3.12 \(\int x^2 \cot ^3(a+b x) \, dx\)

Optimal. Leaf size=126 \[ -\frac {\text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}+\frac {\log (\sin (a+b x))}{b^3}+\frac {i x \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {x \cot (a+b x)}{b^2}-\frac {x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {x^2 \cot ^2(a+b x)}{2 b}-\frac {x^2}{2 b}+\frac {i x^3}{3} \]

[Out]

-1/2*x^2/b+1/3*I*x^3-x*cot(b*x+a)/b^2-1/2*x^2*cot(b*x+a)^2/b-x^2*ln(1-exp(2*I*(b*x+a)))/b+ln(sin(b*x+a))/b^3+I
*x*polylog(2,exp(2*I*(b*x+a)))/b^2-1/2*polylog(3,exp(2*I*(b*x+a)))/b^3

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Rubi [A]  time = 0.19, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3720, 3475, 30, 3717, 2190, 2531, 2282, 6589} \[ \frac {i x \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{b^2}-\frac {\text {PolyLog}\left (3,e^{2 i (a+b x)}\right )}{2 b^3}-\frac {x \cot (a+b x)}{b^2}+\frac {\log (\sin (a+b x))}{b^3}-\frac {x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {x^2 \cot ^2(a+b x)}{2 b}-\frac {x^2}{2 b}+\frac {i x^3}{3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Cot[a + b*x]^3,x]

[Out]

-x^2/(2*b) + (I/3)*x^3 - (x*Cot[a + b*x])/b^2 - (x^2*Cot[a + b*x]^2)/(2*b) - (x^2*Log[1 - E^((2*I)*(a + b*x))]
)/b + Log[Sin[a + b*x]]/b^3 + (I*x*PolyLog[2, E^((2*I)*(a + b*x))])/b^2 - PolyLog[3, E^((2*I)*(a + b*x))]/(2*b
^3)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 3475

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

Rule 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 3720

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(c + d*x)^m*(b*Tan[e
+ f*x])^(n - 1))/(f*(n - 1)), x] + (-Dist[(b*d*m)/(f*(n - 1)), Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1)
, x], x] - Dist[b^2, Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n,
1] && GtQ[m, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int x^2 \cot ^3(a+b x) \, dx &=-\frac {x^2 \cot ^2(a+b x)}{2 b}+\frac {\int x \cot ^2(a+b x) \, dx}{b}-\int x^2 \cot (a+b x) \, dx\\ &=\frac {i x^3}{3}-\frac {x \cot (a+b x)}{b^2}-\frac {x^2 \cot ^2(a+b x)}{2 b}+2 i \int \frac {e^{2 i (a+b x)} x^2}{1-e^{2 i (a+b x)}} \, dx+\frac {\int \cot (a+b x) \, dx}{b^2}-\frac {\int x \, dx}{b}\\ &=-\frac {x^2}{2 b}+\frac {i x^3}{3}-\frac {x \cot (a+b x)}{b^2}-\frac {x^2 \cot ^2(a+b x)}{2 b}-\frac {x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}+\frac {\log (\sin (a+b x))}{b^3}+\frac {2 \int x \log \left (1-e^{2 i (a+b x)}\right ) \, dx}{b}\\ &=-\frac {x^2}{2 b}+\frac {i x^3}{3}-\frac {x \cot (a+b x)}{b^2}-\frac {x^2 \cot ^2(a+b x)}{2 b}-\frac {x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}+\frac {\log (\sin (a+b x))}{b^3}+\frac {i x \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {i \int \text {Li}_2\left (e^{2 i (a+b x)}\right ) \, dx}{b^2}\\ &=-\frac {x^2}{2 b}+\frac {i x^3}{3}-\frac {x \cot (a+b x)}{b^2}-\frac {x^2 \cot ^2(a+b x)}{2 b}-\frac {x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}+\frac {\log (\sin (a+b x))}{b^3}+\frac {i x \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_2(x)}{x} \, dx,x,e^{2 i (a+b x)}\right )}{2 b^3}\\ &=-\frac {x^2}{2 b}+\frac {i x^3}{3}-\frac {x \cot (a+b x)}{b^2}-\frac {x^2 \cot ^2(a+b x)}{2 b}-\frac {x^2 \log \left (1-e^{2 i (a+b x)}\right )}{b}+\frac {\log (\sin (a+b x))}{b^3}+\frac {i x \text {Li}_2\left (e^{2 i (a+b x)}\right )}{b^2}-\frac {\text {Li}_3\left (e^{2 i (a+b x)}\right )}{2 b^3}\\ \end {align*}

