\(\int \frac {\tan (x)}{\sqrt [3]{a^3+b^3 \tan ^2(x)}} \, dx\) [442]

Optimal result

Integrand size = 19, antiderivative size = 133 \[ \int \frac {\tan (x)}{\sqrt [3]{a^3+b^3 \tan ^2(x)}} \, dx=\frac {\sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a^3+b^3 \tan ^2(x)}}{\sqrt [3]{a^3-b^3}}}{\sqrt {3}}\right )}{2 \sqrt [3]{a^3-b^3}}+\frac {\log (\cos (x))}{2 \sqrt [3]{a^3-b^3}}+\frac {3 \log \left (\sqrt [3]{a^3-b^3}-\sqrt [3]{a^3+b^3 \tan ^2(x)}\right )}{4 \sqrt [3]{a^3-b^3}} \] Output:

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

Mathematica [A] (verified)

Time = 0.10 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.79 \[ \int \frac {\tan (x)}{\sqrt [3]{a^3+b^3 \tan ^2(x)}} \, dx=\frac {2 \sqrt {3} \arctan \left (\frac {1+\frac {2 \sqrt [3]{a^3+b^3 \tan ^2(x)}}{\sqrt [3]{a^3-b^3}}}{\sqrt {3}}\right )+2 \log (\cos (x))+3 \log \left (\sqrt [3]{a^3-b^3}-\sqrt [3]{a^3+b^3 \tan ^2(x)}\right )}{4 \sqrt [3]{a^3-b^3}} \] Input:

Integrate[Tan[x]/(a^3 + b^3*Tan[x]^2)^(1/3),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 4153, 353, 67, 16, 1082, 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.

\(\displaystyle \int \frac {\tan (x)}{\sqrt [3]{a^3+b^3 \tan ^2(x)}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4153

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

\(\Big \downarrow \) 353

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

\(\Big \downarrow \) 67

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

\(\Big \downarrow \) 16

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

\(\Big \downarrow \) 1082

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

\(\Big \downarrow \) 217

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

Input:

Int[Tan[x]/(a^3 + b^3*Tan[x]^2)^(1/3),x]
 

Output:

((Sqrt[3]*ArcTan[(1 + (2*(a^3 + b^3*Tan[x]^2)^(1/3))/(a^3 - b^3)^(1/3))/Sq 
rt[3]])/(a^3 - b^3)^(1/3) - Log[1 + Tan[x]^2]/(2*(a^3 - b^3)^(1/3)) + (3*L 
og[(a^3 - b^3)^(1/3) - (a^3 + b^3*Tan[x]^2)^(1/3)])/(2*(a^3 - b^3)^(1/3))) 
/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 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 1082

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S 
implify[a*(c/b^2)]}, Simp[-2/b   Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b 
)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /; Fre 
eQ[{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]

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

Input:

int(tan(x)/(a^3+b^3*tan(x)^2)^(1/3),x)
 

Output:

int(tan(x)/(a^3+b^3*tan(x)^2)^(1/3),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\tan (x)}{\sqrt [3]{a^3+b^3 \tan ^2(x)}} \, dx=\text {Timed out} \] Input:

integrate(tan(x)/(a^3+b^3*tan(x)^2)^(1/3),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate(tan(x)/(a**3+b**3*tan(x)**2)**(1/3),x)
 

Output:

Integral(tan(x)/(a**3 + b**3*tan(x)**2)**(1/3), x)
 

Maxima [F]

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

integrate(tan(x)/(a^3+b^3*tan(x)^2)^(1/3),x, algorithm="maxima")
 

