3.3.53 \(\int x^2 \coth ^{-1}(c+d \cot (a+b x)) \, dx\) [253]
Optimal. Leaf size=391 \[ \frac {1}{3} x^3 \coth ^{-1}(c+d \cot (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )-\frac {i x^2 \text {PolyLog}\left (2,\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b}+\frac {i x^2 \text {PolyLog}\left (2,\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b}+\frac {x \text {PolyLog}\left (3,\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b^2}-\frac {x \text {PolyLog}\left (3,\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b^2}+\frac {i \text {PolyLog}\left (4,\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{8 b^3}-\frac {i \text {PolyLog}\left (4,\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{8 b^3} \]
[Out]
1/3*x^3*arccoth(c+d*cot(b*x+a))+1/6*x^3*ln(1-(1-c-I*d)*exp(2*I*a+2*I*b*x)/(1-c+I*d))-1/6*x^3*ln(1-(1+c+I*d)*ex
p(2*I*a+2*I*b*x)/(1+c-I*d))-1/4*I*x^2*polylog(2,(1-c-I*d)*exp(2*I*a+2*I*b*x)/(1-c+I*d))/b+1/4*I*x^2*polylog(2,
(1+c+I*d)*exp(2*I*a+2*I*b*x)/(1+c-I*d))/b+1/4*x*polylog(3,(1-c-I*d)*exp(2*I*a+2*I*b*x)/(1-c+I*d))/b^2-1/4*x*po
lylog(3,(1+c+I*d)*exp(2*I*a+2*I*b*x)/(1+c-I*d))/b^2+1/8*I*polylog(4,(1-c-I*d)*exp(2*I*a+2*I*b*x)/(1-c+I*d))/b^
3-1/8*I*polylog(4,(1+c+I*d)*exp(2*I*a+2*I*b*x)/(1+c-I*d))/b^3
________________________________________________________________________________________
Rubi [A]
time = 0.35, antiderivative size = 391, 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 = {6405, 2221,
2611, 6744, 2320, 6724} \begin {gather*} \frac {i \text {Li}_4\left (\frac {(-c-i d+1) e^{2 i a+2 i b x}}{-c+i d+1}\right )}{8 b^3}-\frac {i \text {Li}_4\left (\frac {(c+i d+1) e^{2 i a+2 i b x}}{c-i d+1}\right )}{8 b^3}+\frac {x \text {Li}_3\left (\frac {(-c-i d+1) e^{2 i a+2 i b x}}{-c+i d+1}\right )}{4 b^2}-\frac {x \text {Li}_3\left (\frac {(c+i d+1) e^{2 i a+2 i b x}}{c-i d+1}\right )}{4 b^2}-\frac {i x^2 \text {Li}_2\left (\frac {(-c-i d+1) e^{2 i a+2 i b x}}{-c+i d+1}\right )}{4 b}+\frac {i x^2 \text {Li}_2\left (\frac {(c+i d+1) e^{2 i a+2 i b x}}{c-i d+1}\right )}{4 b}+\frac {1}{6} x^3 \log \left (1-\frac {(-c-i d+1) e^{2 i a+2 i b x}}{-c+i d+1}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(c+i d+1) e^{2 i a+2 i b x}}{c-i d+1}\right )+\frac {1}{3} x^3 \coth ^{-1}(d \cot (a+b x)+c) \end {gather*}
Antiderivative was successfully verified.
