\(\int \tan (x) (1-7 \tan ^2(x))^{2/3} \, dx\) [443]

Optimal result

Integrand size = 15, antiderivative size = 69 \[ \int \tan (x) \left (1-7 \tan ^2(x)\right )^{2/3} \, dx=2 \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{1-7 \tan ^2(x)}}{\sqrt {3}}\right )+2 \log (\cos (x))+3 \log \left (2-\sqrt [3]{1-7 \tan ^2(x)}\right )+\frac {3}{4} \left (1-7 \tan ^2(x)\right )^{2/3} \] Output:

2*ln(cos(x))+3*ln(2-(1-7*tan(x)^2)^(1/3))+2*arctan(1/3*(1+(1-7*tan(x)^2)^( 
1/3))*3^(1/2))*3^(1/2)+3/4*(1-7*tan(x)^2)^(2/3)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \tan (x) \left (1-7 \tan ^2(x)\right )^{2/3} \, dx=2 \sqrt {3} \arctan \left (\frac {1+\sqrt [3]{1-7 \tan ^2(x)}}{\sqrt {3}}\right )+2 \log (\cos (x))+3 \log \left (2-\sqrt [3]{1-7 \tan ^2(x)}\right )+\frac {3}{4} \left (1-7 \tan ^2(x)\right )^{2/3} \] Input:

Integrate[Tan[x]*(1 - 7*Tan[x]^2)^(2/3),x]
 

Output:

2*Sqrt[3]*ArcTan[(1 + (1 - 7*Tan[x]^2)^(1/3))/Sqrt[3]] + 2*Log[Cos[x]] + 3 
*Log[2 - (1 - 7*Tan[x]^2)^(1/3)] + (3*(1 - 7*Tan[x]^2)^(2/3))/4
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.32, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {3042, 4153, 353, 60, 67, 16, 1083, 217}

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.

\(\displaystyle \int \tan (x) \left (1-7 \tan ^2(x)\right )^{2/3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \tan (x) \left (1-7 \tan (x)^2\right )^{2/3}dx\)

\(\Big \downarrow \) 4153

\(\displaystyle \int \frac {\tan (x) \left (1-7 \tan ^2(x)\right )^{2/3}}{\tan ^2(x)+1}d\tan (x)\)

\(\Big \downarrow \) 353

\(\displaystyle \frac {1}{2} \int \frac {\left (1-7 \tan ^2(x)\right )^{2/3}}{\tan ^2(x)+1}d\tan ^2(x)\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {1}{2} \left (8 \int \frac {1}{\sqrt [3]{1-7 \tan ^2(x)} \left (\tan ^2(x)+1\right )}d\tan ^2(x)+\frac {3}{2} \left (1-7 \tan ^2(x)\right )^{2/3}\right )\)

\(\Big \downarrow \) 67

\(\displaystyle \frac {1}{2} \left (8 \left (-\frac {3}{4} \int \frac {1}{2-\sqrt [3]{1-7 \tan ^2(x)}}d\sqrt [3]{1-7 \tan ^2(x)}+\frac {3}{2} \int \frac {1}{\tan ^4(x)+2 \sqrt [3]{1-7 \tan ^2(x)}+4}d\sqrt [3]{1-7 \tan ^2(x)}-\frac {1}{4} \log \left (\tan ^2(x)+1\right )\right )+\frac {3}{2} \left (1-7 \tan ^2(x)\right )^{2/3}\right )\)

\(\Big \downarrow \) 16

\(\displaystyle \frac {1}{2} \left (8 \left (\frac {3}{2} \int \frac {1}{\tan ^4(x)+2 \sqrt [3]{1-7 \tan ^2(x)}+4}d\sqrt [3]{1-7 \tan ^2(x)}-\frac {1}{4} \log \left (\tan ^2(x)+1\right )+\frac {3}{4} \log \left (2-\sqrt [3]{1-7 \tan ^2(x)}\right )\right )+\frac {3}{2} \left (1-7 \tan ^2(x)\right )^{2/3}\right )\)

