Integrand size = 22, antiderivative size = 145 \[ \int x^2 \cot ^{-1}(c-(i-c) \tanh (a+b x)) \, dx=-\frac {1}{12} i b x^4+\frac {1}{3} x^3 \cot ^{-1}(c-(i-c) \tanh (a+b x))+\frac {1}{6} i x^3 \log \left (1-i c e^{2 a+2 b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{4 b}-\frac {i x \operatorname {PolyLog}\left (3,i c e^{2 a+2 b x}\right )}{4 b^2}+\frac {i \operatorname {PolyLog}\left (4,i c e^{2 a+2 b x}\right )}{8 b^3} \]
-1/12*I*b*x^4+1/3*x^3*arccot(c-(I-c)*tanh(b*x+a))+1/6*I*x^3*ln(1-I*c*exp(2 *b*x+2*a))+1/4*I*x^2*polylog(2,I*c*exp(2*b*x+2*a))/b-1/4*I*x*polylog(3,I*c *exp(2*b*x+2*a))/b^2+1/8*I*polylog(4,I*c*exp(2*b*x+2*a))/b^3
Time = 0.16 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.92 \[ \int x^2 \cot ^{-1}(c-(i-c) \tanh (a+b x)) \, dx=\frac {8 b^3 x^3 \cot ^{-1}(c+(-i+c) \tanh (a+b x))+4 i b^3 x^3 \log \left (1+\frac {i e^{-2 (a+b x)}}{c}\right )-6 i b^2 x^2 \operatorname {PolyLog}\left (2,-\frac {i e^{-2 (a+b x)}}{c}\right )-6 i b x \operatorname {PolyLog}\left (3,-\frac {i e^{-2 (a+b x)}}{c}\right )-3 i \operatorname {PolyLog}\left (4,-\frac {i e^{-2 (a+b x)}}{c}\right )}{24 b^3} \]
(8*b^3*x^3*ArcCot[c + (-I + c)*Tanh[a + b*x]] + (4*I)*b^3*x^3*Log[1 + I/(c *E^(2*(a + b*x)))] - (6*I)*b^2*x^2*PolyLog[2, (-I)/(c*E^(2*(a + b*x)))] - (6*I)*b*x*PolyLog[3, (-I)/(c*E^(2*(a + b*x)))] - (3*I)*PolyLog[4, (-I)/(c* E^(2*(a + b*x)))])/(24*b^3)
Time = 0.78 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {5719, 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.
| \(\displaystyle \int x^2 \cot ^{-1}(c-(-c+i) \tanh (a+b x)) \, dx\) |
| \(\Big \downarrow \) 5719 |
| \(\displaystyle \frac {1}{3} b \int \frac {x^3}{e^{2 a+2 b x} c+i}dx+\frac {1}{3} x^3 \cot ^{-1}(c-(-c+i) \tanh (a+b x))\) |
| \(\Big \downarrow \) 2615 |
| \(\displaystyle \frac {1}{3} b \left (i c \int \frac {e^{2 a+2 b x} x^3}{e^{2 a+2 b x} c+i}dx-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(-c+i) \tanh (a+b x))\) |
| \(\Big \downarrow \) 2620 |
| \(\displaystyle \frac {1}{3} b \left (i c \left (\frac {x^3 \log \left (1-i c e^{2 a+2 b x}\right )}{2 b c}-\frac {3 \int x^2 \log \left (1-i c e^{2 a+2 b x}\right )dx}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(-c+i) \tanh (a+b x))\) |
| \(\Big \downarrow \) 3011 |
| \(\displaystyle \frac {1}{3} b \left (i c \left (\frac {x^3 \log \left (1-i c e^{2 a+2 b x}\right )}{2 b c}-\frac {3 \left (\frac {\int x \operatorname {PolyLog}\left (2,i c e^{2 a+2 b x}\right )dx}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{2 b}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(-c+i) \tanh (a+b x))\) |
| \(\Big \downarrow \) 7163 |
| \(\displaystyle \frac {1}{3} b \left (i c \left (\frac {x^3 \log \left (1-i c e^{2 a+2 b x}\right )}{2 b c}-\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,i c e^{2 a+2 b x}\right )}{2 b}-\frac {\int \operatorname {PolyLog}\left (3,i c e^{2 a+2 b x}\right )dx}{2 b}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{2 b}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(-c+i) \tanh (a+b x))\) |
| \(\Big \downarrow \) 2720 |
