3.6.52 \(\int \frac {\cot ^3(x)}{a+b \sin ^3(x)} \, dx\) [552]

Optimal. Leaf size=153 \[ \frac {b^{2/3} \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}-\frac {\csc ^2(x)}{2 a}-\frac {\log (\sin (x))}{a}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (x)\right )}{3 a^{5/3}}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (x)+b^{2/3} \sin ^2(x)\right )}{6 a^{5/3}}+\frac {\log \left (a+b \sin ^3(x)\right )}{3 a} \]

[Out]

-1/2*csc(x)^2/a-ln(sin(x))/a-1/3*b^(2/3)*ln(a^(1/3)+b^(1/3)*sin(x))/a^(5/3)+1/6*b^(2/3)*ln(a^(2/3)-a^(1/3)*b^(
1/3)*sin(x)+b^(2/3)*sin(x)^2)/a^(5/3)+1/3*ln(a+b*sin(x)^3)/a+1/3*b^(2/3)*arctan(1/3*(a^(1/3)-2*b^(1/3)*sin(x))
/a^(1/3)*3^(1/2))/a^(5/3)*3^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3309, 1848, 1885, 206, 31, 648, 631, 210, 642, 266} \begin {gather*} \frac {b^{2/3} \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \sin (x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3}}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (x)+b^{2/3} \sin ^2(x)\right )}{6 a^{5/3}}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (x)\right )}{3 a^{5/3}}+\frac {\log \left (a+b \sin ^3(x)\right )}{3 a}-\frac {\csc ^2(x)}{2 a}-\frac {\log (\sin (x))}{a} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cot[x]^3/(a + b*Sin[x]^3),x]

[Out]

(b^(2/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*Sin[x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(5/3)) - Csc[x]^2/(2*a) - Log[Sin[
x]]/a - (b^(2/3)*Log[a^(1/3) + b^(1/3)*Sin[x]])/(3*a^(5/3)) + (b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Sin[x] +
b^(2/3)*Sin[x]^2])/(6*a^(5/3)) + Log[a + b*Sin[x]^3]/(3*a)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

Rule 210

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

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1848

Int[((Pq_)*((c_.)*(x_))^(m_.))/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(Pq/(a + b*x
^n)), x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IntegerQ[n] &&  !IGtQ[m, 0]

Rule 1885

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 3309

Int[((a_) + (b_.)*((c_.)*sin[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[x^m*((a + b*(c*ff*x)^n)^p/(1 - ff^2*x^2)^((
m + 1)/2)), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && ILtQ[(m - 1)/2, 0]

Rubi steps

\begin {align*} \int \frac {\cot ^3(x)}{a+b \sin ^3(x)} \, dx &=\text {Subst}\left (\int \frac {1-x^2}{x^3 \left (a+b x^3\right )} \, dx,x,\sin (x)\right )\\ &=\text {Subst}\left (\int \left (\frac {1}{a x^3}-\frac {1}{a x}+\frac {b \left (-1+x^2\right )}{a \left (a+b x^3\right )}\right ) \, dx,x,\sin (x)\right )\\ &=-\frac {\csc ^2(x)}{2 a}-\frac {\log (\sin (x))}{a}+\frac {b \text {Subst}\left (\int \frac {-1+x^2}{a+b x^3} \, dx,x,\sin (x)\right )}{a}\\ &=-\frac {\csc ^2(x)}{2 a}-\frac {\log (\sin (x))}{a}-\frac {b \text {Subst}\left (\int \frac {1}{a+b x^3} \, dx,x,\sin (x)\right )}{a}+\frac {b \text {Subst}\left (\int \frac {x^2}{a+b x^3} \, dx,x,\sin (x)\right )}{a}\\ &=-\frac {\csc ^2(x)}{2 a}-\frac {\log (\sin (x))}{a}+\frac {\log \left (a+b \sin ^3(x)\right )}{3 a}-\frac {b \text {Subst}\left (\int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,\sin (x)\right )}{3 a^{5/3}}-\frac {b \text {Subst}\left (\int \frac {2 \sqrt [3]{a}-\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (x)\right )}{3 a^{5/3}}\\ &=-\frac {\csc ^2(x)}{2 a}-\frac {\log (\sin (x))}{a}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (x)\right )}{3 a^{5/3}}+\frac {\log \left (a+b \sin ^3(x)\right )}{3 a}+\frac {b^{2/3} \text {Subst}\left (\int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (x)\right )}{6 a^{5/3}}-\frac {b \text {Subst}\left (\int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,\sin (x)\right )}{2 a^{4/3}}\\ &=-\frac {\csc ^2(x)}{2 a}-\frac {\log (\sin (x))}{a}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (x)\right )}{3 a^{5/3}}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (x)+b^{2/3} \sin ^2(x)\right )}{6 a^{5/3}}+\frac {\log \left (a+b \sin ^3(x)\right )}{3 a}-\frac {b^{2/3} \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} \sin (x)}{\sqrt [3]{a}}\right )}{a^{5/3}}\\ &=\frac {b^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} \sin (x)}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt {3} a^{5/3}}-\frac {\csc ^2(x)}{2 a}-\frac {\log (\sin (x))}{a}-\frac {b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (x)\right )}{3 a^{5/3}}+\frac {b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \sin (x)+b^{2/3} \sin ^2(x)\right )}{6 a^{5/3}}+\frac {\log \left (a+b \sin ^3(x)\right )}{3 a}\\ \end {align*}

