3.1.96 \(\int \frac {\tan (x)}{\sqrt {a+b \cos ^4(x)}} \, dx\) [96]

Optimal. Leaf size=28 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^4(x)}}{\sqrt {a}}\right )}{2 \sqrt {a}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 28, 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 = {3308, 272, 65, 214} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^4(x)}}{\sqrt {a}}\right )}{2 \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[x]/Sqrt[a + b*Cos[x]^4],x]

[Out]

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 214

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 3308

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[x^((m - 1)/2)*((a + b*ff^(n/2)*x^(n/2))^p
/(1 - ff*x)^((m + 1)/2)), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2] &
& IntegerQ[n/2]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 28, normalized size = 1.00 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^4(x)}}{\sqrt {a}}\right )}{2 \sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.06, size = 31, normalized size = 1.11

method	result	size
derivativedivides	\(\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\cos ^{4}\left (x \right )\right )}}{\cos \left (x \right )^{2}}\right )}{2 \sqrt {a}}\)	\(31\)
default	\(\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\cos ^{4}\left (x \right )\right )}}{\cos \left (x \right )^{2}}\right )}{2 \sqrt {a}}\)	\(31\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a^(1/2)*ln((2*a+2*a^(1/2)*(a+b*cos(x)^4)^(1/2))/cos(x)^2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.45, size = 67, normalized size = 2.39 \begin {gather*} \left [\frac {\log \left (\frac {b \cos \left (x\right )^{4} + 2 \, \sqrt {b \cos \left (x\right )^{4} + a} \sqrt {a} + 2 \, a}{\cos \left (x\right )^{4}}\right )}{4 \, \sqrt {a}}, -\frac {\sqrt {-a} \arctan \left (\frac {\sqrt {b \cos \left (x\right )^{4} + a} \sqrt {-a}}{a}\right )}{2 \, a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\tan {\left (x \right )}}{\sqrt {a + b \cos ^{4}{\left (x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.42, size = 24, normalized size = 0.86 \begin {gather*} -\frac {\arctan \left (\frac {\sqrt {b \cos \left (x\right )^{4} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*arctan(sqrt(b*cos(x)^4 + a)/sqrt(-a))/sqrt(-a)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int \frac {\mathrm {tan}\left (x\right )}{\sqrt {b\,{\cos \left (x\right )}^4+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a + b*cos(x)^4)^(1/2),x)

[Out]

int(tan(x)/(a + b*cos(x)^4)^(1/2), x)

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