Integrand size = 22, antiderivative size = 155 \[ \int x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx=\frac {b x^4}{12}+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))+\frac {1}{6} i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )+\frac {x^2 \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{4 b}+\frac {i x \operatorname {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {\operatorname {PolyLog}\left (4,-i c e^{2 i a+2 i b x}\right )}{8 b^3} \] Output:
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*e xp(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*po lylog(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
Time = 0.16 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.88 \[ \int x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx=\frac {1}{24} \left (8 x^3 \cot ^{-1}(c+(-1-i c) \cot (a+b x))+4 i x^3 \log \left (1-\frac {i e^{-2 i (a+b x)}}{c}\right )-\frac {6 x^2 \operatorname {PolyLog}\left (2,\frac {i e^{-2 i (a+b x)}}{c}\right )}{b}+\frac {6 i x \operatorname {PolyLog}\left (3,\frac {i e^{-2 i (a+b x)}}{c}\right )}{b^2}+\frac {3 \operatorname {PolyLog}\left (4,\frac {i e^{-2 i (a+b x)}}{c}\right )}{b^3}\right ) \] Input:
Integrate[x^2*ArcCot[c - (1 + I*c)*Cot[a + b*x]],x]
Output:
(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
Time = 0.82 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.28, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {5697, 2615, 2620, 3011, 7163, 2720, 7143}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx\) |
\(\Big \downarrow \) 5697 |
\(\displaystyle \frac {1}{3} i b \int \frac {x^3}{i-c e^{2 i a+2 i b x}}dx+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle \frac {1}{3} i b \left (-i c \int \frac {e^{2 i a+2 i b x} x^3}{i-c e^{2 i a+2 i b x}}dx-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {1}{3} i b \left (-i c \left (\frac {i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )}{2 b c}-\frac {3 i \int x^2 \log \left (i e^{2 i a+2 i b x} c+1\right )dx}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {1}{3} i b \left (-i c \left (\frac {i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )}{2 b c}-\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{2 b}-\frac {i \int x \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )dx}{b}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {1}{3} i b \left (-i c \left (\frac {i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )}{2 b c}-\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{2 b}-\frac {i \left (\frac {i \int \operatorname {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )dx}{2 b}-\frac {i x \operatorname {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{2 b}\right )}{b}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{3} i b \left (-i c \left (\frac {i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )}{2 b c}-\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{2 b}-\frac {i \left (\frac {\int e^{-2 i a-2 i b x} \operatorname {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )de^{2 i a+2 i b x}}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{2 b}\right )}{b}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{3} i b \left (-i c \left (\frac {i x^3 \log \left (1+i c e^{2 i a+2 i b x}\right )}{2 b c}-\frac {3 i \left (\frac {i x^2 \operatorname {PolyLog}\left (2,-i c e^{2 i a+2 i b x}\right )}{2 b}-\frac {i \left (\frac {\operatorname {PolyLog}\left (4,-i c e^{2 i a+2 i b x}\right )}{4 b^2}-\frac {i x \operatorname {PolyLog}\left (3,-i c e^{2 i a+2 i b x}\right )}{2 b}\right )}{b}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(1+i c) \cot (a+b x))\) |
Input:
Int[x^2*ArcCot[c - (1 + I*c)*Cot[a + b*x]],x]
Output:
(x^3*ArcCot[c - (1 + I*c)*Cot[a + b*x]])/3 + (I/3)*b*((-1/4*I)*x^4 - I*c*( ((I/2)*x^3*Log[1 + I*c*E^((2*I)*a + (2*I)*b*x)])/(b*c) - (((3*I)/2)*(((I/2 )*x^2*PolyLog[2, (-I)*c*E^((2*I)*a + (2*I)*b*x)])/b - (I*(((-1/2*I)*x*Poly Log[3, (-I)*c*E^((2*I)*a + (2*I)*b*x)])/b + PolyLog[4, (-I)*c*E^((2*I)*a + (2*I)*b*x)]/(4*b^2)))/b))/(b*c)))
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] - Simp[ 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]
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] - Si mp[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]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
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] + Simp[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]
