\(\int \frac {\sin ^3(x)}{a+b \cot (x)} \, dx\) [21]

Optimal result

Integrand size = 13, antiderivative size = 121 \[ \int \frac {\sin ^3(x)}{a+b \cot (x)} \, dx=\frac {b^4 \text {arctanh}\left (\frac {(b-a \cot (x)) \sin (x)}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {a b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a \cos (x)}{a^2+b^2}+\frac {a \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {b^3 \sin (x)}{\left (a^2+b^2\right )^2}-\frac {b \sin ^3(x)}{3 \left (a^2+b^2\right )} \]

[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)

Rubi [A] (verified)

Time = 0.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.462, Rules used = {3592, 3567, 2713, 2718, 3590, 212} \[ \int \frac {\sin ^3(x)}{a+b \cot (x)} \, dx=\frac {b^4 \text {arctanh}\left (\frac {\sin (x) (b-a \cot (x))}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}-\frac {b \sin ^3(x)}{3 \left (a^2+b^2\right )}+\frac {a \cos ^3(x)}{3 \left (a^2+b^2\right )}-\frac {a b^2 \cos (x)}{\left (a^2+b^2\right )^2}-\frac {a \cos (x)}{a^2+b^2}-\frac {b^3 \sin (x)}{\left (a^2+b^2\right )^2} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2713

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 2718

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

Rule 3567

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 3590

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 3592

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*} \text {integral}& = \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 \text {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 \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (x)\right )}{a^2+b^2} \\ & = \frac {b^4 \text {arctanh}\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*}

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.93 \[ \int \frac {\sin ^3(x)}{a+b \cot (x)} \, dx=\frac {2 b^4 \text {arctanh}\left (\frac {-a+b \tan \left (\frac {x}{2}\right )}{\sqrt {a^2+b^2}}\right )}{\left (a^2+b^2\right )^{5/2}}+\frac {-3 a \left (3 a^2+7 b^2\right ) \cos (x)+a \left (a^2+b^2\right ) \cos (3 x)+2 b \left (-a^2-7 b^2+\left (a^2+b^2\right ) \cos (2 x)\right ) \sin (x)}{12 \left (a^2+b^2\right )^2} \]

[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)

Maple [A] (verified)

Time = 1.80 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.35

method	result	size
default	\(\frac {-2 b^{3} \tan \left (\frac {x}{2}\right )^{5}-2 a \,b^{2} \tan \left (\frac {x}{2}\right )^{4}+2 \left (-\frac {4}{3} a^{2} b -\frac {10}{3} b^{3}\right ) \tan \left (\frac {x}{2}\right )^{3}+2 \left (-2 a^{3}-4 a \,b^{2}\right ) \tan \left (\frac {x}{2}\right )^{2}-2 b^{3} \tan \left (\frac {x}{2}\right )-\frac {4 a^{3}}{3}-\frac {10 a \,b^{2}}{3}}{\left (a^{2}+b^{2}\right )^{2} \left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{3}}-\frac {32 b^{4} \operatorname {arctanh}\left (\frac {-2 b \tan \left (\frac {x}{2}\right )+2 a}{2 \sqrt {a^{2}+b^{2}}}\right )}{\left (16 a^{4}+32 a^{2} b^{2}+16 b^{4}\right ) \sqrt {a^{2}+b^{2}}}\)	\(163\)
risch	\(-\frac {5 i {\mathrm e}^{i x} b}{8 \left (2 i a b +a^{2}-b^{2}\right )}-\frac {3 \,{\mathrm e}^{i x} a}{8 \left (2 i a b +a^{2}-b^{2}\right )}+\frac {5 i {\mathrm e}^{-i x} b}{8 \left (-i b +a \right )^{2}}-\frac {3 \,{\mathrm e}^{-i x} a}{8 \left (-i b +a \right )^{2}}-\frac {i b^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a +b}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{2}}+\frac {i b^{4} \ln \left ({\mathrm e}^{i x}-\frac {i a +b}{\sqrt {-a^{2}-b^{2}}}\right )}{\sqrt {-a^{2}-b^{2}}\, \left (a^{2}+b^{2}\right )^{2}}-\frac {a \cos \left (3 x \right )}{12 \left (-a^{2}-b^{2}\right )}-\frac {b \sin \left (3 x \right )}{12 \left (-a^{2}-b^{2}\right )}\)	\(235\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.82 \[ \int \frac {\sin ^3(x)}{a+b \cot (x)} \, dx=\frac {3 \, \sqrt {a^{2} + b^{2}} b^{4} \log \left (-\frac {2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - a^{2} - 2 \, b^{2} + 2 \, \sqrt {a^{2} + b^{2}} {\left (a \cos \left (x\right ) - b \sin \left (x\right )\right )}}{2 \, a b \cos \left (x\right ) \sin \left (x\right ) - {\left (a^{2} - b^{2}\right )} \cos \left (x\right )^{2} + a^{2}}\right ) + 2 \, {\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (x\right )^{3} - 6 \, {\left (a^{5} + 3 \, a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (x\right ) - 2 \, {\left (a^{4} b + 5 \, a^{2} b^{3} + 4 \, b^{5} - {\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}{6 \, {\left (a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}\right )}} \]