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Mathematica [A]  time = 5.49, size = 221, normalized size = 1.75 \[ -\frac {2 b^3 x^3 \cot (a)+3 b^2 x^2 \csc ^2(a+b x)+2 e^{-i a} \sin (a) (\cot (a)+i) \left (-b^3 x^3 \cot (a)+3 b^2 x^2 \log \left (1-e^{-i (a+b x)}\right )+3 b^2 x^2 \log \left (1+e^{-i (a+b x)}\right )+6 i b x \text {Li}_2\left (-e^{-i (a+b x)}\right )+6 i b x \text {Li}_2\left (e^{-i (a+b x)}\right )+6 \text {Li}_3\left (-e^{-i (a+b x)}\right )+6 \text {Li}_3\left (e^{-i (a+b x)}\right )+i b^3 x^3\right )+6 b x \cot (a)-6 \log (\sin (a+b x))-6 b x \csc (a) \sin (b x) \csc (a+b x)}{6 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Cot[a + b*x]^3,x]

[Out]

-1/6*(6*b*x*Cot[a] + 2*b^3*x^3*Cot[a] + 3*b^2*x^2*Csc[a + b*x]^2 - 6*Log[Sin[a + b*x]] + (2*(I + Cot[a])*(I*b^
3*x^3 - b^3*x^3*Cot[a] + 3*b^2*x^2*Log[1 - E^((-I)*(a + b*x))] + 3*b^2*x^2*Log[1 + E^((-I)*(a + b*x))] + (6*I)
*b*x*PolyLog[2, -E^((-I)*(a + b*x))] + (6*I)*b*x*PolyLog[2, E^((-I)*(a + b*x))] + 6*PolyLog[3, -E^((-I)*(a + b
*x))] + 6*PolyLog[3, E^((-I)*(a + b*x))])*Sin[a])/E^(I*a) - 6*b*x*Csc[a]*Csc[a + b*x]*Sin[b*x])/b^3

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fricas [C]  time = 0.66, size = 423, normalized size = 3.36 \[ \frac {4 \, b^{2} x^{2} + 4 \, b x \sin \left (2 \, b x + 2 \, a\right ) + {\left (2 i \, b x \cos \left (2 \, b x + 2 \, a\right ) - 2 i \, b x\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + {\left (-2 i \, b x \cos \left (2 \, b x + 2 \, a\right ) + 2 i \, b x\right )} {\rm Li}_2\left (\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right ) + 2 \, {\left (a^{2} - {\left (a^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) + \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (a^{2} - {\left (a^{2} - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 1\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, b x + 2 \, a\right ) - \frac {1}{2} i \, \sin \left (2 \, b x + 2 \, a\right ) + \frac {1}{2}\right ) + 2 \, {\left (b^{2} x^{2} - a^{2} - {\left (b^{2} x^{2} - a^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) + 2 \, {\left (b^{2} x^{2} - a^{2} - {\left (b^{2} x^{2} - a^{2}\right )} \cos \left (2 \, b x + 2 \, a\right )\right )} \log \left (-\cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right ) + 1\right ) - {\left (\cos \left (2 \, b x + 2 \, a\right ) - 1\right )} {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) + i \, \sin \left (2 \, b x + 2 \, a\right )\right ) - {\left (\cos \left (2 \, b x + 2 \, a\right ) - 1\right )} {\rm polylog}\left (3, \cos \left (2 \, b x + 2 \, a\right ) - i \, \sin \left (2 \, b x + 2 \, a\right )\right )}{4 \, {\left (b^{3} \cos \left (2 \, b x + 2 \, a\right ) - b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cot(b*x+a)^3,x, algorithm="fricas")

[Out]