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Mathematica [A]
time = 0.23, size = 143, normalized size = 0.93 \begin {gather*} \frac {-3 a^{2/3} \csc ^2(x)-6 a^{2/3} \log (\sin (x))+2 \left (a^{2/3}-(-1)^{2/3} b^{2/3}\right ) \log \left (-(-1)^{2/3} \sqrt [3]{a}-\sqrt [3]{b} \sin (x)\right )+2 \left (a^{2/3}-b^{2/3}\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \sin (x)\right )+2 \left (a^{2/3}+\sqrt [3]{-1} b^{2/3}\right ) \log \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} \sin (x)\right )}{6 a^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cot[x]^3/(a + b*Sin[x]^3),x]

[Out]

(-3*a^(2/3)*Csc[x]^2 - 6*a^(2/3)*Log[Sin[x]] + 2*(a^(2/3) - (-1)^(2/3)*b^(2/3))*Log[-((-1)^(2/3)*a^(1/3)) - b^
(1/3)*Sin[x]] + 2*(a^(2/3) - b^(2/3))*Log[a^(1/3) + b^(1/3)*Sin[x]] + 2*(a^(2/3) + (-1)^(1/3)*b^(2/3))*Log[a^(
1/3) + (-1)^(2/3)*b^(1/3)*Sin[x]])/(6*a^(5/3))

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Maple [A]
time = 0.77, size = 132, normalized size = 0.86

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{2 i x}}{\left ({\mathrm e}^{2 i x}-1\right )^{2} a}-i \left (\munderset {\textit {\_R} =\RootOf \left (27 \textit {\_Z}^{3} a^{5}-27 i a^{4} \textit {\_Z}^{2}-9 \textit {\_Z} \,a^{3}+i a^{2}-i b^{2}\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}+\left (-\frac {6 a^{2} \textit {\_R}}{b}+\frac {2 i a}{b}\right ) {\mathrm e}^{i x}-1\right )\right )-\frac {\ln \left ({\mathrm e}^{2 i x}-1\right )}{a}\)	\(110\)
default	\(\frac {\left (-\frac {\ln \left (\sin \left (x \right )+\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\ln \left (\sin ^{2}\left (x \right )-\left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (x \right )+\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 \sin \left (x \right )}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\ln \left (a +b \left (\sin ^{3}\left (x \right )\right )\right )}{3 b}\right ) b}{a}-\frac {1}{2 a \sin \left (x \right )^{2}}-\frac {\ln \left (\sin \left (x \right )\right )}{a}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/3/b/(1/b*a)^(2/3)*ln(sin(x)+(1/b*a)^(1/3))+1/6/b/(1/b*a)^(2/3)*ln(sin(x)^2-(1/b*a)^(1/3)*sin(x)+(1/b*a)^(2
/3))-1/3/b/(1/b*a)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*sin(x)-1))+1/3/b*ln(a+b*sin(x)^3))/a*b-1/
2/a/sin(x)^2-ln(sin(x))/a

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Maxima [A]
time = 0.53, size = 152, normalized size = 0.99 \begin {gather*} -\frac {\sqrt {3} {\left (b {\left (3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}} - \frac {2 \, a}{b}\right )} + 2 \, a\right )} \arctan \left (-\frac {\sqrt {3} {\left (\left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, \sin \left (x\right )\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, a^{2}} + \frac {{\left (2 \, \left (\frac {a}{b}\right )^{\frac {2}{3}} + 1\right )} \log \left (\sin \left (x\right )^{2} - \left (\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (x\right ) + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (\left (\frac {a}{b}\right )^{\frac {2}{3}} - 1\right )} \log \left (\left (\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (x\right )\right )}{3 \, a \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {\log \left (\sin \left (x\right )\right )}{a} - \frac {1}{2 \, a \sin \left (x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*sqrt(3)*(b*(3*(a/b)^(1/3) - 2*a/b) + 2*a)*arctan(-1/3*sqrt(3)*((a/b)^(1/3) - 2*sin(x))/(a/b)^(1/3))/a^2 +
1/6*(2*(a/b)^(2/3) + 1)*log(sin(x)^2 - (a/b)^(1/3)*sin(x) + (a/b)^(2/3))/(a*(a/b)^(2/3)) + 1/3*((a/b)^(2/3) -
1)*log((a/b)^(1/3) + sin(x))/(a*(a/b)^(2/3)) - log(sin(x))/a - 1/2/(a*sin(x)^2)

