3.4.32 \(\int \frac {\cos ^{\frac {5}{3}}(a+b x)}{\sin ^{\frac {5}{3}}(a+b x)} \, dx\) [332]

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

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 155, normalized size of antiderivative = 1.00, 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 = {2647, 2654, 281, 298, 31, 648, 632, 210, 642} \begin {gather*} \frac {\sqrt {3} \text {ArcTan}\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 {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}+\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[a + b*x]^(5/3)/Sin[a + b*x]^(5/3),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)/C
os[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) - (3*Cos[a + b*x]^(2/3))/(2*b*Sin[a + b*x]^(2/3))

Rule 31

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

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 281

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 298

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 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 2647

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

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 {\cos ^{\frac {5}{3}}(a+b x)}{\sin ^{\frac {5}{3}}(a+b x)} \, dx &=-\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}-\int \frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}} \, dx\\ &=-\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}-\frac {3 \text {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 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}-\frac {3 \text {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 {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}+\frac {\text {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 {\text {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 {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}-\frac {\text {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 \text {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 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}+\frac {3 \text {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}-\frac {3 \cos ^{\frac {2}{3}}(a+b x)}{2 b \sin ^{\frac {2}{3}}(a+b x)}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.02, size = 57, normalized size = 0.37 \begin {gather*} -\frac {3 \cos ^2(a+b x)^{2/3} \, _2F_1\left (-\frac {1}{3},-\frac {1}{3};\frac {2}{3};\sin ^2(a+b x)\right )}{2 b \cos ^{\frac {4}{3}}(a+b x) \sin ^{\frac {2}{3}}(a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\cos ^{\frac {5}{3}}\left (b x +a \right )}{\sin \left (b x +a \right )^{\frac {5}{3}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [A]
time = 0.73, size = 189, normalized size = 1.22 \begin {gather*} -\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 ) \sin \left (b x + a\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 ) \sin \left (b x + a\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 ) \sin \left (b x + a\right ) + 6 \, \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}}}{4 \, b \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(b*x+a)^(5/3)/sin(b*x+a)^(5/3),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))*sin(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))*sin(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)*sin(b*x + a) + 6*cos(b*x + a)^(2/3)*sin(b*x + a)^(1/3))/(b*sin(b*x + a))

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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

[Out]

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

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Mupad [B]
time = 1.20, size = 44, normalized size = 0.28 \begin {gather*} -\frac {3\,{\cos \left (a+b\,x\right )}^{8/3}\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{1/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {4}{3},\frac {4}{3};\ \frac {7}{3};\ {\cos \left (a+b\,x\right )}^2\right )}{8\,b\,{\sin \left (a+b\,x\right )}^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*cos(a + b*x)^(8/3)*(sin(a + b*x)^2)^(1/3)*hypergeom([4/3, 4/3], 7/3, cos(a + b*x)^2))/(8*b*sin(a + b*x)^(2
/3))

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