3.45 \(\int \frac {\sin (x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=61 \[ -\frac {2 b^2 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2}}-\frac {b x}{a^2}-\frac {\cos (x)}{a} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.546, Rules used = {3853, 12, 3783, 2660, 618, 206} \[ -\frac {2 b^2 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^2 \sqrt {a^2-b^2}}-\frac {b x}{a^2}-\frac {\cos (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]/(a + b*Csc[x]),x]

[Out]

-((b*x)/a^2) - (2*b^2*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2*Sqrt[a^2 - b^2]) - Cos[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 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 3783

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(-1), x_Symbol] :> Simp[x/a, x] - Dist[1/a, Int[1/(1 + (a*Sin[c + d
*x])/b), x], x] /; FreeQ[{a, b, c, d}, 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]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 56, normalized size = 0.92 \[ -\frac {-\frac {2 b^2 \tan ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+a \cos (x)+b x}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]/(a + b*Csc[x]),x]

[Out]

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

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fricas [A]  time = 0.49, size = 235, normalized size = 3.85 \[ \left [\frac {\sqrt {a^{2} - b^{2}} b^{2} \log \left (-\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} - 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) - 2 \, {\left (a^{2} b - b^{3}\right )} x - 2 \, {\left (a^{3} - a b^{2}\right )} \cos \relax (x)}{2 \, {\left (a^{4} - a^{2} b^{2}\right )}}, -\frac {\sqrt {-a^{2} + b^{2}} b^{2} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right ) + {\left (a^{2} b - b^{3}\right )} x + {\left (a^{3} - a b^{2}\right )} \cos \relax (x)}{a^{4} - a^{2} b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.52, size = 77, normalized size = 1.26 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b^{2}}{\sqrt {-a^{2} + b^{2}} a^{2}} - \frac {b x}{a^{2}} - \frac {2}{{\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)/(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^2/(sqrt(-a^2 + b^2)*a^2) -
b*x/a^2 - 2/((tan(1/2*x)^2 + 1)*a)

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maple [A]  time = 0.46, size = 72, normalized size = 1.18 \[ \frac {2 b^{2} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{2} \sqrt {-a^{2}+b^{2}}}-\frac {2}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )}-\frac {2 b \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b^2/a^2/(-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/(tan(1/2*x)^2+1)-2/a^2*b*arct
an(tan(1/2*x))

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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)/(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 = 0.65, size = 766, normalized size = 12.56 \[ -\frac {2}{a\,\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}-\frac {b\,x}{a^2}-\frac {b^2\,\mathrm {atan}\left (\frac {\frac {b^2\,\sqrt {a^2-b^2}\,\left (\frac {32\,b^4}{a}-\frac {32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a\,b^5-2\,a^3\,b^3\right )}{a^3}+\frac {b^2\,\sqrt {a^2-b^2}\,\left (32\,a^2\,b^2+64\,a\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {b^2\,\sqrt {a^2-b^2}\,\left (32\,a^3\,b^2+\frac {32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^7\,b-2\,a^5\,b^3\right )}{a^3}\right )}{a^4-a^2\,b^2}\right )}{a^4-a^2\,b^2}\right )\,1{}\mathrm {i}}{a^4-a^2\,b^2}-\frac {b^2\,\sqrt {a^2-b^2}\,\left (\frac {32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a\,b^5-2\,a^3\,b^3\right )}{a^3}-\frac {32\,b^4}{a}+\frac {b^2\,\sqrt {a^2-b^2}\,\left (32\,a^2\,b^2+64\,a\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {b^2\,\sqrt {a^2-b^2}\,\left (32\,a^3\,b^2+\frac {32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^7\,b-2\,a^5\,b^3\right )}{a^3}\right )}{a^4-a^2\,b^2}\right )}{a^4-a^2\,b^2}\right )\,1{}\mathrm {i}}{a^4-a^2\,b^2}}{\frac {128\,b^5\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^3}+\frac {b^2\,\sqrt {a^2-b^2}\,\left (\frac {32\,b^4}{a}-\frac {32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a\,b^5-2\,a^3\,b^3\right )}{a^3}+\frac {b^2\,\sqrt {a^2-b^2}\,\left (32\,a^2\,b^2+64\,a\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )+\frac {b^2\,\sqrt {a^2-b^2}\,\left (32\,a^3\,b^2+\frac {32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^7\,b-2\,a^5\,b^3\right )}{a^3}\right )}{a^4-a^2\,b^2}\right )}{a^4-a^2\,b^2}\right )}{a^4-a^2\,b^2}+\frac {b^2\,\sqrt {a^2-b^2}\,\left (\frac {32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (2\,a\,b^5-2\,a^3\,b^3\right )}{a^3}-\frac {32\,b^4}{a}+\frac {b^2\,\sqrt {a^2-b^2}\,\left (32\,a^2\,b^2+64\,a\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )-\frac {b^2\,\sqrt {a^2-b^2}\,\left (32\,a^3\,b^2+\frac {32\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (3\,a^7\,b-2\,a^5\,b^3\right )}{a^3}\right )}{a^4-a^2\,b^2}\right )}{a^4-a^2\,b^2}\right )}{a^4-a^2\,b^2}}\right )\,\sqrt {a^2-b^2}\,2{}\mathrm {i}}{a^4-a^2\,b^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)/(a + b/sin(x)),x)

