\(\int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx\) [327]

Optimal result

Integrand size = 21, antiderivative size = 155 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}}{\sqrt {3}}\right )}{2 b}+\frac {\log \left (1+\frac {\cos ^{\frac {4}{3}}(a+b x)}{\sin ^{\frac {4}{3}}(a+b x)}-\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{4 b}-\frac {\log \left (1+\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right )}{2 b}+\frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2646, 2655, 281, 298, 31, 648, 632, 210, 642} \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}}{\sqrt {3}}\right )}{2 b}+\frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)}+\frac {\log \left (\frac {\cos ^{\frac {4}{3}}(a+b x)}{\sin ^{\frac {4}{3}}(a+b x)}-\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}+1\right )}{4 b}-\frac {\log \left (\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}+1\right )}{2 b} \]

[In]

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

[Out]

-1/2*(Sqrt[3]*ArcTan[(1 - (2*Cos[a + b*x]^(2/3))/Sin[a + b*x]^(2/3))/Sqrt[3]])/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 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 2646

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] && (Int
egersQ[2*m, 2*n] || EqQ[m + n, 0])

Rule 2655

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*
Sin[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*} \text {integral}& = \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 \text {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 \text {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 {\text {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 {\text {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 {\text {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 \text {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 \text {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} \arctan \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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.37 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\frac {3 \sqrt [3]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {4}{3},\frac {7}{3},\sin ^2(a+b x)\right ) \sin ^{\frac {8}{3}}(a+b x)}{8 b \cos ^{\frac {2}{3}}(a+b x)} \]

[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))

Maple [F]

\[\int \frac {\sin ^{\frac {5}{3}}\left (b x +a \right )}{\cos \left (b x +a \right )^{\frac {5}{3}}}d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.27 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\frac {2 \, \sqrt {3} \arctan \left (\frac {2 \, \sqrt {3} \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}} - \sqrt {3} \sin \left (b x + a\right )}{3 \, \sin \left (b x + a\right )}\right ) \cos \left (b x + a\right ) + \cos \left (b x + a\right ) \log \left (\frac {4 \, {\left (\cos \left (b x + a\right )^{2} - \cos \left (b x + a\right )^{\frac {4}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} + \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {4}{3}} - 1\right )}}{\cos \left (b x + a\right )^{2} - 1}\right ) - 2 \, \cos \left (b x + a\right ) \log \left (-\frac {2 \, {\left (\cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}} + \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )}\right ) + 6 \, \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}}}{4 \, b \cos \left (b x + a\right )} \]

[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))

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{\frac {5}{3}}}{\cos \left (b x + a\right )^{\frac {5}{3}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{\frac {5}{3}}}{\cos \left (b x + a\right )^{\frac {5}{3}}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 1.01 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.28 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\frac {3\,{\sin \left (a+b\,x\right )}^{8/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},-\frac {1}{3};\ \frac {2}{3};\ {\cos \left (a+b\,x\right )}^2\right )}{2\,b\,{\cos \left (a+b\,x\right )}^{2/3}\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{4/3}} \]

[In]

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

[Out]

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