3.324 \(\int \frac{\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}} \, dx\)

Optimal. Leaf size=128 \[ -\frac{\log \left (\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}+1\right )}{2 b}+\frac{\log \left (\frac{\sin ^{\frac{4}{3}}(a+b x)}{\cos ^{\frac{4}{3}}(a+b x)}-\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}+1\right )}{4 b}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}}{\sqrt{3}}\right )}{2 b} \]

[Out]

-(Sqrt[3]*ArcTan[(1 - (2*Sin[a + b*x]^(2/3))/Cos[a + b*x]^(2/3))/Sqrt[3]])/(2*b) - Log[1 + Sin[a + b*x]^(2/3)/
Cos[a + b*x]^(2/3)]/(2*b) + Log[1 - Sin[a + b*x]^(2/3)/Cos[a + b*x]^(2/3) + Sin[a + b*x]^(4/3)/Cos[a + b*x]^(4
/3)]/(4*b)

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Rubi [A]  time = 0.15127, antiderivative size = 128, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.381, Rules used = {2574, 275, 292, 31, 634, 618, 204, 628} \[ -\frac{\log \left (\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}+1\right )}{2 b}+\frac{\log \left (\frac{\sin ^{\frac{4}{3}}(a+b x)}{\cos ^{\frac{4}{3}}(a+b x)}-\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}+1\right )}{4 b}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}}{\sqrt{3}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^(1/3)/Cos[a + b*x]^(1/3),x]

[Out]

-(Sqrt[3]*ArcTan[(1 - (2*Sin[a + b*x]^(2/3))/Cos[a + b*x]^(2/3))/Sqrt[3]])/(2*b) - Log[1 + Sin[a + b*x]^(2/3)/
Cos[a + b*x]^(2/3)]/(2*b) + Log[1 - Sin[a + b*x]^(2/3)/Cos[a + b*x]^(2/3) + Sin[a + b*x]^(4/3)/Cos[a + b*x]^(4
/3)]/(4*b)

Rule 2574

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[{k = Denomina
tor[m]}, Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Sin[e + f*x])^(1/k)/(b*Cos
[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}} \, dx &=\frac{3 \operatorname{Subst}\left (\int \frac{x^3}{1+x^6} \, dx,x,\frac{\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{b}\\ &=\frac{3 \operatorname{Subst}\left (\int \frac{x}{1+x^3} \, dx,x,\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}\right )}{2 b}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{1+x}{1-x+x^2} \, dx,x,\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}\right )}{2 b}\\ &=-\frac{\log \left (1+\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}\right )}{4 b}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}\right )}{4 b}\\ &=-\frac{\log \left (1+\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}\right )}{2 b}+\frac{\log \left (1-\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}+\frac{\sin ^{\frac{4}{3}}(a+b x)}{\cos ^{\frac{4}{3}}(a+b x)}\right )}{4 b}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+\frac{2 \sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}\right )}{2 b}\\ &=-\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}}{\sqrt{3}}\right )}{2 b}-\frac{\log \left (1+\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}\right )}{2 b}+\frac{\log \left (1-\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}+\frac{\sin ^{\frac{4}{3}}(a+b x)}{\cos ^{\frac{4}{3}}(a+b x)}\right )}{4 b}\\ \end{align*}

Mathematica [C]  time = 0.040584, size = 57, normalized size = 0.45 \[ \frac{3 \sin ^{\frac{4}{3}}(a+b x) \cos ^2(a+b x)^{2/3} \, _2F_1\left (\frac{2}{3},\frac{2}{3};\frac{5}{3};\sin ^2(a+b x)\right )}{4 b \cos ^{\frac{4}{3}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^(1/3)/Cos[a + b*x]^(1/3),x]

[Out]

(3*(Cos[a + b*x]^2)^(2/3)*Hypergeometric2F1[2/3, 2/3, 5/3, Sin[a + b*x]^2]*Sin[a + b*x]^(4/3))/(4*b*Cos[a + b*
x]^(4/3))

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Maple [F]  time = 0.096, size = 0, normalized size = 0. \begin{align*} \int{\sqrt [3]{\sin \left ( bx+a \right ) }{\frac{1}{\sqrt [3]{\cos \left ( bx+a \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{\frac{1}{3}}}{\cos \left (b x + a\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(1/3)/cos(b*x + a)^(1/3), x)

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Fricas [A]  time = 2.24578, size = 424, normalized size = 3.31 \begin{align*} \frac{2 \, \sqrt{3} \arctan \left (-\frac{\sqrt{3} \cos \left (b x + a\right ) - 2 \, \sqrt{3} \cos \left (b x + a\right )^{\frac{1}{3}} \sin \left (b x + a\right )^{\frac{2}{3}}}{3 \, \cos \left (b x + a\right )}\right ) - 2 \, \log \left (\frac{\cos \left (b x + a\right )^{\frac{1}{3}} \sin \left (b x + a\right )^{\frac{2}{3}} + \cos \left (b x + a\right )}{\cos \left (b x + a\right )}\right ) + \log \left (\frac{\cos \left (b x + a\right )^{2} - \cos \left (b x + a\right )^{\frac{4}{3}} \sin \left (b x + a\right )^{\frac{2}{3}} + \cos \left (b x + a\right )^{\frac{2}{3}} \sin \left (b x + a\right )^{\frac{4}{3}}}{\cos \left (b x + a\right )^{2}}\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*sqrt(3)*arctan(-1/3*(sqrt(3)*cos(b*x + a) - 2*sqrt(3)*cos(b*x + a)^(1/3)*sin(b*x + a)^(2/3))/cos(b*x +
a)) - 2*log((cos(b*x + a)^(1/3)*sin(b*x + a)^(2/3) + cos(b*x + a))/cos(b*x + a)) + log((cos(b*x + a)^2 - cos(b
*x + a)^(4/3)*sin(b*x + a)^(2/3) + cos(b*x + a)^(2/3)*sin(b*x + a)^(4/3))/cos(b*x + a)^2))/b

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x)**(1/3)/cos(a + b*x)**(1/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{\frac{1}{3}}}{\cos \left (b x + a\right )^{\frac{1}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(1/3)/cos(b*x + a)^(1/3), x)