[In]
Int[x^2*ArcCoth[c + d*Cot[a + b*x]],x]
[Out]
(x^3*ArcCoth[c + d*Cot[a + b*x]])/3 + (x^3*Log[1 - ((1 - c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c + I*d)])/6 -
(x^3*Log[1 - ((1 + c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c - I*d)])/6 - ((I/4)*x^2*PolyLog[2, ((1 - c - I*d)
*E^((2*I)*a + (2*I)*b*x))/(1 - c + I*d)])/b + ((I/4)*x^2*PolyLog[2, ((1 + c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1
+ c - I*d)])/b + (x*PolyLog[3, ((1 - c - I*d)*E^((2*I)*a + (2*I)*b*x))/(1 - c + I*d)])/(4*b^2) - (x*PolyLog[3
, ((1 + c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c - I*d)])/(4*b^2) + ((I/8)*PolyLog[4, ((1 - c - I*d)*E^((2*I)*
a + (2*I)*b*x))/(1 - c + I*d)])/b^3 - ((I/8)*PolyLog[4, ((1 + c + I*d)*E^((2*I)*a + (2*I)*b*x))/(1 + c - I*d)]
)/b^3
Rule 2221
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/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], 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 2320
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 2611
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 6405
Int[ArcCoth[(c_.) + Cot[(a_.) + (b_.)*(x_)]*(d_.)]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[(e + f*x)^(m
+ 1)*(ArcCoth[c + d*Cot[a + b*x]]/(f*(m + 1))), x] + (-Dist[I*b*((1 - c - I*d)/(f*(m + 1))), Int[(e + f*x)^(m
+ 1)*(E^(2*I*a + 2*I*b*x)/(1 - c + I*d - (1 - c - I*d)*E^(2*I*a + 2*I*b*x))), x], x] + Dist[I*b*((1 + c + I*d)
/(f*(m + 1))), Int[(e + f*x)^(m + 1)*(E^(2*I*a + 2*I*b*x)/(1 + c - I*d - (1 + c + I*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 6724
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 6744
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 \coth ^{-1}(c+d \cot (a+b x)) \, dx &=\frac {1}{3} x^3 \coth ^{-1}(c+d \cot (a+b x))+\frac {1}{3} (b (i (1+c)-d)) \int \frac {e^{2 i a+2 i b x} x^3}{1+c-i d+(-1-c-i d) e^{2 i a+2 i b x}} \, dx-\frac {1}{3} (b (i-i c+d)) \int \frac {e^{2 i a+2 i b x} x^3}{1-c+i d+(-1+c+i d) e^{2 i a+2 i b x}} \, dx\\ &=\frac {1}{3} x^3 \coth ^{-1}(c+d \cot (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )+\frac {1}{2} \int x^2 \log \left (1+\frac {(-1-c-i d) e^{2 i a+2 i b x}}{1+c-i d}\right ) \, dx-\frac {1}{2} \int x^2 \log \left (1+\frac {(-1+c+i d) e^{2 i a+2 i b x}}{1-c+i d}\right ) \, dx\\ &=\frac {1}{3} x^3 \coth ^{-1}(c+d \cot (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )-\frac {i x^2 \text {Li}_2\left (\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b}+\frac {i x^2 \text {Li}_2\left (\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b}-\frac {i \int x \text {Li}_2\left (-\frac {(-1-c-i d) e^{2 i a+2 i b x}}{1+c-i d}\right ) \, dx}{2 b}+\frac {i \int x \text {Li}_2\left (-\frac {(-1+c+i d) e^{2 i a+2 i b x}}{1-c+i d}\right ) \, dx}{2 b}\\ &=\frac {1}{3} x^3 \coth ^{-1}(c+d \cot (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )-\frac {i x^2 \text {Li}_2\left (\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b}+\frac {i x^2 \text {Li}_2\left (\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b}+\frac {x \text {Li}_3\left (\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b^2}-\frac {x \text {Li}_3\left (\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b^2}+\frac {\int \text {Li}_3\left (-\frac {(-1-c-i d) e^{2 i a+2 i b x}}{1+c-i d}\right ) \, dx}{4 b^2}-\frac {\int \text {Li}_3\left (-\frac {(-1+c+i d) e^{2 i a+2 i b x}}{1-c+i d}\right ) \, dx}{4 b^2}\\ &=\frac {1}{3} x^3 \coth ^{-1}(c+d \cot (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )-\frac {i x^2 \text {Li}_2\left (\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b}+\frac {i x^2 \text {Li}_2\left (\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b}+\frac {x \text {Li}_3\left (\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b^2}-\frac {x \text {Li}_3\left (\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b^2}+\frac {i \text {Subst}\left (\int \frac {\text {Li}_3\left (\frac {(-1+c+i d) x}{-1+c-i d}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}-\frac {i \text {Subst}\left (\int \frac {\text {Li}_3\left (\frac {(1+c+i d) x}{1+c-i d}\right )}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=\frac {1}{3} x^3 \coth ^{-1}(c+d \cot (a+b x))+\frac {1}{6} x^3 \log \left (1-\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )-\frac {1}{6} x^3 \log \left (1-\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )-\frac {i x^2 \text {Li}_2\left (\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b}+\frac {i x^2 \text {Li}_2\left (\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b}+\frac {x \text {Li}_3\left (\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{4 b^2}-\frac {x \text {Li}_3\left (\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{4 b^2}+\frac {i \text {Li}_4\left (\frac {(1-c-i d) e^{2 i a+2 i b x}}{1-c+i d}\right )}{8 b^3}-\frac {i \text {Li}_4\left (\frac {(1+c+i d) e^{2 i a+2 i b x}}{1+c-i d}\right )}{8 b^3}\\ \end {align*}
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Mathematica [A]
time = 0.23, size = 339, normalized size = 0.87 \begin {gather*} \frac {1}{3} x^3 \coth ^{-1}(c+d \cot (a+b x))+\frac {4 b^3 x^3 \log \left (1-\frac {(-1+c+i d) e^{2 i (a+b x)}}{-1+c-i d}\right )-4 b^3 x^3 \log \left (1-\frac {(1+c+i d) e^{2 i (a+b x)}}{1+c-i d}\right )-6 i b^2 x^2 \text {PolyLog}\left (2,\frac {(-1+c+i d) e^{2 i (a+b x)}}{-1+c-i d}\right )+6 i b^2 x^2 \text {PolyLog}\left (2,\frac {(1+c+i d) e^{2 i (a+b x)}}{1+c-i d}\right )+6 b x \text {PolyLog}\left (3,\frac {(-1+c+i d) e^{2 i (a+b x)}}{-1+c-i d}\right )-6 b x \text {PolyLog}\left (3,\frac {(1+c+i d) e^{2 i (a+b x)}}{1+c-i d}\right )+3 i \text {PolyLog}\left (4,\frac {(-1+c+i d) e^{2 i (a+b x)}}{-1+c-i d}\right )-3 i \text {PolyLog}\left (4,\frac {(1+c+i d) e^{2 i (a+b x)}}{1+c-i d}\right )}{24 b^3} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[x^2*ArcCoth[c + d*Cot[a + b*x]],x]
[Out]
(x^3*ArcCoth[c + d*Cot[a + b*x]])/3 + (4*b^3*x^3*Log[1 - ((-1 + c + I*d)*E^((2*I)*(a + b*x)))/(-1 + c - I*d)]
- 4*b^3*x^3*Log[1 - ((1 + c + I*d)*E^((2*I)*(a + b*x)))/(1 + c - I*d)] - (6*I)*b^2*x^2*PolyLog[2, ((-1 + c + I
*d)*E^((2*I)*(a + b*x)))/(-1 + c - I*d)] + (6*I)*b^2*x^2*PolyLog[2, ((1 + c + I*d)*E^((2*I)*(a + b*x)))/(1 + c
- I*d)] + 6*b*x*PolyLog[3, ((-1 + c + I*d)*E^((2*I)*(a + b*x)))/(-1 + c - I*d)] - 6*b*x*PolyLog[3, ((1 + c +
I*d)*E^((2*I)*(a + b*x)))/(1 + c - I*d)] + (3*I)*PolyLog[4, ((-1 + c + I*d)*E^((2*I)*(a + b*x)))/(-1 + c - I*d
)] - (3*I)*PolyLog[4, ((1 + c + I*d)*E^((2*I)*(a + b*x)))/(1 + c - I*d)])/(24*b^3)
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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order
4.
time = 18.16, size = 6698, normalized size = 17.13
| | |
method |
result |
size |
| | |
risch |
\(\text {Expression too large to display}\) |
\(6698\) |
| | |
|
|
|
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*arccoth(c+d*cot(b*x+a)),x,method=_RETURNVERBOSE)
[Out]
result too large to display
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arccoth(c+d*cot(b*x+a)),x, algorithm="maxima")
[Out]
1/12*x^3*log((c^2 + d^2 + 2*c + 1)*cos(2*b*x + 2*a)^2 + 4*(c + 1)*d*sin(2*b*x + 2*a) + (c^2 + d^2 + 2*c + 1)*s
in(2*b*x + 2*a)^2 + c^2 + d^2 - 2*(c^2 - d^2 + 2*c + 1)*cos(2*b*x + 2*a) + 2*c + 1) - 1/12*x^3*log((c^2 + d^2
- 2*c + 1)*cos(2*b*x + 2*a)^2 + 4*(c - 1)*d*sin(2*b*x + 2*a) + (c^2 + d^2 - 2*c + 1)*sin(2*b*x + 2*a)^2 + c^2
+ d^2 - 2*(c^2 - d^2 - 2*c + 1)*cos(2*b*x + 2*a) - 2*c + 1) - 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*co
s(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)*co
s(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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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 1798 vs. \(2 (275) = 550\).
