3.93 \(\int \sqrt{a+b \cos ^3(x)} \tan (x) \, dx\)

Optimal. Leaf size=45 \[ \frac{2}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \cos ^3(x)}}{\sqrt{a}}\right )-\frac{2}{3} \sqrt{a+b \cos ^3(x)} \]

[Out]

(2*Sqrt[a]*ArcTanh[Sqrt[a + b*Cos[x]^3]/Sqrt[a]])/3 - (2*Sqrt[a + b*Cos[x]^3])/3

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Rubi [A]  time = 0.0729281, antiderivative size = 45, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3230, 266, 50, 63, 208} \[ \frac{2}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \cos ^3(x)}}{\sqrt{a}}\right )-\frac{2}{3} \sqrt{a+b \cos ^3(x)} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Cos[x]^3]*Tan[x],x]

[Out]

(2*Sqrt[a]*ArcTanh[Sqrt[a + b*Cos[x]^3]/Sqrt[a]])/3 - (2*Sqrt[a + b*Cos[x]^3])/3

Rule 3230

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + b*(c*ff*x)^n)^p)/(1 - ff^2*x^2)^(
(m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 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 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 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 \sqrt{a+b \cos ^3(x)} \tan (x) \, dx &=-\operatorname{Subst}\left (\int \frac{\sqrt{a+b x^3}}{x} \, dx,x,\cos (x)\right )\\ &=-\left (\frac{1}{3} \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{x} \, dx,x,\cos ^3(x)\right )\right )\\ &=-\frac{2}{3} \sqrt{a+b \cos ^3(x)}-\frac{1}{3} a \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,\cos ^3(x)\right )\\ &=-\frac{2}{3} \sqrt{a+b \cos ^3(x)}-\frac{(2 a) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \cos ^3(x)}\right )}{3 b}\\ &=\frac{2}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \cos ^3(x)}}{\sqrt{a}}\right )-\frac{2}{3} \sqrt{a+b \cos ^3(x)}\\ \end{align*}

Mathematica [A]  time = 0.0249099, size = 45, normalized size = 1. \[ \frac{2}{3} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+b \cos ^3(x)}}{\sqrt{a}}\right )-\frac{2}{3} \sqrt{a+b \cos ^3(x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Cos[x]^3]*Tan[x],x]

[Out]

(2*Sqrt[a]*ArcTanh[Sqrt[a + b*Cos[x]^3]/Sqrt[a]])/3 - (2*Sqrt[a + b*Cos[x]^3])/3

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Maple [A]  time = 0.179, size = 34, normalized size = 0.8 \begin{align*}{\frac{2}{3}{\it Artanh} \left ({\sqrt{a+b \left ( \cos \left ( x \right ) \right ) ^{3}}{\frac{1}{\sqrt{a}}}} \right ) \sqrt{a}}-{\frac{2}{3}\sqrt{a+b \left ( \cos \left ( x \right ) \right ) ^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 26.9559, size = 340, normalized size = 7.56 \begin{align*} \left [\frac{1}{6} \, \sqrt{a} \log \left (-\frac{b^{2} \cos \left (x\right )^{6} + 8 \, a b \cos \left (x\right )^{3} + 4 \,{\left (b \cos \left (x\right )^{3} + 2 \, a\right )} \sqrt{b \cos \left (x\right )^{3} + a} \sqrt{a} + 8 \, a^{2}}{\cos \left (x\right )^{6}}\right ) - \frac{2}{3} \, \sqrt{b \cos \left (x\right )^{3} + a}, -\frac{1}{3} \, \sqrt{-a} \arctan \left (\frac{2 \, \sqrt{b \cos \left (x\right )^{3} + a} \sqrt{-a}}{b \cos \left (x\right )^{3} + 2 \, a}\right ) - \frac{2}{3} \, \sqrt{b \cos \left (x\right )^{3} + a}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*sqrt(a)*log(-(b^2*cos(x)^6 + 8*a*b*cos(x)^3 + 4*(b*cos(x)^3 + 2*a)*sqrt(b*cos(x)^3 + a)*sqrt(a) + 8*a^2)/
cos(x)^6) - 2/3*sqrt(b*cos(x)^3 + a), -1/3*sqrt(-a)*arctan(2*sqrt(b*cos(x)^3 + a)*sqrt(-a)/(b*cos(x)^3 + 2*a))
- 2/3*sqrt(b*cos(x)^3 + a)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \cos ^{3}{\left (x \right )}} \tan{\left (x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a + b*cos(x)**3)*tan(x), x)

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Giac [A]  time = 1.17278, size = 51, normalized size = 1.13 \begin{align*} -\frac{2 \, a \arctan \left (\frac{\sqrt{b \cos \left (x\right )^{3} + a}}{\sqrt{-a}}\right )}{3 \, \sqrt{-a}} - \frac{2}{3} \, \sqrt{b \cos \left (x\right )^{3} + a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*a*arctan(sqrt(b*cos(x)^3 + a)/sqrt(-a))/sqrt(-a) - 2/3*sqrt(b*cos(x)^3 + a)