[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)

Sympy [F]

\[ \int \frac {\sin ^3(x)}{a+b \cot (x)} \, dx=\int \frac {\sin ^{3}{\left (x \right )}}{a + b \cot {\left (x \right )}}\, dx \]

[In]

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

[Out]

Integral(sin(x)**3/(a + b*cot(x)), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 283 vs. \(2 (115) = 230\).

Time = 0.35 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.34 \[ \int \frac {\sin ^3(x)}{a+b \cot (x)} \, dx=-\frac {b^{4} \log \left (\frac {a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} + \sqrt {a^{2} + b^{2}}}{a - \frac {b \sin \left (x\right )}{\cos \left (x\right ) + 1} - \sqrt {a^{2} + b^{2}}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (2 \, a^{3} + 5 \, a b^{2} + \frac {3 \, b^{3} \sin \left (x\right )}{\cos \left (x\right ) + 1} + \frac {3 \, a b^{2} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {3 \, b^{3} \sin \left (x\right )^{5}}{{\left (\cos \left (x\right ) + 1\right )}^{5}} + \frac {6 \, {\left (a^{3} + 2 \, a b^{2}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {2 \, {\left (2 \, a^{2} b + 5 \, b^{3}\right )} \sin \left (x\right )^{3}}{{\left (\cos \left (x\right ) + 1\right )}^{3}}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{2}}{{\left (\cos \left (x\right ) + 1\right )}^{2}} + \frac {3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{4}}{{\left (\cos \left (x\right ) + 1\right )}^{4}} + \frac {{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (x\right )^{6}}{{\left (\cos \left (x\right ) + 1\right )}^{6}}\right )}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.66 \[ \int \frac {\sin ^3(x)}{a+b \cot (x)} \, dx=-\frac {b^{4} \log \left (\frac {{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a - 2 \, \sqrt {a^{2} + b^{2}} \right |}}{{\left | 2 \, b \tan \left (\frac {1}{2} \, x\right ) - 2 \, a + 2 \, \sqrt {a^{2} + b^{2}} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} \sqrt {a^{2} + b^{2}}} - \frac {2 \, {\left (3 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{5} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{4} + 4 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{3} + 10 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} + 6 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{2} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, b^{3} \tan \left (\frac {1}{2} \, x\right ) + 2 \, a^{3} + 5 \, a b^{2}\right )}}{3 \, {\left (a^{4} + 2 \, a^{2} b^{2} + b^{4}\right )} {\left (\tan \left (\frac {1}{2} \, x\right )^{2} + 1\right )}^{3}} \]

[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)

Mupad [B] (verification not implemented)

Time = 12.85 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.31 \[ \int \frac {\sin ^3(x)}{a+b \cot (x)} \, dx=-\frac {\frac {4\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3\,\left (2\,a^2\,b+5\,b^3\right )}{3\,\left (a^4+2\,a^2\,b^2+b^4\right )}+\frac {2\,b^3\,\mathrm {tan}\left (\frac {x}{2}\right )}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,\left (2\,a^2+5\,b^2\right )}{3\,{\left (a^2+b^2\right )}^2}+\frac {2\,b^3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{a^4+2\,a^2\,b^2+b^4}+\frac {2\,a\,b^2\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4}{a^4+2\,a^2\,b^2+b^4}+\frac {4\,a\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2\,\left (a^2+2\,b^2\right )}{a^4+2\,a^2\,b^2+b^4}}{{\mathrm {tan}\left (\frac {x}{2}\right )}^6+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^4+3\,{\mathrm {tan}\left (\frac {x}{2}\right )}^2+1}-\frac {2\,b^4\,\mathrm {atanh}\left (\frac {2\,a\,b^4+2\,a^5+4\,a^3\,b^2-2\,b\,\mathrm {tan}\left (\frac {x}{2}\right )\,\left (a^4+2\,a^2\,b^2+b^4\right )}{2\,{\left (a^2+b^2\right )}^{5/2}}\right )}{{\left (a^2+b^2\right )}^{5/2}} \]

[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)