3.1.97 \(\int \sqrt {a+b \cos ^n(x)} \tan (x) \, dx\) [97]

Optimal. Leaf size=47 \[ \frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^n(x)}}{\sqrt {a}}\right )}{n}-\frac {2 \sqrt {a+b \cos ^n(x)}}{n} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 47, 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 = {3309, 272, 52, 65, 214} \begin {gather*} \frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^n(x)}}{\sqrt {a}}\right )}{n}-\frac {2 \sqrt {a+b \cos ^n(x)}}{n} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Cos[x]^n]*Tan[x],x]

[Out]

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

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

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Mathematica [A]
time = 0.04, size = 46, normalized size = 0.98 \begin {gather*} -\frac {-2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \cos ^n(x)}}{\sqrt {a}}\right )+2 \sqrt {a+b \cos ^n(x)}}{n} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.57, size = 39, normalized size = 0.83

method	result	size
derivativedivides	\(-\frac {2 \sqrt {a +b \left (\cos ^{n}\left (x \right )\right )}-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {a +b \left (\cos ^{n}\left (x \right )\right )}}{\sqrt {a}}\right )}{n}\)	\(39\)
default	\(-\frac {2 \sqrt {a +b \left (\cos ^{n}\left (x \right )\right )}-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {a +b \left (\cos ^{n}\left (x \right )\right )}}{\sqrt {a}}\right )}{n}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.49, size = 58, normalized size = 1.23 \begin {gather*} -\frac {\sqrt {a} \log \left (\frac {\sqrt {b \cos \left (x\right )^{n} + a} - \sqrt {a}}{\sqrt {b \cos \left (x\right )^{n} + a} + \sqrt {a}}\right )}{n} - \frac {2 \, \sqrt {b \cos \left (x\right )^{n} + a}}{n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.38, size = 97, normalized size = 2.06 \begin {gather*} \left [\frac {\sqrt {a} \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 ) - 2 \, \sqrt {b \cos \left (x\right )^{n} + a}}{n}, -\frac {2 \, {\left (\sqrt {-a} \arctan \left (\frac {\sqrt {b \cos \left (x\right )^{n} + a} \sqrt {-a}}{a}\right ) + \sqrt {b \cos \left (x\right )^{n} + a}\right )}}{n}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a + b \cos ^{n}{\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)**n)**(1/2)*tan(x),x)

[Out]

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

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Giac [A]
time = 0.49, size = 46, normalized size = 0.98 \begin {gather*} -\frac {2 \, {\left (\frac {a b \arctan \left (\frac {\sqrt {b \cos \left (x\right )^{n} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \sqrt {b \cos \left (x\right )^{n} + a} b\right )}}{b n} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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