ^2(x)}}\right )}{\sqrt{a-b}}+\frac{\tan (x) \sqrt{a+b \cot ^2(x)}}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^2/Sqrt[a + b*Cot[x]^2],x]

[Out]

ArcTan[(Sqrt[a - b]*Cot[x])/Sqrt[a + b*Cot[x]^2]]/Sqrt[a - b] + (Sqrt[a + b*Cot[x]^2]*Tan[x])/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 480

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

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

Rule 203

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

Rubi steps

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

Mathematica [C]  time = 1.55225, size = 134, normalized size = 2.48 \[ \frac{\sin ^2(x) \tan (x) \left (\frac{b \cot ^2(x)}{a}+1\right ) \left (4 (a-b) \cos ^2(x) \left (a+b \cot ^2(x)\right ) \text{Hypergeometric2F1}\left (2,2,\frac{5}{2},\frac{(a-b) \cos ^2(x)}{a}\right )+\frac{3 a \left (a+2 b \cot ^2(x)\right ) \sin ^{-1}\left (\sqrt{\frac{(a-b) \cos ^2(x)}{a}}\right )}{\sqrt{\frac{(a-b) \sin ^2(x) \cos ^2(x) \left (a+b \cot ^2(x)\right )}{a^2}}}\right )}{3 a^2 \sqrt{a+b \cot ^2(x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 + (b*Cot[x]^2)/a)*Sin[x]^2*(4*(a - b)*Cos[x]^2*(a + b*Cot[x]^2)*Hypergeometric2F1[2, 2, 5/2, ((a - b)*Cos[
x]^2)/a] + (3*a*ArcSin[Sqrt[((a - b)*Cos[x]^2)/a]]*(a + 2*b*Cot[x]^2))/Sqrt[((a - b)*Cos[x]^2*(a + b*Cot[x]^2)
*Sin[x]^2)/a^2])*Tan[x])/(3*a^2*Sqrt[a + b*Cot[x]^2])