time = 0.56, size = 1798, normalized size = 4.60 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arccoth(c+d*cot(b*x+a)),x, algorithm="fricas")
[Out]
1/48*(8*b^3*x^3*log((d*cos(2*b*x + 2*a) + (c + 1)*sin(2*b*x + 2*a) + d)/(d*cos(2*b*x + 2*a) + (c - 1)*sin(2*b*
x + 2*a) + d)) + 6*I*b^2*x^2*dilog(-(c^2 + d^2 - (c^2 + 2*I*(c + 1)*d - d^2 + 2*c + 1)*cos(2*b*x + 2*a) + (-I*
c^2 + 2*(c + 1)*d + I*d^2 - 2*I*c - I)*sin(2*b*x + 2*a) + 2*c + 1)/(c^2 + d^2 + 2*c + 1) + 1) - 6*I*b^2*x^2*di
log(-(c^2 + d^2 - (c^2 - 2*I*(c + 1)*d - d^2 + 2*c + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*(c + 1)*d - I*d^2 + 2*I*
c + I)*sin(2*b*x + 2*a) + 2*c + 1)/(c^2 + d^2 + 2*c + 1) + 1) - 6*I*b^2*x^2*dilog(-(c^2 + d^2 - (c^2 + 2*I*(c
- 1)*d - d^2 - 2*c + 1)*cos(2*b*x + 2*a) + (-I*c^2 + 2*(c - 1)*d + I*d^2 + 2*I*c - I)*sin(2*b*x + 2*a) - 2*c +
1)/(c^2 + d^2 - 2*c + 1) + 1) + 6*I*b^2*x^2*dilog(-(c^2 + d^2 - (c^2 - 2*I*(c - 1)*d - d^2 - 2*c + 1)*cos(2*b
*x + 2*a) + (I*c^2 + 2*(c - 1)*d - I*d^2 - 2*I*c + I)*sin(2*b*x + 2*a) - 2*c + 1)/(c^2 + d^2 - 2*c + 1) + 1) +
4*a^3*log(1/2*c^2 + I*(c + 1)*d - 1/2*d^2 - 1/2*(c^2 + d^2 + 2*c + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 +
2*I*c + I)*sin(2*b*x + 2*a) + c + 1/2) - 4*a^3*log(1/2*c^2 + I*(c - 1)*d - 1/2*d^2 - 1/2*(c^2 + d^2 - 2*c + 1
)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 - 2*I*c + I)*sin(2*b*x + 2*a) - c + 1/2) + 4*a^3*log(-1/2*c^2 + I*(c +
1)*d + 1/2*d^2 + 1/2*(c^2 + d^2 + 2*c + 1)*cos(2*b*x + 2*a) + 1/2*(I*c^2 + I*d^2 + 2*I*c + I)*sin(2*b*x + 2*a
) - c - 1/2) - 4*a^3*log(-1/2*c^2 + I*(c - 1)*d + 1/2*d^2 + 1/2*(c^2 + d^2 - 2*c + 1)*cos(2*b*x + 2*a) + 1/2*(
I*c^2 + I*d^2 - 2*I*c + I)*sin(2*b*x + 2*a) + c - 1/2) - 6*b*x*polylog(3, ((c^2 + 2*I*(c + 1)*d - d^2 + 2*c +
1)*cos(2*b*x + 2*a) + (I*c^2 - 2*(c + 1)*d - I*d^2 + 2*I*c + I)*sin(2*b*x + 2*a))/(c^2 + d^2 + 2*c + 1)) - 6*b
*x*polylog(3, ((c^2 - 2*I*(c + 1)*d - d^2 + 2*c + 1)*cos(2*b*x + 2*a) + (-I*c^2 - 2*(c + 1)*d + I*d^2 - 2*I*c
- I)*sin(2*b*x + 2*a))/(c^2 + d^2 + 2*c + 1)) + 6*b*x*polylog(3, ((c^2 + 2*I*(c - 1)*d - d^2 - 2*c + 1)*cos(2*
b*x + 2*a) + (I*c^2 - 2*(c - 1)*d - I*d^2 - 2*I*c + I)*sin(2*b*x + 2*a))/(c^2 + d^2 - 2*c + 1)) + 6*b*x*polylo
