3.4.25 \(\int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)} \, dx\) [325]

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

[Out]

arctan(sin(b*x+a)^(1/3)/cos(b*x+a)^(1/3))/b+1/2*arctan(2*sin(b*x+a)^(1/3)/cos(b*x+a)^(1/3)-3^(1/2))/b+1/2*arct
an(2*sin(b*x+a)^(1/3)/cos(b*x+a)^(1/3)+3^(1/2))/b+1/4*ln(1+sin(b*x+a)^(2/3)/cos(b*x+a)^(2/3)-sin(b*x+a)^(1/3)*
3^(1/2)/cos(b*x+a)^(1/3))*3^(1/2)/b-1/4*ln(1+sin(b*x+a)^(2/3)/cos(b*x+a)^(2/3)+sin(b*x+a)^(1/3)*3^(1/2)/cos(b*
x+a)^(1/3))*3^(1/2)/b

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Rubi [A]
time = 0.22, antiderivative size = 224, normalized size of antiderivative = 1.00, 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 = {2654, 301, 648, 632, 210, 642, 209} \begin {gather*} -\frac {\text {ArcTan}\left (\sqrt {3}-\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{2 b}+\frac {\text {ArcTan}\left (\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\sqrt {3}\right )}{2 b}+\frac {\text {ArcTan}\left (\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{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 {\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*ArcTan[Sqrt[3] - (2*Sin[a + b*x]^(1/3))/Cos[a + b*x]^(1/3)]/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 209

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

Rule 210

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

Rule 301

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)/(a*n*s^m))*Int[1/(r^2 + s^2*x^2), x] +
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 632

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 642

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 648

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 2654

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]

Rubi steps

\begin {align*} \int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)} \, dx &=\frac {3 \text {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 {\text {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 {\text {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 {\text {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 {\text {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 {\text {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} \text {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} \text {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 {\text {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 {\text {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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.03, size = 57, normalized size = 0.25 \begin {gather*} \frac {3 \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 ) \sin ^{\frac {5}{3}}(a+b x)}{5 b \cos ^{\frac {5}{3}}(a+b x)} \end {gather*}

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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\sin ^{\frac {2}{3}}\left (b x +a \right )}{\cos \left (b x +a \right )^{\frac {2}{3}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{\frac {2}{3}}{\left (a + b x \right )}}{\cos ^{\frac {2}{3}}{\left (a + b x \right )}}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [B]
time = 1.05, size = 44, normalized size = 0.20 \begin {gather*} -\frac {3\,{\cos \left (a+b\,x\right )}^{1/3}\,{\sin \left (a+b\,x\right )}^{5/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{6},\frac {1}{6};\ \frac {7}{6};\ {\cos \left (a+b\,x\right )}^2\right )}{b\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{5/6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*cos(a + b*x)^(1/3)*sin(a + b*x)^(5/3)*hypergeom([1/6, 1/6], 7/6, cos(a + b*x)^2))/(b*(sin(a + b*x)^2)^(5/6
))

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