3.1.98 \(\int \frac {\tan (x)}{\sqrt {a+b \cos ^n(x)}} \, dx\) [98]

Optimal. Leaf size=29 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^n(x)}}{\sqrt {a}}\right )}{\sqrt {a} n} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 29, 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 = {3309, 272, 65, 214} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^n(x)}}{\sqrt {a}}\right )}{\sqrt {a} n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[x]/Sqrt[a + b*Cos[x]^n],x]

[Out]

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

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 3309

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*((a + b*(c*ff*x)^n)^p/(1 - ff^2*x^2)^((
m + 1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 29, normalized size = 1.00 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^n(x)}}{\sqrt {a}}\right )}{\sqrt {a} n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.05, size = 24, normalized size = 0.83

method	result	size
derivativedivides	\(\frac {2 \arctanh \left (\frac {\sqrt {a +b \left (\cos ^{n}\left (x \right )\right )}}{\sqrt {a}}\right )}{n \sqrt {a}}\)	\(24\)
default	\(\frac {2 \arctanh \left (\frac {\sqrt {a +b \left (\cos ^{n}\left (x \right )\right )}}{\sqrt {a}}\right )}{n \sqrt {a}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.47, size = 42, normalized size = 1.45 \begin {gather*} -\frac {\log \left (\frac {\sqrt {b \cos \left (x\right )^{n} + a} - \sqrt {a}}{\sqrt {b \cos \left (x\right )^{n} + a} + \sqrt {a}}\right )}{\sqrt {a} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-log((sqrt(b*cos(x)^n + a) - sqrt(a))/(sqrt(b*cos(x)^n + a) + sqrt(a)))/(sqrt(a)*n)

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Fricas [A]
time = 0.44, size = 74, normalized size = 2.55 \begin {gather*} \left [\frac {\log \left (\frac {b \cos \left (x\right )^{n} + 2 \, \sqrt {b \cos \left (x\right )^{n} + a} \sqrt {a} + 2 \, a}{\cos \left (x\right )^{n}}\right )}{\sqrt {a} n}, -\frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {b \cos \left (x\right )^{n} + a} \sqrt {-a}}{a}\right )}{a n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 ^{n}{\left (x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.51, size = 27, normalized size = 0.93 \begin {gather*} -\frac {2 \, \arctan \left (\frac {\sqrt {b \cos \left (x\right )^{n} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\mathrm {tan}\left (x\right )}{\sqrt {a+b\,{\cos \left (x\right )}^n}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(x)/(a + b*cos(x)^n)^(1/2),x)

[Out]

int(tan(x)/(a + b*cos(x)^n)^(1/2), x)

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