3.325 \(\int \frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)} \, dx\)

Optimal. Leaf size=224 \[ \frac{\sqrt{3} \log \left (\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}-\frac{\sqrt{3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1\right )}{4 b}-\frac{\sqrt{3} \log \left (\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}+\frac{\sqrt{3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1\right )}{4 b}-\frac{\tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{2 b}+\frac{\tan ^{-1}\left (\frac{2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\sqrt{3}\right )}{2 b}+\frac{\tan ^{-1}\left (\frac{\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.329462, antiderivative size = 224, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2574, 295, 634, 618, 204, 628, 203} \[ \frac{\sqrt{3} \log \left (\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}-\frac{\sqrt{3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1\right )}{4 b}-\frac{\sqrt{3} \log \left (\frac{\sin ^{\frac{2}{3}}(a+b x)}{\cos ^{\frac{2}{3}}(a+b x)}+\frac{\sqrt{3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1\right )}{4 b}-\frac{\tan ^{-1}\left (\sqrt{3}-\frac{2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{2 b}+\frac{\tan ^{-1}\left (\frac{2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\sqrt{3}\right )}{2 b}+\frac{\tan ^{-1}\left (\frac{\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-ArcTan[Sqrt[3] - (2*Sin[a + b*x]^(1/3))/Cos[a + b*x]^(1/3)]/(2*b) + ArcTan[Sqrt[3] + (2*Sin[a + b*x]^(1/3))/C
os[a + b*x]^(1/3)]/(2*b) + ArcTan[Sin[a + b*x]^(1/3)/Cos[a + b*x]^(1/3)]/b + (Sqrt[3]*Log[1 - (Sqrt[3]*Sin[a +
b*x]^(1/3))/Cos[a + b*x]^(1/3) + Sin[a + b*x]^(2/3)/Cos[a + b*x]^(2/3)])/(4*b) - (Sqrt[3]*Log[1 + (Sqrt[3]*Si
n[a + b*x]^(1/3))/Cos[a + b*x]^(1/3) + Sin[a + b*x]^(2/3)/Cos[a + b*x]^(2/3)])/(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 295

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Module[{r = Numerator[Rt[a/b, n]], s = Denominator[Rt[a/
b, n]], k, u}, Simp[u = Int[(r*Cos[((2*k - 1)*m*Pi)/n] - s*Cos[((2*k - 1)*(m + 1)*Pi)/n]*x)/(r^2 - 2*r*s*Cos[(
(2*k - 1)*Pi)/n]*x + s^2*x^2), x] + Int[(r*Cos[((2*k - 1)*m*Pi)/n] + s*Cos[((2*k - 1)*(m + 1)*Pi)/n]*x)/(r^2 +
2*r*s*Cos[((2*k - 1)*Pi)/n]*x + s^2*x^2), x]; (2*(-1)^(m/2)*r^(m + 2)*Int[1/(r^2 + s^2*x^2), x])/(a*n*s^m) +
Dist[(2*r^(m + 1))/(a*n*s^m), Sum[u, {k, 1, (n - 2)/4}], x], x]] /; FreeQ[{a, b}, x] && IGtQ[(n - 2)/4, 0] &&
IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]

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]

Rule 203

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

Rubi steps

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

Mathematica [C]  time = 0.0405218, size = 57, normalized size = 0.25 \[ \frac{3 \sin ^{\frac{5}{3}}(a+b x) \cos ^2(a+b x)^{5/6} \, _2F_1\left (\frac{5}{6},\frac{5}{6};\frac{11}{6};\sin ^2(a+b x)\right )}{5 b \cos ^{\frac{5}{3}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(Cos[a + b*x]^2)^(5/6)*Hypergeometric2F1[5/6, 5/6, 11/6, Sin[a + b*x]^2]*Sin[a + b*x]^(5/3))/(5*b*Cos[a + b
*x]^(5/3))

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Maple [F]  time = 0.07, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sin \left ( bx+a \right ) \right ) ^{{\frac{2}{3}}} \left ( \cos \left ( bx+a \right ) \right ) ^{-{\frac{2}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(sin(b*x+a)^(2/3)/cos(b*x+a)^(2/3),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{2}{3}}}{\cos \left (b x + a\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(2/3)/cos(b*x + a)^(2/3), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin ^{\frac{2}{3}}{\left (a + b x \right )}}{\cos ^{\frac{2}{3}}{\left (a + b x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x)**(2/3)/cos(a + b*x)**(2/3), 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{2}{3}}}{\cos \left (b x + a\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sin(b*x + a)^(2/3)/cos(b*x + a)^(2/3), x)