| \(\displaystyle \frac {1}{3} b \left (i c \left (\frac {x^3 \log \left (1-i c e^{2 a+2 b x}\right )}{2 b c}-\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,i c e^{2 a+2 b x}\right )}{2 b}-\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (3,i c e^{2 a+2 b x}\right )de^{2 a+2 b x}}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{2 b}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(-c+i) \tanh (a+b x))\) |
| \(\Big \downarrow \) 7143 |
| \(\displaystyle \frac {1}{3} b \left (i c \left (\frac {x^3 \log \left (1-i c e^{2 a+2 b x}\right )}{2 b c}-\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,i c e^{2 a+2 b x}\right )}{2 b}-\frac {\operatorname {PolyLog}\left (4,i c e^{2 a+2 b x}\right )}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,i c e^{2 a+2 b x}\right )}{2 b}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \cot ^{-1}(c-(-c+i) \tanh (a+b x))\) |
(x^3*ArcCot[c - (I - c)*Tanh[a + b*x]])/3 + (b*((-1/4*I)*x^4 + I*c*((x^3*L og[1 - I*c*E^(2*a + 2*b*x)])/(2*b*c) - (3*(-1/2*(x^2*PolyLog[2, I*c*E^(2*a + 2*b*x)])/b + ((x*PolyLog[3, I*c*E^(2*a + 2*b*x)])/(2*b) - PolyLog[4, I* c*E^(2*a + 2*b*x)]/(4*b^2))/b))/(2*b*c))))/3
3.2.96.3.1 Defintions of rubi rules used
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_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcCot[c + d*Tanh[a + b*x]]/(f*(m + 1))), x] + Simp[b/(f*(m + 1)) Int[(e + f*x)^(m + 1)/(c - d + c*E^(2*a + 2*b*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && EqQ[(c - 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 = 2.47 (sec) , antiderivative size = 1409, normalized size of antiderivative = 9.72
1/6*I*x^3*ln(2*I*exp(2*b*x+2*a)-2*exp(2*b*x+2*a)*c)-1/12*Pi*(csgn(I/(exp(2 *b*x+2*a)+1))*csgn(I*(2*exp(2*b*x+2*a)*c+2*I))*csgn(I*(2*exp(2*b*x+2*a)*c+ 2*I)/(exp(2*b*x+2*a)+1))-csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(-2*I*exp(2*b*x +2*a)+2*exp(2*b*x+2*a)*c))*csgn(I*(-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c) /(exp(2*b*x+2*a)+1))-csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(2*exp(2*b*x+2*a)*c +2*I)/(exp(2*b*x+2*a)+1))^2+csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(-2*I*exp(2* b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))^2+csgn(I*(2*exp(2*b*x+2*a )*c+2*I))*csgn(I*(2*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)+1))^2-csgn(I*(-2 *I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c))*csgn(I*(-2*I*exp(2*b*x+2*a)+2*exp(2 *b*x+2*a)*c)/(exp(2*b*x+2*a)+1))^2-csgn(I*(2*exp(2*b*x+2*a)*c+2*I)/(exp(2* b*x+2*a)+1))^3+csgn(I*(2*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)+1))*csgn((2 *exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)+1))^2+csgn(I*(2*exp(2*b*x+2*a)*c+2* I)/(exp(2*b*x+2*a)+1))*csgn((2*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x+2*a)+1))+c sgn(I*(-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))^3-csgn( I*(-2*I*exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))*csgn((-2*I* exp(2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))^2-csgn(I*(-2*I*exp( 2*b*x+2*a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))*csgn((-2*I*exp(2*b*x+2* a)+2*exp(2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))+csgn((-2*I*exp(2*b*x+2*a)+2*exp (2*b*x+2*a)*c)/(exp(2*b*x+2*a)+1))^3+csgn((-2*I*exp(2*b*x+2*a)+2*exp(2*b*x +2*a)*c)/(exp(2*b*x+2*a)+1))^2+csgn((2*exp(2*b*x+2*a)*c+2*I)/(exp(2*b*x...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (105) = 210\).
