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

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [F]
Fricas [B] (verification not implemented)
Sympy [F]
Maxima [F]
Giac [F]
Mupad [B] (verification not implemented)
Reduce [F]

Optimal result

Integrand size = 21, antiderivative size = 163 \[ \int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=-\frac {\arctan \left (\sqrt {3}-\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{2 b}+\frac {\arctan \left (\sqrt {3}+\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{2 b}+\frac {\arctan \left (\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{b}-\frac {\sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)} \left (1+\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}\right )}{2 b} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.35 \[ \int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=\frac {3 \cos ^2(a+b x)^{5/6} \operatorname {Hypergeometric2F1}\left (\frac {5}{6},\frac {5}{6},\frac {11}{6},\sin ^2(a+b x)\right ) \sin ^{\frac {5}{3}}(a+b x)}{5 b \cos ^{\frac {5}{3}}(a+b x)} \] Input:

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

Output:

(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))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.36, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 3054, 824, 27, 216, 1142, 25, 1083, 217, 1103}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3054

\(\displaystyle \frac {3 \int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x) \left (\tan ^2(a+b x)+1\right )}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}}{b}\)

\(\Big \downarrow \) 824

\(\displaystyle \frac {3 \left (\frac {1}{3} \int \frac {1}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\frac {1}{3} \int -\frac {1-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}}{2 \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1\right )}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\frac {1}{3} \int -\frac {\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}{2 \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1\right )}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {3 \left (\frac {1}{3} \int \frac {1}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}-\frac {1}{6} \int \frac {1-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}-\frac {1}{6} \int \frac {\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )}{b}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {3 \left (-\frac {1}{6} \int \frac {1-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}-\frac {1}{6} \int \frac {\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\frac {1}{3} \arctan \left (\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )\right )}{b}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {3 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\frac {1}{2} \sqrt {3} \int -\frac {\sqrt {3}-\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}-\frac {1}{2} \sqrt {3} \int \frac {\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\sqrt {3}}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )+\frac {1}{3} \arctan \left (\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )\right )}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3 \left (\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )+\frac {1}{6} \left (\frac {1}{2} \int \frac {1}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}-\frac {1}{2} \sqrt {3} \int \frac {\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\sqrt {3}}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )+\frac {1}{3} \arctan \left (\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )\right )}{b}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {3 \left (\frac {1}{6} \left (-\int \frac {1}{-\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-1}d\left (\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}-\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )+\frac {1}{6} \left (-\int \frac {1}{-\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-1}d\left (\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\sqrt {3}}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )+\frac {1}{3} \arctan \left (\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )\right )}{b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 \left (\frac {1}{6} \left (-\frac {1}{2} \sqrt {3} \int \frac {\sqrt {3}-\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}-\arctan \left (\sqrt {3}-\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )\right )+\frac {1}{6} \left (\arctan \left (\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \int \frac {\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\sqrt {3}}{\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1}d\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )+\frac {1}{3} \arctan \left (\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )\right )}{b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3 \left (\frac {1}{3} \arctan \left (\frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )+\frac {1}{6} \left (\frac {1}{2} \sqrt {3} \log \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1\right )-\arctan \left (\sqrt {3}-\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}\right )\right )+\frac {1}{6} \left (\arctan \left (\frac {2 \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+\sqrt {3}\right )-\frac {1}{2} \sqrt {3} \log \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sqrt {3} \sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}+1\right )\right )\right )}{b}\)

Input:

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

Output:

(3*(ArcTan[Sin[a + b*x]^(1/3)/Cos[a + b*x]^(1/3)]/3 + (-ArcTan[Sqrt[3] - ( 
2*Sin[a + b*x]^(1/3))/Cos[a + b*x]^(1/3)] + (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) 
])/2)/6 + (ArcTan[Sqrt[3] + (2*Sin[a + b*x]^(1/3))/Cos[a + b*x]^(1/3)] - ( 
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)])/2)/6))/b
 

Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 217
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 824
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] + 2*(r^(m 
+ 1)/(a*n*s^m))   Sum[u, {k, 1, (n - 2)/4}], x]] /; FreeQ[{a, b}, x] && IGt 
Q[(n - 2)/4, 0] && IGtQ[m, 0] && LtQ[m, n - 1] && PosQ[a/b]
 

rule 1083
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Simp[-2   Subst[I 
nt[1/Simp[b^2 - 4*a*c - x^2, x], x], x, b + 2*c*x], x] /; FreeQ[{a, b, c}, 
x]
 

rule 1103
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[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 1142
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S 
imp[(2*c*d - b*e)/(2*c)   Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) 
Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3054
Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m 
_), x_Symbol] :> With[{k = Denominator[m]}, Simp[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]
 
Maple [F]

\[\int \frac {\sin \left (b x +a \right )^{\frac {2}{3}}}{\cos \left (b x +a \right )^{\frac {2}{3}}}d x\]

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (131) = 262\).

Time = 0.11 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.64 \[ \int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=-\frac {\sqrt {3} b \sqrt {\frac {1}{b^{2}}} \log \left (\frac {2 \, {\left (\sqrt {3} b \sqrt {\frac {1}{b^{2}}} \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} + \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}} + \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )}\right ) - \sqrt {3} b \sqrt {\frac {1}{b^{2}}} \log \left (-\frac {2 \, {\left (\sqrt {3} b \sqrt {\frac {1}{b^{2}}} \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} - \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}} - \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )}\right ) + 2 \, \arctan \left (\frac {\sqrt {3} b \sqrt {\frac {1}{b^{2}}} \sin \left (b x + a\right ) + 2 \, \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}}}{\sin \left (b x + a\right )}\right ) + 2 \, \arctan \left (-\frac {\sqrt {3} b \sqrt {\frac {1}{b^{2}}} \sin \left (b x + a\right ) - 2 \, \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}}}{\sin \left (b x + a\right )}\right ) + 4 \, \arctan \left (\frac {\cos \left (b x + a\right )^{\frac {1}{3}}}{\sin \left (b x + a\right )^{\frac {1}{3}}}\right )}{4 \, b} \] Input:

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

Output:

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

Sympy [F]

\[ \int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=\int \frac {\sin ^{\frac {2}{3}}{\left (a + b x \right )}}{\cos ^{\frac {2}{3}}{\left (a + b x \right )}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{\frac {2}{3}}}{\cos \left (b x + a\right )^{\frac {2}{3}}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{\frac {2}{3}}}{\cos \left (b x + a\right )^{\frac {2}{3}}} \,d x } \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 26.43 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.27 \[ \int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=-\frac {3\,{\cos \left (a+b\,x\right )}^{1/3}\,{\sin \left (a+b\,x\right )}^{5/3}\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{6},\frac {1}{6};\ \frac {7}{6};\ {\cos \left (a+b\,x\right )}^2\right )}{b\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{5/6}} \] Input:

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

Output:

-(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))
 

Reduce [F]

\[ \int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)} \, dx=\int \frac {\sin \left (b x +a \right )^{\frac {2}{3}}}{\cos \left (b x +a \right )^{\frac {2}{3}}}d x \] Input:

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

Output:

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