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

Optimal. Leaf size=121 \[ -\frac {b \sin ^3(x)}{3 \left (a^2+b^2\right )}+\frac {a \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a \cos (x)}{a^2+b^2}+\frac {b^4 \tanh ^{-1}\left (\frac {\sin (x) (b-a \cot (x))}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {b^3 \sin (x)}{\left (a^2+b^2\right )^2} \]

[Out]

b^4*arctanh((b-a*cot(x))*sin(x)/(a^2+b^2)^(1/2))/(a^2+b^2)^(5/2)-a*b^2*cos(x)/(a^2+b^2)^2-a*cos(x)/(a^2+b^2)+1
/3*a*cos(x)^3/(a^2+b^2)-b^3*sin(x)/(a^2+b^2)^2-1/3*b*sin(x)^3/(a^2+b^2)

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Rubi [A]  time = 0.16, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3511, 3486, 2633, 2638, 3509, 206} \[ -\frac {b \sin ^3(x)}{3 \left (a^2+b^2\right )}-\frac {b^3 \sin (x)}{\left (a^2+b^2\right )^2}+\frac {a \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a \cos (x)}{a^2+b^2}+\frac {b^4 \tanh ^{-1}\left (\frac {\sin (x) (b-a \cot (x))}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]^3/(a + b*Cot[x]),x]

[Out]

(b^4*ArcTanh[((b - a*Cot[x])*Sin[x])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) - (a*b^2*Cos[x])/(a^2 + b^2)^2 - (a*C
os[x])/(a^2 + b^2) + (a*Cos[x]^3)/(3*(a^2 + b^2)) - (b^3*Sin[x])/(a^2 + b^2)^2 - (b*Sin[x]^3)/(3*(a^2 + b^2))

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 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2638

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

Rule 3486

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*(d*Sec[
e + f*x])^m)/(f*m), x] + Dist[a, Int[(d*Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2
*m] || NeQ[a^2 + b^2, 0])

Rule 3509

Int[sec[(e_.) + (f_.)*(x_)]/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Dist[f^(-1), Subst[Int[1/(a^
2 + b^2 - x^2), x], x, (b - a*Tan[e + f*x])/Sec[e + f*x]], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 + b^2, 0]

Rule 3511

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^
2), Int[(d*Sec[e + f*x])^m*(a - b*Tan[e + f*x]), x], x] + Dist[b^2/(d^2*(a^2 + b^2)), Int[(d*Sec[e + f*x])^(m
+ 2)/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {\sin ^3(x)}{a+b \cot (x)} \, dx &=\frac {\int (a-b \cot (x)) \sin ^3(x) \, dx}{a^2+b^2}+\frac {b^2 \int \frac {\sin (x)}{a+b \cot (x)} \, dx}{a^2+b^2}\\ &=-\frac {b \sin ^3(x)}{3 \left (a^2+b^2\right )}+\frac {b^2 \int (a-b \cot (x)) \sin (x) \, dx}{\left (a^2+b^2\right )^2}+\frac {b^4 \int \frac {\csc (x)}{a+b \cot (x)} \, dx}{\left (a^2+b^2\right )^2}+\frac {a \int \sin ^3(x) \, dx}{a^2+b^2}\\ &=-\frac {b^3 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {b \sin ^3(x)}{3 \left (a^2+b^2\right )}+\frac {\left (a b^2\right ) \int \sin (x) \, dx}{\left (a^2+b^2\right )^2}-\frac {b^4 \operatorname {Subst}\left (\int \frac {1}{a^2+b^2-x^2} \, dx,x,(-b+a \cot (x)) \sin (x)\right )}{\left (a^2+b^2\right )^2}-\frac {a \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (x)\right )}{a^2+b^2}\\ &=\frac {b^4 \tanh ^{-1}\left (\frac {(b-a \cot (x)) \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {a b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a \cos (x)}{a^2+b^2}+\frac {a \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {b^3 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {b \sin ^3(x)}{3 \left (a^2+b^2\right )}\\ \end {align*}

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Mathematica [A]  time = 0.61, size = 113, normalized size = 0.93 \[ \frac {-3 a \left (3 a^2+7 b^2\right ) \cos (x)+a \left (a^2+b^2\right ) \cos (3 x)+2 b \sin (x) \left (\left (a^2+b^2\right ) \cos (2 x)-a^2-7 b^2\right )}{12 \left (a^2+b^2\right )^2}+\frac {2 b^4 \tanh ^{-1}\left (\frac {b \tan \left (\frac {x}{2}\right )-a}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]^3/(a + b*Cot[x]),x]