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] + Simp[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] && E qQ[(c - I*d)^2, -1]
Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S ymbol] :> 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]
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] - Simp[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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 4.20 (sec) , antiderivative size = 1449, normalized size of antiderivative = 9.35
Input:
int(x^2*(Pi-arccot(-c+(I*c+1)*cot(b*x+a))),x,method=_RETURNVERBOSE)
Output:
-1/6*I*x^3*ln(exp(2*I*(b*x+a))*c-I)+1/12*(Pi*csgn(I*exp(I*(b*x+a)))^2*csgn (I*exp(2*I*(b*x+a)))-2*Pi*csgn(I*exp(I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a)))^ 2+Pi*csgn(I*exp(2*I*(b*x+a)))^3+Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I*(-I+c)/ (exp(2*I*(b*x+a))-1))*csgn(I*exp(2*I*(b*x+a))*(-I+c)/(exp(2*I*(b*x+a))-1)) -Pi*csgn(I*exp(2*I*(b*x+a)))*csgn(I*exp(2*I*(b*x+a))*(-I+c)/(exp(2*I*(b*x+ a))-1))^2-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))+Pi*csgn((exp(2*I*(b*x+a))*c-I)/(e xp(2*I*(b*x+a))-1))^2+Pi*csgn(I*(-I+c))*csgn(I/(exp(2*I*(b*x+a))-1))*csgn( I*(-I+c)/(exp(2*I*(b*x+a))-1))-Pi*csgn(I*(-I+c))*csgn(I*(-I+c)/(exp(2*I*(b *x+a))-1))^2+Pi*csgn(I*exp(2*I*(b*x+a))*(-I+c)/(exp(2*I*(b*x+a))-1))*csgn( exp(2*I*(b*x+a))*(-I+c)/(exp(2*I*(b*x+a))-1))+Pi*csgn(exp(2*I*(b*x+a))*(-I +c)/(exp(2*I*(b*x+a))-1))^2-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))+Pi*cs gn(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-Pi*csgn(I/(exp(2*I*(b*x+a))-1))*csgn(I*(-I+c)/(exp(2*I*(b*x+a))-1 ))^2+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+Pi*csgn(I*(-I+c)/(exp(2*I*(b*x+a))-1))^3-Pi*csgn(I*(-I+c) /(exp(2*I*(b*x+a))-1))*csgn(I*exp(2*I*(b*x+a))*(-I+c)/(exp(2*I*(b*x+a))-1) )^2-Pi*csgn(I*(exp(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))^3+Pi*csgn(I*(ex p(2*I*(b*x+a))*c-I)/(exp(2*I*(b*x+a))-1))*csgn((exp(2*I*(b*x+a))*c-I)/(...
Time = 0.10 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.12 \[ \int x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx=\frac {2 \, b^{4} x^{4} + 8 \, \pi b^{3} x^{3} + 4 i \, b^{3} x^{3} \log \left (\frac {{\left (c - i\right )} e^{\left (2 i \, b x + 2 i \, a\right )}}{c e^{\left (2 i \, b x + 2 i \, a\right )} - i}\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 2 \, a^{4} - 4 i \, a^{3} \log \left (\frac {c e^{\left (2 i \, b x + 2 i \, a\right )} - i}{c}\right ) + 6 i \, b x {\rm polylog}\left (3, -i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right ) - 4 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (i \, c e^{\left (2 i \, b x + 2 i \, a\right )} + 1\right ) - 3 \, {\rm polylog}\left (4, -i \, c e^{\left (2 i \, b x + 2 i \, a\right )}\right )}{24 \, b^{3}} \] Input:
integrate(x^2*(pi-arccot(-c+(1+I*c)*cot(b*x+a))),x, algorithm="fricas")
Output:
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 - 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
Exception generated. \[ \int x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \] Input:
integrate(x**2*(pi-acot(-c+(1+I*c)*cot(b*x+a))),x)
Output:
Exception raised: CoercionFailed >> Cannot convert 2*_t0**2*c*exp(2*I*a) - _t0**2*I*exp(2*I*a) - I of type <class 'sympy.core.add.Add'> to QQ_I[x,b, c,_t0,exp(I*a)]
Exception generated. \[ \int x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^2*(pi-arccot(-c+(1+I*c)*cot(b*x+a))),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(c-1>0)', see `assume?` for more details)Is
\[ \int x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx=\int { {\left (\pi - \operatorname {arccot}\left ({\left (i \, c + 1\right )} \cot \left (b x + a\right ) - c\right )\right )} x^{2} \,d x } \] Input:
integrate(x^2*(pi-arccot(-c+(1+I*c)*cot(b*x+a))),x, algorithm="giac")
Output:
integrate((pi - arccot((I*c + 1)*cot(b*x + a) - c))*x^2, x)
Timed out. \[ \int x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx=\int x^2\,\left (\Pi +\mathrm {acot}\left (c-\mathrm {cot}\left (a+b\,x\right )\,\left (1+c\,1{}\mathrm {i}\right )\right )\right ) \,d x \] Input:
int(x^2*(Pi + acot(c - cot(a + b*x)*(c*1i + 1))),x)
Output:
int(x^2*(Pi + acot(c - cot(a + b*x)*(c*1i + 1))), x)
\[ \int x^2 \cot ^{-1}(c-(1+i c) \cot (a+b x)) \, dx=-\left (\int \mathit {acot} \left (\cot \left (b x +a \right ) c i +\cot \left (b x +a \right )-c \right ) x^{2}d x \right )+\frac {\pi \,x^{3}}{3} \] Input:
int(x^2*(Pi-acot(-c+(1+I*c)*cot(b*x+a))),x)
Output:
( - 3*int(acot(cot(a + b*x)*c*i + cot(a + b*x) - c)*x**2,x) + pi*x**3)/3