3.171 \(\int x^2 \cot ^{-1}(c+d \cot (a+b x)) \, dx\)

Optimal. Leaf size=399 \[ \frac {\text {Li}_4\left (\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{8 b^3}-\frac {\text {Li}_4\left (\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{8 b^3}-\frac {i x \text {Li}_3\left (\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{4 b^2}+\frac {i x \text {Li}_3\left (\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}-\frac {x^2 \text {Li}_2\left (\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{4 b}+\frac {x^2 \text {Li}_2\left (\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}-\frac {1}{6} i x^3 \log \left (1-\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac {1}{3} x^3 \cot ^{-1}(d \cot (a+b x)+c) \]

[Out]

1/3*x^3*arccot(c+d*cot(b*x+a))-1/6*I*x^3*ln(1-(1+I*c-d)*exp(2*I*a+2*I*b*x)/(1+I*c+d))+1/6*I*x^3*ln(1-(c+I*(1+d
))*exp(2*I*a+2*I*b*x)/(c+I*(1-d)))-1/4*x^2*polylog(2,(1+I*c-d)*exp(2*I*a+2*I*b*x)/(1+I*c+d))/b+1/4*x^2*polylog
(2,(c+I*(1+d))*exp(2*I*a+2*I*b*x)/(c+I*(1-d)))/b-1/4*I*x*polylog(3,(1+I*c-d)*exp(2*I*a+2*I*b*x)/(1+I*c+d))/b^2
+1/4*I*x*polylog(3,(c+I*(1+d))*exp(2*I*a+2*I*b*x)/(c+I*(1-d)))/b^2+1/8*polylog(4,(1+I*c-d)*exp(2*I*a+2*I*b*x)/
(1+I*c+d))/b^3-1/8*polylog(4,(c+I*(1+d))*exp(2*I*a+2*I*b*x)/(c+I*(1-d)))/b^3

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Rubi [A]  time = 0.51, antiderivative size = 399, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 6, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {5178, 2190, 2531, 6609, 2282, 6589} \[ -\frac {i x \text {PolyLog}\left (3,\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{4 b^2}+\frac {i x \text {PolyLog}\left (3,\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}+\frac {\text {PolyLog}\left (4,\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{8 b^3}-\frac {\text {PolyLog}\left (4,\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{8 b^3}-\frac {x^2 \text {PolyLog}\left (2,\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )}{4 b}+\frac {x^2 \text {PolyLog}\left (2,\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}-\frac {1}{6} i x^3 \log \left (1-\frac {(i c-d+1) e^{2 i a+2 i b x}}{i c+d+1}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (d+1)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )+\frac {1}{3} x^3 \cot ^{-1}(d \cot (a+b x)+c) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCot[c + d*Cot[a + b*x]],x]

[Out]

(x^3*ArcCot[c + d*Cot[a + b*x]])/3 - (I/6)*x^3*Log[1 - ((1 + I*c - d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)]
+ (I/6)*x^3*Log[1 - ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 - d))] - (x^2*PolyLog[2, ((1 + I*c - d
)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)])/(4*b) + (x^2*PolyLog[2, ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(
c + I*(1 - d))])/(4*b) - ((I/4)*x*PolyLog[3, ((1 + I*c - d)*E^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)])/b^2 + ((I
/4)*x*PolyLog[3, ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(1 - d))])/b^2 + PolyLog[4, ((1 + I*c - d)*E
^((2*I)*a + (2*I)*b*x))/(1 + I*c + d)]/(8*b^3) - PolyLog[4, ((c + I*(1 + d))*E^((2*I)*a + (2*I)*b*x))/(c + I*(
1 - d))]/(8*b^3)

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 5178

Int[ArcCot[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^(m
+ 1)*ArcCot[c + d*Cot[a + b*x]])/(f*(m + 1)), x] + (-Dist[(b*(1 + I*c - d))/(f*(m + 1)), Int[((e + f*x)^(m + 1
)*E^(2*I*a + 2*I*b*x))/(1 + I*c + d - (1 + I*c - d)*E^(2*I*a + 2*I*b*x)), x], x] + Dist[(b*(1 - I*c + d))/(f*(
m + 1)), Int[((e + f*x)^(m + 1)*E^(2*I*a + 2*I*b*x))/(1 - I*c - d - (1 - I*c + d)*E^(2*I*a + 2*I*b*x)), x], x]
) /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && NeQ[(c - I*d)^2, -1]