[Out]

(2*b^4*ArcTanh[(-a + b*Tan[x/2])/Sqrt[a^2 + b^2]])/(a^2 + b^2)^(5/2) + (-3*a*(3*a^2 + 7*b^2)*Cos[x] + a*(a^2 +
b^2)*Cos[3*x] + 2*b*(-a^2 - 7*b^2 + (a^2 + b^2)*Cos[2*x])*Sin[x])/(12*(a^2 + b^2)^2)

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fricas [A]  time = 0.63, size = 220, normalized size = 1.82 \[ \frac {3 \, \sqrt {a^{2} + b^{2}} b^{4} \log \left (-\frac {2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} - a^{2} - 2 \, b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \relax (x) - b \sin \relax (x)\right )}}{2 \, a b \cos \relax (x) \sin \relax (x) - {\left (a^{2} - b^{2}\right )} \cos \relax (x)^{2} + a^{2}}\right ) + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \relax (x)^{3} - 6 \, {\left (a^{5} + 3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \relax (x) - 2 \, {\left (a^{4} b + 5 \, a^{2} b^{3} + 4 \, b^{5} - {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \relax (x)^{2}\right )} \sin \relax (x)}{6 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*sqrt(a^2 + b^2)*b^4*log(-(2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 - a^2 - 2*b^2 + 2*sqrt(a^2 + b^2)*
(a*cos(x) - b*sin(x)))/(2*a*b*cos(x)*sin(x) - (a^2 - b^2)*cos(x)^2 + a^2)) + 2*(a^5 + 2*a^3*b^2 + a*b^4)*cos(x
)^3 - 6*(a^5 + 3*a^3*b^2 + 2*a*b^4)*cos(x) - 2*(a^4*b + 5*a^2*b^3 + 4*b^5 - (a^4*b + 2*a^2*b^3 + b^5)*cos(x)^2
)*sin(x))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6)

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giac [A]  time = 0.72, size = 201, normalized size = 1.66 \[ -\frac {b^{4} \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (3 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{5} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{4} + 4 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{3} + 10 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, b^{3} \tan \left (\frac {1}{2} \, x\right ) + 2 \, a^{3} + 5 \, a b^{2}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^4*log(abs(2*b*tan(1/2*x) - 2*a - 2*sqrt(a^2 + b^2))/abs(2*b*tan(1/2*x) - 2*a + 2*sqrt(a^2 + b^2)))/((a^4 +
2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2/3*(3*b^3*tan(1/2*x)^5 + 3*a*b^2*tan(1/2*x)^4 + 4*a^2*b*tan(1/2*x)^3 + 10
*b^3*tan(1/2*x)^3 + 6*a^3*tan(1/2*x)^2 + 12*a*b^2*tan(1/2*x)^2 + 3*b^3*tan(1/2*x) + 2*a^3 + 5*a*b^2)/((a^4 + 2
*a^2*b^2 + b^4)*(tan(1/2*x)^2 + 1)^3)

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maple [A]  time = 0.26, size = 163, normalized size = 1.35 \[ \frac {32 b^{4} \arctanh \left (\frac {2 \tan \left (\frac {x}{2}\right ) b -2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (16 a^{4}+32 a^{2} b^{2}+16 b^{4}\right ) \sqrt {a^{2}+b^{2}}}+\frac {-2 b^{3} \left (\tan ^{5}\left (\frac {x}{2}\right )\right )-2 b^{2} a \left (\tan ^{4}\left (\frac {x}{2}\right )\right )+2 \left (-\frac {4}{3} a^{2} b -\frac {10}{3} b^{3}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+2 \left (-2 a^{3}-4 b^{2} a \right ) \left (\tan ^{2}\left (\frac {x}{2}\right )\right )-2 b^{3} \tan \left (\frac {x}{2}\right )-\frac {4 a^{3}}{3}-\frac {10 b^{2} a}{3}}{\left (a^{2}+b^{2}\right )^{2} \left (\tan ^{2}\left (\frac {x}{2}\right )+1\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32*b^4/(16*a^4+32*a^2*b^2+16*b^4)/(a^2+b^2)^(1/2)*arctanh(1/2*(2*tan(1/2*x)*b-2*a)/(a^2+b^2)^(1/2))+2/(a^2+b^2
)^2*(-b^3*tan(1/2*x)^5-b^2*a*tan(1/2*x)^4+(-4/3*a^2*b-10/3*b^3)*tan(1/2*x)^3+(-2*a^3-4*a*b^2)*tan(1/2*x)^2-b^3
*tan(1/2*x)-2/3*a^3-5/3*b^2*a)/(tan(1/2*x)^2+1)^3

