\(\int x \arctan (c+(i+c) \tanh (a+b x)) \, dx\) [86]

Optimal result

Integrand size = 17, antiderivative size = 113 \[ \int x \arctan (c+(i+c) \tanh (a+b x)) \, dx=-\frac {1}{6} i b x^3+\frac {1}{2} x^2 \arctan (c+(i+c) \tanh (a+b x))+\frac {1}{4} i x^2 \log \left (1+i c e^{2 a+2 b x}\right )+\frac {i x \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{4 b}-\frac {i \operatorname {PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{8 b^2} \] Output:

-1/6*I*b*x^3+1/2*x^2*arctan(c+(I+c)*tanh(b*x+a))+1/4*I*x^2*ln(1+I*c*exp(2* 
b*x+2*a))+1/4*I*x*polylog(2,-I*c*exp(2*b*x+2*a))/b-1/8*I*polylog(3,-I*c*ex 
p(2*b*x+2*a))/b^2
 

Mathematica [A] (verified)

Time = 0.07 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.91 \[ \int x \arctan (c+(i+c) \tanh (a+b x)) \, dx=\frac {2 b^2 x^2 \left (2 \arctan (c+(i+c) \tanh (a+b x))+i \log \left (1-\frac {i e^{-2 (a+b x)}}{c}\right )\right )-2 i b x \operatorname {PolyLog}\left (2,\frac {i e^{-2 (a+b x)}}{c}\right )-i \operatorname {PolyLog}\left (3,\frac {i e^{-2 (a+b x)}}{c}\right )}{8 b^2} \] Input:

Integrate[x*ArcTan[c + (I + c)*Tanh[a + b*x]],x]
 

Output:

(2*b^2*x^2*(2*ArcTan[c + (I + c)*Tanh[a + b*x]] + I*Log[1 - I/(c*E^(2*(a + 
 b*x)))]) - (2*I)*b*x*PolyLog[2, I/(c*E^(2*(a + b*x)))] - I*PolyLog[3, I/( 
c*E^(2*(a + b*x)))])/(8*b^2)
 

Rubi [A] (verified)

Time = 0.67 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {5718, 25, 2615, 2620, 3011, 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.

\(\displaystyle \int x \arctan (c+(c+i) \tanh (a+b x)) \, dx\)

\(\Big \downarrow \) 5718

\(\displaystyle \frac {1}{2} x^2 \arctan (c+(c+i) \tanh (a+b x))-\frac {1}{2} b \int -\frac {x^2}{i-c e^{2 a+2 b x}}dx\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {1}{2} b \int \frac {x^2}{i-c e^{2 a+2 b x}}dx+\frac {1}{2} x^2 \arctan (c+(c+i) \tanh (a+b x))\)

\(\Big \downarrow \) 2615

\(\displaystyle \frac {1}{2} b \left (-i c \int \frac {e^{2 a+2 b x} x^2}{i-c e^{2 a+2 b x}}dx-\frac {i x^3}{3}\right )+\frac {1}{2} x^2 \arctan (c+(c+i) \tanh (a+b x))\)

\(\Big \downarrow \) 2620

\(\displaystyle \frac {1}{2} b \left (-i c \left (\frac {\int x \log \left (i e^{2 a+2 b x} c+1\right )dx}{b c}-\frac {x^2 \log \left (1+i c e^{2 a+2 b x}\right )}{2 b c}\right )-\frac {i x^3}{3}\right )+\frac {1}{2} x^2 \arctan (c+(c+i) \tanh (a+b x))\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {1}{2} b \left (-i c \left (\frac {\frac {\int \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )dx}{2 b}-\frac {x \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{2 b}}{b c}-\frac {x^2 \log \left (1+i c e^{2 a+2 b x}\right )}{2 b c}\right )-\frac {i x^3}{3}\right )+\frac {1}{2} x^2 \arctan (c+(c+i) \tanh (a+b x))\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {1}{2} b \left (-i c \left (\frac {\frac {\int e^{-2 a-2 b x} \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )de^{2 a+2 b x}}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{2 b}}{b c}-\frac {x^2 \log \left (1+i c e^{2 a+2 b x}\right )}{2 b c}\right )-\frac {i x^3}{3}\right )+\frac {1}{2} x^2 \arctan (c+(c+i) \tanh (a+b x))\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {1}{2} x^2 \arctan (c+(c+i) \tanh (a+b x))+\frac {1}{2} b \left (-i c \left (\frac {\frac {\operatorname {PolyLog}\left (3,-i c e^{2 a+2 b x}\right )}{4 b^2}-\frac {x \operatorname {PolyLog}\left (2,-i c e^{2 a+2 b x}\right )}{2 b}}{b c}-\frac {x^2 \log \left (1+i c e^{2 a+2 b x}\right )}{2 b c}\right )-\frac {i x^3}{3}\right )\)