1/4*(4*b^2*x^2 + 4*b*x*sin(2*b*x + 2*a) + (2*I*b*x*cos(2*b*x + 2*a) - 2*I*b*x)*dilog(cos(2*b*x + 2*a) + I*sin(
2*b*x + 2*a)) + (-2*I*b*x*cos(2*b*x + 2*a) + 2*I*b*x)*dilog(cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a)) + 2*(a^2 -
(a^2 - 1)*cos(2*b*x + 2*a) - 1)*log(-1/2*cos(2*b*x + 2*a) + 1/2*I*sin(2*b*x + 2*a) + 1/2) + 2*(a^2 - (a^2 - 1)
*cos(2*b*x + 2*a) - 1)*log(-1/2*cos(2*b*x + 2*a) - 1/2*I*sin(2*b*x + 2*a) + 1/2) + 2*(b^2*x^2 - a^2 - (b^2*x^2
- a^2)*cos(2*b*x + 2*a))*log(-cos(2*b*x + 2*a) + I*sin(2*b*x + 2*a) + 1) + 2*(b^2*x^2 - a^2 - (b^2*x^2 - a^2)
*cos(2*b*x + 2*a))*log(-cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a) + 1) - (cos(2*b*x + 2*a) - 1)*polylog(3, cos(2*b
*x + 2*a) + I*sin(2*b*x + 2*a)) - (cos(2*b*x + 2*a) - 1)*polylog(3, cos(2*b*x + 2*a) - I*sin(2*b*x + 2*a)))/(b
^3*cos(2*b*x + 2*a) - b^3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cot \left (b x + a\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cot(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^2*cot(b*x + a)^3, x)

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maple [B]  time = 1.01, size = 293, normalized size = 2.33 \[ \frac {2 i \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}+\frac {2 x \left (b x \,{\mathrm e}^{2 i \left (b x +a \right )}-i {\mathrm e}^{2 i \left (b x +a \right )}+i\right )}{b^{2} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{2}}-\frac {2 \polylog \left (3, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 \polylog \left (3, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}-\frac {2 i a^{2} x}{b^{2}}+\frac {\ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a^{2}}{b^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {2 i \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x^{2}}{b}+\frac {i x^{3}}{3}-\frac {\ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x^{2}}{b}-\frac {4 i a^{3}}{3 b^{3}}-\frac {2 \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{3}}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{3}}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cot(b*x+a)^3,x)

[Out]

-2*I/b^2*a^2*x+2*x*(b*x*exp(2*I*(b*x+a))-I*exp(2*I*(b*x+a))+I)/b^2/(exp(2*I*(b*x+a))-1)^2-2/b^3*polylog(3,exp(
I*(b*x+a)))-2/b^3*polylog(3,-exp(I*(b*x+a)))-4/3*I/b^3*a^3+1/b^3*ln(1-exp(I*(b*x+a)))*a^2+1/b^3*ln(exp(I*(b*x+
a))-1)+2*I/b^2*polylog(2,-exp(I*(b*x+a)))*x-1/b*ln(exp(I*(b*x+a))+1)*x^2+1/3*I*x^3-1/b*ln(1-exp(I*(b*x+a)))*x^
2+2*I/b^2*polylog(2,exp(I*(b*x+a)))*x-2/b^3*ln(exp(I*(b*x+a)))+1/b^3*ln(exp(I*(b*x+a))+1)-1/b^3*a^2*ln(exp(I*(
b*x+a))-1)+2/b^3*a^2*ln(exp(I*(b*x+a)))

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maxima [B]  time = 1.18, size = 1216, normalized size = 9.65 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*cot(b*x+a)^3,x, algorithm="maxima")

[Out]

