3.179 \(\int x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx\)
Optimal. Leaf size=155 \[ -\frac {\text {Li}_4\left (-i c e^{2 i a+2 i b x}\right )}{8 b^3}+\frac {i x \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{4 b^2}+\frac {x^2 \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {b x^4}{12} \]
[Out]
1/12*b*x^4+1/3*x^3*(Pi-arccot(-c+(1+I*c)*cot(b*x+a)))+1/6*I*x^3*ln(1+I*c*exp(2*I*a+2*I*b*x))+1/4*x^2*polylog(2
,-I*c*exp(2*I*a+2*I*b*x))/b+1/4*I*x*polylog(3,-I*c*exp(2*I*a+2*I*b*x))/b^2-1/8*polylog(4,-I*c*exp(2*I*a+2*I*b*
x))/b^3
________________________________________________________________________________________
Rubi [A] time = 0.26, antiderivative size = 155, normalized size of antiderivative = 1.00,
number of steps used = 7, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used =
{5174, 2184, 2190, 2531, 6609, 2282, 6589} \[ \frac {i x \text {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {\text {PolyLog}\left (4,-i c e^{2 i a+2 i b x}\right )}{8 b^3}+\frac {x^2 \text {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {b x^4}{12} \]
Antiderivative was successfully verified.
[In]
Int[x^2*ArcCot[c - (1 + I*c)*Cot[a + b*x]],x]
[Out]
(b*x^4)/12 + (x^3*ArcCot[c - (1 + I*c)*Cot[a + b*x]])/3 + (I/6)*x^3*Log[1 + I*c*E^((2*I)*a + (2*I)*b*x)] + (x^
2*PolyLog[2, (-I)*c*E^((2*I)*a + (2*I)*b*x)])/(4*b) + ((I/4)*x*PolyLog[3, (-I)*c*E^((2*I)*a + (2*I)*b*x)])/b^2
- PolyLog[4, (-I)*c*E^((2*I)*a + (2*I)*b*x)]/(8*b^3)
Rule 2184
Int[((c_.) + (d_.)*(x_))^(m_.)/((a_) + (b_.)*((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[(c
+ d*x)^(m + 1)/(a*d*(m + 1)), x] - Dist[b/a, Int[((c + d*x)^m*(F^(g*(e + f*x)))^n)/(a + b*(F^(g*(e + f*x)))^n)
, x], x] /; FreeQ[{F, a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]
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 5174
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[(I*b)/(f*(m + 1)), Int[(e + f*x)^(m + 1)/(c - I*d - c*
E^(2*I*a + 2*I*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(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-(1+i c) \cot (a+b x)) \, dx &=\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{3} (i b) \int \frac {x^3}{-i (-1-i c)+c-c e^{2 i a+2 i b x}} \, dx\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{3} (b c) \int \frac {e^{2 i a+2 i b x} x^3}{-i (-1-i c)+c-c e^{2 i a+2 i b x}} \, dx\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )-\frac {1}{2} i \int x^2 \log \left (1-\frac {c e^{2 i a+2 i b x}}{-i (-1-i c)+c}\right ) \, dx\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {x^2 \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}-\frac {\int x \text {Li}_2\left (\frac {c e^{2 i a+2 i b x}}{-i (-1-i c)+c}\right ) \, dx}{2 b}\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {x^2 \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i x \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i \int \text {Li}_3\left (\frac {c e^{2 i a+2 i b x}}{-i (-1-i c)+c}\right ) \, dx}{4 b^2}\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {x^2 \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i x \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {\operatorname {Subst}\left (\int \frac {\text {Li}_3(-i c x)}{x} \, dx,x,e^{2 i a+2 i b x}\right )}{8 b^3}\\ &=\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {x^2 \text {Li}_2\left (-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i x \text {Li}_3\left (-i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {\text {Li}_4\left (-i c e^{2 i a+2 i b x}\right )}{8 b^3}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.23, size = 136, normalized size = 0.88 \[ \frac {1}{24} \left (\frac {3 \text {Li}_4\left (\frac {i e^{-2 i (a+b x)}}{c}\right )}{b^3}+\frac {6 i x \text {Li}_3\left (\frac {i e^{-2 i (a+b x)}}{c}\right )}{b^2}-\frac {6 x^2 \text {Li}_2\left (\frac {i e^{-2 i (a+b x)}}{c}\right )}{b}+4 i x^3 \log \left (1-\frac {i e^{-2 i (a+b x)}}{c}\right )+8 x^3 \cot ^{-1}(c+(-1-i c) \cot (a+b x))\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[x^2*ArcCot[c - (1 + I*c)*Cot[a + b*x]],x]
[Out]
(8*x^3*ArcCot[c + (-1 - I*c)*Cot[a + b*x]] + (4*I)*x^3*Log[1 - I/(c*E^((2*I)*(a + b*x)))] - (6*x^2*PolyLog[2,
I/(c*E^((2*I)*(a + b*x)))])/b + ((6*I)*x*PolyLog[3, I/(c*E^((2*I)*(a + b*x)))])/b^2 + (3*PolyLog[4, I/(c*E^((2
*I)*(a + b*x)))])/b^3)/24
________________________________________________________________________________________
fricas [C] time = 0.81, size = 173, normalized size = 1.12 \[ \frac {2 \, b^{4} x^{4} + 8 \, \pi b^{3} x^{3} + 4 i \, b^{3} x^{3} \log \left (\frac {{\left (c - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )}}{c e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 2 \, a^{4} - 4 i \, a^{3} \log \left (\frac {c e^{\left (2 i \, b x + 2 i \, a\right )} - i}{c}\right ) + 6 i \, b x {\rm polylog}\left (3, -i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) + {\left (4 i \, b^{3} x^{3} + 4 i \, a^{3}\right )} \log \left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) - 3 \, {\rm polylog}\left (4, -i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{24 \, b^{3}} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*(pi-arccot(-c+(1+I*c)*cot(b*x+a))),x, algorithm="fricas")
[Out]
1/24*(2*b^4*x^4 + 8*pi*b^3*x^3 + 4*I*b^3*x^3*log((c - I)*e^(2*I*b*x + 2*I*a)/(c*e^(2*I*b*x + 2*I*a) - I)) + 6*
b^2*x^2*dilog(-I*c*e^(2*I*b*x + 2*I*a)) - 2*a^4 - 4*I*a^3*log((c*e^(2*I*b*x + 2*I*a) - I)/c) + 6*I*b*x*polylog
(3, -I*c*e^(2*I*b*x + 2*I*a)) + (4*I*b^3*x^3 + 4*I*a^3)*log(I*c*e^(2*I*b*x + 2*I*a) + 1) - 3*polylog(4, -I*c*e
^(2*I*b*x + 2*I*a)))/b^3
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (\pi - \operatorname {arccot}\left ({\left (i \, c + 1\right )} \cot \left (b x + a\right ) - c\right )\right )} x^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*(pi-arccot(-c+(1+I*c)*cot(b*x+a))),x, algorithm="giac")
[Out]
integrate((pi - arccot((I*c + 1)*cot(b*x + a) - c))*x^2, x)
________________________________________________________________________________________
maple [C] time = 6.80, size = 1527, normalized size = 9.85 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*(Pi-arccot(-c+(1+I*c)*cot(b*x+a))),x)
[Out]
1/12*x^3*Pi*csgn(I*(c-I)/(exp(2*I*(b*x+a))-1))^3+1/12*x^3*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I*(c-I)/(exp(2*I*(b
*x+a))-1))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))+1/4*x^2*polylog(2,-I*exp(2*I*(b*x+a))*c)/b+1/12
*b*x^4-1/12*x^3*Pi*csgn(I*(exp(2*I*(b*x+a))*c-I))*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(exp(2*I*(b*x+a))*c-I)/(
exp(2*I*(b*x+a))-1))-1/12*x^3*Pi*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^3+1/12*x^3*Pi*csgn(exp(2*
