3.359 \(\int \sqrt {2+3 \cos (x)} \tan (x) \, dx\)

Optimal. Leaf size=37 \[ 2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {3 \cos (x)+2}}{\sqrt {2}}\right )-2 \sqrt {3 \cos (x)+2} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 37, 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 = {2721, 50, 63, 207} \[ 2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {3 \cos (x)+2}}{\sqrt {2}}\right )-2 \sqrt {3 \cos (x)+2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[2 + 3*Cos[x]]*Tan[x],x]

[Out]

2*Sqrt[2]*ArcTanh[Sqrt[2 + 3*Cos[x]]/Sqrt[2]] - 2*Sqrt[2 + 3*Cos[x]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 2721

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_.), x_Symbol] :> Dist[1/f, Subst[I
nt[(x^p*(a + x)^m)/(b^2 - x^2)^((p + 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2
- b^2, 0] && IntegerQ[(p + 1)/2]

Rubi steps

\begin {align*} \int \sqrt {2+3 \cos (x)} \tan (x) \, dx &=-\operatorname {Subst}\left (\int \frac {\sqrt {2+x}}{x} \, dx,x,3 \cos (x)\right )\\ &=-2 \sqrt {2+3 \cos (x)}-2 \operatorname {Subst}\left (\int \frac {1}{x \sqrt {2+x}} \, dx,x,3 \cos (x)\right )\\ &=-2 \sqrt {2+3 \cos (x)}-4 \operatorname {Subst}\left (\int \frac {1}{-2+x^2} \, dx,x,\sqrt {2+3 \cos (x)}\right )\\ &=2 \sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {2+3 \cos (x)}}{\sqrt {2}}\right )-2 \sqrt {2+3 \cos (x)}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 33, normalized size = 0.89 \[ 2 \sqrt {2} \tanh ^{-1}\left (\sqrt {\frac {3 \cos (x)}{2}+1}\right )-2 \sqrt {3 \cos (x)+2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[2 + 3*Cos[x]]*Tan[x],x]

[Out]

2*Sqrt[2]*ArcTanh[Sqrt[1 + (3*Cos[x])/2]] - 2*Sqrt[2 + 3*Cos[x]]

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fricas [A]  time = 0.56, size = 58, normalized size = 1.57 \[ \frac {1}{2} \, \sqrt {2} \log \left (-\frac {9 \, \cos \relax (x)^{2} + 4 \, {\left (3 \, \sqrt {2} \cos \relax (x) + 4 \, \sqrt {2}\right )} \sqrt {3 \, \cos \relax (x) + 2} + 48 \, \cos \relax (x) + 32}{\cos \relax (x)^{2}}\right ) - 2 \, \sqrt {3 \, \cos \relax (x) + 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(2)*log(-(9*cos(x)^2 + 4*(3*sqrt(2)*cos(x) + 4*sqrt(2))*sqrt(3*cos(x) + 2) + 48*cos(x) + 32)/cos(x)^2)
- 2*sqrt(3*cos(x) + 2)

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giac [A]  time = 1.05, size = 50, normalized size = 1.35 \[ -\sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 2 \, \sqrt {3 \, \cos \relax (x) + 2} \right |}}{2 \, {\left (\sqrt {2} + \sqrt {3 \, \cos \relax (x) + 2}\right )}}\right ) - 2 \, \sqrt {3 \, \cos \relax (x) + 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 31, normalized size = 0.84 \[ 2 \sqrt {2}\, \arctanh \left (\frac {\sqrt {3 \cos \relax (x )+2}\, \sqrt {2}}{2}\right )-2 \sqrt {3 \cos \relax (x )+2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.37, size = 47, normalized size = 1.27 \[ -\sqrt {2} \log \left (-\frac {\sqrt {2} - \sqrt {3 \, \cos \relax (x) + 2}}{\sqrt {2} + \sqrt {3 \, \cos \relax (x) + 2}}\right ) - 2 \, \sqrt {3 \, \cos \relax (x) + 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \mathrm {tan}\relax (x)\,\sqrt {3\,\cos \relax (x)+2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)*(3*cos(x) + 2)^(1/2),x)

[Out]

int(tan(x)*(3*cos(x) + 2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {3 \cos {\relax (x )} + 2} \tan {\relax (x )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(3*cos(x) + 2)*tan(x), x)

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