3.47 \(\int \frac {\sin ^3(x)}{a+b \csc (x)} \, dx\)

Optimal. Leaf size=110 \[ \frac {b \sin (x) \cos (x)}{2 a^2}-\frac {b x \left (a^2+2 b^2\right )}{2 a^4}-\frac {2 b^4 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \sqrt {a^2-b^2}}-\frac {\left (2 a^2+3 b^2\right ) \cos (x)}{3 a^3}-\frac {\sin ^2(x) \cos (x)}{3 a} \]

[Out]

-1/2*b*(a^2+2*b^2)*x/a^4-1/3*(2*a^2+3*b^2)*cos(x)/a^3+1/2*b*cos(x)*sin(x)/a^2-1/3*cos(x)*sin(x)^2/a-2*b^4*arct
anh((a+b*tan(1/2*x))/(a^2-b^2)^(1/2))/a^4/(a^2-b^2)^(1/2)

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Rubi [A]  time = 0.40, antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \[ -\frac {b x \left (a^2+2 b^2\right )}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cos (x)}{3 a^3}-\frac {2 b^4 \tanh ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \sqrt {a^2-b^2}}+\frac {b \sin (x) \cos (x)}{2 a^2}-\frac {\sin ^2(x) \cos (x)}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3/(a + b*Csc[x]),x]

[Out]

-(b*(a^2 + 2*b^2)*x)/(2*a^4) - (2*b^4*ArcTanh[(a + b*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^4*Sqrt[a^2 - b^2]) - ((2*a
^2 + 3*b^2)*Cos[x])/(3*a^3) + (b*Cos[x]*Sin[x])/(2*a^2) - (Cos[x]*Sin[x]^2)/(3*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 ^3(x)}{a+b \csc (x)} \, dx &=-\frac {\cos (x) \sin ^2(x)}{3 a}+\frac {\int \frac {\left (-3 b+2 a \csc (x)+2 b \csc ^2(x)\right ) \sin ^2(x)}{a+b \csc (x)} \, dx}{3 a}\\ &=\frac {b \cos (x) \sin (x)}{2 a^2}-\frac {\cos (x) \sin ^2(x)}{3 a}-\frac {\int \frac {\left (-2 \left (2 a^2+3 b^2\right )-a b \csc (x)+3 b^2 \csc ^2(x)\right ) \sin (x)}{a+b \csc (x)} \, dx}{6 a^2}\\ &=-\frac {\left (2 a^2+3 b^2\right ) \cos (x)}{3 a^3}+\frac {b \cos (x) \sin (x)}{2 a^2}-\frac {\cos (x) \sin ^2(x)}{3 a}+\frac {\int \frac {-3 b \left (a^2+2 b^2\right )-3 a b^2 \csc (x)}{a+b \csc (x)} \, dx}{6 a^3}\\ &=-\frac {b \left (a^2+2 b^2\right ) x}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cos (x)}{3 a^3}+\frac {b \cos (x) \sin (x)}{2 a^2}-\frac {\cos (x) \sin ^2(x)}{3 a}+\frac {b^4 \int \frac {\csc (x)}{a+b \csc (x)} \, dx}{a^4}\\ &=-\frac {b \left (a^2+2 b^2\right ) x}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cos (x)}{3 a^3}+\frac {b \cos (x) \sin (x)}{2 a^2}-\frac {\cos (x) \sin ^2(x)}{3 a}+\frac {b^3 \int \frac {1}{1+\frac {a \sin (x)}{b}} \, dx}{a^4}\\ &=-\frac {b \left (a^2+2 b^2\right ) x}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cos (x)}{3 a^3}+\frac {b \cos (x) \sin (x)}{2 a^2}-\frac {\cos (x) \sin ^2(x)}{3 a}+\frac {\left (2 b^3\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^4}\\ &=-\frac {b \left (a^2+2 b^2\right ) x}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cos (x)}{3 a^3}+\frac {b \cos (x) \sin (x)}{2 a^2}-\frac {\cos (x) \sin ^2(x)}{3 a}-\frac {\left (4 b^3\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^4}\\ &=-\frac {b \left (a^2+2 b^2\right ) x}{2 a^4}-\frac {2 b^4 \tanh ^{-1}\left (\frac {b \left (\frac {a}{b}+\tan \left (\frac {x}{2}\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \sqrt {a^2-b^2}}-\frac {\left (2 a^2+3 b^2\right ) \cos (x)}{3 a^3}+\frac {b \cos (x) \sin (x)}{2 a^2}-\frac {\cos (x) \sin ^2(x)}{3 a}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 98, normalized size = 0.89 \[ \frac {a^3 \cos (3 x)-6 b x \left (a^2+2 b^2\right )-3 a \left (3 a^2+4 b^2\right ) \cos (x)+\frac {24 b^4 \tan ^{-1}\left (\frac {a+b \tan \left (\frac {x}{2}\right )}{\sqrt {b^2-a^2}}\right )}{\sqrt {b^2-a^2}}+3 a^2 b \sin (2 x)}{12 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^3/(a + b*Csc[x]),x]

[Out]

