3.7 \(\int \frac{\tan (x)}{\sqrt{1+\sec ^3(x)}} \, dx\)

Optimal. Leaf size=15 \[ -\frac{2}{3} \tanh ^{-1}\left (\sqrt{\sec ^3(x)+1}\right ) \]

[Out]

(-2*ArcTanh[Sqrt[1 + Sec[x]^3]])/3

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Rubi [A]  time = 0.0309224, antiderivative size = 15, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {4139, 266, 63, 207} \[ -\frac{2}{3} \tanh ^{-1}\left (\sqrt{\sec ^3(x)+1}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/Sqrt[1 + Sec[x]^3],x]

[Out]

(-2*ArcTanh[Sqrt[1 + Sec[x]^3]])/3

Rule 4139

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sec[e + f*x], x]}, Dist[1/f, Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^n)^p)/x
, x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[
n, 2] || EqQ[n, 4] || IGtQ[p, 0] || IntegersQ[2*n, p])

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 207

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

Rubi steps

\begin{align*} \int \frac{\tan (x)}{\sqrt{1+\sec ^3(x)}} \, dx &=\operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+x^3}} \, dx,x,\sec (x)\right )\\ &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+x}} \, dx,x,\sec ^3(x)\right )\\ &=\frac{2}{3} \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\sqrt{1+\sec ^3(x)}\right )\\ &=-\frac{2}{3} \tanh ^{-1}\left (\sqrt{1+\sec ^3(x)}\right )\\ \end{align*}

Mathematica [A]  time = 0.013665, size = 15, normalized size = 1. \[ -\frac{2}{3} \tanh ^{-1}\left (\sqrt{\sec ^3(x)+1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/Sqrt[1 + Sec[x]^3],x]

[Out]

(-2*ArcTanh[Sqrt[1 + Sec[x]^3]])/3

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Maple [A]  time = 0.043, size = 12, normalized size = 0.8 \begin{align*} -{\frac{2}{3}{\it Artanh} \left ( \sqrt{1+ \left ( \sec \left ( x \right ) \right ) ^{3}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(1+sec(x)^3)^(1/2),x)

[Out]

-2/3*arctanh((1+sec(x)^3)^(1/2))

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Maxima [B]  time = 0.961202, size = 36, normalized size = 2.4 \begin{align*} -\frac{1}{3} \, \log \left (\sqrt{\frac{1}{\cos \left (x\right )^{3}} + 1} + 1\right ) + \frac{1}{3} \, \log \left (\sqrt{\frac{1}{\cos \left (x\right )^{3}} + 1} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+sec(x)^3)^(1/2),x, algorithm="maxima")

[Out]

-1/3*log(sqrt(1/cos(x)^3 + 1) + 1) + 1/3*log(sqrt(1/cos(x)^3 + 1) - 1)

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Fricas [B]  time = 3.13441, size = 92, normalized size = 6.13 \begin{align*} \frac{1}{3} \, \log \left (2 \, \sqrt{\frac{\cos \left (x\right )^{3} + 1}{\cos \left (x\right )^{3}}} \cos \left (x\right )^{3} - 2 \, \cos \left (x\right )^{3} - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+sec(x)^3)^(1/2),x, algorithm="fricas")

[Out]

1/3*log(2*sqrt((cos(x)^3 + 1)/cos(x)^3)*cos(x)^3 - 2*cos(x)^3 - 1)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan{\left (x \right )}}{\sqrt{\left (\sec{\left (x \right )} + 1\right ) \left (\sec ^{2}{\left (x \right )} - \sec{\left (x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+sec(x)**3)**(1/2),x)

[Out]

Integral(tan(x)/sqrt((sec(x) + 1)*(sec(x)**2 - sec(x) + 1)), x)

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Giac [B]  time = 1.23377, size = 38, normalized size = 2.53 \begin{align*} -\frac{1}{3} \, \log \left (\sqrt{\frac{1}{\cos \left (x\right )^{3}} + 1} + 1\right ) + \frac{1}{3} \, \log \left ({\left | \sqrt{\frac{1}{\cos \left (x\right )^{3}} + 1} - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+sec(x)^3)^(1/2),x, algorithm="giac")

[Out]

-1/3*log(sqrt(1/cos(x)^3 + 1) + 1) + 1/3*log(abs(sqrt(1/cos(x)^3 + 1) - 1))