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

Optimal result

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

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

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 = 0.37 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\frac {3 \sqrt [3]{\cos ^2(a+b x)} \operatorname {Hypergeometric2F1}\left (\frac {4}{3},\frac {4}{3},\frac {7}{3},\sin ^2(a+b x)\right ) \sin ^{\frac {8}{3}}(a+b x)}{8 b \cos ^{\frac {2}{3}}(a+b x)} \] Input:

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

Output:

(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))
 

Rubi [A] (warning: unable to verify)

Time = 0.39 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.65, 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, 807, 821, 16, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3046

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3055

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

\(\Big \downarrow \) 807

\(\displaystyle \frac {3 \int \frac {\cos ^{\frac {2}{3}}(a+b x)}{(\cot (a+b x)+1) \sin ^{\frac {2}{3}}(a+b x)}d\frac {\cos ^{\frac {2}{3}}(a+b x)}{\sin ^{\frac {2}{3}}(a+b x)}}{2 b}+\frac {3 \sin ^{\frac {2}{3}}(a+b x)}{2 b \cos ^{\frac {2}{3}}(a+b x)}\)

\(\Big \downarrow \) 821

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1142

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 1083

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

\(\Big \downarrow \) 217

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

\(\Big \downarrow \) 1103

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 16

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

rule 25

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

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 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[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 821

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 
1)   Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[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 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]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.27 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\frac {2 \, \sqrt {3} \arctan \left (\frac {2 \, \sqrt {3} \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}} - \sqrt {3} \sin \left (b x + a\right )}{3 \, \sin \left (b x + a\right )}\right ) \cos \left (b x + a\right ) + \cos \left (b x + a\right ) \log \left (\frac {4 \, {\left (\cos \left (b x + a\right )^{2} - \cos \left (b x + a\right )^{\frac {4}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} + \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {4}{3}} - 1\right )}}{\cos \left (b x + a\right )^{2} - 1}\right ) - 2 \, \cos \left (b x + a\right ) \log \left (-\frac {2 \, {\left (\cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {1}{3}} + \sin \left (b x + a\right )\right )}}{\sin \left (b x + a\right )}\right ) + 6 \, \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}}}{4 \, b \cos \left (b x + a\right )} \] Input:

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

Output:

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 26.55 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.28 \[ \int \frac {\sin ^{\frac {5}{3}}(a+b x)}{\cos ^{\frac {5}{3}}(a+b x)} \, dx=\frac {3\,{\sin \left (a+b\,x\right )}^{8/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{3},-\frac {1}{3};\ \frac {2}{3};\ {\cos \left (a+b\,x\right )}^2\right )}{2\,b\,{\cos \left (a+b\,x\right )}^{2/3}\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{4/3}} \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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