I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))^2-1/12*x^3*Pi*csgn((exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^3+1/6*I
*x^3*ln(1+I*c*exp(2*I*(b*x+a)))-1/4/b^3*polylog(2,-I*exp(2*I*(b*x+a))*c)*a^2+1/12*x^3*Pi*csgn(I*exp(2*I*(b*x+a
)))^3-1/6*I*x^3*ln(exp(2*I*(b*x+a))*c-I)-1/12*x^3*Pi*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))*csgn(
(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))-1/12*x^3*Pi*csgn(I*(c-I))*csgn(I*(c-I)/(exp(2*I*(b*x+a))-1))^2-1/
12*x^3*Pi*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(c-I)/(exp(2*I*(b*x+a))-1))^2+1/12*x^3*Pi*csgn(I*(exp(2*I*(b*x+a
))*c-I))*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^2+1/4*I*x*polylog(3,-I*exp(2*I*(b*x+a))*c)/b^2-1/
12*x^3*Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))^2-1/12*x^3*Pi*csgn(I*(c
-I)/(exp(2*I*(b*x+a))-1))*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))^2+1/12*x^3*Pi*csgn(I*(exp(2*I*(b
*x+a))*c-I)/(exp(2*I*(b*x+a))-1))*csgn((exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^2+1/12*x^3*Pi*csgn(I*(c-I)
)*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(c-I)/(exp(2*I*(b*x+a))-1))-1/6*x^3*Pi*csgn(I*exp(I*(b*x+a)))*csgn(I*exp
(2*I*(b*x+a)))^2+1/12*x^3*Pi*csgn((exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^2-1/12*x^3*Pi*csgn(I*exp(2*I*(b
*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))*csgn(exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))^2-1/12*x^3*Pi*csgn(exp(2*
I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))^3+1/2/b^3*a^2*dilog(1+I*exp(I*(b*x+a))*(I*c)^(1/2))+1/2/b^3*a^2*dilog(1
-I*exp(I*(b*x+a))*(I*c)^(1/2))-1/2*I/b^2*ln(1+I*c*exp(2*I*(b*x+a)))*x*a^2+1/2*I/b^2*a^2*ln(1+I*exp(I*(b*x+a))*
(I*c)^(1/2))*x+1/2*I/b^2*a^2*ln(1-I*exp(I*(b*x+a))*(I*c)^(1/2))*x+1/12*x^3*Pi*csgn(I*exp(2*I*(b*x+a))*(c-I)/(e
xp(2*I*(b*x+a))-1))*csgn(exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+a))-1))+1/12*x^3*Pi*csgn(I/(exp(2*I*(b*x+a))-1))
*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^2-1/8*polylog(4,-I*exp(2*I*(b*x+a))*c)/b^3+1/3*I*x^3*ln(e
xp(I*(b*x+a)))+1/12*x^3*Pi*csgn(I*exp(I*(b*x+a)))^2*csgn(I*exp(2*I*(b*x+a)))+1/6*I*x^3*ln(c-I)+1/2*I/b^3*a^3*l
n(1+I*exp(I*(b*x+a))*(I*c)^(1/2))+1/2*I/b^3*a^3*ln(1-I*exp(I*(b*x+a))*(I*c)^(1/2))-1/3*I/b^3*ln(1+I*c*exp(2*I*
(b*x+a)))*a^3-1/6*I/b^3*a^3*ln(-exp(2*I*(b*x+a))*c+I)+1/12*x^3*Pi*csgn(I*exp(2*I*(b*x+a))*(c-I)/(exp(2*I*(b*x+
a))-1))^3
________________________________________________________________________________________
maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^2*(pi-arccot(-c+(1+I*c)*cot(b*x+a))),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c-1>0)', see `assume?` for mor
e details)Is c-1 zero or nonzero?
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\left (\Pi +\mathrm {acot}\left (c-\mathrm {cot}\left (a+b\,x\right )\,\left (1+c\,1{}\mathrm {i}\right )\right )\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^2*(Pi + acot(c - cot(a + b*x)*(c*1i + 1))),x)
[Out]
int(x^2*(Pi + acot(c - cot(a + b*x)*(c*1i + 1))), x)
________________________________________________________________________________________
sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: CoercionFailed} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x**2*(pi-acot(-c+(1+I*c)*cot(b*x+a))),x)
[Out]
Exception raised: CoercionFailed
________________________________________________________________________________________