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

Optimal. Leaf size=84 \[ \frac{a \tan (x)}{a^2-b^2}-\frac{b \sec (x)}{a^2-b^2}+\frac{2 b^3 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{3/2}}-\frac{x}{a} \]

[Out]

-(x/a) + (2*b^3*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b^2)^(3/2)) - (b*Sec[x])/(a^2 - b^2) + (a
*Tan[x])/(a^2 - b^2)

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Rubi [A]  time = 0.187451, antiderivative size = 112, normalized size of antiderivative = 1.33, number of steps used = 10, number of rules used = 9, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.692, Rules used = {3898, 2902, 2606, 8, 3473, 2735, 2660, 618, 206} \[ \frac{b^2 x}{a \left (a^2-b^2\right )}-\frac{a x}{a^2-b^2}+\frac{a \tan (x)}{a^2-b^2}-\frac{b \sec (x)}{a^2-b^2}+\frac{2 b^3 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[x]^2/(a + b*Csc[x]),x]

[Out]

-((a*x)/(a^2 - b^2)) + (b^2*x)/(a*(a^2 - b^2)) + (2*b^3*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a*(a^2 - b
^2)^(3/2)) - (b*Sec[x])/(a^2 - b^2) + (a*Tan[x])/(a^2 - b^2)

Rule 3898

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[(Cos[c + d*x]^
m*(b + a*Sin[c + d*x])^n)/Sin[c + d*x]^(m + n), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[
n] && IntegerQ[m] && (IntegerQ[m/2] || LeQ[m, 1])

Rule 2902

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.
)*(x_)]), x_Symbol] :> Dist[(a*d^2)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-D
ist[(b*d)/(a^2 - b^2), Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Dist[(a^2*d^2)/(g^2*(a^2 - b^
2)), Int[((g*Cos[e + f*x])^(p + 2)*(d*Sin[e + f*x])^(n - 2))/(a + b*Sin[e + f*x]), x], x]) /; FreeQ[{a, b, d,
e, f, g}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 2735

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

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^2(x)}{a+b \csc (x)} \, dx &=\int \frac{\sin (x) \tan ^2(x)}{b+a \sin (x)} \, dx\\ &=\frac{a \int \tan ^2(x) \, dx}{a^2-b^2}-\frac{b \int \sec (x) \tan (x) \, dx}{a^2-b^2}+\frac{b^2 \int \frac{\sin (x)}{b+a \sin (x)} \, dx}{a^2-b^2}\\ &=\frac{b^2 x}{a \left (a^2-b^2\right )}+\frac{a \tan (x)}{a^2-b^2}-\frac{a \int 1 \, dx}{a^2-b^2}-\frac{b \operatorname{Subst}(\int 1 \, dx,x,\sec (x))}{a^2-b^2}-\frac{b^3 \int \frac{1}{b+a \sin (x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{a x}{a^2-b^2}+\frac{b^2 x}{a \left (a^2-b^2\right )}-\frac{b \sec (x)}{a^2-b^2}+\frac{a \tan (x)}{a^2-b^2}-\frac{\left (2 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{b+2 a x+b x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )}\\ &=-\frac{a x}{a^2-b^2}+\frac{b^2 x}{a \left (a^2-b^2\right )}-\frac{b \sec (x)}{a^2-b^2}+\frac{a \tan (x)}{a^2-b^2}+\frac{\left (4 b^3\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (a^2-b^2\right )-x^2} \, dx,x,2 a+2 b \tan \left (\frac{x}{2}\right )\right )}{a \left (a^2-b^2\right )}\\ &=-\frac{a x}{a^2-b^2}+\frac{b^2 x}{a \left (a^2-b^2\right )}+\frac{2 b^3 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a \left (a^2-b^2\right )^{3/2}}-\frac{b \sec (x)}{a^2-b^2}+\frac{a \tan (x)}{a^2-b^2}\\ \end{align*}

Mathematica [A]  time = 0.247568, size = 87, normalized size = 1.04 \[ -\frac{\sqrt{b^2-a^2} \left (a^2 (-x)+a^2 \tan (x)-a b \sec (x)+b^2 x\right )-2 b^3 \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{a \left (b^2-a^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[x]^2/(a + b*Csc[x]),x]

[Out]

