3.12 \(\int \frac{\tan ^2(x)}{a+b \cos (x)} \, dx\)

Optimal. Leaf size=61 \[ -\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^2}-\frac{b \tanh ^{-1}(\sin (x))}{a^2}+\frac{\tan (x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.224174, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {2723, 3056, 3001, 3770, 2659, 205} \[ -\frac{2 \sqrt{a-b} \sqrt{a+b} \tan ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{x}{2}\right )}{\sqrt{a+b}}\right )}{a^2}-\frac{b \tanh ^{-1}(\sin (x))}{a^2}+\frac{\tan (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^2/(a + b*Cos[x]),x]

[Out]

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

Rule 2723

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)/tan[(e_.) + (f_.)*(x_)]^2, x_Symbol] :> Int[((a + b*Sin[e + f*
x])^m*(1 - Sin[e + f*x]^2))/Sin[e + f*x]^2, x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0]

Rule 3056

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
&& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/(a + b*Sin[e + f*x]), x], x] + Dist[(B*c - A
*d)/(b*c - a*d), Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 205

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

Rubi steps

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

Mathematica [A]  time = 0.170191, size = 85, normalized size = 1.39 \[ \frac{-2 \sqrt{b^2-a^2} \tanh ^{-1}\left (\frac{(a-b) \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )+a \tan (x)+b \left (\log \left (\cos \left (\frac{x}{2}\right )-\sin \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )+\cos \left (\frac{x}{2}\right )\right )\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^2/(a + b*Cos[x]),x]

[Out]

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

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Maple [B]  time = 0.098, size = 129, normalized size = 2.1 \begin{align*} -{\frac{1}{a} \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) ^{-1}}+{\frac{b}{{a}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) -1 \right ) }-{\frac{1}{a} \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) ^{-1}}-{\frac{b}{{a}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) +1 \right ) }-2\,{\frac{1}{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) }+2\,{\frac{{b}^{2}}{{a}^{2}\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}\arctan \left ({\frac{\tan \left ( x/2 \right ) \left ( a-b \right ) }{\sqrt{ \left ( a-b \right ) \left ( a+b \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65146, size = 564, normalized size = 9.25 \begin{align*} \left [-\frac{b \cos \left (x\right ) \log \left (\sin \left (x\right ) + 1\right ) - b \cos \left (x\right ) \log \left (-\sin \left (x\right ) + 1\right ) - \sqrt{-a^{2} + b^{2}} \cos \left (x\right ) \log \left (\frac{2 \, a b \cos \left (x\right ) +{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + 2 \, \sqrt{-a^{2} + b^{2}}{\left (a \cos \left (x\right ) + b\right )} \sin \left (x\right ) - a^{2} + 2 \, b^{2}}{b^{2} \cos \left (x\right )^{2} + 2 \, a b \cos \left (x\right ) + a^{2}}\right ) - 2 \, a \sin \left (x\right )}{2 \, a^{2} \cos \left (x\right )}, -\frac{b \cos \left (x\right ) \log \left (\sin \left (x\right ) + 1\right ) - b \cos \left (x\right ) \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \cos \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \sin \left (x\right )}\right ) \cos \left (x\right ) - 2 \, a \sin \left (x\right )}{2 \, a^{2} \cos \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (x \right )}}{a + b \cos{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(x)**2/(a + b*cos(x)), x)

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Giac [B]  time = 1.43594, size = 150, normalized size = 2.46 \begin{align*} -\frac{b \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) + 1 \right |}\right )}{a^{2}} + \frac{b \log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) - 1 \right |}\right )}{a^{2}} + \frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac{a \tan \left (\frac{1}{2} \, x\right ) - b \tan \left (\frac{1}{2} \, x\right )}{\sqrt{a^{2} - b^{2}}}\right )\right )} \sqrt{a^{2} - b^{2}}}{a^{2}} - \frac{2 \, \tan \left (\frac{1}{2} \, x\right )}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} - 1\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b*log(abs(tan(1/2*x) + 1))/a^2 + b*log(abs(tan(1/2*x) - 1))/a^2 + 2*(pi*floor(1/2*x/pi + 1/2)*sgn(-2*a + 2*b)
+ arctan(-(a*tan(1/2*x) - b*tan(1/2*x))/sqrt(a^2 - b^2)))*sqrt(a^2 - b^2)/a^2 - 2*tan(1/2*x)/((tan(1/2*x)^2 -
1)*a)