\(\Big \downarrow \) 1083

\(\displaystyle \frac {1}{2} \left (8 \left (-3 \int \frac {1}{-\tan ^4(x)-12}d\left (2 \sqrt [3]{1-7 \tan ^2(x)}+2\right )-\frac {1}{4} \log \left (\tan ^2(x)+1\right )+\frac {3}{4} \log \left (2-\sqrt [3]{1-7 \tan ^2(x)}\right )\right )+\frac {3}{2} \left (1-7 \tan ^2(x)\right )^{2/3}\right )\)

\(\Big \downarrow \) 217

\(\displaystyle \frac {1}{2} \left (8 \left (\frac {1}{2} \sqrt {3} \arctan \left (\frac {2 \sqrt [3]{1-7 \tan ^2(x)}+2}{2 \sqrt {3}}\right )-\frac {1}{4} \log \left (\tan ^2(x)+1\right )+\frac {3}{4} \log \left (2-\sqrt [3]{1-7 \tan ^2(x)}\right )\right )+\frac {3}{2} \left (1-7 \tan ^2(x)\right )^{2/3}\right )\)

Input:

Int[Tan[x]*(1 - 7*Tan[x]^2)^(2/3),x]
 

Output:

(8*((Sqrt[3]*ArcTan[(2 + 2*(1 - 7*Tan[x]^2)^(1/3))/(2*Sqrt[3])])/2 - Log[1 
 + Tan[x]^2]/4 + (3*Log[2 - (1 - 7*Tan[x]^2)^(1/3)])/4) + (3*(1 - 7*Tan[x] 
^2)^(2/3))/2)/2
 

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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 67

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(1/3)), x_Symbol] :> With[ 
{q = Rt[(b*c - a*d)/b, 3]}, Simp[-Log[RemoveContent[a + b*x, x]]/(2*b*q), x 
] + (Simp[3/(2*b)   Subst[Int[1/(q^2 + q*x + x^2), x], x, (c + d*x)^(1/3)], 
 x] - Simp[3/(2*b*q)   Subst[Int[1/(q - x), x], x, (c + d*x)^(1/3)], x])] / 
; FreeQ[{a, b, c, d}, x] && PosQ[(b*c - a*d)/b]
 

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 353

Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_Symbol] 
 :> Simp[1/2   Subst[Int[(a + b*x)^p*(c + d*x)^q, x], x, x^2], x] /; FreeQ[ 
{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0]
 

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 3042

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

rule 4153

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], 
 x]}, Simp[c*(ff/f)   Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2 + f 
f^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, 
n, p}, x] && (IGtQ[p, 0] || EqQ[n, 2] || EqQ[n, 4] || (IntegerQ[p] && Ratio 
nalQ[n]))
 

Maple [F]

\[\int \tan \left (x \right ) \left (1-7 \tan \left (x \right )^{2}\right )^{\frac {2}{3}}d x\]

Input:

int(tan(x)*(1-7*tan(x)^2)^(2/3),x)
 

Output:

int(tan(x)*(1-7*tan(x)^2)^(2/3),x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (58) = 116\).

Time = 0.40 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.70 \[ \int \tan (x) \left (1-7 \tan ^2(x)\right )^{2/3} \, dx=2 \, \sqrt {3} \arctan \left (\frac {7 \, \sqrt {3} \tan \left (x\right )^{2} + 4 \, \sqrt {3} {\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {2}{3}} - 16 \, \sqrt {3} {\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {1}{3}} - \sqrt {3}}{7 \, \tan \left (x\right )^{2} - 65}\right ) + \frac {3}{4} \, {\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {2}{3}} + \log \left (\frac {7 \, \tan \left (x\right )^{2} + 6 \, {\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {2}{3}} - 12 \, {\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {1}{3}} + 7}{\tan \left (x\right )^{2} + 1}\right ) \] Input:

integrate(tan(x)*(1-7*tan(x)^2)^(2/3),x, algorithm="fricas")
 

Output:

2*sqrt(3)*arctan((7*sqrt(3)*tan(x)^2 + 4*sqrt(3)*(-7*tan(x)^2 + 1)^(2/3) - 
 16*sqrt(3)*(-7*tan(x)^2 + 1)^(1/3) - sqrt(3))/(7*tan(x)^2 - 65)) + 3/4*(- 
7*tan(x)^2 + 1)^(2/3) + log((7*tan(x)^2 + 6*(-7*tan(x)^2 + 1)^(2/3) - 12*( 
-7*tan(x)^2 + 1)^(1/3) + 7)/(tan(x)^2 + 1))
 