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Maple [B]  time = 0.147, size = 330, normalized size = 6.1 \begin{align*}{\frac{\sin \left ( x \right ) }{a\cos \left ( x \right ) \left ( \left ( \cos \left ( x \right ) \right ) ^{2}-1 \right ) } \left ( - \left ( \cos \left ( x \right ) \right ) ^{2}\sqrt{-{\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}a-b \left ( \cos \left ( x \right ) \right ) ^{2}-a}{ \left ( \cos \left ( x \right ) +1 \right ) ^{2}}}}\ln \left ( 4\,\cos \left ( x \right ) \sqrt{-a+b}\sqrt{-{\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}a-b \left ( \cos \left ( x \right ) \right ) ^{2}-a}{ \left ( \cos \left ( x \right ) +1 \right ) ^{2}}}}-4\,a\cos \left ( x \right ) +4\,b\cos \left ( x \right ) +4\,\sqrt{-a+b}\sqrt{-{\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}a-b \left ( \cos \left ( x \right ) \right ) ^{2}-a}{ \left ( \cos \left ( x \right ) +1 \right ) ^{2}}}} \right ) a+ \left ( \cos \left ( x \right ) \right ) ^{2}\sqrt{-a+b}a- \left ( \cos \left ( x \right ) \right ) ^{2}\sqrt{-a+b}b-\cos \left ( x \right ) \sqrt{-{\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}a-b \left ( \cos \left ( x \right ) \right ) ^{2}-a}{ \left ( \cos \left ( x \right ) +1 \right ) ^{2}}}}\ln \left ( 4\,\cos \left ( x \right ) \sqrt{-a+b}\sqrt{-{\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}a-b \left ( \cos \left ( x \right ) \right ) ^{2}-a}{ \left ( \cos \left ( x \right ) +1 \right ) ^{2}}}}-4\,a\cos \left ( x \right ) +4\,b\cos \left ( x \right ) +4\,\sqrt{-a+b}\sqrt{-{\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}a-b \left ( \cos \left ( x \right ) \right ) ^{2}-a}{ \left ( \cos \left ( x \right ) +1 \right ) ^{2}}}} \right ) a-\sqrt{-a+b}a \right ){\frac{1}{\sqrt{-a+b}}}{\frac{1}{\sqrt{{\frac{ \left ( \cos \left ( x \right ) \right ) ^{2}a-b \left ( \cos \left ( x \right ) \right ) ^{2}-a}{ \left ( \cos \left ( x \right ) \right ) ^{2}-1}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/(-a+b)^(1/2)/a*sin(x)*(-cos(x)^2*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)*ln(4*cos(x)*(-a+b)^(1/2)*(-
(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x)+4*b*cos(x)+4*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)
/(cos(x)+1)^2)^(1/2))*a+cos(x)^2*(-a+b)^(1/2)*a-cos(x)^2*(-a+b)^(1/2)*b-cos(x)*(-(cos(x)^2*a-b*cos(x)^2-a)/(co
s(x)+1)^2)^(1/2)*ln(4*cos(x)*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2)-4*a*cos(x)+4*b*cos(x
)+4*(-a+b)^(1/2)*(-(cos(x)^2*a-b*cos(x)^2-a)/(cos(x)+1)^2)^(1/2))*a-(-a+b)^(1/2)*a)/cos(x)/((cos(x)^2*a-b*cos(
x)^2-a)/(cos(x)^2-1))^(1/2)/(cos(x)^2-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.40979, size = 593, normalized size = 10.98 \begin{align*} \left [-\frac{a \sqrt{-a + b} \log \left (-\frac{a^{2} \tan \left (x\right )^{4} - 2 \,{\left (3 \, a^{2} - 4 \, a b\right )} \tan \left (x\right )^{2} + a^{2} - 8 \, a b + 8 \, b^{2} + 4 \,{\left (a \tan \left (x\right )^{3} -{\left (a - 2 \, b\right )} \tan \left (x\right )\right )} \sqrt{-a + b} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}}}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right ) - 4 \,{\left (a - b\right )} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}} \tan \left (x\right )}{4 \,{\left (a^{2} - a b\right )}}, \frac{\sqrt{a - b} a \arctan \left (\frac{2 \, \sqrt{a - b} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}} \tan \left (x\right )}{a \tan \left (x\right )^{2} - a + 2 \, b}\right ) + 2 \,{\left (a - b\right )} \sqrt{\frac{a \tan \left (x\right )^{2} + b}{\tan \left (x\right )^{2}}} \tan \left (x\right )}{2 \,{\left (a^{2} - a b\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*(a*sqrt(-a + b)*log(-(a^2*tan(x)^4 - 2*(3*a^2 - 4*a*b)*tan(x)^2 + a^2 - 8*a*b + 8*b^2 + 4*(a*tan(x)^3 -
(a - 2*b)*tan(x))*sqrt(-a + b)*sqrt((a*tan(x)^2 + b)/tan(x)^2))/(tan(x)^4 + 2*tan(x)^2 + 1)) - 4*(a - b)*sqrt(
(a*tan(x)^2 + b)/tan(x)^2)*tan(x))/(a^2 - a*b), 1/2*(sqrt(a - b)*a*arctan(2*sqrt(a - b)*sqrt((a*tan(x)^2 + b)/
tan(x)^2)*tan(x)/(a*tan(x)^2 - a + 2*b)) + 2*(a - b)*sqrt((a*tan(x)^2 + b)/tan(x)^2)*tan(x))/(a^2 - a*b)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.51555, size = 132, normalized size = 2.44 \begin{align*} \frac{{\left (a \arctan \left (\frac{\sqrt{b}}{\sqrt{a - b}}\right ) - \sqrt{a - b} \sqrt{b}\right )} \mathrm{sgn}\left (\tan \left (x\right )\right )}{\sqrt{a - b} a} - \frac{\frac{a \arctan \left (\frac{\sqrt{a \tan \left (x\right )^{2} + b}}{\sqrt{a - b}}\right )}{\sqrt{a - b} \mathrm{sgn}\left (\tan \left (x\right )\right )} - \frac{\sqrt{a \tan \left (x\right )^{2} + b}}{\mathrm{sgn}\left (\tan \left (x\right )\right )}}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*arctan(sqrt(b)/sqrt(a - b)) - sqrt(a - b)*sqrt(b))*sgn(tan(x))/(sqrt(a - b)*a) - (a*arctan(sqrt(a*tan(x)^2
+ b)/sqrt(a - b))/(sqrt(a - b)*sgn(tan(x))) - sqrt(a*tan(x)^2 + b)/sgn(tan(x)))/a