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

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

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 82, normalized size of antiderivative = 1.00, 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [A]  time = 0.68, size = 285, normalized size = 3.48 \[ \left [\frac {\sqrt {a^{2} - b^{2}} b^{3} \log \left (\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} + 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - {\left (a^{4} - a^{2} b^{2}\right )} \cos \relax (x) \sin \relax (x) + {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} x + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x)}{2 \, {\left (a^{5} - a^{3} b^{2}\right )}}, \frac {2 \, \sqrt {-a^{2} + b^{2}} b^{3} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right ) - {\left (a^{4} - a^{2} b^{2}\right )} \cos \relax (x) \sin \relax (x) + {\left (a^{4} + a^{2} b^{2} - 2 \, b^{4}\right )} x + 2 \, {\left (a^{3} b - a b^{3}\right )} \cos \relax (x)}{2 \, {\left (a^{5} - a^{3} b^{2}\right )}}\right ] \]

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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giac [A]  time = 0.64, size = 112, normalized size = 1.37 \[ -\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b^{3}}{\sqrt {-a^{2} + b^{2}} a^{3}} + \frac {{\left (a^{2} + 2 \, b^{2}\right )} x}{2 \, a^{3}} + \frac {a \tan \left (\frac {1}{2} \, x\right )^{3} + 2 \, b \tan \left (\frac {1}{2} \, x\right )^{2} - a \tan \left (\frac {1}{2} \, x\right ) + 2 \, b}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{2} a^{2}} \]

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
)

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maple [A]  time = 0.49, size = 142, normalized size = 1.73 \[ -\frac {2 b^{3} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{3} \sqrt {-a^{2}+b^{2}}}+\frac {\tan ^{3}\left (\frac {x}{2}\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2 b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}-\frac {\tan \left (\frac {x}{2}\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2 b}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{2}}+\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right ) b^{2}}{a^{3}}+\frac {x}{2 a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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

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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(sin(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?

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mupad [B]  time = 0.84, size = 1147, normalized size = 13.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(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((40*b^3
*tan(x/2))/(8*a^2*b + 40*b^3 + (48*b^5)/a^2) + (48*b^5*tan(x/2))/(8*a^4*b + 48*b^5 + 40*a^2*b^3) + (8*a*b*tan(
x/2))/(8*a*b + (40*b^3)/a + (48*b^5)/a^3))*(a^2*1i + b^2*2i)*1i)/a^3 + (b^3*atan(((b^3*(a^2 - b^2)^(1/2)*((8*(
4*a^2*b^6 + 4*a^4*b^4 + a^6*b^2))/a^5 + (8*tan(x/2)*(2*a^8*b - 8*a^2*b^7 + 4*a^4*b^5 + 7*a^6*b^3))/a^6 + (b^3*
(a^2 - b^2)^(1/2)*(64*b^4*tan(x/2) + (8*(2*a^8*b + 2*a^6*b^3))/a^5 + (b^3*(a^2 - b^2)^(1/2)*(32*a^3*b^2 + (8*t
an(x/2)*(12*a^10*b - 8*a^8*b^3))/a^6))/(a^5 - a^3*b^2)))/(a^5 - a^3*b^2))*1i)/(a^5 - a^3*b^2) + (b^3*(a^2 - b^
2)^(1/2)*((8*(4*a^2*b^6 + 4*a^4*b^4 + a^6*b^2))/a^5 + (8*tan(x/2)*(2*a^8*b - 8*a^2*b^7 + 4*a^4*b^5 + 7*a^6*b^3
))/a^6 - (b^3*(a^2 - b^2)^(1/2)*(64*b^4*tan(x/2) + (8*(2*a^8*b + 2*a^6*b^3))/a^5 - (b^3*(a^2 - b^2)^(1/2)*(32*
a^3*b^2 + (8*tan(x/2)*(12*a^10*b - 8*a^8*b^3))/a^6))/(a^5 - a^3*b^2)))/(a^5 - a^3*b^2))*1i)/(a^5 - a^3*b^2))/(
(16*(2*b^7 + a^2*b^5))/a^5 + (16*tan(x/2)*(8*b^8 + 8*a^2*b^6 + 2*a^4*b^4))/a^6 + (b^3*(a^2 - b^2)^(1/2)*((8*(4
*a^2*b^6 + 4*a^4*b^4 + a^6*b^2))/a^5 + (8*tan(x/2)*(2*a^8*b - 8*a^2*b^7 + 4*a^4*b^5 + 7*a^6*b^3))/a^6 + (b^3*(
a^2 - b^2)^(1/2)*(64*b^4*tan(x/2) + (8*(2*a^8*b + 2*a^6*b^3))/a^5 + (b^3*(a^2 - b^2)^(1/2)*(32*a^3*b^2 + (8*ta
n(x/2)*(12*a^10*b - 8*a^8*b^3))/a^6))/(a^5 - a^3*b^2)))/(a^5 - a^3*b^2)))/(a^5 - a^3*b^2) - (b^3*(a^2 - b^2)^(
1/2)*((8*(4*a^2*b^6 + 4*a^4*b^4 + a^6*b^2))/a^5 + (8*tan(x/2)*(2*a^8*b - 8*a^2*b^7 + 4*a^4*b^5 + 7*a^6*b^3))/a
^6 - (b^3*(a^2 - b^2)^(1/2)*(64*b^4*tan(x/2) + (8*(2*a^8*b + 2*a^6*b^3))/a^5 - (b^3*(a^2 - b^2)^(1/2)*(32*a^3*
b^2 + (8*tan(x/2)*(12*a^10*b - 8*a^8*b^3))/a^6))/(a^5 - a^3*b^2)))/(a^5 - a^3*b^2)))/(a^5 - a^3*b^2)))*(a^2 -
b^2)^(1/2)*2i)/(a^5 - a^3*b^2)

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

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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