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

Optimal. Leaf size=25 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cos ^2(x)}}{\sqrt{a}}\right )}{\sqrt{a}} \]

[Out]

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

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Rubi [A]  time = 0.0599249, antiderivative size = 25, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3194, 63, 208} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cos ^2(x)}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/Sqrt[a + b*Cos[x]^2],x]

[Out]

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

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^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]

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

Mathematica [A]  time = 0.0123528, size = 25, normalized size = 1. \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{a+b \cos ^2(x)}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.02, size = 30, normalized size = 1.2 \begin{align*}{\ln \left ({\frac{1}{\cos \left ( x \right ) } \left ( 2\,a+2\,\sqrt{a}\sqrt{a+b \left ( \cos \left ( x \right ) \right ) ^{2}} \right ) } \right ){\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.12165, size = 184, normalized size = 7.36 \begin{align*} \left [\frac{\log \left (\frac{b \cos \left (x\right )^{2} + 2 \, \sqrt{b \cos \left (x\right )^{2} + a} \sqrt{a} + 2 \, a}{\cos \left (x\right )^{2}}\right )}{2 \, \sqrt{a}}, -\frac{\sqrt{-a} \arctan \left (\frac{\sqrt{b \cos \left (x\right )^{2} + a} \sqrt{-a}}{a}\right )}{a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.15484, size = 32, normalized size = 1.28 \begin{align*} -\frac{\arctan \left (\frac{\sqrt{b \cos \left (x\right )^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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