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]

Rule 6609

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

Rubi steps

\begin {align*} \int x^2 \cot ^{-1}(c+d \cot (a+b x)) \, dx &=\frac {1}{3} x^3 \cot ^{-1}(c+d \cot (a+b x))-\frac {1}{3} (b (1+i c-d)) \int \frac {e^{2 i a+2 i b x} x^3}{1+i c+d+(-1-i c+d) e^{2 i a+2 i b x}} \, dx+\frac {1}{3} (b (1-i c+d)) \int \frac {e^{2 i a+2 i b x} x^3}{1-i c-d+(-1+i c-d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{3} x^3 \cot ^{-1}(c+d \cot (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )-\frac {1}{2} i \int x^2 \log \left (1+\frac {(-1+i c-d) e^{2 i a+2 i b x}}{1-i c-d}\right ) \, dx+\frac {1}{2} i \int x^2 \log \left (1+\frac {(-1-i c+d) e^{2 i a+2 i b x}}{1+i c+d}\right ) \, dx\\ &=\frac {1}{3} x^3 \cot ^{-1}(c+d \cot (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )-\frac {x^2 \text {Li}_2\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}+\frac {x^2 \text {Li}_2\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}-\frac {\int x \text {Li}_2\left (-\frac {(-1+i c-d) e^{2 i a+2 i b x}}{1-i c-d}\right ) \, dx}{2 b}+\frac {\int x \text {Li}_2\left (-\frac {(-1-i c+d) e^{2 i a+2 i b x}}{1+i c+d}\right ) \, dx}{2 b}\\ &=\frac {1}{3} x^3 \cot ^{-1}(c+d \cot (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )-\frac {x^2 \text {Li}_2\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}+\frac {x^2 \text {Li}_2\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}-\frac {i x \text {Li}_3\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b^2}+\frac {i x \text {Li}_3\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}-\frac {i \int \text {Li}_3\left (-\frac {(-1+i c-d) e^{2 i a+2 i b x}}{1-i c-d}\right ) \, dx}{4 b^2}+\frac {i \int \text {Li}_3\left (-\frac {(-1-i c+d) e^{2 i a+2 i b x}}{1+i c+d}\right ) \, dx}{4 b^2}\\ &=\frac {1}{3} x^3 \cot ^{-1}(c+d \cot (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )-\frac {x^2 \text {Li}_2\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}+\frac {x^2 \text {Li}_2\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}-\frac {i x \text {Li}_3\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b^2}+\frac {i x \text {Li}_3\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}+\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-\frac {(-1-i c+d) x}{1+i c+d}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3\left (\frac {(c+i (1+d)) x}{c-i (-1+d)}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=\frac {1}{3} x^3 \cot ^{-1}(c+d \cot (a+b x))-\frac {1}{6} i x^3 \log \left (1-\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )+\frac {1}{6} i x^3 \log \left (1-\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )-\frac {x^2 \text {Li}_2\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b}+\frac {x^2 \text {Li}_2\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b}-\frac {i x \text {Li}_3\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{4 b^2}+\frac {i x \text {Li}_3\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{4 b^2}+\frac {\text {Li}_4\left (\frac {(1+i c-d) e^{2 i a+2 i b x}}{1+i c+d}\right )}{8 b^3}-\frac {\text {Li}_4\left (\frac {(c+i (1+d)) e^{2 i a+2 i b x}}{c+i (1-d)}\right )}{8 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.91, size = 359, normalized size = 0.90 \[ \frac {1}{3} x^3 \cot ^{-1}(d \cot (a+b x)+c)+\frac {-4 i b^3 x^3 \log \left (1-\frac {(c+i (d-1)) e^{2 i (a+b x)}}{c-i (d+1)}\right )+4 i b^3 x^3 \log \left (1-\frac {(c+i (d+1)) e^{2 i (a+b x)}}{c-i d+i}\right )-6 b^2 x^2 \text {Li}_2\left (\frac {(c+i (d-1)) e^{2 i (a+b x)}}{c-i (d+1)}\right )+6 b^2 x^2 \text {Li}_2\left (\frac {(c+i (d+1)) e^{2 i (a+b x)}}{c-i d+i}\right )-6 i b x \text {Li}_3\left (\frac {(c+i (d-1)) e^{2 i (a+b x)}}{c-i (d+1)}\right )+6 i b x \text {Li}_3\left (\frac {(c+i (d+1)) e^{2 i (a+b x)}}{c-i d+i}\right )+3 \text {Li}_4\left (\frac {(c+i (d-1)) e^{2 i (a+b x)}}{c-i (d+1)}\right )-3 \text {Li}_4\left (\frac {(c+i (d+1)) e^{2 i (a+b x)}}{c-i d+i}\right )}{24 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCot[c + d*Cot[a + b*x]],x]