Sympy [F]

\[ \int \tan (x) \left (1-7 \tan ^2(x)\right )^{2/3} \, dx=\int \left (1 - 7 \tan ^{2}{\left (x \right )}\right )^{\frac {2}{3}} \tan {\left (x \right )}\, dx \] Input:

integrate(tan(x)*(1-7*tan(x)**2)**(2/3),x)
 

Output:

Integral((1 - 7*tan(x)**2)**(2/3)*tan(x), x)
 

Maxima [F]

\[ \int \tan (x) \left (1-7 \tan ^2(x)\right )^{2/3} \, dx=\int { {\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {2}{3}} \tan \left (x\right ) \,d x } \] Input:

integrate(tan(x)*(1-7*tan(x)^2)^(2/3),x, algorithm="maxima")
 

Output:

integrate((-7*tan(x)^2 + 1)^(2/3)*tan(x), x)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14 \[ \int \tan (x) \left (1-7 \tan ^2(x)\right )^{2/3} \, dx=2 \, \sqrt {3} \arctan \left (\frac {1}{3} \, \sqrt {3} {\left ({\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {1}{3}} + 1\right )}\right ) + \frac {3}{4} \, {\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {2}{3}} - \log \left ({\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {2}{3}} + 2 \, {\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {1}{3}} + 4\right ) + 2 \, \log \left ({\left | {\left (-7 \, \tan \left (x\right )^{2} + 1\right )}^{\frac {1}{3}} - 2 \right |}\right ) \] Input:

integrate(tan(x)*(1-7*tan(x)^2)^(2/3),x, algorithm="giac")
 

Output:

2*sqrt(3)*arctan(1/3*sqrt(3)*((-7*tan(x)^2 + 1)^(1/3) + 1)) + 3/4*(-7*tan( 
x)^2 + 1)^(2/3) - log((-7*tan(x)^2 + 1)^(2/3) + 2*(-7*tan(x)^2 + 1)^(1/3) 
+ 4) + 2*log(abs((-7*tan(x)^2 + 1)^(1/3) - 2))
 

Mupad [B] (verification not implemented)

Time = 0.41 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.46 \[ \int \tan (x) \left (1-7 \tan ^2(x)\right )^{2/3} \, dx=2\,\ln \left (144\,{\left (1-7\,{\mathrm {tan}\left (x\right )}^2\right )}^{1/3}-288\right )+\frac {3\,{\left (1-7\,{\mathrm {tan}\left (x\right )}^2\right )}^{2/3}}{4}+\ln \left (144\,{\left (1-7\,{\mathrm {tan}\left (x\right )}^2\right )}^{1/3}-72\,{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )-\ln \left (144\,{\left (1-7\,{\mathrm {tan}\left (x\right )}^2\right )}^{1/3}-72\,{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right ) \] Input:

int(tan(x)*(1 - 7*tan(x)^2)^(2/3),x)
 

Output:

2*log(144*(1 - 7*tan(x)^2)^(1/3) - 288) + (3*(1 - 7*tan(x)^2)^(2/3))/4 + l 
og(144*(1 - 7*tan(x)^2)^(1/3) - 72*(3^(1/2)*1i - 1)^2)*(3^(1/2)*1i - 1) - 
log(144*(1 - 7*tan(x)^2)^(1/3) - 72*(3^(1/2)*1i + 1)^2)*(3^(1/2)*1i + 1)
 

Reduce [F]

\[ \int \tan (x) \left (1-7 \tan ^2(x)\right )^{2/3} \, dx=-\frac {3 \left (-7 \tan \left (x \right )^{2}+1\right )^{\frac {2}{3}}}{28}+8 \left (\int \frac {\left (-7 \tan \left (x \right )^{2}+1\right )^{\frac {2}{3}} \tan \left (x \right )^{3}}{7 \tan \left (x \right )^{2}-1}d x \right ) \] Input:

int(tan(x)*(1-7*tan(x)^2)^(2/3),x)
 

Output:

( - 3*( - 7*tan(x)**2 + 1)**(2/3) + 224*int((( - 7*tan(x)**2 + 1)**(2/3)*t 
an(x)**3)/(7*tan(x)**2 - 1),x))/28