-((-2*b^3*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]] + Sqrt[-a^2 + b^2]*(-(a^2*x) + b^2*x - a*b*Sec[x] + a^2*Ta
n[x]))/(a*(-a^2 + b^2)^(3/2)))

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Maple [A]  time = 0.06, size = 106, normalized size = 1.3 \begin{align*} -2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ) }{a}}-16\,{\frac{1}{ \left ( 16\,a+16\,b \right ) \left ( \tan \left ( x/2 \right ) -1 \right ) }}-16\,{\frac{1}{ \left ( 16\,a-16\,b \right ) \left ( \tan \left ( x/2 \right ) +1 \right ) }}-2\,{\frac{{b}^{3}}{ \left ( a-b \right ) \left ( a+b \right ) a\sqrt{-{a}^{2}+{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,b\tan \left ( x/2 \right ) +2\,a}{\sqrt{-{a}^{2}+{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/a*arctan(tan(1/2*x))-16/(16*a+16*b)/(tan(1/2*x)-1)-16/(16*a-16*b)/(tan(1/2*x)+1)-2/(a-b)/(a+b)*b^3/a/(-a^2+
b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*x)+2*a)/(-a^2+b^2)^(1/2))

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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*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.544479, size = 690, normalized size = 8.21 \begin{align*} \left [-\frac{\sqrt{a^{2} - b^{2}} b^{3} \cos \left (x\right ) \log \left (-\frac{{\left (a^{2} - 2 \, b^{2}\right )} \cos \left (x\right )^{2} + 2 \, a b \sin \left (x\right ) + a^{2} + b^{2} - 2 \,{\left (b \cos \left (x\right ) \sin \left (x\right ) + a \cos \left (x\right )\right )} \sqrt{a^{2} - b^{2}}}{a^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 2 \, a^{3} b - 2 \, a b^{3} + 2 \,{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x \cos \left (x\right ) - 2 \,{\left (a^{4} - a^{2} b^{2}\right )} \sin \left (x\right )}{2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )}, \frac{\sqrt{-a^{2} + b^{2}} b^{3} \arctan \left (-\frac{\sqrt{-a^{2} + b^{2}}{\left (b \sin \left (x\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \left (x\right )}\right ) \cos \left (x\right ) - a^{3} b + a b^{3} -{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} x \cos \left (x\right ) +{\left (a^{4} - a^{2} b^{2}\right )} \sin \left (x\right )}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(sqrt(a^2 - b^2)*b^3*cos(x)*log(-((a^2 - 2*b^2)*cos(x)^2 + 2*a*b*sin(x) + a^2 + b^2 - 2*(b*cos(x)*sin(x)
+ a*cos(x))*sqrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + 2*a^3*b - 2*a*b^3 + 2*(a^4 - 2*a^2*
b^2 + b^4)*x*cos(x) - 2*(a^4 - a^2*b^2)*sin(x))/((a^5 - 2*a^3*b^2 + a*b^4)*cos(x)), (sqrt(-a^2 + b^2)*b^3*arct
an(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x)))*cos(x) - a^3*b + a*b^3 - (a^4 - 2*a^2*b^2 + b^4)*x*c
os(x) + (a^4 - a^2*b^2)*sin(x))/((a^5 - 2*a^3*b^2 + a*b^4)*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 \csc{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(tan(x)**2/(a + b*csc(x)), x)

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Giac [A]  time = 1.33368, size = 139, normalized size = 1.65 \begin{align*} -\frac{2 \,{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (\frac{1}{2} \, x\right ) + a}{\sqrt{-a^{2} + b^{2}}}\right )\right )} b^{3}}{{\left (a^{3} - a b^{2}\right )} \sqrt{-a^{2} + b^{2}}} - \frac{x}{a} - \frac{2 \,{\left (a \tan \left (\frac{1}{2} \, x\right ) - b\right )}}{{\left (a^{2} - b^{2}\right )}{\left (\tan \left (\frac{1}{2} \, x\right )^{2} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*(pi*floor(1/2*x/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*x) + a)/sqrt(-a^2 + b^2)))*b^3/((a^3 - a*b^2)*sqrt(-a^
2 + b^2)) - x/a - 2*(a*tan(1/2*x) - b)/((a^2 - b^2)*(tan(1/2*x)^2 - 1))