Integrand size = 19, antiderivative size = 142 \[ \int x^2 \arctan (c+(i+c) \tanh (a+b x)) \, dx=-\frac {1}{12} i b x^4+\frac {1}{3} x^3 \arctan (c+(i+c) \tanh (a+b x))+\frac {1}{6} i x^3 \log \left (1+i c e^{2 a+2 b x}\right )+\frac {i x^2 \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}-\frac {i x \operatorname {PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{4 b^2}+\frac {i \operatorname {PolyLog}\left (4,-i c e^{2 a+2 b x}\right )}{8 b^3} \]
-1/12*I*b*x^4+1/3*x^3*arctan(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 = 0.17 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.94 \[ \int x^2 \arctan (c+(i+c) \tanh (a+b x)) \, dx=\frac {8 b^3 x^3 \arctan (c+(i+c) \tanh (a+b x))+4 i b^3 x^3 \log \left (1-\frac {i e^{-2 (a+b x)}}{c}\right )-6 i b^2 x^2 \operatorname {PolyLog}\left (2,\frac {i e^{-2 (a+b x)}}{c}\right )-6 i b x \operatorname {PolyLog}\left (3,\frac {i e^{-2 (a+b x)}}{c}\right )-3 i \operatorname {PolyLog}\left (4,\frac {i e^{-2 (a+b x)}}{c}\right )}{24 b^3} \]
(8*b^3*x^3*ArcTan[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 = 0.81 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.18, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {5718, 25, 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.
\(\displaystyle \int x^2 \arctan (c+(c+i) \tanh (a+b x)) \, dx\) |
\(\Big \downarrow \) 5718 |
\(\displaystyle \frac {1}{3} x^3 \arctan (c+(c+i) \tanh (a+b x))-\frac {1}{3} b \int -\frac {x^3}{i-c e^{2 a+2 b x}}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{3} b \int \frac {x^3}{i-c e^{2 a+2 b x}}dx+\frac {1}{3} x^3 \arctan (c+(c+i) \tanh (a+b x))\) |
\(\Big \downarrow \) 2615 |
\(\displaystyle \frac {1}{3} b \left (-i c \int \frac {e^{2 a+2 b x} x^3}{i-c e^{2 a+2 b x}}dx-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \arctan (c+(c+i) \tanh (a+b x))\) |
\(\Big \downarrow \) 2620 |
\(\displaystyle \frac {1}{3} b \left (-i c \left (\frac {3 \int x^2 \log \left (i e^{2 a+2 b x} c+1\right )dx}{2 b c}-\frac {x^3 \log \left (1+i c e^{2 a+2 b x}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \arctan (c+(c+i) \tanh (a+b x))\) |
\(\Big \downarrow \) 3011 |
\(\displaystyle \frac {1}{3} b \left (-i c \left (\frac {3 \left (\frac {\int x \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )dx}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{2 b}\right )}{2 b c}-\frac {x^3 \log \left (1+i c e^{2 a+2 b x}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \arctan (c+(c+i) \tanh (a+b x))\) |
\(\Big \downarrow \) 7163 |
\(\displaystyle \frac {1}{3} b \left (-i c \left (\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{2 b}-\frac {\int \operatorname {PolyLog}\left (3,-i c e^{2 a+2 b x}\right )dx}{2 b}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{2 b}\right )}{2 b c}-\frac {x^3 \log \left (1+i c e^{2 a+2 b x}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \arctan (c+(c+i) \tanh (a+b x))\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {1}{3} b \left (-i c \left (\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{2 b}-\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (3,-i c e^{2 a+2 b x}\right )de^{2 a+2 b x}}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{2 b}\right )}{2 b c}-\frac {x^3 \log \left (1+i c e^{2 a+2 b x}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )+\frac {1}{3} x^3 \arctan (c+(c+i) \tanh (a+b x))\) |
\(\Big \downarrow \) 7143 |
\(\displaystyle \frac {1}{3} x^3 \arctan (c+(c+i) \tanh (a+b x))+\frac {1}{3} b \left (-i c \left (\frac {3 \left (\frac {\frac {x \operatorname {PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{2 b}-\frac {\operatorname {PolyLog}\left (4,-i c e^{2 a+2 b x}\right )}{4 b^2}}{b}-\frac {x^2 \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{2 b}\right )}{2 b c}-\frac {x^3 \log \left (1+i c e^{2 a+2 b x}\right )}{2 b c}\right )-\frac {i x^4}{4}\right )\) |