[Out]

(x^3*ArcCot[c + d*Cot[a + b*x]])/3 + ((-4*I)*b^3*x^3*Log[1 - ((c + I*(-1 + d))*E^((2*I)*(a + b*x)))/(c - I*(1
+ d))] + (4*I)*b^3*x^3*Log[1 - ((c + I*(1 + d))*E^((2*I)*(a + b*x)))/(I + c - I*d)] - 6*b^2*x^2*PolyLog[2, ((c
+ I*(-1 + d))*E^((2*I)*(a + b*x)))/(c - I*(1 + d))] + 6*b^2*x^2*PolyLog[2, ((c + I*(1 + d))*E^((2*I)*(a + b*x
)))/(I + c - I*d)] - (6*I)*b*x*PolyLog[3, ((c + I*(-1 + d))*E^((2*I)*(a + b*x)))/(c - I*(1 + d))] + (6*I)*b*x*
PolyLog[3, ((c + I*(1 + d))*E^((2*I)*(a + b*x)))/(I + c - I*d)] + 3*PolyLog[4, ((c + I*(-1 + d))*E^((2*I)*(a +
b*x)))/(c - I*(1 + d))] - 3*PolyLog[4, ((c + I*(1 + d))*E^((2*I)*(a + b*x)))/(I + c - I*d)])/(24*b^3)

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fricas [C]  time = 0.89, size = 1585, normalized size = 3.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*b^3*x^3*arccot(d*cot(b*x + a) + c) - 6*b^2*x^2*dilog(-(c^2 + d^2 - (c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*
x + 2*a) + (-I*c^2 + 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a) + 2*d + 1)/(c^2 + d^2 + 2*d + 1) + 1) - 6*b^2*x^2*dil
og(-(c^2 + d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a) + 2
*d + 1)/(c^2 + d^2 + 2*d + 1) + 1) + 6*b^2*x^2*dilog(-(c^2 + d^2 - (c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a)
+ (-I*c^2 + 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a) - 2*d + 1)/(c^2 + d^2 - 2*d + 1) + 1) + 6*b^2*x^2*dilog(-(c^2
+ d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a) - 2*d + 1)/(
c^2 + d^2 - 2*d + 1) + 1) + 4*I*a^3*log(1/2*c^2 + I*c*d - 1/2*d^2 - 1/2*(c^2 + d^2 + 2*d + 1)*cos(2*b*x + 2*a)
+ 1/2*(I*c^2 + I*d^2 + 2*I*d + I)*sin(2*b*x + 2*a) + 1/2) - 4*I*a^3*log(1/2*c^2 + I*c*d - 1/2*d^2 - 1/2*(c^2
+ d^2 - 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 - 2*I*d + I)*sin(2*b*x + 2*a) + 1/2) - 4*I*a^3*log(-1/2
*c^2 + I*c*d + 1/2*d^2 + 1/2*(c^2 + d^2 + 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 + 2*I*d + I)*sin(2*b*
x + 2*a) - 1/2) + 4*I*a^3*log(-1/2*c^2 + I*c*d + 1/2*d^2 + 1/2*(c^2 + d^2 - 2*d + 1)*cos(2*b*x + 2*a) + 1/2*(I
*c^2 + I*d^2 - 2*I*d + I)*sin(2*b*x + 2*a) - 1/2) - 6*I*b*x*polylog(3, ((c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x +
2*a) + (I*c^2 - 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a))/(c^2 + d^2 + 2*d + 1)) + 6*I*b*x*polylog(3, ((c^2 + 2*I*c
*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 - 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a))/(c^2 + d^2 - 2*d + 1)) + 6*I*b*
x*polylog(3, ((c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 - 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a))/(c^2
+ d^2 + 2*d + 1)) - 6*I*b*x*polylog(3, ((c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 - 2*c*d + I*d^2
- I)*sin(2*b*x + 2*a))/(c^2 + d^2 - 2*d + 1)) + (-4*I*b^3*x^3 - 4*I*a^3)*log((c^2 + d^2 - (c^2 + 2*I*c*d - d^2
+ 1)*cos(2*b*x + 2*a) + (-I*c^2 + 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a) + 2*d + 1)/(c^2 + d^2 + 2*d + 1)) + (4*
I*b^3*x^3 + 4*I*a^3)*log((c^2 + d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d - I*d^2 + I)
*sin(2*b*x + 2*a) + 2*d + 1)/(c^2 + d^2 + 2*d + 1)) + (4*I*b^3*x^3 + 4*I*a^3)*log((c^2 + d^2 - (c^2 + 2*I*c*d
- d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 + 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a) - 2*d + 1)/(c^2 + d^2 - 2*d + 1))
+ (-4*I*b^3*x^3 - 4*I*a^3)*log((c^2 + d^2 - (c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*c*d - I*d^
2 + I)*sin(2*b*x + 2*a) - 2*d + 1)/(c^2 + d^2 - 2*d + 1)) + 3*polylog(4, ((c^2 + 2*I*c*d - d^2 + 1)*cos(2*b*x
+ 2*a) + (I*c^2 - 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a))/(c^2 + d^2 + 2*d + 1)) - 3*polylog(4, ((c^2 + 2*I*c*d -
d^2 + 1)*cos(2*b*x + 2*a) + (I*c^2 - 2*c*d - I*d^2 + I)*sin(2*b*x + 2*a))/(c^2 + d^2 - 2*d + 1)) + 3*polylog(
4, ((c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 - 2*c*d + I*d^2 - I)*sin(2*b*x + 2*a))/(c^2 + d^2 + 2
*d + 1)) - 3*polylog(4, ((c^2 - 2*I*c*d - d^2 + 1)*cos(2*b*x + 2*a) + (-I*c^2 - 2*c*d + I*d^2 - I)*sin(2*b*x +
2*a))/(c^2 + d^2 - 2*d + 1)))/b^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^2*arccot(d*cot(b*x + a) + c), x)