Time = 0.28 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.01 \[ \int x^2 \cot ^{-1}(c-(i-c) \tanh (a+b x)) \, dx=\frac {-i \, b^{4} x^{4} + 2 i \, b^{3} x^{3} \log \left (\frac {{\left (c - i\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c e^{\left (2 \, b x + 2 \, a\right )} + i}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + i \, a^{4} - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (b x + a\right )} + i \, \sqrt {4 i \, c}}{2 \, c}\right ) - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (b x + a\right )} - i \, \sqrt {4 i \, c}}{2 \, c}\right ) - 12 i \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) - 12 i \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) - 2 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) - 2 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )} + 1\right ) + 12 i \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right ) + 12 i \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {4 i \, c} e^{\left (b x + a\right )}\right )}{12 \, b^{3}} \]
1/12*(-I*b^4*x^4 + 2*I*b^3*x^3*log((c - I)*e^(2*b*x + 2*a)/(c*e^(2*b*x + 2 *a) + I)) + 6*I*b^2*x^2*dilog(1/2*sqrt(4*I*c)*e^(b*x + a)) + 6*I*b^2*x^2*d ilog(-1/2*sqrt(4*I*c)*e^(b*x + a)) + I*a^4 - 2*I*a^3*log(1/2*(2*c*e^(b*x + a) + I*sqrt(4*I*c))/c) - 2*I*a^3*log(1/2*(2*c*e^(b*x + a) - I*sqrt(4*I*c) )/c) - 12*I*b*x*polylog(3, 1/2*sqrt(4*I*c)*e^(b*x + a)) - 12*I*b*x*polylog (3, -1/2*sqrt(4*I*c)*e^(b*x + a)) - 2*(-I*b^3*x^3 - I*a^3)*log(1/2*sqrt(4* I*c)*e^(b*x + a) + 1) - 2*(-I*b^3*x^3 - I*a^3)*log(-1/2*sqrt(4*I*c)*e^(b*x + a) + 1) + 12*I*polylog(4, 1/2*sqrt(4*I*c)*e^(b*x + a)) + 12*I*polylog(4 , -1/2*sqrt(4*I*c)*e^(b*x + a)))/b^3
Exception generated. \[ \int x^2 \cot ^{-1}(c-(i-c) \tanh (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \]
Exception raised: CoercionFailed >> Cannot convert 2*_t0**2*c*exp(2*a) - _ t0**2*I*exp(2*a) + I of type <class 'sympy.core.add.Add'> to QQ_I[x,b,c,_t 0,exp(a)]
Time = 1.44 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.89 \[ \int x^2 \cot ^{-1}(c-(i-c) \tanh (a+b x)) \, dx=\frac {1}{3} \, x^{3} \operatorname {arccot}\left ({\left (c - i\right )} \tanh \left (b x + a\right ) + c\right ) + \frac {4}{9} \, {\left (\frac {3 \, x^{4}}{4 i \, c + 4} - \frac {4 \, b^{3} x^{3} \log \left (-i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (i \, c e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(i \, c e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(i \, c e^{\left (2 \, b x + 2 \, a\right )})}{-2 \, b^{4} {\left (-i \, c - 1\right )}}\right )} b {\left (c - i\right )} \]
1/3*x^3*arccot((c - I)*tanh(b*x + a) + c) + 4/9*(3*x^4/(4*I*c + 4) - (4*b^ 3*x^3*log(-I*c*e^(2*b*x + 2*a) + 1) + 6*b^2*x^2*dilog(I*c*e^(2*b*x + 2*a)) - 6*b*x*polylog(3, I*c*e^(2*b*x + 2*a)) + 3*polylog(4, I*c*e^(2*b*x + 2*a )))/(b^4*(2*I*c + 2)))*b*(c - I)
\[ \int x^2 \cot ^{-1}(c-(i-c) \tanh (a+b x)) \, dx=\int { x^{2} \operatorname {arccot}\left ({\left (c - i\right )} \tanh \left (b x + a\right ) + c\right ) \,d x } \]
Timed out. \[ \int x^2 \cot ^{-1}(c-(i-c) \tanh (a+b x)) \, dx=\int x^2\,\mathrm {acot}\left (c+\mathrm {tanh}\left (a+b\,x\right )\,\left (c-\mathrm {i}\right )\right ) \,d x \]