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

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

[Out]

1/2*(a^2-2*b^2)*x/a^3-1/2*cos(x)*(2*b-a*sin(x))/a^2+2*b*arctanh((a+b*tan(1/2*x))/(a^2-b^2)^(1/2))*(a^2-b^2)^(1
/2)/a^3

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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 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]

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 = 0.12, 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)

________________________________________________________________________________________

fricas [A]  time = 0.53, size = 205, normalized size = 2.56 \[ \left [\frac {a^{2} \cos \relax (x) \sin \relax (x) - 2 \, a b \cos \relax (x) + \sqrt {a^{2} - b^{2}} b \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} + 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) + {\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}}, \frac {a^{2} \cos \relax (x) \sin \relax (x) - 2 \, a b \cos \relax (x) + 2 \, \sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right ) + {\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}}\right ] \]

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]

________________________________________________________________________________________

giac [A]  time = 0.36, size = 121, normalized size = 1.51 \[ \frac {{\left (a^{2} - 2 \, b^{2}\right )} x}{2 \, a^{3}} - \frac {2 \, {\left (a^{2} b - b^{3}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{3}} - \frac {a \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, x\right )^{2} - a \tan \left (\frac {1}{2} \, x\right ) + 2 \, b}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{2} a^{2}} \]

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)

________________________________________________________________________________________

maple [B]  time = 0.27, size = 184, normalized size = 2.30 \[ -\frac {2 b \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a \sqrt {-a^{2}+b^{2}}}+\frac {2 b^{3} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{3} \sqrt {-a^{2}+b^{2}}}-\frac {\tan ^{3}\left (\frac {x}{2}\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {\tan \left (\frac {x}{2}\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 b}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right ) b^{2}}{a^{3}}+\frac {x}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 0.54, size = 362, normalized size = 4.52 \[ -\frac {\frac {2\,b}{a^2}-\frac {\mathrm {tan}\left (\frac {x}{2}\right )}{a}+\frac {{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{a}+\frac {2\,b\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2}{a^2}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {2\,b\,\mathrm {atanh}\left (\frac {32\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}}{16\,b^3-\frac {16\,b^5}{a^2}-\frac {32\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )}{a}+32\,a\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}-\frac {16\,b^3\,\sqrt {a^2-b^2}}{32\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )-16\,a\,b^3+\frac {16\,b^5}{a}-32\,a^2\,b^2\,\mathrm {tan}\left (\frac {x}{2}\right )}+\frac {16\,b^4\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {a^2-b^2}}{-32\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^3\,b^2-16\,a^2\,b^3+32\,\mathrm {tan}\left (\frac {x}{2}\right )\,a\,b^4+16\,b^5}\right )\,\sqrt {a^2-b^2}}{a^3}-\frac {\mathrm {atan}\left (\frac {16\,b^5\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a^4\,b-24\,a^2\,b^3+16\,b^5}-\frac {24\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a^2\,b-24\,b^3+\frac {16\,b^5}{a^2}}+\frac {8\,a\,b\,\mathrm {tan}\left (\frac {x}{2}\right )}{8\,a\,b-\frac {24\,b^3}{a}+\frac {16\,b^5}{a^3}}\right )\,\left (a^2\,1{}\mathrm {i}-b^2\,2{}\mathrm {i}\right )\,1{}\mathrm {i}}{a^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(x)^2/(a + b/sin(x)),x)

[Out]

- ((2*b)/a^2 - tan(x/2)/a + tan(x/2)^3/a + (2*b*tan(x/2)^2)/a^2)/(2*tan(x/2)^2 + tan(x/2)^4 + 1) - (atan((16*b
^5*tan(x/2))/(8*a^4*b + 16*b^5 - 24*a^2*b^3) - (24*b^3*tan(x/2))/(8*a^2*b - 24*b^3 + (16*b^5)/a^2) + (8*a*b*ta
n(x/2))/(8*a*b - (24*b^3)/a + (16*b^5)/a^3))*(a^2*1i - b^2*2i)*1i)/a^3 - (2*b*atanh((32*b^2*tan(x/2)*(a^2 - b^
2)^(1/2))/(16*b^3 - (16*b^5)/a^2 - (32*b^4*tan(x/2))/a + 32*a*b^2*tan(x/2)) - (16*b^3*(a^2 - b^2)^(1/2))/(32*b
^4*tan(x/2) - 16*a*b^3 + (16*b^5)/a - 32*a^2*b^2*tan(x/2)) + (16*b^4*tan(x/2)*(a^2 - b^2)^(1/2))/(16*b^5 - 16*
a^2*b^3 - 32*a^3*b^2*tan(x/2) + 32*a*b^4*tan(x/2)))*(a^2 - b^2)^(1/2))/a^3

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos ^{2}{\relax (x )}}{a + b \csc {\relax (x )}}\, dx \]

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)

________________________________________________________________________________________