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

Optimal. Leaf size=82 \[ \frac{x \left (a^2+2 b^2\right )}{2 a^3}+\frac{2 b^3 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a^3 \sqrt{a^2-b^2}}+\frac{b \cos (x)}{a^2}-\frac{\sin (x) \cos (x)}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.261108, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.538, Rules used = {3853, 4104, 3919, 3831, 2660, 618, 206} \[ \frac{x \left (a^2+2 b^2\right )}{2 a^3}+\frac{2 b^3 \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a^3 \sqrt{a^2-b^2}}+\frac{b \cos (x)}{a^2}-\frac{\sin (x) \cos (x)}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^2/(a + b*Csc[x]),x]

[Out]

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

Rule 3853

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(Cot[e + f*
x]*(d*Csc[e + f*x])^n)/(a*f*n), x] - Dist[1/(a*d*n), Int[((d*Csc[e + f*x])^(n + 1)*Simp[b*n - a*(n + 1)*Csc[e
+ f*x] - b*(n + 1)*Csc[e + f*x]^2, x])/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 -
b^2, 0] && LeQ[n, -1] && IntegerQ[2*n]

Rule 4104

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]

Rule 3919

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

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 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{\sin ^2(x)}{a+b \csc (x)} \, dx &=-\frac{\cos (x) \sin (x)}{2 a}+\frac{\int \frac{\left (-2 b+a \csc (x)+b \csc ^2(x)\right ) \sin (x)}{a+b \csc (x)} \, dx}{2 a}\\ &=\frac{b \cos (x)}{a^2}-\frac{\cos (x) \sin (x)}{2 a}-\frac{\int \frac{-a^2-2 b^2-a b \csc (x)}{a+b \csc (x)} \, dx}{2 a^2}\\ &=\frac{\left (a^2+2 b^2\right ) x}{2 a^3}+\frac{b \cos (x)}{a^2}-\frac{\cos (x) \sin (x)}{2 a}-\frac{b^3 \int \frac{\csc (x)}{a+b \csc (x)} \, dx}{a^3}\\ &=\frac{\left (a^2+2 b^2\right ) x}{2 a^3}+\frac{b \cos (x)}{a^2}-\frac{\cos (x) \sin (x)}{2 a}-\frac{b^2 \int \frac{1}{1+\frac{a \sin (x)}{b}} \, dx}{a^3}\\ &=\frac{\left (a^2+2 b^2\right ) x}{2 a^3}+\frac{b \cos (x)}{a^2}-\frac{\cos (x) \sin (x)}{2 a}-\frac{\left (2 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{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{b \cos (x)}{a^2}-\frac{\cos (x) \sin (x)}{2 a}+\frac{\left (4 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (1-\frac{a^2}{b^2}\right )-x^2} \, dx,x,\frac{2 a}{b}+2 \tan \left (\frac{x}{2}\right )\right )}{a^3}\\ &=\frac{\left (a^2+2 b^2\right ) x}{2 a^3}+\frac{2 b^3 \tanh ^{-1}\left (\frac{b \left (\frac{a}{b}+\tan \left (\frac{x}{2}\right )\right )}{\sqrt{a^2-b^2}}\right )}{a^3 \sqrt{a^2-b^2}}+\frac{b \cos (x)}{a^2}-\frac{\cos (x) \sin (x)}{2 a}\\ \end{align*}

Mathematica [A]  time = 0.118575, size = 78, normalized size = 0.95 \[ \frac{-\frac{8 b^3 \tan ^{-1}\left (\frac{a+b \tan \left (\frac{x}{2}\right )}{\sqrt{b^2-a^2}}\right )}{\sqrt{b^2-a^2}}+2 a^2 x-a^2 \sin (2 x)+4 a b \cos (x)+4 b^2 x}{4 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^2/(a + b*Csc[x]),x]

[Out]

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

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Maple [A]  time = 0.072, size = 142, normalized size = 1.7 \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}^{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(sin(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^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(sin(x)^2/(a+b*csc(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.540342, size = 644, normalized size = 7.85 \begin{align*} \left [\frac{\sqrt{a^{2} - b^{2}} b^{3} \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^{4} - a^{2} b^{2}\right )} \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} x + 2 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (x\right )}{2 \,{\left (a^{5} - a^{3} b^{2}\right )}}, \frac{2 \, \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 ) -{\left (a^{4} - a^{2} b^{2}\right )} \cos \left (x\right ) \sin \left (x\right ) +{\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} x + 2 \,{\left (a^{3} b - a b^{3}\right )} \cos \left (x\right )}{2 \,{\left (a^{5} - a^{3} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(sqrt(a^2 - b^2)*b^3*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^4 - a^2*b^2)*cos(x)*sin(x) + (a^4 + a^2*b
^2 - 2*b^4)*x + 2*(a^3*b - a*b^3)*cos(x))/(a^5 - a^3*b^2), 1/2*(2*sqrt(-a^2 + b^2)*b^3*arctan(-sqrt(-a^2 + b^2
)*(b*sin(x) + a)/((a^2 - b^2)*cos(x))) - (a^4 - a^2*b^2)*cos(x)*sin(x) + (a^4 + a^2*b^2 - 2*b^4)*x + 2*(a^3*b
- a*b^3)*cos(x))/(a^5 - a^3*b^2)]

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

[Out]

Integral(sin(x)**2/(a + b*csc(x)), x)

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Giac [A]  time = 1.38052, size = 151, normalized size = 1.84 \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}}{\sqrt{-a^{2} + b^{2}} a^{3}} + \frac{{\left (a^{2} + 2 \, b^{2}\right )} x}{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(sin(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/(sqrt(-a^2 + b^2)*a^3)
+ 1/2*(a^2 + 2*b^2)*x/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
)