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

Optimal. Leaf size=41 \[ -\frac {\tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {3670, 1248, 725, 206} \[ -\frac {\tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]/Sqrt[a + b*Tan[x]^4],x]

[Out]

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

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

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Mathematica [A]  time = 0.02, size = 41, normalized size = 1.00 \[ -\frac {\tanh ^{-1}\left (\frac {a-b \tan ^2(x)}{\sqrt {a+b} \sqrt {a+b \tan ^4(x)}}\right )}{2 \sqrt {a+b}} \]

Antiderivative was successfully verified.

[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]

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fricas [A]  time = 0.64, size = 150, normalized size = 3.66 \[ \left [\frac {\log \left (\frac {{\left (a b + 2 \, b^{2}\right )} \tan \relax (x)^{4} - 2 \, a b \tan \relax (x)^{2} + 2 \, \sqrt {b \tan \relax (x)^{4} + a} {\left (b \tan \relax (x)^{2} - a\right )} \sqrt {a + b} + 2 \, a^{2} + a b}{\tan \relax (x)^{4} + 2 \, \tan \relax (x)^{2} + 1}\right )}{4 \, \sqrt {a + b}}, -\frac {\sqrt {-a - b} \arctan \left (\frac {\sqrt {b \tan \relax (x)^{4} + a} {\left (b \tan \relax (x)^{2} - a\right )} \sqrt {-a - b}}{{\left (a b + b^{2}\right )} \tan \relax (x)^{4} + a^{2} + a b}\right )}{2 \, {\left (a + b\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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giac [A]  time = 0.46, size = 46, normalized size = 1.12 \[ \frac {\arctan \left (-\frac {\sqrt {b} \tan \relax (x)^{2} - \sqrt {b \tan \relax (x)^{4} + a} + \sqrt {b}}{\sqrt {-a - b}}\right )}{\sqrt {-a - b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.23, size = 65, normalized size = 1.59 \[ -\frac {\ln \left (\frac {2 a +2 b -2 \left (1+\tan ^{2}\relax (x )\right ) b +2 \sqrt {a +b}\, \sqrt {\left (1+\tan ^{2}\relax (x )\right )^{2} b -2 \left (1+\tan ^{2}\relax (x )\right ) b +a +b}}{1+\tan ^{2}\relax (x )}\right )}{2 \sqrt {a +b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \relax (x)}{\sqrt {b \tan \relax (x)^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\mathrm {tan}\relax (x)}{\sqrt {b\,{\mathrm {tan}\relax (x)}^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a + b*tan(x)^4)^(1/2),x)

[Out]

int(tan(x)/(a + b*tan(x)^4)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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