(x^3*ArcTan[c + (I + c)*Tanh[a + b*x]])/3 + (b*((-1/4*I)*x^4 - I*c*(-1/2*( x^3*Log[1 + I*c*E^(2*a + 2*b*x)])/(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) - Poly Log[4, (-I)*c*E^(2*a + 2*b*x)]/(4*b^2))/b))/(2*b*c))))/3
3.1.85.3.1 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[ArcTan[(c_.) + (d_.)*Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e + f*x)^(m + 1)*(ArcTan[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 = 1.60 (sec) , antiderivative size = 1406, normalized size of antiderivative = 9.90
1/6*I*x^3*ln(1+I*c*exp(2*b*x+2*a))-1/2*I/b^2*ln(1+I*c*exp(2*b*x+2*a))*a^2* x-1/6*I*x^3*ln(2*exp(2*b*x+2*a)*c-2*I)+1/2*I/b^2*a^2*ln(1+I*exp(b*x+a)*(I* c)^(1/2))*x+1/2*I/b^2*a^2*ln(1-I*exp(b*x+a)*(I*c)^(1/2))*x-1/3*I/b^3*ln(1+ I*c*exp(2*b*x+2*a))*a^3-1/12*I*b*c/(I+c)*x^4+1/12*I/b^3*c/(I+c)*a^4-1/4*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+1/2*I/b^3*a^3*ln(1+I*exp(b*x+a)*(I*c)^(1/2))+1/2*I/b^3*a^2*dilog(1+I*e xp(b*x+a)*(I*c)^(1/2))+1/2*I/b^3*a^2*dilog(1-I*exp(b*x+a)*(I*c)^(1/2))+1/4 *I*x^2*polylog(2,-I*c*exp(2*b*x+2*a))/b-1/4*I/b^3*polylog(2,-I*c*exp(2*b*x +2*a))*a^2-1/6*I/b^3*a^3*ln(-exp(2*b*x+2*a)*c+I)-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)/(e...
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 293 vs. \(2 (105) = 210\).
Time = 0.27 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.06 \[ \int x^2 \arctan (c+(i+c) \tanh (a+b x)) \, dx=\frac {-i \, b^{4} x^{4} + 2 i \, b^{3} x^{3} \log \left (-\frac {{\left (c + i\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c e^{\left (2 \, b x + 2 \, a\right )} - i}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) + 6 i \, b^{2} x^{2} {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) + i \, a^{4} - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (b x + a\right )} + i \, \sqrt {-4 i \, c}}{2 \, c}\right ) - 2 i \, a^{3} \log \left (\frac {2 \, c e^{\left (b x + a\right )} - i \, \sqrt {-4 i \, c}}{2 \, c}\right ) - 12 i \, b x {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) - 12 i \, b x {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) - 2 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )} + 1\right ) - 2 \, {\left (-i \, b^{3} x^{3} - i \, a^{3}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )} + 1\right ) + 12 i \, {\rm polylog}\left (4, \frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) + 12 i \, {\rm polylog}\left (4, -\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right )}{12 \, b^{3}} \]
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 *dilog(-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*p olylog(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. \[ \int x^2 \arctan (c+(i+c) \tanh (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \]
Exception raised: CoercionFailed >> Cannot convert _t0**2*exp(2*a) + 1 of type <class 'sympy.core.add.Add'> to QQ_I[x,b,_t0,exp(a)]
Time = 1.32 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.91 \[ \int x^2 \arctan (c+(i+c) \tanh (a+b x)) \, dx=\frac {1}{3} \, x^{3} \arctan \left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right ) + \frac {4}{9} \, {\left (\frac {3 \, x^{4}}{4 i \, c - 4} - \frac {4 \, b^{3} x^{3} \log \left (i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 6 \, b^{2} x^{2} {\rm Li}_2\left (-i \, c e^{\left (2 \, b x + 2 \, a\right )}\right ) - 6 \, b x {\rm Li}_{3}(-i \, c e^{\left (2 \, b x + 2 \, a\right )}) + 3 \, {\rm Li}_{4}(-i \, c e^{\left (2 \, b x + 2 \, a\right )})}{-2 \, b^{4} {\left (-i \, c + 1\right )}}\right )} b {\left (c + i\right )} \]
1/3*x^3*arctan((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)
\[ \int x^2 \arctan (c+(i+c) \tanh (a+b x)) \, dx=\int { x^{2} \arctan \left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right ) \,d x } \]
Timed out. \[ \int x^2 \arctan (c+(i+c) \tanh (a+b x)) \, dx=\int x^2\,\mathrm {atan}\left (c+\mathrm {tanh}\left (a+b\,x\right )\,\left (c+1{}\mathrm {i}\right )\right ) \,d x \]