3.1.95 \(\int \sqrt {a+b \cos ^4(x)} \tan (x) \, dx\) [95]

Optimal. Leaf size=45 \[ \frac {1}{2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^4(x)}}{\sqrt {a}}\right )-\frac {1}{2} \sqrt {a+b \cos ^4(x)} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3308, 272, 52, 65, 214} \begin {gather*} \frac {1}{2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^4(x)}}{\sqrt {a}}\right )-\frac {1}{2} \sqrt {a+b \cos ^4(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Cos[x]^4]*Tan[x],x]

[Out]

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

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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

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Mathematica [A]
time = 0.04, size = 45, normalized size = 1.00 \begin {gather*} \frac {1}{2} \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^4(x)}}{\sqrt {a}}\right )-\frac {1}{2} \sqrt {a+b \cos ^4(x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.07, size = 44, normalized size = 0.98

method	result	size
derivativedivides	\(-\frac {\sqrt {a +b \left (\cos ^{4}\left (x \right )\right )}}{2}+\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\cos ^{4}\left (x \right )\right )}}{\cos \left (x \right )^{2}}\right )}{2}\)	\(44\)
default	\(-\frac {\sqrt {a +b \left (\cos ^{4}\left (x \right )\right )}}{2}+\frac {\sqrt {a}\, \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {a +b \left (\cos ^{4}\left (x \right )\right )}}{\cos \left (x \right )^{2}}\right )}{2}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a+b*cos(x)^4)^(1/2)+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 [A]
time = 0.27, size = 43, normalized size = 0.96 \begin {gather*} \frac {1}{2} \, \sqrt {a} \operatorname {arsinh}\left (-\frac {a}{\sqrt {a b} {\left (\sin \left (x\right )^{2} - 1\right )}}\right ) - \frac {1}{2} \, \sqrt {b \sin \left (x\right )^{4} - 2 \, b \sin \left (x\right )^{2} + a + b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(a)*arcsinh(-a/(sqrt(a*b)*(sin(x)^2 - 1))) - 1/2*sqrt(b*sin(x)^4 - 2*b*sin(x)^2 + a + b)

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Fricas [A]
time = 0.48, size = 90, normalized size = 2.00 \begin {gather*} \left [\frac {1}{4} \, \sqrt {a} \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 ) - \frac {1}{2} \, \sqrt {b \cos \left (x\right )^{4} + a}, -\frac {1}{2} \, \sqrt {-a} \arctan \left (\frac {\sqrt {b \cos \left (x\right )^{4} + a} \sqrt {-a}}{a}\right ) - \frac {1}{2} \, \sqrt {b \cos \left (x\right )^{4} + a}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.48, size = 38, normalized size = 0.84 \begin {gather*} -\frac {a \arctan \left (\frac {\sqrt {b \cos \left (x\right )^{4} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a}} - \frac {1}{2} \, \sqrt {b \cos \left (x\right )^{4} + a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \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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