3.43 \(\int \frac {\tan (x)}{\sqrt {1+\tan ^4(x)}} \, dx\)

Optimal. Leaf size=34 \[ -\frac {\tanh ^{-1}\left (\frac {1-\tan ^2(x)}{\sqrt {2} \sqrt {\tan ^4(x)+1}}\right )}{2 \sqrt {2}} \]

[Out]

-1/4*arctanh(1/2*(1-tan(x)^2)*2^(1/2)/(1+tan(x)^4)^(1/2))*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.05, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {3670, 1248, 725, 206} \[ -\frac {\tanh ^{-1}\left (\frac {1-\tan ^2(x)}{\sqrt {2} \sqrt {\tan ^4(x)+1}}\right )}{2 \sqrt {2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/Sqrt[1 + Tan[x]^4],x]

[Out]

-ArcTanh[(1 - Tan[x]^2)/(Sqrt[2]*Sqrt[1 + Tan[x]^4])]/(2*Sqrt[2])

Rule 206

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

Rule 725

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

Rule 1248

Int[(x_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[(d + e*x)^q
*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x]

Rule 3670

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]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^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] && RationalQ[n]))

Rubi steps

\begin {align*} \int \frac {\tan (x)}{\sqrt {1+\tan ^4(x)}} \, dx &=\operatorname {Subst}\left (\int \frac {x}{\left (1+x^2\right ) \sqrt {1+x^4}} \, dx,x,\tan (x)\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{(1+x) \sqrt {1+x^2}} \, dx,x,\tan ^2(x)\right )\\ &=-\left (\frac {1}{2} \operatorname {Subst}\left (\int \frac {1}{2-x^2} \, dx,x,\frac {1-\tan ^2(x)}{\sqrt {1+\tan ^4(x)}}\right )\right )\\ &=-\frac {\tanh ^{-1}\left (\frac {1-\tan ^2(x)}{\sqrt {2} \sqrt {1+\tan ^4(x)}}\right )}{2 \sqrt {2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.07, size = 55, normalized size = 1.62 \[ -\frac {\sqrt {\cos (4 x)+3} \sec ^2(x) \log \left (\sqrt {2} \cos (2 x)+\sqrt {\cos (4 x)+3}\right )}{4 \sqrt {2} \sqrt {\tan ^4(x)+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]/Sqrt[1 + Tan[x]^4],x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.44, size = 186, normalized size = 5.47 \[ \frac {1}{32} \, \sqrt {2} \log \left (\frac {577 \, \tan \relax (x)^{16} - 1912 \, \tan \relax (x)^{14} + 4124 \, \tan \relax (x)^{12} - 6216 \, \tan \relax (x)^{10} + 7110 \, \tan \relax (x)^{8} - 6216 \, \tan \relax (x)^{6} + 4124 \, \tan \relax (x)^{4} - 1912 \, \tan \relax (x)^{2} + 8 \, {\left (51 \, \sqrt {2} \tan \relax (x)^{14} - 169 \, \sqrt {2} \tan \relax (x)^{12} + 339 \, \sqrt {2} \tan \relax (x)^{10} - 465 \, \sqrt {2} \tan \relax (x)^{8} + 465 \, \sqrt {2} \tan \relax (x)^{6} - 339 \, \sqrt {2} \tan \relax (x)^{4} + 169 \, \sqrt {2} \tan \relax (x)^{2} - 51 \, \sqrt {2}\right )} \sqrt {\tan \relax (x)^{4} + 1} + 577}{\tan \relax (x)^{16} + 8 \, \tan \relax (x)^{14} + 28 \, \tan \relax (x)^{12} + 56 \, \tan \relax (x)^{10} + 70 \, \tan \relax (x)^{8} + 56 \, \tan \relax (x)^{6} + 28 \, \tan \relax (x)^{4} + 8 \, \tan \relax (x)^{2} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+tan(x)^4)^(1/2),x, algorithm="fricas")

[Out]

1/32*sqrt(2)*log((577*tan(x)^16 - 1912*tan(x)^14 + 4124*tan(x)^12 - 6216*tan(x)^10 + 7110*tan(x)^8 - 6216*tan(
x)^6 + 4124*tan(x)^4 - 1912*tan(x)^2 + 8*(51*sqrt(2)*tan(x)^14 - 169*sqrt(2)*tan(x)^12 + 339*sqrt(2)*tan(x)^10
- 465*sqrt(2)*tan(x)^8 + 465*sqrt(2)*tan(x)^6 - 339*sqrt(2)*tan(x)^4 + 169*sqrt(2)*tan(x)^2 - 51*sqrt(2))*sqr
t(tan(x)^4 + 1) + 577)/(tan(x)^16 + 8*tan(x)^14 + 28*tan(x)^12 + 56*tan(x)^10 + 70*tan(x)^8 + 56*tan(x)^6 + 28
*tan(x)^4 + 8*tan(x)^2 + 1))

