\(\int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)} \, dx\) [326]

Optimal result

Integrand size = 21, antiderivative size = 188 \[ \int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)} \, dx=-\frac {\arctan \left (\sqrt {3}-\frac {2 \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{2 b}+\frac {\arctan \left (\sqrt {3}+\frac {2 \sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{2 b}+\frac {\arctan \left (\frac {\sqrt [3]{\cos (a+b x)}}{\sqrt [3]{\sin (a+b x)}}\right )}{b}-\frac {\sqrt {3} \text {arctanh}\left (\frac {\sqrt {3} \sqrt [3]{\cos (a+b x)}}{\left (1+\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}\right ) \sqrt [3]{\sin (a+b x)}}\right )}{2 b}+\frac {3 \sqrt [3]{\sin (a+b x)}}{b \sqrt [3]{\cos (a+b x)}} \] Output:

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

Mathematica [C] (verified)

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

Time = 0.06 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.30 \[ \int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)} \, dx=\frac {3 \sqrt [6]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {7}{6},\frac {7}{6},\frac {13}{6},\sin ^2(a+b x)\right ) \sin ^{\frac {7}{3}}(a+b x)}{7 b \sqrt [3]{\cos (a+b x)}} \] Input:

Integrate[Sin[a + b*x]^(4/3)/Cos[a + b*x]^(4/3),x]
 

Output:

(3*(Cos[a + b*x]^2)^(1/6)*Hypergeometric2F1[7/6, 7/6, 13/6, Sin[a + b*x]^2 
]*Sin[a + b*x]^(7/3))/(7*b*Cos[a + b*x]^(1/3))
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.31, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 3046, 3042, 3055, 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.

\(\displaystyle \int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (a+b x)^{4/3}}{\cos (a+b x)^{4/3}}dx\)

\(\Big \downarrow \) 3046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3055

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

\(\Big \downarrow \) 824

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 216

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

Int[Sin[a + b*x]^(4/3)/Cos[a + b*x]^(4/3),x]
 

Output:

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

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 3046

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] + Simp[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] && (IntegersQ[2*m, 2*n] || EqQ[m + n, 0])
 

rule 3055

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n 
_), 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*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]
 

Maple [F]

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

Input:

int(sin(b*x+a)^(4/3)/cos(b*x+a)^(4/3),x)
 

Output:

int(sin(b*x+a)^(4/3)/cos(b*x+a)^(4/3),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (152) = 304\).

Time = 0.13 (sec) , antiderivative size = 322, normalized size of antiderivative = 1.71 \[ \int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)} \, dx=-\frac {\sqrt {3} b \sqrt {\frac {1}{b^{2}}} \cos \left (b x + a\right ) \log \left (\frac {\sqrt {3} b \sqrt {\frac {1}{b^{2}}} \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}} + \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} + \cos \left (b x + a\right )}{\cos \left (b x + a\right )}\right ) - \sqrt {3} b \sqrt {\frac {1}{b^{2}}} \cos \left (b x + a\right ) \log \left (-\frac {\sqrt {3} b \sqrt {\frac {1}{b^{2}}} \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}} - \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} - \cos \left (b x + a\right )}{\cos \left (b x + a\right )}\right ) + 2 \, \arctan \left (\frac {\sqrt {3} b \sqrt {\frac {1}{b^{2}}} \cos \left (b x + a\right ) + 2 \, \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}}}{\cos \left (b x + a\right )}\right ) \cos \left (b x + a\right ) + 2 \, \arctan \left (-\frac {\sqrt {3} b \sqrt {\frac {1}{b^{2}}} \cos \left (b x + a\right ) - 2 \, \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}}}{\cos \left (b x + a\right )}\right ) \cos \left (b x + a\right ) + 4 \, \arctan \left (\frac {\sin \left (b x + a\right )^{\frac {1}{3}}}{\cos \left (b x + a\right )^{\frac {1}{3}}}\right ) \cos \left (b x + a\right ) - 12 \, \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}}}{4 \, b \cos \left (b x + a\right )} \] Input:

integrate(sin(b*x+a)^(4/3)/cos(b*x+a)^(4/3),x, algorithm="fricas")
 