Output:

integrate(tan(x)/(b^3*tan(x)^2 + a^3)^(1/3), x)
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.40 \[ \int \frac {\tan (x)}{\sqrt [3]{a^3+b^3 \tan ^2(x)}} \, dx=\frac {3 \, {\left (a^{3} - b^{3}\right )}^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, {\left (b^{3} \tan \left (x\right )^{2} + a^{3}\right )}^{\frac {1}{3}} + {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}}\right )}}{3 \, {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}}}\right )}{2 \, {\left (\sqrt {3} a^{3} - \sqrt {3} b^{3}\right )}} - \frac {\log \left ({\left (b^{3} \tan \left (x\right )^{2} + a^{3}\right )}^{\frac {2}{3}} + {\left (b^{3} \tan \left (x\right )^{2} + a^{3}\right )}^{\frac {1}{3}} {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}} + {\left (a^{3} - b^{3}\right )}^{\frac {2}{3}}\right )}{4 \, {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}}} + \frac {\log \left ({\left | {\left (b^{3} \tan \left (x\right )^{2} + a^{3}\right )}^{\frac {1}{3}} - {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}} \right |}\right )}{2 \, {\left (a^{3} - b^{3}\right )}^{\frac {1}{3}}} \] Input:

integrate(tan(x)/(a^3+b^3*tan(x)^2)^(1/3),x, algorithm="giac")
 

Output:

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

Mupad [B] (verification not implemented)

Time = 0.75 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.88 \[ \int \frac {\tan (x)}{\sqrt [3]{a^3+b^3 \tan ^2(x)}} \, dx=\frac {\ln \left (\frac {9\,{\left (a^3+b^3\,{\mathrm {tan}\left (x\right )}^2\right )}^{1/3}}{4}-\frac {9\,a^3-9\,b^3}{4\,{\left (a-b\right )}^{2/3}\,{\left (a^2+a\,b+b^2\right )}^{2/3}}\right )}{2\,{\left (a-b\right )}^{1/3}\,{\left (a^2+a\,b+b^2\right )}^{1/3}}+\frac {\ln \left (\frac {9\,{\left (a^3+b^3\,{\mathrm {tan}\left (x\right )}^2\right )}^{1/3}}{4}-\frac {{\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (9\,a^3-9\,b^3\right )}{16\,{\left (a-b\right )}^{2/3}\,{\left (a^2+a\,b+b^2\right )}^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,{\left (a-b\right )}^{1/3}\,{\left (a^2+a\,b+b^2\right )}^{1/3}}-\frac {\ln \left (\frac {9\,{\left (a^3+b^3\,{\mathrm {tan}\left (x\right )}^2\right )}^{1/3}}{4}-\frac {{\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}^2\,\left (9\,a^3-9\,b^3\right )}{16\,{\left (a-b\right )}^{2/3}\,{\left (a^2+a\,b+b^2\right )}^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{4\,{\left (a-b\right )}^{1/3}\,{\left (a^2+a\,b+b^2\right )}^{1/3}} \] Input:

int(tan(x)/(b^3*tan(x)^2 + a^3)^(1/3),x)
 

Output:

log((9*(b^3*tan(x)^2 + a^3)^(1/3))/4 - (9*a^3 - 9*b^3)/(4*(a - b)^(2/3)*(a 
*b + a^2 + b^2)^(2/3)))/(2*(a - b)^(1/3)*(a*b + a^2 + b^2)^(1/3)) + (log(( 
9*(b^3*tan(x)^2 + a^3)^(1/3))/4 - ((3^(1/2)*1i - 1)^2*(9*a^3 - 9*b^3))/(16 
*(a - b)^(2/3)*(a*b + a^2 + b^2)^(2/3)))*(3^(1/2)*1i - 1))/(4*(a - b)^(1/3 
)*(a*b + a^2 + b^2)^(1/3)) - (log((9*(b^3*tan(x)^2 + a^3)^(1/3))/4 - ((3^( 
1/2)*1i + 1)^2*(9*a^3 - 9*b^3))/(16*(a - b)^(2/3)*(a*b + a^2 + b^2)^(2/3)) 
)*(3^(1/2)*1i + 1))/(4*(a - b)^(1/3)*(a*b + a^2 + b^2)^(1/3))
 

Reduce [F]

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

int(tan(x)/(a^3+b^3*tan(x)^2)^(1/3),x)
 

Output:

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