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Fricas [C] Result contains complex when optimal does not.
time = 38.73, size = 1764, normalized size = 11.53 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(6*sqrt(1/3)*(a*cos(x)^2 - a)*sqrt(((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)
^(1/3) - 2/a)^2*a^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a)*a +
4)/a^2)*arctan(-1/8*(2*sqrt(1/3)*sqrt((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/
3) - 2/a)^2*a^4 - 4*b^2*cos(x)^2 - 4*a*b*sin(x) - 2*(a^2*b*sin(x) - 2*a^3)*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/5
4*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a) + 4*a^2 + 4*b^2)*((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5
+ 1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a)*a^3 + 2*a^2)*sqrt(((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a
^2 - b^2)/a^5)^(1/3) - 2/a)^2*a^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/
3) - 2/a)*a + 4)/a^2) + sqrt(1/3)*((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3)
- 2/a)^2*a^5 - 8*a^2*b*sin(x) + 4*a^3 - 4*(a^3*b*sin(x) - a^4)*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 +
1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a))*sqrt(((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)
^(1/3) - 2/a)^2*a^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a)*a +
4)/a^2))/b^2) - 6*sqrt(1/3)*(a*cos(x)^2 - a)*sqrt(((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 -
b^2)/a^5)^(1/3) - 2/a)^2*a^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3) -
2/a)*a + 4)/a^2)*arctan(-1/8*(2*sqrt(1/3)*sqrt((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)
/a^5)^(1/3) - 2/a)^2*a^4 - 4*b^2*cos(x)^2 - 4*a*b*sin(x) - 2*(a^2*b*sin(x) - 2*a^3)*(3*(I*sqrt(3) + 1)*(-1/54/
a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a) + 4*a^2 + 4*b^2)*((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54
*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a)*a^3 + 2*a^2)*sqrt(((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5
+ 1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a)^2*a^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)
/a^5)^(1/3) - 2/a)*a + 4)/a^2) - sqrt(1/3)*((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^
5)^(1/3) - 2/a)^2*a^5 - 8*a^2*b*sin(x) + 4*a^3 - 4*(a^3*b*sin(x) - a^4)*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b
^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a))*sqrt(((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 -
b^2)/a^5)^(1/3) - 2/a)^2*a^2 + 4*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3) -
2/a)*a + 4)/a^2))/b^2) + (a*cos(x)^2 - a)*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)
^(1/3) - 2/a)*log(1/4*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a)^2*a^4
- b^2*cos(x)^2 + 2*a*b*sin(x) + (a^2*b*sin(x) + a^3)*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2
- b^2)/a^5)^(1/3) - 2/a) + a^2 + b^2) - ((a*cos(x)^2 - a)*(3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*
(a^2 - b^2)/a^5)^(1/3) - 2/a) + 6*cos(x)^2 - 6)*log((3*(I*sqrt(3) + 1)*(-1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 -
b^2)/a^5)^(1/3) - 2/a)^2*a^4 - 4*b^2*cos(x)^2 - 4*a*b*sin(x) - 2*(a^2*b*sin(x) - 2*a^3)*(3*(I*sqrt(3) + 1)*(-
1/54/a^3 + 1/54*b^2/a^5 + 1/54*(a^2 - b^2)/a^5)^(1/3) - 2/a) + 4*a^2 + 4*b^2) + 12*(cos(x)^2 - 1)*log(-1/2*sin
(x)) - 6)/(a*cos(x)^2 - a)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cot ^{3}{\left (x \right )}}{a + b \sin ^{3}{\left (x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.50, size = 144, normalized size = 0.94 \begin {gather*} \frac {b \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | -\left (-\frac {a}{b}\right )^{\frac {1}{3}} + \sin \left (x\right ) \right |}\right )}{3 \, a^{2}} + \frac {\log \left ({\left | b \sin \left (x\right )^{3} + a \right |}\right )}{3 \, a} - \frac {\log \left ({\left | \sin \left (x\right ) \right |}\right )}{a} - \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (\left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, \sin \left (x\right )\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{3 \, a^{2}} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (\sin \left (x\right )^{2} + \left (-\frac {a}{b}\right )^{\frac {1}{3}} \sin \left (x\right ) + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{2}} - \frac {1}{2 \, a \sin \left (x\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*(-a/b)^(1/3)*log(abs(-(-a/b)^(1/3) + sin(x)))/a^2 + 1/3*log(abs(b*sin(x)^3 + a))/a - log(abs(sin(x)))/a
- 1/3*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*((-a/b)^(1/3) + 2*sin(x))/(-a/b)^(1/3))/a^2 - 1/6*(-a*b^2)^(1/
3)*log(sin(x)^2 + (-a/b)^(1/3)*sin(x) + (-a/b)^(2/3))/a^2 - 1/2/(a*sin(x)^2)