(-6*b*(a^2 + 2*b^2)*x + (24*b^4*ArcTan[(a + b*Tan[x/2])/Sqrt[-a^2 + b^2]])/Sqrt[-a^2 + b^2] - 3*a*(3*a^2 + 4*b
^2)*Cos[x] + a^3*Cos[3*x] + 3*a^2*b*Sin[2*x])/(12*a^4)

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fricas [A]  time = 0.62, size = 329, normalized size = 2.99 \[ \left [\frac {3 \, \sqrt {a^{2} - b^{2}} b^{4} \log \left (-\frac {{\left (a^{2} - 2 \, b^{2}\right )} \cos \relax (x)^{2} + 2 \, a b \sin \relax (x) + a^{2} + b^{2} - 2 \, {\left (b \cos \relax (x) \sin \relax (x) + a \cos \relax (x)\right )} \sqrt {a^{2} - b^{2}}}{a^{2} \cos \relax (x)^{2} - 2 \, a b \sin \relax (x) - a^{2} - b^{2}}\right ) + 2 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \relax (x)^{3} + 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \relax (x) \sin \relax (x) - 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} x - 6 \, {\left (a^{5} - a b^{4}\right )} \cos \relax (x)}{6 \, {\left (a^{6} - a^{4} b^{2}\right )}}, -\frac {6 \, \sqrt {-a^{2} + b^{2}} b^{4} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \sin \relax (x) + a\right )}}{{\left (a^{2} - b^{2}\right )} \cos \relax (x)}\right ) - 2 \, {\left (a^{5} - a^{3} b^{2}\right )} \cos \relax (x)^{3} - 3 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \relax (x) \sin \relax (x) + 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} x + 6 \, {\left (a^{5} - a b^{4}\right )} \cos \relax (x)}{6 \, {\left (a^{6} - a^{4} b^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3/(a+b*csc(x)),x, algorithm="fricas")

[Out]

[1/6*(3*sqrt(a^2 - b^2)*b^4*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*c
os(x))*sqrt(a^2 - b^2))/(a^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) + 2*(a^5 - a^3*b^2)*cos(x)^3 + 3*(a^4*b - a
^2*b^3)*cos(x)*sin(x) - 3*(a^4*b + a^2*b^3 - 2*b^5)*x - 6*(a^5 - a*b^4)*cos(x))/(a^6 - a^4*b^2), -1/6*(6*sqrt(
-a^2 + b^2)*b^4*arctan(-sqrt(-a^2 + b^2)*(b*sin(x) + a)/((a^2 - b^2)*cos(x))) - 2*(a^5 - a^3*b^2)*cos(x)^3 - 3
*(a^4*b - a^2*b^3)*cos(x)*sin(x) + 3*(a^4*b + a^2*b^3 - 2*b^5)*x + 6*(a^5 - a*b^4)*cos(x))/(a^6 - a^4*b^2)]

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giac [A]  time = 0.61, size = 149, normalized size = 1.35 \[ \frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (\frac {1}{2} \, x\right ) + a}{\sqrt {-a^{2} + b^{2}}}\right )\right )} b^{4}}{\sqrt {-a^{2} + b^{2}} a^{4}} - \frac {{\left (a^{2} b + 2 \, b^{3}\right )} x}{2 \, a^{4}} - \frac {3 \, a b \tan \left (\frac {1}{2} \, x\right )^{5} + 6 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{4} + 12 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 12 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, x\right ) + 4 \, a^{2} + 6 \, b^{2}}{3 \, {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)^3/(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^4/(sqrt(-a^2 + b^2)*a^4) -
1/2*(a^2*b + 2*b^3)*x/a^4 - 1/3*(3*a*b*tan(1/2*x)^5 + 6*b^2*tan(1/2*x)^4 + 12*a^2*tan(1/2*x)^2 + 12*b^2*tan(1
/2*x)^2 - 3*a*b*tan(1/2*x) + 4*a^2 + 6*b^2)/((tan(1/2*x)^2 + 1)^3*a^3)

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maple [B]  time = 0.44, size = 213, normalized size = 1.94 \[ \frac {2 b^{4} \arctan \left (\frac {2 \tan \left (\frac {x}{2}\right ) b +2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{a^{4} \sqrt {-a^{2}+b^{2}}}-\frac {\left (\tan ^{5}\left (\frac {x}{2}\right )\right ) b}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 b^{2} \left (\tan ^{4}\left (\frac {x}{2}\right )\right )}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {4 \left (\tan ^{2}\left (\frac {x}{2}\right )\right )}{a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {4 \left (\tan ^{2}\left (\frac {x}{2}\right )\right ) b^{2}}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}+\frac {b \tan \left (\frac {x}{2}\right )}{a^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {4}{3 a \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {2 b^{2}}{a^{3} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}}-\frac {b \arctan \left (\tan \left (\frac {x}{2}\right )\right )}{a^{2}}-\frac {2 \arctan \left (\tan \left (\frac {x}{2}\right )\right ) b^{3}}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(a+b*csc(x)),x)

[Out]

