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

Optimal. Leaf size=62 \[ \frac {2 b^2 \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2}}+\frac {b \tanh ^{-1}(\cos (x))}{a^2}-\frac {\cot (x)}{a} \]

[Out]

b*arctanh(cos(x))/a^2-cot(x)/a+2*b^2*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/a^2/(a^2-b^2)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.11, antiderivative size = 62, 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 = {2802, 12, 2747, 3770, 2660, 618, 204} \[ \frac {2 b^2 \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2}}+\frac {b \tanh ^{-1}(\cos (x))}{a^2}-\frac {\cot (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^2/(a + b*Sin[x]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[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 2747

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

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 91, normalized size = 1.47 \[ \frac {\csc \left (\frac {x}{2}\right ) \sec \left (\frac {x}{2}\right ) \left (\frac {2 b^2 \sin (x) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-a \cos (x)+b \sin (x) \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )\right )}{2 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^2/(a + b*Sin[x]),x]

[Out]

(Csc[x/2]*Sec[x/2]*(-(a*Cos[x]) + (2*b^2*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]]*Sin[x])/Sqrt[a^2 - b^2] + b*
(Log[Cos[x/2]] - Log[Sin[x/2]])*Sin[x]))/(2*a^2)

________________________________________________________________________________________

fricas [B]  time = 0.56, size = 302, normalized size = 4.87 \[ \left [-\frac {\sqrt {-a^{2} + b^{2}} b^{2} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2} + 2 \, {\left (a \cos \relax (x) \sin \relax (x) + b \cos \relax (x)\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) \sin \relax (x) - {\left (a^{2} b - b^{3}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + {\left (a^{2} b - b^{3}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + 2 \, {\left (a^{3} - a b^{2}\right )} \cos \relax (x)}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} \sin \relax (x)}, -\frac {2 \, \sqrt {a^{2} - b^{2}} b^{2} \arctan \left (-\frac {a \sin \relax (x) + b}{\sqrt {a^{2} - b^{2}} \cos \relax (x)}\right ) \sin \relax (x) - {\left (a^{2} b - b^{3}\right )} \log \left (\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + {\left (a^{2} b - b^{3}\right )} \log \left (-\frac {1}{2} \, \cos \relax (x) + \frac {1}{2}\right ) \sin \relax (x) + 2 \, {\left (a^{3} - a b^{2}\right )} \cos \relax (x)}{2 \, {\left (a^{4} - a^{2} b^{2}\right )} \sin \relax (x)}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.16, size = 98, normalized size = 1.58 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{2}}{\sqrt {a^{2} - b^{2}} a^{2}} - \frac {b \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{a^{2}} + \frac {\tan \left (\frac {1}{2} \, x\right )}{2 \, a} + \frac {2 \, b \tan \left (\frac {1}{2} \, x\right ) - a}{2 \, a^{2} \tan \left (\frac {1}{2} \, x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.10, size = 77, normalized size = 1.24 \[ \frac {\tan \left (\frac {x}{2}\right )}{2 a}-\frac {1}{2 a \tan \left (\frac {x}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}+\frac {2 b^{2} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{2} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(csc(x)^2/(a+b*sin(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*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 7.01, size = 179, normalized size = 2.89 \[ \frac {b^3\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )-a^2\,b\,\ln \left (\mathrm {tan}\left (\frac {x}{2}\right )\right )+b^2\,\mathrm {atan}\left (\frac {-a^2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^2\,\mathrm {tan}\left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}+a\,b\,\sqrt {b^2-a^2}\,2{}\mathrm {i}}{-a^3-3\,\mathrm {tan}\left (\frac {x}{2}\right )\,a^2\,b+2\,a\,b^2+4\,\mathrm {tan}\left (\frac {x}{2}\right )\,b^3}\right )\,\sqrt {b^2-a^2}\,2{}\mathrm {i}}{a^4-a^2\,b^2}+\frac {a\,b^2-a^3}{a^4\,\mathrm {tan}\relax (x)-a^2\,b^2\,\mathrm {tan}\relax (x)} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^2*(a + b*sin(x))),x)

[Out]

(b^3*log(tan(x/2)) - a^2*b*log(tan(x/2)) + b^2*atan((b^2*tan(x/2)*(b^2 - a^2)^(1/2)*4i - a^2*tan(x/2)*(b^2 - a
^2)^(1/2)*1i + a*b*(b^2 - a^2)^(1/2)*2i)/(4*b^3*tan(x/2) + 2*a*b^2 - a^3 - 3*a^2*b*tan(x/2)))*(b^2 - a^2)^(1/2
)*2i)/(a^4 - a^2*b^2) + (a*b^2 - a^3)/(a^4*tan(x) - a^2*b^2*tan(x))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**2/(a + b*sin(x)), x)

________________________________________________________________________________________