Output:

-1/4*(sqrt(3)*b*sqrt(b^(-2))*cos(b*x + a)*log((sqrt(3)*b*sqrt(b^(-2))*cos( 
b*x + a)^(2/3)*sin(b*x + a)^(1/3) + cos(b*x + a)^(1/3)*sin(b*x + a)^(2/3) 
+ cos(b*x + a))/cos(b*x + a)) - sqrt(3)*b*sqrt(b^(-2))*cos(b*x + a)*log(-( 
sqrt(3)*b*sqrt(b^(-2))*cos(b*x + a)^(2/3)*sin(b*x + a)^(1/3) - cos(b*x + a 
)^(1/3)*sin(b*x + a)^(2/3) - cos(b*x + a))/cos(b*x + a)) + 2*arctan((sqrt( 
3)*b*sqrt(b^(-2))*cos(b*x + a) + 2*cos(b*x + a)^(2/3)*sin(b*x + a)^(1/3))/ 
cos(b*x + a))*cos(b*x + a) + 2*arctan(-(sqrt(3)*b*sqrt(b^(-2))*cos(b*x + a 
) - 2*cos(b*x + a)^(2/3)*sin(b*x + a)^(1/3))/cos(b*x + a))*cos(b*x + a) + 
4*arctan(sin(b*x + a)^(1/3)/cos(b*x + a)^(1/3))*cos(b*x + a) - 12*cos(b*x 
+ a)^(2/3)*sin(b*x + a)^(1/3))/(b*cos(b*x + a))
 

Sympy [F]

\[ \int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)} \, dx=\int \frac {\sin ^{\frac {4}{3}}{\left (a + b x \right )}}{\cos ^{\frac {4}{3}}{\left (a + b x \right )}}\, dx \] Input:

integrate(sin(b*x+a)**(4/3)/cos(b*x+a)**(4/3),x)
 

Output:

Integral(sin(a + b*x)**(4/3)/cos(a + b*x)**(4/3), x)
 

Maxima [F]

\[ \int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{\frac {4}{3}}}{\cos \left (b x + a\right )^{\frac {4}{3}}} \,d x } \] Input:

integrate(sin(b*x+a)^(4/3)/cos(b*x+a)^(4/3),x, algorithm="maxima")
 

Output:

integrate(sin(b*x + a)^(4/3)/cos(b*x + a)^(4/3), x)
 

Giac [F]

\[ \int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{\frac {4}{3}}}{\cos \left (b x + a\right )^{\frac {4}{3}}} \,d x } \] Input:

integrate(sin(b*x+a)^(4/3)/cos(b*x+a)^(4/3),x, algorithm="giac")
 

Output:

integrate(sin(b*x + a)^(4/3)/cos(b*x + a)^(4/3), x)
 

Mupad [B] (verification not implemented)

Time = 27.00 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.23 \[ \int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)} \, dx=\frac {3\,{\sin \left (a+b\,x\right )}^{7/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{6},-\frac {1}{6};\ \frac {5}{6};\ {\cos \left (a+b\,x\right )}^2\right )}{b\,{\cos \left (a+b\,x\right )}^{1/3}\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{7/6}} \] Input:

int(sin(a + b*x)^(4/3)/cos(a + b*x)^(4/3),x)
 

Output:

(3*sin(a + b*x)^(7/3)*hypergeom([-1/6, -1/6], 5/6, cos(a + b*x)^2))/(b*cos 
(a + b*x)^(1/3)*(sin(a + b*x)^2)^(7/6))
 

Reduce [F]

\[ \int \frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)} \, dx=\int \frac {\sin \left (b x +a \right )^{\frac {4}{3}}}{\cos \left (b x +a \right )^{\frac {4}{3}}}d x \] Input:

int(sin(b*x+a)^(4/3)/cos(b*x+a)^(4/3),x)
 

Output:

int((sin(a + b*x)**(1/3)*sin(a + b*x))/(cos(a + b*x)**(1/3)*cos(a + b*x)), 
x)