-1/2*(a^2*(1/sin(b*x + a)^2 + log(sin(b*x + a)^2)) - 2*(2*(b*x + a)^3 - 6*(b*x + a)^2*a - (6*(b*x + a)^2 - 12*
(b*x + a)*a + 6*((b*x + a)^2 - 2*(b*x + a)*a - 1)*cos(4*b*x + 4*a) - 12*((b*x + a)^2 - 2*(b*x + a)*a - 1)*cos(
2*b*x + 2*a) + (6*I*(b*x + a)^2 - 12*I*(b*x + a)*a - 6*I)*sin(4*b*x + 4*a) + (-12*I*(b*x + a)^2 + 24*I*(b*x +
a)*a + 12*I)*sin(2*b*x + 2*a) - 6)*arctan2(sin(b*x + a), cos(b*x + a) + 1) + (6*cos(4*b*x + 4*a) - 12*cos(2*b*
x + 2*a) + 6*I*sin(4*b*x + 4*a) - 12*I*sin(2*b*x + 2*a) + 6)*arctan2(sin(b*x + a), cos(b*x + a) - 1) + (6*(b*x
+ a)^2 - 12*(b*x + a)*a + 6*((b*x + a)^2 - 2*(b*x + a)*a)*cos(4*b*x + 4*a) - 12*((b*x + a)^2 - 2*(b*x + a)*a)
*cos(2*b*x + 2*a) - (-6*I*(b*x + a)^2 + 12*I*(b*x + a)*a)*sin(4*b*x + 4*a) - (12*I*(b*x + a)^2 - 24*I*(b*x + a
)*a)*sin(2*b*x + 2*a))*arctan2(sin(b*x + a), -cos(b*x + a) + 1) + 2*((b*x + a)^3 - 3*(b*x + a)^2*a - 6*b*x - 6
*a)*cos(4*b*x + 4*a) - (4*(b*x + a)^3 - (b*x + a)^2*(12*a - 12*I) - 12*(b*x + a)*(2*I*a + 1) - 12*a)*cos(2*b*x
+ 2*a) + (12*b*x*cos(4*b*x + 4*a) - 24*b*x*cos(2*b*x + 2*a) + 12*I*b*x*sin(4*b*x + 4*a) - 24*I*b*x*sin(2*b*x
+ 2*a) + 12*b*x)*dilog(-e^(I*b*x + I*a)) + (12*b*x*cos(4*b*x + 4*a) - 24*b*x*cos(2*b*x + 2*a) + 12*I*b*x*sin(4
*b*x + 4*a) - 24*I*b*x*sin(2*b*x + 2*a) + 12*b*x)*dilog(e^(I*b*x + I*a)) - (-3*I*(b*x + a)^2 + 6*I*(b*x + a)*a
+ (-3*I*(b*x + a)^2 + 6*I*(b*x + a)*a + 3*I)*cos(4*b*x + 4*a) + (6*I*(b*x + a)^2 - 12*I*(b*x + a)*a - 6*I)*co
s(2*b*x + 2*a) + 3*((b*x + a)^2 - 2*(b*x + a)*a - 1)*sin(4*b*x + 4*a) - 6*((b*x + a)^2 - 2*(b*x + a)*a - 1)*si
n(2*b*x + 2*a) + 3*I)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x + a) + 1) - (-3*I*(b*x + a)^2 + 6*I*(b*x
+ a)*a + (-3*I*(b*x + a)^2 + 6*I*(b*x + a)*a + 3*I)*cos(4*b*x + 4*a) + (6*I*(b*x + a)^2 - 12*I*(b*x + a)*a -
6*I)*cos(2*b*x + 2*a) + 3*((b*x + a)^2 - 2*(b*x + a)*a - 1)*sin(4*b*x + 4*a) - 6*((b*x + a)^2 - 2*(b*x + a)*a
- 1)*sin(2*b*x + 2*a) + 3*I)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) - (-12*I*cos(4*b*x + 4*
a) + 24*I*cos(2*b*x + 2*a) + 12*sin(4*b*x + 4*a) - 24*sin(2*b*x + 2*a) - 12*I)*polylog(3, -e^(I*b*x + I*a)) -
(-12*I*cos(4*b*x + 4*a) + 24*I*cos(2*b*x + 2*a) + 12*sin(4*b*x + 4*a) - 24*sin(2*b*x + 2*a) - 12*I)*polylog(3,
e^(I*b*x + I*a)) - (-2*I*(b*x + a)^3 + 6*I*(b*x + a)^2*a + 12*I*b*x + 12*I*a)*sin(4*b*x + 4*a) - (4*I*(b*x +
a)^3 - 12*(b*x + a)^2*(I*a + 1) + (b*x + a)*(24*a - 12*I) - 12*I*a)*sin(2*b*x + 2*a) - 12*a)/(-6*I*cos(4*b*x +
4*a) + 12*I*cos(2*b*x + 2*a) + 6*sin(4*b*x + 4*a) - 12*sin(2*b*x + 2*a) - 6*I))/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,{\mathrm {cot}\left (a+b\,x\right )}^3 \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*cot(a + b*x)^3,x)

[Out]

int(x^2*cot(a + b*x)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \cot ^{3}{\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*cot(b*x+a)**3,x)

[Out]

Integral(x**2*cot(a + b*x)**3, x)

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