3.855 \(\int \frac {\sin ^3(x)}{\cos ^3(x)+\sin ^3(x)} \, dx\)

Optimal. Leaf size=29 \[ \frac {x}{2}+\frac {1}{3} \log (2-\sin (2 x))-\frac {1}{6} \log (\sin (x)+\cos (x)) \]

[Out]

1/2*x-1/6*ln(cos(x)+sin(x))+1/3*ln(2-sin(2*x))

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Rubi [A]  time = 0.13, antiderivative size = 37, normalized size of antiderivative = 1.28, number of steps used = 7, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2074, 635, 203, 260, 628} \[ \frac {x}{2}+\frac {1}{3} \log \left (\tan ^2(x)-\tan (x)+1\right )-\frac {1}{6} \log (\tan (x)+1)+\frac {1}{2} \log (\cos (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3/(Cos[x]^3 + Sin[x]^3),x]

[Out]

x/2 + Log[Cos[x]]/2 - Log[1 + Tan[x]]/6 + Log[1 - Tan[x] + Tan[x]^2]/3

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(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 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 2074

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Q^q, x], x] /;  !SumQ[N
onfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x] && PolyQ[Q, x] && IntegerQ[p] && NeQ[P, x]

Rubi steps

\begin {align*} \int \frac {\sin ^3(x)}{\cos ^3(x)+\sin ^3(x)} \, dx &=\operatorname {Subst}\left (\int \frac {x^3}{1+x^2+x^3+x^5} \, dx,x,\tan (x)\right )\\ &=\operatorname {Subst}\left (\int \left (-\frac {1}{6 (1+x)}+\frac {1-x}{2 \left (1+x^2\right )}+\frac {-1+2 x}{3 \left (1-x+x^2\right )}\right ) \, dx,x,\tan (x)\right )\\ &=-\frac {1}{6} \log (1+\tan (x))+\frac {1}{3} \operatorname {Subst}\left (\int \frac {-1+2 x}{1-x+x^2} \, dx,x,\tan (x)\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1-x}{1+x^2} \, dx,x,\tan (x)\right )\\ &=-\frac {1}{6} \log (1+\tan (x))+\frac {1}{3} \log \left (1-\tan (x)+\tan ^2(x)\right )+\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (x)\right )-\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{1+x^2} \, dx,x,\tan (x)\right )\\ &=\frac {x}{2}+\frac {1}{2} \log (\cos (x))-\frac {1}{6} \log (1+\tan (x))+\frac {1}{3} \log \left (1-\tan (x)+\tan ^2(x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 29, normalized size = 1.00 \[ \frac {x}{2}+\frac {1}{3} \log (2-\sin (2 x))-\frac {1}{6} \log (\sin (x)+\cos (x)) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^3/(Cos[x]^3 + Sin[x]^3),x]

[Out]

x/2 - Log[Cos[x] + Sin[x]]/6 + Log[2 - Sin[2*x]]/3

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fricas [A]  time = 2.01, size = 26, normalized size = 0.90 \[ \frac {1}{2} \, x - \frac {1}{12} \, \log \left (2 \, \cos \relax (x) \sin \relax (x) + 1\right ) + \frac {1}{3} \, \log \left (-\cos \relax (x) \sin \relax (x) + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3/(cos(x)^3+sin(x)^3),x, algorithm="fricas")

[Out]

1/2*x - 1/12*log(2*cos(x)*sin(x) + 1) + 1/3*log(-cos(x)*sin(x) + 1)

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giac [A]  time = 0.17, size = 34, normalized size = 1.17 \[ \frac {1}{2} \, x + \frac {1}{3} \, \log \left (\tan \relax (x)^{2} - \tan \relax (x) + 1\right ) - \frac {1}{4} \, \log \left (\tan \relax (x)^{2} + 1\right ) - \frac {1}{6} \, \log \left ({\left | \tan \relax (x) + 1 \right |}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3/(cos(x)^3+sin(x)^3),x, algorithm="giac")

[Out]

1/2*x + 1/3*log(tan(x)^2 - tan(x) + 1) - 1/4*log(tan(x)^2 + 1) - 1/6*log(abs(tan(x) + 1))

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maple [A]  time = 0.14, size = 34, normalized size = 1.17 \[ \frac {\ln \left (1-\tan \relax (x )+\tan ^{2}\relax (x )\right )}{3}-\frac {\ln \left (1+\tan ^{2}\relax (x )\right )}{4}-\frac {\ln \left (1+\tan \relax (x )\right )}{6}+\frac {x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(cos(x)^3+sin(x)^3),x)

[Out]

1/3*ln(1-tan(x)+tan(x)^2)-1/4*ln(1+tan(x)^2)-1/6*ln(1+tan(x))+1/2*x

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maxima [B]  time = 0.41, size = 103, normalized size = 3.55 \[ \arctan \left (\frac {\sin \relax (x)}{\cos \relax (x) + 1}\right ) + \frac {1}{3} \, \log \left (-\frac {2 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {2 \, \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {2 \, \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}} + \frac {\sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + 1\right ) - \frac {1}{6} \, \log \left (-\frac {2 \, \sin \relax (x)}{\cos \relax (x) + 1} + \frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} - 1\right ) - \frac {1}{2} \, \log \left (\frac {\sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + 1\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3/(cos(x)^3+sin(x)^3),x, algorithm="maxima")

[Out]

arctan(sin(x)/(cos(x) + 1)) + 1/3*log(-2*sin(x)/(cos(x) + 1) + 2*sin(x)^2/(cos(x) + 1)^2 + 2*sin(x)^3/(cos(x)
+ 1)^3 + sin(x)^4/(cos(x) + 1)^4 + 1) - 1/6*log(-2*sin(x)/(cos(x) + 1) + sin(x)^2/(cos(x) + 1)^2 - 1) - 1/2*lo
g(sin(x)^2/(cos(x) + 1)^2 + 1)

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mupad [B]  time = 3.31, size = 45, normalized size = 1.55 \[ \frac {x}{2}-\frac {\ln \left (\frac {1}{{\cos \left (\frac {x}{2}\right )}^2}\right )}{2}+\frac {\ln \left (\frac {\sin \left (2\,x\right )-2}{{\cos \left (\frac {x}{2}\right )}^4}\right )}{3}-\frac {\ln \left (\frac {\sin \left (x+\frac {\pi }{4}\right )}{{\cos \left (\frac {x}{2}\right )}^2}\right )}{6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(cos(x)^3 + sin(x)^3),x)

[Out]

x/2 - log(1/cos(x/2)^2)/2 + log((sin(2*x) - 2)/cos(x/2)^4)/3 - log(sin(x + pi/4)/cos(x/2)^2)/6

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sympy [A]  time = 0.37, size = 32, normalized size = 1.10 \[ \frac {x}{2} - \frac {\log {\left (\sin {\relax (x )} + \cos {\relax (x )} \right )}}{6} + \frac {\log {\left (\sin ^{2}{\relax (x )} - \sin {\relax (x )} \cos {\relax (x )} + \cos ^{2}{\relax (x )} \right )}}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**3/(cos(x)**3+sin(x)**3),x)

[Out]

x/2 - log(sin(x) + cos(x))/6 + log(sin(x)**2 - sin(x)*cos(x) + cos(x)**2)/3

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