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maple [C]  time = 75.34, size = 7900, normalized size = 19.80 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccot(c+d*cot(b*x+a)),x)

[Out]

result too large to display

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{6} \, x^{3} \arctan \left ({\left (d + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + c \sin \left (2 \, b x + 2 \, a\right ) + d - 1, c \cos \left (2 \, b x + 2 \, a\right ) - {\left (d + 1\right )} \sin \left (2 \, b x + 2 \, a\right ) - c\right ) - \frac {1}{6} \, x^{3} \arctan \left ({\left (d - 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + c \sin \left (2 \, b x + 2 \, a\right ) + d + 1, c \cos \left (2 \, b x + 2 \, a\right ) - {\left (d - 1\right )} \sin \left (2 \, b x + 2 \, a\right ) - c\right ) - 4 \, b d \int \frac {2 \, {\left (c^{2} + d^{2} + 1\right )} x^{3} \cos \left (2 \, b x + 2 \, a\right )^{2} + 2 \, c d x^{3} \sin \left (2 \, b x + 2 \, a\right ) + 2 \, {\left (c^{2} + d^{2} + 1\right )} x^{3} \sin \left (2 \, b x + 2 \, a\right )^{2} - {\left (c^{2} - d^{2} + 1\right )} x^{3} \cos \left (2 \, b x + 2 \, a\right ) - {\left (2 \, c d x^{3} \sin \left (2 \, b x + 2 \, a\right ) + {\left (c^{2} - d^{2} + 1\right )} x^{3} \cos \left (2 \, b x + 2 \, a\right )\right )} \cos \left (4 \, b x + 4 \, a\right ) + {\left (2 \, c d x^{3} \cos \left (2 \, b x + 2 \, a\right ) - {\left (c^{2} - d^{2} + 1\right )} x^{3} \sin \left (2 \, b x + 2 \, a\right )\right )} \sin \left (4 \, b x + 4 \, a\right )}{3 \, {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} - 1\right )} d^{2} + {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} - 1\right )} d^{2} + 2 \, c^{2} + 1\right )} \cos \left (4 \, b x + 4 \, a\right )^{2} + 4 \, {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} + 1\right )} d^{2} + 2 \, c^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right )^{2} + {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} - 1\right )} d^{2} + 2 \, c^{2} + 1\right )} \sin \left (4 \, b x + 4 \, a\right )^{2} + 4 \, {\left (c^{4} + d^{4} + 2 \, {\left (c^{2} + 1\right )} d^{2} + 2 \, c^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )^{2} + 2 \, c^{2} + 2 \, {\left (c^{4} + d^{4} - 2 \, {\left (3 \, c^{2} + 1\right )} d^{2} + 2 \, c^{2} - 2 \, {\left (c^{4} - d^{4} + 2 \, c^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) - 4 \, {\left (c d^{3} + {\left (c^{3} + c\right )} d\right )} \sin \left (2 \, b x + 2 \, a\right ) + 1\right )} \cos \left (4 \, b x + 4 \, a\right ) - 4 \, {\left (c^{4} - d^{4} + 2 \, c^{2} + 1\right )} \cos \left (2 \, b x + 2 \, a\right ) + 4 \, {\left (2 \, c d^{3} - 2 \, {\left (c^{3} + c\right )} d + 2 \, {\left (c d^{3} + {\left (c^{3} + c\right )} d\right )} \cos \left (2 \, b x + 2 \, a\right ) - {\left (c^{4} - d^{4} + 2 \, c^{2} + 1\right )} \sin \left (2 \, b x + 2 \, a\right )\right )} \sin \left (4 \, b x + 4 \, a\right ) + 8 \, {\left (c d^{3} + {\left (c^{3} + c\right )} d\right )} \sin \left (2 \, b x + 2 \, a\right ) + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x^3*arctan2((d + 1)*cos(2*b*x + 2*a) + c*sin(2*b*x + 2*a) + d - 1, c*cos(2*b*x + 2*a) - (d + 1)*sin(2*b*x
+ 2*a) - c) - 1/6*x^3*arctan2((d - 1)*cos(2*b*x + 2*a) + c*sin(2*b*x + 2*a) + d + 1, c*cos(2*b*x + 2*a) - (d -
1)*sin(2*b*x + 2*a) - c) - 4*b*d*integrate(1/3*(2*(c^2 + d^2 + 1)*x^3*cos(2*b*x + 2*a)^2 + 2*c*d*x^3*sin(2*b*
x + 2*a) + 2*(c^2 + d^2 + 1)*x^3*sin(2*b*x + 2*a)^2 - (c^2 - d^2 + 1)*x^3*cos(2*b*x + 2*a) - (2*c*d*x^3*sin(2*
b*x + 2*a) + (c^2 - d^2 + 1)*x^3*cos(2*b*x + 2*a))*cos(4*b*x + 4*a) + (2*c*d*x^3*cos(2*b*x + 2*a) - (c^2 - d^2
+ 1)*x^3*sin(2*b*x + 2*a))*sin(4*b*x + 4*a))/(c^4 + d^4 + 2*(c^2 - 1)*d^2 + (c^4 + d^4 + 2*(c^2 - 1)*d^2 + 2*
c^2 + 1)*cos(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 + 1)*d^2 + 2*c^2 + 1)*cos(2*b*x + 2*a)^2 + (c^4 + d^4 + 2*
(c^2 - 1)*d^2 + 2*c^2 + 1)*sin(4*b*x + 4*a)^2 + 4*(c^4 + d^4 + 2*(c^2 + 1)*d^2 + 2*c^2 + 1)*sin(2*b*x + 2*a)^2
+ 2*c^2 + 2*(c^4 + d^4 - 2*(3*c^2 + 1)*d^2 + 2*c^2 - 2*(c^4 - d^4 + 2*c^2 + 1)*cos(2*b*x + 2*a) - 4*(c*d^3 +
(c^3 + c)*d)*sin(2*b*x + 2*a) + 1)*cos(4*b*x + 4*a) - 4*(c^4 - d^4 + 2*c^2 + 1)*cos(2*b*x + 2*a) + 4*(2*c*d^3
- 2*(c^3 + c)*d + 2*(c*d^3 + (c^3 + c)*d)*cos(2*b*x + 2*a) - (c^4 - d^4 + 2*c^2 + 1)*sin(2*b*x + 2*a))*sin(4*b
*x + 4*a) + 8*(c*d^3 + (c^3 + c)*d)*sin(2*b*x + 2*a) + 1), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*acot(c + d*cot(a + b*x)),x)

[Out]

int(x^2*acot(c + d*cot(a + b*x)), x)

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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(x**2*acot(c+d*cot(b*x+a)),x)

[Out]

Timed out

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