Input:

Int[x*ArcTan[c + (I + c)*Tanh[a + b*x]],x]
 

Output:

(x^2*ArcTan[c + (I + c)*Tanh[a + b*x]])/2 + (b*((-1/3*I)*x^3 - I*c*(-1/2*( 
x^2*Log[1 + I*c*E^(2*a + 2*b*x)])/(b*c) + (-1/2*(x*PolyLog[2, (-I)*c*E^(2* 
a + 2*b*x)])/b + PolyLog[3, (-I)*c*E^(2*a + 2*b*x)]/(4*b^2))/(b*c))))/2
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 2615

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]
 

rule 2620

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]
 

rule 2720

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]]]
 

rule 3011

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]
 

rule 5718

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]
 

rule 7143

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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.54 (sec) , antiderivative size = 1370, normalized size of antiderivative = 12.12

method	result	size
risch	\(\text {Expression too large to display}\)	\(1370\)

Input:

int(x*arctan(c+(I+c)*tanh(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

-1/2*I/b^2*ln(1+I*exp(b*x+a)*(I*c)^(1/2))*a^2+1/8*Pi*(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*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))^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))*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))^3+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*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))+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))^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))^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))*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)/(ex 
p(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+2*a)+1))^3-csgn((2...
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (83) = 166\).

Time = 0.13 (sec) , antiderivative size = 247, normalized size of antiderivative = 2.19 \[ \int x \arctan (c+(i+c) \tanh (a+b x)) \, dx=\frac {-2 i \, b^{3} x^{3} + 3 i \, b^{2} x^{2} \log \left (-\frac {{\left (c + i\right )} e^{\left (2 \, b x + 2 \, a\right )}}{c e^{\left (2 \, b x + 2 \, a\right )} - i}\right ) - 2 i \, a^{3} + 6 i \, b x {\rm Li}_2\left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) + 6 i \, b x {\rm Li}_2\left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) + 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (b x + a\right )} + i \, \sqrt {-4 i \, c}}{2 \, c}\right ) + 3 i \, a^{2} \log \left (\frac {2 \, c e^{\left (b x + a\right )} - i \, \sqrt {-4 i \, c}}{2 \, c}\right ) - 3 \, {\left (-i \, b^{2} x^{2} + i \, a^{2}\right )} \log \left (\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )} + 1\right ) - 3 \, {\left (-i \, b^{2} x^{2} + i \, a^{2}\right )} \log \left (-\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )} + 1\right ) - 6 i \, {\rm polylog}\left (3, \frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right ) - 6 i \, {\rm polylog}\left (3, -\frac {1}{2} \, \sqrt {-4 i \, c} e^{\left (b x + a\right )}\right )}{12 \, b^{2}} \] Input:

integrate(x*arctan(c+(I+c)*tanh(b*x+a)),x, algorithm="fricas")
 

Output:

1/12*(-2*I*b^3*x^3 + 3*I*b^2*x^2*log(-(c + I)*e^(2*b*x + 2*a)/(c*e^(2*b*x 
+ 2*a) - I)) - 2*I*a^3 + 6*I*b*x*dilog(1/2*sqrt(-4*I*c)*e^(b*x + a)) + 6*I 
*b*x*dilog(-1/2*sqrt(-4*I*c)*e^(b*x + a)) + 3*I*a^2*log(1/2*(2*c*e^(b*x + 
a) + I*sqrt(-4*I*c))/c) + 3*I*a^2*log(1/2*(2*c*e^(b*x + a) - I*sqrt(-4*I*c 
))/c) - 3*(-I*b^2*x^2 + I*a^2)*log(1/2*sqrt(-4*I*c)*e^(b*x + a) + 1) - 3*( 
-I*b^2*x^2 + I*a^2)*log(-1/2*sqrt(-4*I*c)*e^(b*x + a) + 1) - 6*I*polylog(3 
, 1/2*sqrt(-4*I*c)*e^(b*x + a)) - 6*I*polylog(3, -1/2*sqrt(-4*I*c)*e^(b*x 
+ a)))/b^2
 

Sympy [F(-2)]

Exception generated. \[ \int x \arctan (c+(i+c) \tanh (a+b x)) \, dx=\text {Exception raised: CoercionFailed} \] Input:

integrate(x*atan(c+(I+c)*tanh(b*x+a)),x)
 

Output:

Exception raised: CoercionFailed >> Cannot convert _t0**2 + exp(2*a) of ty 
pe <class 'sympy.core.add.Add'> to QQ_I[x,b,_t0,exp(a)]
 

Maxima [A] (verification not implemented)

Time = 1.46 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.94 \[ \int x \arctan (c+(i+c) \tanh (a+b x)) \, dx={\left (\frac {2 \, x^{3}}{3 i \, c - 3} - \frac {2 \, b^{2} x^{2} \log \left (i \, c e^{\left (2 \, b x + 2 \, a\right )} + 1\right ) + 2 \, b x {\rm Li}_2\left (-i \, c e^{\left (2 \, b x + 2 \, a\right )}\right ) - {\rm Li}_{3}(-i \, c e^{\left (2 \, b x + 2 \, a\right )})}{-2 \, b^{3} {\left (-i \, c + 1\right )}}\right )} b {\left (c + i\right )} + \frac {1}{2} \, x^{2} \arctan \left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right ) \] Input:

integrate(x*arctan(c+(I+c)*tanh(b*x+a)),x, algorithm="maxima")
 

Output:

(2*x^3/(3*I*c - 3) - (2*b^2*x^2*log(I*c*e^(2*b*x + 2*a) + 1) + 2*b*x*dilog 
(-I*c*e^(2*b*x + 2*a)) - polylog(3, -I*c*e^(2*b*x + 2*a)))/(b^3*(2*I*c - 2 
)))*b*(c + I) + 1/2*x^2*arctan((c + I)*tanh(b*x + a) + c)
 

Giac [F]

\[ \int x \arctan (c+(i+c) \tanh (a+b x)) \, dx=\int { x \arctan \left ({\left (c + i\right )} \tanh \left (b x + a\right ) + c\right ) \,d x } \] Input:

integrate(x*arctan(c+(I+c)*tanh(b*x+a)),x, algorithm="giac")
 

Output:

integrate(x*arctan((c + I)*tanh(b*x + a) + c), x)
 

Mupad [F(-1)]

Timed out. \[ \int x \arctan (c+(i+c) \tanh (a+b x)) \, dx=\int x\,\mathrm {atan}\left (c+\mathrm {tanh}\left (a+b\,x\right )\,\left (c+1{}\mathrm {i}\right )\right ) \,d x \] Input:

int(x*atan(c + tanh(a + b*x)*(c + 1i)),x)
 

Output:

int(x*atan(c + tanh(a + b*x)*(c + 1i)), x)
 

Reduce [F]

\[ \int x \arctan (c+(i+c) \tanh (a+b x)) \, dx=\int \mathit {atan} \left (\tanh \left (b x +a \right ) c +\tanh \left (b x +a \right ) i +c \right ) x d x \] Input:

int(x*atan(c+(I+c)*tanh(b*x+a)),x)
 

Output:

int(atan(tanh(a + b*x)*c + tanh(a + b*x)*i + c)*x,x)