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

Optimal. Leaf size=80 \[ \frac{x \left (a^2-2 b^2\right )}{2 a^3}+\frac{2 b \sqrt{a^2-b^2} \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a^3}-\frac{\cos (x) (2 b-a \sin (x))}{2 a^2} \]

[Out]

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

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Rubi [A]  time = 0.189825, antiderivative size = 80, 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 = {3872, 2865, 2735, 2660, 618, 206} \[ \frac{x \left (a^2-2 b^2\right )}{2 a^3}+\frac{2 b \sqrt{a^2-b^2} \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a^3}-\frac{\cos (x) (2 b-a \sin (x))}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[x]^2/(a + b*Csc[x]),x]

[Out]

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

Rule 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

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

Mathematica [A]  time = 0.134177, size = 75, normalized size = 0.94 \[ \frac{8 b \sqrt{b^2-a^2} \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )+2 a^2 x+a^2 \sin (2 x)-4 a b \cos (x)-4 b^2 x}{4 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[x]^2/(a + b*Csc[x]),x]

[Out]

(2*a^2*x - 4*b^2*x + 8*b*Sqrt[-a^2 + b^2]*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]] - 4*a*b*Cos[x] + a^2*Sin[2
*x])/(4*a^3)

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Maple [B]  time = 0.053, size = 184, normalized size = 2.3 \begin{align*} -{\frac{1}{a} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3} \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-2\,{\frac{b \left ( \tan \left ( x/2 \right ) \right ) ^{2}}{{a}^{2} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+{\frac{1}{a}\tan \left ({\frac{x}{2}} \right ) \left ( \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) ^{-2}}-2\,{\frac{b}{{a}^{2} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-2\,{\frac{\arctan \left ( \tan \left ( x/2 \right ) \right ){b}^{2}}{{a}^{3}}}-2\,{\frac{b}{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 ) }+2\,{\frac{{b}^{3}}{{a}^{3}\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 ) }+{\frac{x}{2\,a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.536135, size = 514, normalized size = 6.42 \begin{align*} \left [\frac{a^{2} \cos \left (x\right ) \sin \left (x\right ) - 2 \, a b \cos \left (x\right ) + \sqrt{a^{2} - b^{2}} b \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 ) +{\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}}, \frac{a^{2} \cos \left (x\right ) \sin \left (x\right ) - 2 \, a b \cos \left (x\right ) + 2 \, \sqrt{-a^{2} + b^{2}} b \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 ) +{\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(a^2*cos(x)*sin(x) - 2*a*b*cos(x) + sqrt(a^2 - b^2)*b*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)) + (a^2 - 2*b^
2)*x)/a^3, 1/2*(a^2*cos(x)*sin(x) - 2*a*b*cos(x) + 2*sqrt(-a^2 + b^2)*b*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a
)/((a^2 - b^2)*cos(x))) + (a^2 - 2*b^2)*x)/a^3]

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

[Out]

Integral(cos(x)**2/(a + b*csc(x)), x)

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Giac [A]  time = 1.45519, size = 163, normalized size = 2.04 \begin{align*} \frac{{\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}} - \frac{2 \,{\left (a^{2} b - b^{3}\right )}{\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 )}}{\sqrt{-a^{2} + b^{2}} a^{3}} - \frac{a \tan \left (\frac{1}{2} \, x\right )^{3} + 2 \, b \tan \left (\frac{1}{2} \, x\right )^{2} - a \tan \left (\frac{1}{2} \, x\right ) + 2 \, b}{{\left (\tan \left (\frac{1}{2} \, x\right )^{2} + 1\right )}^{2} a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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