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Mupad [B]
time = 17.54, size = 2003, normalized size = 13.09 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(x)^3/(a + b*sin(x)^3),x)

[Out]

symsum(log(-(256*(64*b^7*tan(x/2) + 32*a*b^6 - 44*a^3*b^4 + 15*a^5*b^2 - 1024*root(27*a^5*e^3 - 27*a^4*e^2 + 9
*a^3*e - a^2 + b^2, e, k)*b^8*tan(x/2)^2 - 84*a^2*b^5*tan(x/2) + 26*a^4*b^3*tan(x/2) + 48*a*b^6*tan(x/2)^2 - 1
6*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)*a^2*b^6 + 328*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3
*e - a^2 + b^2, e, k)*a^4*b^4 - 165*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)*a^6*b^2 - 70*a^3
*b^4*tan(x/2)^2 + 25*a^5*b^2*tan(x/2)^2 - 48*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^2*a^3*b
^6 - 915*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^2*a^5*b^4 + 630*root(27*a^5*e^3 - 27*a^4*e^
2 + 9*a^3*e - a^2 + b^2, e, k)^2*a^7*b^2 + 873*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^3*a^6
*b^4 - 810*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^3*a^8*b^2 + 864*root(27*a^5*e^3 - 27*a^4*
e^2 + 9*a^3*e - a^2 + b^2, e, k)^4*a^7*b^4 - 405*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^4*a
^9*b^2 - 1296*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^5*a^8*b^4 + 1215*root(27*a^5*e^3 - 27*
a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^5*a^10*b^2 - 608*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k
)*a*b^7*tan(x/2) - 8880*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^2*a^3*b^6*tan(x/2)^2 + 5067*
root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^2*a^5*b^4*tan(x/2)^2 + 1050*root(27*a^5*e^3 - 27*a^4
*e^2 + 9*a^3*e - a^2 + b^2, e, k)^2*a^7*b^2*tan(x/2)^2 + 27648*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 +
b^2, e, k)^3*a^4*b^6*tan(x/2)^2 - 15543*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^3*a^6*b^4*ta
n(x/2)^2 - 1350*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^3*a^8*b^2*tan(x/2)^2 - 27648*root(27
*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^4*a^5*b^6*tan(x/2)^2 + 10800*root(27*a^5*e^3 - 27*a^4*e^2 +
9*a^3*e - a^2 + b^2, e, k)^4*a^7*b^4*tan(x/2)^2 - 675*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e,
k)^4*a^9*b^2*tan(x/2)^2 + 9072*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^5*a^8*b^4*tan(x/2)^2
+ 2025*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^5*a^10*b^2*tan(x/2)^2 + 1566*root(27*a^5*e^3
- 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)*a^3*b^5*tan(x/2) - 610*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2
+ b^2, e, k)*a^5*b^3*tan(x/2) + 1760*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)*a^2*b^6*tan(x/2
)^2 - 260*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)*a^4*b^4*tan(x/2)^2 - 275*root(27*a^5*e^3 -
27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)*a^6*b^2*tan(x/2)^2 + 1536*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^
2 + b^2, e, k)^2*a^2*b^7*tan(x/2) - 9870*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^2*a^4*b^5*t
an(x/2) + 5238*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^2*a^6*b^3*tan(x/2) + 31968*root(27*a^
5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^3*a^5*b^5*tan(x/2) - 21150*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^
3*e - a^2 + b^2, e, k)^3*a^7*b^3*tan(x/2) - 57888*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^4*
a^6*b^5*tan(x/2) + 40824*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^4*a^8*b^3*tan(x/2) + 41472*
root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2, e, k)^5*a^7*b^5*tan(x/2) - 30456*root(27*a^5*e^3 - 27*a^4*
e^2 + 9*a^3*e - a^2 + b^2, e, k)^5*a^9*b^3*tan(x/2)))/a^3)*root(27*a^5*e^3 - 27*a^4*e^2 + 9*a^3*e - a^2 + b^2,
e, k), k, 1, 3) - 1/(8*a*tan(x/2)^2) - tan(x/2)^2/(8*a) - log(tan(x/2))/a

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