[Out]

- 2/(a*(tan(x/2)^2 + 1)) - (b*x)/a^2 - (b^2*atan(((b^2*(a^2 - b^2)^(1/2)*((32*b^4)/a - (32*tan(x/2)*(2*a*b^5 -
2*a^3*b^3))/a^3 + (b^2*(a^2 - b^2)^(1/2)*(32*a^2*b^2 + 64*a*b^3*tan(x/2) + (b^2*(a^2 - b^2)^(1/2)*(32*a^3*b^2
+ (32*tan(x/2)*(3*a^7*b - 2*a^5*b^3))/a^3))/(a^4 - a^2*b^2)))/(a^4 - a^2*b^2))*1i)/(a^4 - a^2*b^2) - (b^2*(a^
2 - b^2)^(1/2)*((32*tan(x/2)*(2*a*b^5 - 2*a^3*b^3))/a^3 - (32*b^4)/a + (b^2*(a^2 - b^2)^(1/2)*(32*a^2*b^2 + 64
*a*b^3*tan(x/2) - (b^2*(a^2 - b^2)^(1/2)*(32*a^3*b^2 + (32*tan(x/2)*(3*a^7*b - 2*a^5*b^3))/a^3))/(a^4 - a^2*b^
2)))/(a^4 - a^2*b^2))*1i)/(a^4 - a^2*b^2))/((128*b^5*tan(x/2))/a^3 + (b^2*(a^2 - b^2)^(1/2)*((32*b^4)/a - (32*
tan(x/2)*(2*a*b^5 - 2*a^3*b^3))/a^3 + (b^2*(a^2 - b^2)^(1/2)*(32*a^2*b^2 + 64*a*b^3*tan(x/2) + (b^2*(a^2 - b^2
)^(1/2)*(32*a^3*b^2 + (32*tan(x/2)*(3*a^7*b - 2*a^5*b^3))/a^3))/(a^4 - a^2*b^2)))/(a^4 - a^2*b^2)))/(a^4 - a^2
*b^2) + (b^2*(a^2 - b^2)^(1/2)*((32*tan(x/2)*(2*a*b^5 - 2*a^3*b^3))/a^3 - (32*b^4)/a + (b^2*(a^2 - b^2)^(1/2)*
(32*a^2*b^2 + 64*a*b^3*tan(x/2) - (b^2*(a^2 - b^2)^(1/2)*(32*a^3*b^2 + (32*tan(x/2)*(3*a^7*b - 2*a^5*b^3))/a^3
))/(a^4 - a^2*b^2)))/(a^4 - a^2*b^2)))/(a^4 - a^2*b^2)))*(a^2 - b^2)^(1/2)*2i)/(a^4 - a^2*b^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(x)/(a + b*csc(x)), x)

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