\(\int \frac {\tan (x)}{\sqrt {a+b \tan ^4(x)}} \, dx\) [397]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [F]
   Giac [A] (verification not implemented)
   Mupad [F(-1)]

Optimal result

Integrand size = 15, antiderivative size = 41 \[ \int \frac {\tan (x)}{\sqrt {a+b \tan ^4(x)}} \, dx=-\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3751, 1262, 739, 212} \[ \int \frac {\tan (x)}{\sqrt {a+b \tan ^4(x)}} \, dx=-\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \]

[In]

Int[Tan[x]/Sqrt[a + b*Tan[x]^4],x]

[Out]

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

Rule 212

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

Rule 739

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 1262

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 3751

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*} \text {integral}& = \text {Subst}\left (\int \frac {x}{\left (1+x^2\right ) \sqrt {a+b x^4}} \, dx,x,\tan (x)\right ) \\ & = \frac {1}{2} \text {Subst}\left (\int \frac {1}{(1+x) \sqrt {a+b x^2}} \, dx,x,\tan ^2(x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {1}{a+b-x^2} \, dx,x,\frac {a-b \tan ^2(x)}{\sqrt {a+b \tan ^4(x)}}\right )\right ) \\ & = -\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (x)}{\sqrt {a+b \tan ^4(x)}} \, dx=-\frac {\text {arctanh}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \]

[In]

Integrate[Tan[x]/Sqrt[a + b*Tan[x]^4],x]

[Out]

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

Maple [A] (verified)

Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.59

method result size
derivativedivides \(-\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \sqrt {a +b}}\) \(65\)
default \(-\frac {\ln \left (\frac {2 a +2 b -2 b \left (1+\tan \left (x \right )^{2}\right )+2 \sqrt {a +b}\, \sqrt {b \left (1+\tan \left (x \right )^{2}\right )^{2}-2 b \left (1+\tan \left (x \right )^{2}\right )+a +b}}{1+\tan \left (x \right )^{2}}\right )}{2 \sqrt {a +b}}\) \(65\)

[In]

int(tan(x)/(a+b*tan(x)^4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2/(a+b)^(1/2)*ln((2*a+2*b-2*b*(1+tan(x)^2)+2*(a+b)^(1/2)*(b*(1+tan(x)^2)^2-2*b*(1+tan(x)^2)+a+b)^(1/2))/(1+
tan(x)^2))

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 150, normalized size of antiderivative = 3.66 \[ \int \frac {\tan (x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\left [\frac {\log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \tan \left (x\right )^{4} - 2 \, a b \tan \left (x\right )^{2} + 2 \, \sqrt {b \tan \left (x\right )^{4} + a} {\left (b \tan \left (x\right )^{2} - a\right )} \sqrt {a + b} + 2 \, a^{2} + a b}{\tan \left (x\right )^{4} + 2 \, \tan \left (x\right )^{2} + 1}\right )}{4 \, \sqrt {a + b}}, -\frac {\sqrt {-a - b} \arctan \left (\frac {\sqrt {b \tan \left (x\right )^{4} + a} {\left (b \tan \left (x\right )^{2} - a\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \tan \left (x\right )^{4} + a^{2} + a b}\right )}{2 \, {\left (a + b\right )}}\right ] \]

[In]

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

[Out]

[1/4*log(((a*b + 2*b^2)*tan(x)^4 - 2*a*b*tan(x)^2 + 2*sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 - a)*sqrt(a + b) + 2*a^
2 + a*b)/(tan(x)^4 + 2*tan(x)^2 + 1))/sqrt(a + b), -1/2*sqrt(-a - b)*arctan(sqrt(b*tan(x)^4 + a)*(b*tan(x)^2 -
 a)*sqrt(-a - b)/((a*b + b^2)*tan(x)^4 + a^2 + a*b))/(a + b)]

Sympy [F]

\[ \int \frac {\tan (x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\int \frac {\tan {\left (x \right )}}{\sqrt {a + b \tan ^{4}{\left (x \right )}}}\, dx \]

[In]

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

[Out]

Integral(tan(x)/sqrt(a + b*tan(x)**4), x)

Maxima [F]

\[ \int \frac {\tan (x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\int { \frac {\tan \left (x\right )}{\sqrt {b \tan \left (x\right )^{4} + a}} \,d x } \]

[In]

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

[Out]

integrate(tan(x)/sqrt(b*tan(x)^4 + a), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.12 \[ \int \frac {\tan (x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\frac {\arctan \left (-\frac {\sqrt {b} \tan \left (x\right )^{2} - \sqrt {b \tan \left (x\right )^{4} + a} + \sqrt {b}}{\sqrt {-a - b}}\right )}{\sqrt {-a - b}} \]

[In]

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

[Out]

arctan(-(sqrt(b)*tan(x)^2 - sqrt(b*tan(x)^4 + a) + sqrt(b))/sqrt(-a - b))/sqrt(-a - b)

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan (x)}{\sqrt {a+b \tan ^4(x)}} \, dx=\int \frac {\mathrm {tan}\left (x\right )}{\sqrt {b\,{\mathrm {tan}\left (x\right )}^4+a}} \,d x \]

[In]

int(tan(x)/(a + b*tan(x)^4)^(1/2),x)

[Out]

int(tan(x)/(a + b*tan(x)^4)^(1/2), x)