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

Optimal. Leaf size=225 \[ -\frac {\sqrt {3} \log \left (\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}+1\right )}{4 b}+\frac {\sqrt {3} \log \left (\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}+1\right )}{4 b}+\frac {\tan ^{-1}\left (\sqrt {3}-\frac {2 \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{2 b}-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}+\sqrt {3}\right )}{2 b}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{b} \]

[Out]

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

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Rubi [A]  time = 0.30, antiderivative size = 225, 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 = {2575, 295, 634, 618, 204, 628, 203} \[ -\frac {\sqrt {3} \log \left (\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}+1\right )}{4 b}+\frac {\sqrt {3} \log \left (\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}+1\right )}{4 b}+\frac {\tan ^{-1}\left (\sqrt {3}-\frac {2 \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{2 b}-\frac {\tan ^{-1}\left (\frac {2 \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}+\sqrt {3}\right )}{2 b}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 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 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 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 2575

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 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*Cos[e + f*x])^(1/k)/(b*Si
n[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 {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)} \, dx &=-\frac {3 \operatorname {Subst}\left (\int \frac {x^4}{1+x^6} \, dx,x,\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{b}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (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]{\cos (a+b x)}}{\sqrt [3]{\sin (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]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{b}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{b}-\frac {\operatorname {Subst}\left (\int \frac {1}{1-\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{4 b}-\frac {\operatorname {Subst}\left (\int \frac {1}{1+\sqrt {3} x+x^2} \, dx,x,\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (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]{\cos (a+b x)}}{\sqrt [3]{\sin (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]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{4 b}\\ &=-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{b}-\frac {\sqrt {3} \log \left (1+\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{4 b}+\frac {\sqrt {3} \log \left (1+\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{4 b}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,-\sqrt {3}+\frac {2 \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{2 b}+\frac {\operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,\sqrt {3}+\frac {2 \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{2 b}\\ &=\frac {\tan ^{-1}\left (\sqrt {3}-\frac {2 \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{2 b}-\frac {\tan ^{-1}\left (\sqrt {3}+\frac {2 \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{2 b}-\frac {\tan ^{-1}\left (\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{b}-\frac {\sqrt {3} \log \left (1+\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{4 b}+\frac {\sqrt {3} \log \left (1+\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{4 b}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 55, normalized size = 0.24 \[ \frac {3 \sqrt [3]{\sin (a+b x)} \sqrt [6]{\cos ^2(a+b x)} \, _2F_1\left (\frac {1}{6},\frac {1}{6};\frac {7}{6};\sin ^2(a+b x)\right )}{b \sqrt [3]{\cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(Cos[a + b*x]^2)^(1/6)*Hypergeometric2F1[1/6, 1/6, 7/6, Sin[a + b*x]^2]*Sin[a + b*x]^(1/3))/(b*Cos[a + b*x]
^(1/3))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 0.08, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{\frac {2}{3}}\left (b x +a \right )}{\sin \left (b x +a \right )^{\frac {2}{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.12, size = 44, normalized size = 0.20 \[ -\frac {3\,{\cos \left (a+b\,x\right )}^{5/3}\,{\sin \left (a+b\,x\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {5}{6},\frac {5}{6};\ \frac {11}{6};\ {\cos \left (a+b\,x\right )}^2\right )}{5\,b\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{1/6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*cos(a + b*x)^(5/3)*sin(a + b*x)^(1/3)*hypergeom([5/6, 5/6], 11/6, cos(a + b*x)^2))/(5*b*(sin(a + b*x)^2)^(
1/6))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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