________________________________________________________________________________________

giac [A]  time = 1.06, size = 50, normalized size = 1.47 \[ \frac {1}{4} \, \sqrt {2} \log \left (-\frac {\tan \relax (x)^{2} + \sqrt {2} - \sqrt {\tan \relax (x)^{4} + 1} + 1}{\tan \relax (x)^{2} - \sqrt {2} - \sqrt {\tan \relax (x)^{4} + 1} + 1}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+tan(x)^4)^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt(2)*log(-(tan(x)^2 + sqrt(2) - sqrt(tan(x)^4 + 1) + 1)/(tan(x)^2 - sqrt(2) - sqrt(tan(x)^4 + 1) + 1))

________________________________________________________________________________________

maple [A]  time = 0.06, size = 37, normalized size = 1.09 \[ -\frac {\sqrt {2}\, \arctanh \left (\frac {\left (-2 \left (\tan ^{2}\relax (x )\right )+2\right ) \sqrt {2}}{4 \sqrt {-2 \left (\tan ^{2}\relax (x )\right )+\left (\tan ^{2}\relax (x )+1\right )^{2}}}\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(1+tan(x)^4)^(1/2),x)

[Out]

-1/4*2^(1/2)*arctanh(1/4*(-2*tan(x)^2+2)*2^(1/2)/(-2*tan(x)^2+(tan(x)^2+1)^2)^(1/2))

________________________________________________________________________________________

maxima [B]  time = 1.35, size = 565, normalized size = 16.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+tan(x)^4)^(1/2),x, algorithm="maxima")

[Out]

-1/16*sqrt(2)*(log(4*sqrt(2*(6*cos(4*x) + 1)*cos(8*x) + cos(8*x)^2 + 36*cos(4*x)^2 + sin(8*x)^2 + 12*sin(8*x)*
sin(4*x) + 36*sin(4*x)^2 + 12*cos(4*x) + 1)*cos(1/2*arctan2(sin(8*x) + 6*sin(4*x), cos(8*x) + 6*cos(4*x) + 1))
^2 + 4*sqrt(2*(6*cos(4*x) + 1)*cos(8*x) + cos(8*x)^2 + 36*cos(4*x)^2 + sin(8*x)^2 + 12*sin(8*x)*sin(4*x) + 36*
sin(4*x)^2 + 12*cos(4*x) + 1)*sin(1/2*arctan2(sin(8*x) + 6*sin(4*x), cos(8*x) + 6*cos(4*x) + 1))^2 + 32*(2*(6*
cos(4*x) + 1)*cos(8*x) + cos(8*x)^2 + 36*cos(4*x)^2 + sin(8*x)^2 + 12*sin(8*x)*sin(4*x) + 36*sin(4*x)^2 + 12*c
os(4*x) + 1)^(1/4)*cos(1/2*arctan2(sin(8*x) + 6*sin(4*x), cos(8*x) + 6*cos(4*x) + 1)) + 64) + log(4*cos(4*x)^2
+ 4*sin(4*x)^2 + 4*sqrt(2*(6*cos(4*x) + 1)*cos(8*x) + cos(8*x)^2 + 36*cos(4*x)^2 + sin(8*x)^2 + 12*sin(8*x)*s
in(4*x) + 36*sin(4*x)^2 + 12*cos(4*x) + 1)*(cos(1/2*arctan2(sin(8*x) + 6*sin(4*x), cos(8*x) + 6*cos(4*x) + 1))
^2 + sin(1/2*arctan2(sin(8*x) + 6*sin(4*x), cos(8*x) + 6*cos(4*x) + 1))^2) + 8*(2*(6*cos(4*x) + 1)*cos(8*x) +
cos(8*x)^2 + 36*cos(4*x)^2 + sin(8*x)^2 + 12*sin(8*x)*sin(4*x) + 36*sin(4*x)^2 + 12*cos(4*x) + 1)^(1/4)*((cos(
4*x) + 3)*cos(1/2*arctan2(sin(8*x) + 6*sin(4*x), cos(8*x) + 6*cos(4*x) + 1)) + sin(4*x)*sin(1/2*arctan2(sin(8*
x) + 6*sin(4*x), cos(8*x) + 6*cos(4*x) + 1))) + 24*cos(4*x) + 36))

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\mathrm {tan}\relax (x)}{\sqrt {{\mathrm {tan}\relax (x)}^4+1}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(tan(x)^4 + 1)^(1/2),x)

[Out]

int(tan(x)/(tan(x)^4 + 1)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan {\relax (x )}}{\sqrt {\tan ^{4}{\relax (x )} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(x)/(1+tan(x)**4)**(1/2),x)

[Out]

Integral(tan(x)/sqrt(tan(x)**4 + 1), x)

________________________________________________________________________________________