2/a^4*b^4/(-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^2/(tan(1/2*x)^2+1)^3*tan(1/2*
x)^5*b-2/a^3/(tan(1/2*x)^2+1)^3*b^2*tan(1/2*x)^4-4/a/(tan(1/2*x)^2+1)^3*tan(1/2*x)^2-4/a^3/(tan(1/2*x)^2+1)^3*
tan(1/2*x)^2*b^2+1/a^2/(tan(1/2*x)^2+1)^3*b*tan(1/2*x)-4/3/a/(tan(1/2*x)^2+1)^3-2/a^3/(tan(1/2*x)^2+1)^3*b^2-1
/a^2*b*arctan(tan(1/2*x))-2/a^4*arctan(tan(1/2*x))*b^3

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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)^3/(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 = 0.95, size = 1218, normalized size = 11.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(a + b/sin(x)),x)

[Out]

- ((2*(2*a^2 + 3*b^2))/(3*a^3) + (b*tan(x/2)^5)/a^2 + (2*b^2*tan(x/2)^4)/a^3 + (4*tan(x/2)^2*(a^2 + b^2))/a^3
- (b*tan(x/2))/a^2)/(3*tan(x/2)^2 + 3*tan(x/2)^4 + tan(x/2)^6 + 1) - (b^4*atan(((b^4*(a^2 - b^2)^(1/2)*((8*(4*
a^3*b^8 + 4*a^5*b^6 + a^7*b^4))/a^8 + (8*tan(x/2)*(4*a^5*b^7 - 8*a^3*b^9 + 7*a^7*b^5 + 2*a^9*b^3))/a^9 + (b^4*
(a^2 - b^2)^(1/2)*((8*(2*a^8*b^4 + 2*a^10*b^2))/a^8 + (64*b^5*tan(x/2))/a + (b^4*(a^2 - b^2)^(1/2)*(32*a^3*b^2
+ (8*tan(x/2)*(12*a^13*b - 8*a^11*b^3))/a^9))/(a^6 - a^4*b^2)))/(a^6 - a^4*b^2))*1i)/(a^6 - a^4*b^2) + (b^4*(
a^2 - b^2)^(1/2)*((8*(4*a^3*b^8 + 4*a^5*b^6 + a^7*b^4))/a^8 + (8*tan(x/2)*(4*a^5*b^7 - 8*a^3*b^9 + 7*a^7*b^5 +
2*a^9*b^3))/a^9 - (b^4*(a^2 - b^2)^(1/2)*((8*(2*a^8*b^4 + 2*a^10*b^2))/a^8 + (64*b^5*tan(x/2))/a - (b^4*(a^2
- b^2)^(1/2)*(32*a^3*b^2 + (8*tan(x/2)*(12*a^13*b - 8*a^11*b^3))/a^9))/(a^6 - a^4*b^2)))/(a^6 - a^4*b^2))*1i)/
(a^6 - a^4*b^2))/((16*(2*b^10 + a^2*b^8))/a^8 + (16*tan(x/2)*(8*b^11 + 8*a^2*b^9 + 2*a^4*b^7))/a^9 + (b^4*(a^2
- b^2)^(1/2)*((8*(4*a^3*b^8 + 4*a^5*b^6 + a^7*b^4))/a^8 + (8*tan(x/2)*(4*a^5*b^7 - 8*a^3*b^9 + 7*a^7*b^5 + 2*
a^9*b^3))/a^9 + (b^4*(a^2 - b^2)^(1/2)*((8*(2*a^8*b^4 + 2*a^10*b^2))/a^8 + (64*b^5*tan(x/2))/a + (b^4*(a^2 - b
^2)^(1/2)*(32*a^3*b^2 + (8*tan(x/2)*(12*a^13*b - 8*a^11*b^3))/a^9))/(a^6 - a^4*b^2)))/(a^6 - a^4*b^2)))/(a^6 -
a^4*b^2) - (b^4*(a^2 - b^2)^(1/2)*((8*(4*a^3*b^8 + 4*a^5*b^6 + a^7*b^4))/a^8 + (8*tan(x/2)*(4*a^5*b^7 - 8*a^3
*b^9 + 7*a^7*b^5 + 2*a^9*b^3))/a^9 - (b^4*(a^2 - b^2)^(1/2)*((8*(2*a^8*b^4 + 2*a^10*b^2))/a^8 + (64*b^5*tan(x/
2))/a - (b^4*(a^2 - b^2)^(1/2)*(32*a^3*b^2 + (8*tan(x/2)*(12*a^13*b - 8*a^11*b^3))/a^9))/(a^6 - a^4*b^2)))/(a^
6 - a^4*b^2)))/(a^6 - a^4*b^2)))*(a^2 - b^2)^(1/2)*2i)/(a^6 - a^4*b^2) - (b*atan((8*b^4*tan(x/2))/(8*b^4 + (40
*b^6)/a^2 + (48*b^8)/a^4) + (40*b^6*tan(x/2))/(40*b^6 + 8*a^2*b^4 + (48*b^8)/a^2) + (48*b^8*tan(x/2))/(48*b^8
+ 40*a^2*b^6 + 8*a^4*b^4))*(a^2 + 2*b^2))/a^4

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)**3/(a+b*csc(x)),x)

[Out]

Timed out

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