3.327 \(\int \frac{\sin ^{\frac{5}{3}}(a+b x)}{\cos ^{\frac{5}{3}}(a+b x)} \, dx\)

Optimal. Leaf size=155 \[ \frac{3 \sin ^{\frac{2}{3}}(a+b x)}{2 b \cos ^{\frac{2}{3}}(a+b x)}+\frac{\log \left (\frac{\cos ^{\frac{4}{3}}(a+b x)}{\sin ^{\frac{4}{3}}(a+b x)}-\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}+1\right )}{4 b}-\frac{\log \left (\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}+1\right )}{2 b}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}}{\sqrt{3}}\right )}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.174671, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2566, 2575, 275, 292, 31, 634, 618, 204, 628} \[ \frac{3 \sin ^{\frac{2}{3}}(a+b x)}{2 b \cos ^{\frac{2}{3}}(a+b x)}+\frac{\log \left (\frac{\cos ^{\frac{4}{3}}(a+b x)}{\sin ^{\frac{4}{3}}(a+b x)}-\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}+1\right )}{4 b}-\frac{\log \left (\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}+1\right )}{2 b}-\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}}{\sqrt{3}}\right )}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2566

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

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]

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{\sin ^{\frac{5}{3}}(a+b x)}{\cos ^{\frac{5}{3}}(a+b x)} \, dx &=\frac{3 \sin ^{\frac{2}{3}}(a+b x)}{2 b \cos ^{\frac{2}{3}}(a+b x)}-\int \frac{\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}} \, dx\\ &=\frac{3 \sin ^{\frac{2}{3}}(a+b x)}{2 b \cos ^{\frac{2}{3}}(a+b x)}+\frac{3 \operatorname{Subst}\left (\int \frac{x^3}{1+x^6} \, dx,x,\frac{\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{b}\\ &=\frac{3 \sin ^{\frac{2}{3}}(a+b x)}{2 b \cos ^{\frac{2}{3}}(a+b x)}+\frac{3 \operatorname{Subst}\left (\int \frac{x}{1+x^3} \, dx,x,\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{2 b}\\ &=\frac{3 \sin ^{\frac{2}{3}}(a+b x)}{2 b \cos ^{\frac{2}{3}}(a+b x)}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x} \, dx,x,\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{2 b}+\frac{\operatorname{Subst}\left (\int \frac{1+x}{1-x+x^2} \, dx,x,\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{2 b}\\ &=-\frac{\log \left (1+\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{2 b}+\frac{3 \sin ^{\frac{2}{3}}(a+b x)}{2 b \cos ^{\frac{2}{3}}(a+b x)}+\frac{\operatorname{Subst}\left (\int \frac{-1+2 x}{1-x+x^2} \, dx,x,\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{4 b}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{1-x+x^2} \, dx,x,\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{4 b}\\ &=\frac{\log \left (1+\frac{\cos ^{\frac{4}{3}}(a+b x)}{\sin ^{\frac{4}{3}}(a+b x)}-\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{4 b}-\frac{\log \left (1+\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{2 b}+\frac{3 \sin ^{\frac{2}{3}}(a+b x)}{2 b \cos ^{\frac{2}{3}}(a+b x)}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,-1+\frac{2 \cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{2 b}\\ &=-\frac{\sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}}{\sqrt{3}}\right )}{2 b}+\frac{\log \left (1+\frac{\cos ^{\frac{4}{3}}(a+b x)}{\sin ^{\frac{4}{3}}(a+b x)}-\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{4 b}-\frac{\log \left (1+\frac{\cos ^{\frac{2}{3}}(a+b x)}{\sin ^{\frac{2}{3}}(a+b x)}\right )}{2 b}+\frac{3 \sin ^{\frac{2}{3}}(a+b x)}{2 b \cos ^{\frac{2}{3}}(a+b x)}\\ \end{align*}

Mathematica [C]  time = 0.0558447, size = 57, normalized size = 0.37 \[ \frac{3 \sin ^{\frac{8}{3}}(a+b x) \sqrt [3]{\cos ^2(a+b x)} \, _2F_1\left (\frac{4}{3},\frac{4}{3};\frac{7}{3};\sin ^2(a+b x)\right )}{8 b \cos ^{\frac{2}{3}}(a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(Cos[a + b*x]^2)^(1/3)*Hypergeometric2F1[4/3, 4/3, 7/3, Sin[a + b*x]^2]*Sin[a + b*x]^(8/3))/(8*b*Cos[a + b*
x]^(2/3))

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Maple [F]  time = 0.052, size = 0, normalized size = 0. \begin{align*} \int{ \left ( \sin \left ( bx+a \right ) \right ) ^{{\frac{5}{3}}} \left ( \cos \left ( bx+a \right ) \right ) ^{-{\frac{5}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^(5/3)/cos(b*x+a)^(5/3),x)

[Out]

int(sin(b*x+a)^(5/3)/cos(b*x+a)^(5/3),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{5}{3}}}{\cos \left (b x + a\right )^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^(5/3)/cos(b*x+a)^(5/3),x, algorithm="maxima")

[Out]

integrate(sin(b*x + a)^(5/3)/cos(b*x + a)^(5/3), x)

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Fricas [A]  time = 3.23847, size = 572, normalized size = 3.69 \begin{align*} \frac{2 \, \sqrt{3} \arctan \left (\frac{2 \, \sqrt{3} \cos \left (b x + a\right )^{\frac{2}{3}} \sin \left (b x + a\right )^{\frac{1}{3}} - \sqrt{3} \sin \left (b x + a\right )}{3 \, \sin \left (b x + a\right )}\right ) \cos \left (b x + a\right ) + \cos \left (b x + a\right ) \log \left (\frac{4 \,{\left (\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}} - 1\right )}}{\cos \left (b x + a\right )^{2} - 1}\right ) - 2 \, \cos \left (b x + a\right ) \log \left (-\frac{2 \,{\left (\cos \left (b x + a\right )^{\frac{2}{3}} \sin \left (b x + a\right )^{\frac{1}{3}} + \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )}\right ) + 6 \, \cos \left (b x + a\right )^{\frac{1}{3}} \sin \left (b x + a\right )^{\frac{2}{3}}}{4 \, b \cos \left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^(5/3)/cos(b*x+a)^(5/3),x, algorithm="fricas")

[Out]

1/4*(2*sqrt(3)*arctan(1/3*(2*sqrt(3)*cos(b*x + a)^(2/3)*sin(b*x + a)^(1/3) - sqrt(3)*sin(b*x + a))/sin(b*x + a
))*cos(b*x + a) + cos(b*x + a)*log(4*(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) - 1)/(cos(b*x + a)^2 - 1)) - 2*cos(b*x + a)*log(-2*(cos(b*x + a)^(2/3)*sin(b*x + a)^(1/
3) + sin(b*x + a))/sin(b*x + a)) + 6*cos(b*x + a)^(1/3)*sin(b*x + a)^(2/3))/(b*cos(b*x + a))

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Sympy [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)**(5/3)/cos(b*x+a)**(5/3),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (b x + a\right )^{\frac{5}{3}}}{\cos \left (b x + a\right )^{\frac{5}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^(5/3)/cos(b*x+a)^(5/3),x, algorithm="giac")

[Out]

integrate(sin(b*x + a)^(5/3)/cos(b*x + a)^(5/3), x)