\(\int \frac {\sin ^{\frac {7}{3}}(a+b x)}{\cos ^{\frac {7}{3}}(a+b x)} \, dx\) [328]

Optimal result

Integrand size = 21, antiderivative size = 155 \[ \int \frac {\sin ^{\frac {7}{3}}(a+b x)}{\cos ^{\frac {7}{3}}(a+b x)} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}}{\sqrt {3}}\right )}{2 b}+\frac {\log \left (1+\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )}{2 b}-\frac {\log \left (1-\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {\sin ^{\frac {4}{3}}(a+b x)}{\cos ^{\frac {4}{3}}(a+b x)}\right )}{4 b}+\frac {3 \sin ^{\frac {4}{3}}(a+b x)}{4 b \cos ^{\frac {4}{3}}(a+b x)} \] Output:

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

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 \[ \int \frac {\sin ^{\frac {7}{3}}(a+b x)}{\cos ^{\frac {7}{3}}(a+b x)} \, dx=\frac {3 \cos ^2(a+b x)^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {5}{3},\frac {5}{3},\frac {8}{3},\sin ^2(a+b x)\right ) \sin ^{\frac {10}{3}}(a+b x)}{10 b \cos ^{\frac {4}{3}}(a+b x)} \] Input:

Integrate[Sin[a + b*x]^(7/3)/Cos[a + b*x]^(7/3),x]
 

Output:

(3*(Cos[a + b*x]^2)^(2/3)*Hypergeometric2F1[5/3, 5/3, 8/3, Sin[a + b*x]^2] 
*Sin[a + b*x]^(10/3))/(10*b*Cos[a + b*x]^(4/3))
 

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, 3054, 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.

\(\displaystyle \int \frac {\sin ^{\frac {7}{3}}(a+b x)}{\cos ^{\frac {7}{3}}(a+b x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (a+b x)^{7/3}}{\cos (a+b x)^{7/3}}dx\)

\(\Big \downarrow \) 3046

\(\displaystyle \frac {3 \sin ^{\frac {4}{3}}(a+b x)}{4 b \cos ^{\frac {4}{3}}(a+b x)}-\int \frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 \sin ^{\frac {4}{3}}(a+b x)}{4 b \cos ^{\frac {4}{3}}(a+b x)}-\int \frac {\sqrt [3]{\sin (a+b x)}}{\sqrt [3]{\cos (a+b x)}}dx\)

\(\Big \downarrow \) 3054

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

\(\Big \downarrow \) 807

\(\displaystyle \frac {3 \sin ^{\frac {4}{3}}(a+b x)}{4 b \cos ^{\frac {4}{3}}(a+b x)}-\frac {3 \int \frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x) (\tan (a+b x)+1)}d\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}}{2 b}\)

\(\Big \downarrow \) 821

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

\(\Big \downarrow \) 16

\(\displaystyle \frac {3 \sin ^{\frac {4}{3}}(a+b x)}{4 b \cos ^{\frac {4}{3}}(a+b x)}-\frac {3 \left (\frac {1}{3} \int \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1\right )d\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {1}{3} \log \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1\right )\right )}{2 b}\)

\(\Big \downarrow \) 1142

\(\displaystyle \frac {3 \sin ^{\frac {4}{3}}(a+b x)}{4 b \cos ^{\frac {4}{3}}(a+b x)}-\frac {3 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+\frac {1}{2} \int \left (\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-1\right )d\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )-\frac {1}{3} \log \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1\right )\right )}{2 b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3 \sin ^{\frac {4}{3}}(a+b x)}{4 b \cos ^{\frac {4}{3}}(a+b x)}-\frac {3 \left (\frac {1}{3} \left (\frac {3}{2} \int 1d\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-\frac {1}{2} \int \left (1-\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )d\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )-\frac {1}{3} \log \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1\right )\right )}{2 b}\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {3 \sin ^{\frac {4}{3}}(a+b x)}{4 b \cos ^{\frac {4}{3}}(a+b x)}-\frac {3 \left (\frac {1}{3} \left (-3 \int \frac {1}{-\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-2}d\left (\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-1\right )-\frac {1}{2} \int \left (1-\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )d\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )-\frac {1}{3} \log \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1\right )\right )}{2 b}\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {3 \sin ^{\frac {4}{3}}(a+b x)}{4 b \cos ^{\frac {4}{3}}(a+b x)}-\frac {3 \left (\frac {1}{3} \left (\sqrt {3} \arctan \left (\frac {\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-1}{\sqrt {3}}\right )-\frac {1}{2} \int \left (1-\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )d\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}\right )-\frac {1}{3} \log \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1\right )\right )}{2 b}\)

\(\Big \downarrow \) 1103

\(\displaystyle \frac {3 \sin ^{\frac {4}{3}}(a+b x)}{4 b \cos ^{\frac {4}{3}}(a+b x)}-\frac {3 \left (\frac {\arctan \left (\frac {\frac {2 \sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log \left (\frac {\sin ^{\frac {2}{3}}(a+b x)}{\cos ^{\frac {2}{3}}(a+b x)}+1\right )\right )}{2 b}\)

Input:

Int[Sin[a + b*x]^(7/3)/Cos[a + b*x]^(7/3),x]
 

Output:

(-3*(ArcTan[(-1 + (2*Sin[a + b*x]^(2/3))/Cos[a + b*x]^(2/3))/Sqrt[3]]/Sqrt 
[3] - Log[1 + Sin[a + b*x]^(2/3)/Cos[a + b*x]^(2/3)]/3))/(2*b) + (3*Sin[a 
+ b*x]^(4/3))/(4*b*Cos[a + b*x]^(4/3))
 

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 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]

\[\int \frac {\sin \left (b x +a \right )^{\frac {7}{3}}}{\cos \left (b x +a \right )^{\frac {7}{3}}}d x\]

Input:

int(sin(b*x+a)^(7/3)/cos(b*x+a)^(7/3),x)
 

Output:

int(sin(b*x+a)^(7/3)/cos(b*x+a)^(7/3),x)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.26 \[ \int \frac {\sin ^{\frac {7}{3}}(a+b x)}{\cos ^{\frac {7}{3}}(a+b x)} \, dx=-\frac {2 \, \sqrt {3} \arctan \left (-\frac {\sqrt {3} \cos \left (b x + a\right ) - 2 \, \sqrt {3} \cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}}}{3 \, \cos \left (b x + a\right )}\right ) \cos \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right )^{2} \log \left (\frac {\cos \left (b x + a\right )^{\frac {1}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} + \cos \left (b x + a\right )}{\cos \left (b x + a\right )}\right ) + \cos \left (b x + a\right )^{2} \log \left (\frac {\cos \left (b x + a\right )^{2} - \cos \left (b x + a\right )^{\frac {4}{3}} \sin \left (b x + a\right )^{\frac {2}{3}} + \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {4}{3}}}{\cos \left (b x + a\right )^{2}}\right ) - 3 \, \cos \left (b x + a\right )^{\frac {2}{3}} \sin \left (b x + a\right )^{\frac {4}{3}}}{4 \, b \cos \left (b x + a\right )^{2}} \] Input:

integrate(sin(b*x+a)^(7/3)/cos(b*x+a)^(7/3),x, algorithm="fricas")
 

Output:

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

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin ^{\frac {7}{3}}(a+b x)}{\cos ^{\frac {7}{3}}(a+b x)} \, dx=\text {Timed out} \] Input:

integrate(sin(b*x+a)**(7/3)/cos(b*x+a)**(7/3),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\sin ^{\frac {7}{3}}(a+b x)}{\cos ^{\frac {7}{3}}(a+b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{\frac {7}{3}}}{\cos \left (b x + a\right )^{\frac {7}{3}}} \,d x } \] Input:

integrate(sin(b*x+a)^(7/3)/cos(b*x+a)^(7/3),x, algorithm="maxima")
 

Output:

integrate(sin(b*x + a)^(7/3)/cos(b*x + a)^(7/3), x)
 

Giac [F]

\[ \int \frac {\sin ^{\frac {7}{3}}(a+b x)}{\cos ^{\frac {7}{3}}(a+b x)} \, dx=\int { \frac {\sin \left (b x + a\right )^{\frac {7}{3}}}{\cos \left (b x + a\right )^{\frac {7}{3}}} \,d x } \] Input:

integrate(sin(b*x+a)^(7/3)/cos(b*x+a)^(7/3),x, algorithm="giac")
 

Output:

integrate(sin(b*x + a)^(7/3)/cos(b*x + a)^(7/3), x)
 

Mupad [B] (verification not implemented)

Time = 27.03 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.28 \[ \int \frac {\sin ^{\frac {7}{3}}(a+b x)}{\cos ^{\frac {7}{3}}(a+b x)} \, dx=\frac {3\,{\sin \left (a+b\,x\right )}^{10/3}\,{{}}_2{\mathrm {F}}_1\left (-\frac {2}{3},-\frac {2}{3};\ \frac {1}{3};\ {\cos \left (a+b\,x\right )}^2\right )}{4\,b\,{\cos \left (a+b\,x\right )}^{4/3}\,{\left ({\sin \left (a+b\,x\right )}^2\right )}^{5/3}} \] Input:

int(sin(a + b*x)^(7/3)/cos(a + b*x)^(7/3),x)
 

Output:

(3*sin(a + b*x)^(10/3)*hypergeom([-2/3, -2/3], 1/3, cos(a + b*x)^2))/(4*b* 
cos(a + b*x)^(4/3)*(sin(a + b*x)^2)^(5/3))
 

Reduce [F]

\[ \int \frac {\sin ^{\frac {7}{3}}(a+b x)}{\cos ^{\frac {7}{3}}(a+b x)} \, dx=\int \frac {\sin \left (b x +a \right )^{\frac {7}{3}}}{\cos \left (b x +a \right )^{\frac {7}{3}}}d x \] Input:

int(sin(b*x+a)^(7/3)/cos(b*x+a)^(7/3),x)
 

Output:

int((sin(a + b*x)**(1/3)*sin(a + b*x)**2)/(cos(a + b*x)**(1/3)*cos(a + b*x 
)**2),x)