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]

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

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Rubi [A]  time = 0.03, antiderivative size = 15, normalized size of antiderivative = 1.00, 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 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])

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 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])

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*}

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Mathematica [A]  time = 0.02, size = 15, normalized size = 1.00 \[ -\frac {2}{3} \tanh ^{-1}\left (\sqrt {\sec ^3(x)+1}\right ) \]

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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fricas [B]  time = 0.50, size = 30, normalized size = 2.00 \[ \frac {1}{3} \, \log \left (2 \, \sqrt {\frac {\cos \relax (x)^{3} + 1}{\cos \relax (x)^{3}}} \cos \relax (x)^{3} - 2 \, \cos \relax (x)^{3} - 1\right ) \]

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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giac [B]  time = 1.23, size = 28, normalized size = 1.87 \[ -\frac {1}{3} \, \log \left (\sqrt {\frac {1}{\cos \relax (x)^{3}} + 1} + 1\right ) + \frac {1}{3} \, \log \left ({\left | \sqrt {\frac {1}{\cos \relax (x)^{3}} + 1} - 1 \right |}\right ) \]

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))

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maple [A]  time = 0.08, size = 12, normalized size = 0.80 \[ -\frac {2 \arctanh \left (\sqrt {\sec ^{3}\relax (x )+1}\right )}{3} \]

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 = 0.42, size = 27, normalized size = 1.80 \[ -\frac {1}{3} \, \log \left (\sqrt {\frac {1}{\cos \relax (x)^{3}} + 1} + 1\right ) + \frac {1}{3} \, \log \left (\sqrt {\frac {1}{\cos \relax (x)^{3}} + 1} - 1\right ) \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.07 \[ \int \frac {\mathrm {tan}\relax (x)}{\sqrt {\frac {1}{{\cos \relax (x)}^3}+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(1/cos(x)^3 + 1)^(1/2),x)

[Out]

int(tan(x)/(1/cos(x)^3 + 1)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan {\relax (x )}}{\sqrt {\left (\sec {\relax (x )} + 1\right ) \left (\sec ^{2}{\relax (x )} - \sec {\relax (x )} + 1\right )}}\, dx \]

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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