g(3, ((c^2 - 2*I*(c - 1)*d - d^2 - 2*c + 1)*cos(2*b*x + 2*a) + (-I*c^2 - 2*(c - 1)*d + I*d^2 + 2*I*c - I)*sin(
2*b*x + 2*a))/(c^2 + d^2 - 2*c + 1)) - 4*(b^3*x^3 + a^3)*log((c^2 + d^2 - (c^2 + 2*I*(c + 1)*d - d^2 + 2*c + 1
)*cos(2*b*x + 2*a) + (-I*c^2 + 2*(c + 1)*d + I*d^2 - 2*I*c - I)*sin(2*b*x + 2*a) + 2*c + 1)/(c^2 + d^2 + 2*c +
1)) - 4*(b^3*x^3 + a^3)*log((c^2 + d^2 - (c^2 - 2*I*(c + 1)*d - d^2 + 2*c + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*
(c + 1)*d - I*d^2 + 2*I*c + I)*sin(2*b*x + 2*a) + 2*c + 1)/(c^2 + d^2 + 2*c + 1)) + 4*(b^3*x^3 + a^3)*log((c^2
+ d^2 - (c^2 + 2*I*(c - 1)*d - d^2 - 2*c + 1)*cos(2*b*x + 2*a) + (-I*c^2 + 2*(c - 1)*d + I*d^2 + 2*I*c - I)*s
in(2*b*x + 2*a) - 2*c + 1)/(c^2 + d^2 - 2*c + 1)) + 4*(b^3*x^3 + a^3)*log((c^2 + d^2 - (c^2 - 2*I*(c - 1)*d -
d^2 - 2*c + 1)*cos(2*b*x + 2*a) + (I*c^2 + 2*(c - 1)*d - I*d^2 - 2*I*c + I)*sin(2*b*x + 2*a) - 2*c + 1)/(c^2 +
d^2 - 2*c + 1)) - 3*I*polylog(4, ((c^2 + 2*I*(c + 1)*d - d^2 + 2*c + 1)*cos(2*b*x + 2*a) + (I*c^2 - 2*(c + 1)
*d - I*d^2 + 2*I*c + I)*sin(2*b*x + 2*a))/(c^2 + d^2 + 2*c + 1)) + 3*I*polylog(4, ((c^2 - 2*I*(c + 1)*d - d^2
+ 2*c + 1)*cos(2*b*x + 2*a) + (-I*c^2 - 2*(c + 1)*d + I*d^2 - 2*I*c - I)*sin(2*b*x + 2*a))/(c^2 + d^2 + 2*c +
1)) + 3*I*polylog(4, ((c^2 + 2*I*(c - 1)*d - d^2 - 2*c + 1)*cos(2*b*x + 2*a) + (I*c^2 - 2*(c - 1)*d - I*d^2 -
2*I*c + I)*sin(2*b*x + 2*a))/(c^2 + d^2 - 2*c + 1)) - 3*I*polylog(4, ((c^2 - 2*I*(c - 1)*d - d^2 - 2*c + 1)*co
s(2*b*x + 2*a) + (-I*c^2 - 2*(c - 1)*d + I*d^2 + 2*I*c - I)*sin(2*b*x + 2*a))/(c^2 + d^2 - 2*c + 1)))/b^3
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{2} \operatorname {acoth}{\left (c + d \cot {\left (a + b x \right )} \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2*acoth(c+d*cot(b*x+a)),x)
[Out]
Integral(x**2*acoth(c + d*cot(a + b*x)), x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*arccoth(c+d*cot(b*x+a)),x, algorithm="giac")
[Out]
integrate(x^2*arccoth(d*cot(b*x + a) + c), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^2\,\mathrm {acoth}\left (c+d\,\mathrm {cot}\left (a+b\,x\right )\right ) \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*acoth(c + d*cot(a + b*x)),x)
[Out]
int(x^2*acoth(c + d*cot(a + b*x)), x)
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