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maxima [B]  time = 0.79, size = 283, normalized size = 2.34 \[ -\frac {b^{4} \log \left (\frac {a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \relax (x)}{\cos \relax (x) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (2 \, a^{3} + 5 \, a b^{2} + \frac {3 \, b^{3} \sin \relax (x)}{\cos \relax (x) + 1} + \frac {3 \, a b^{2} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {3 \, b^{3} \sin \relax (x)^{5}}{{\left (\cos \relax (x) + 1\right )}^{5}} + \frac {6 \, {\left (a^{3} + 2 \, a b^{2}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {2 \, {\left (2 \, a^{2} b + 5 \, b^{3}\right )} \sin \relax (x)^{3}}{{\left (\cos \relax (x) + 1\right )}^{3}}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \relax (x)^{2}}{{\left (\cos \relax (x) + 1\right )}^{2}} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \relax (x)^{4}}{{\left (\cos \relax (x) + 1\right )}^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \relax (x)^{6}}{{\left (\cos \relax (x) + 1\right )}^{6}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-b^4*log((a - b*sin(x)/(cos(x) + 1) + sqrt(a^2 + b^2))/(a - b*sin(x)/(cos(x) + 1) - sqrt(a^2 + b^2)))/((a^4 +
2*a^2*b^2 + b^4)*sqrt(a^2 + b^2)) - 2/3*(2*a^3 + 5*a*b^2 + 3*b^3*sin(x)/(cos(x) + 1) + 3*a*b^2*sin(x)^4/(cos(x
) + 1)^4 + 3*b^3*sin(x)^5/(cos(x) + 1)^5 + 6*(a^3 + 2*a*b^2)*sin(x)^2/(cos(x) + 1)^2 + 2*(2*a^2*b + 5*b^3)*sin
(x)^3/(cos(x) + 1)^3)/(a^4 + 2*a^2*b^2 + b^4 + 3*(a^4 + 2*a^2*b^2 + b^4)*sin(x)^2/(cos(x) + 1)^2 + 3*(a^4 + 2*
a^2*b^2 + b^4)*sin(x)^4/(cos(x) + 1)^4 + (a^4 + 2*a^2*b^2 + b^4)*sin(x)^6/(cos(x) + 1)^6)

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mupad [B]  time = 0.65, size = 280, normalized size = 2.31 \[ -\frac {\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (2\,a^2\,b+5\,b^3\right )}{3\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {2\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,\left (2\,a^2+5\,b^2\right )}{3\,{\left (a^2+b^2\right )}^2}+\frac {2\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{a^4+2\,a^2\,b^2+b^4}+\frac {4\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a^2+2\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {2\,b^4\,\mathrm {atanh}\left (\frac {2\,a\,b^4+2\,a^5+4\,a^3\,b^2-2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{2\,{\left (a^2+b^2\right )}^{5/2}}\right )}{{\left (a^2+b^2\right )}^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)^3/(a + b*cot(x)),x)

[Out]

- ((4*tan(x/2)^3*(2*a^2*b + 5*b^3))/(3*(a^4 + b^4 + 2*a^2*b^2)) + (2*b^3*tan(x/2))/(a^4 + b^4 + 2*a^2*b^2) + (
2*a*(2*a^2 + 5*b^2))/(3*(a^2 + b^2)^2) + (2*b^3*tan(x/2)^5)/(a^4 + b^4 + 2*a^2*b^2) + (2*a*b^2*tan(x/2)^4)/(a^
4 + b^4 + 2*a^2*b^2) + (4*a*tan(x/2)^2*(a^2 + 2*b^2))/(a^4 + b^4 + 2*a^2*b^2))/(3*tan(x/2)^2 + 3*tan(x/2)^4 +
tan(x/2)^6 + 1) - (2*b^4*atanh((2*a*b^4 + 2*a^5 + 4*a^3*b^2 - 2*b*tan(x/2)*(a^4 + b^4 + 2*a^2*b^2))/(2*(a^2 +
b^2)^(5/2))))/(a^2 + b^2)^(5/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{3}{\relax (x )}}{a + b \cot {\relax (x )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(x)**3/(a + b*cot(x)), x)

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