\(\int (e+f x)^3 \arctan (\tanh (a+b x)) \, dx\) [76]

Optimal result

Integrand size = 15, antiderivative size = 299 \[ \int (e+f x)^3 \arctan (\tanh (a+b x)) \, dx=-\frac {(e+f x)^4 \arctan \left (e^{2 a+2 b x}\right )}{4 f}+\frac {(e+f x)^4 \arctan (\tanh (a+b x))}{4 f}+\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{4 b}-\frac {i (e+f x)^3 \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{4 b}-\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{8 b^2}+\frac {3 i f (e+f x)^2 \operatorname {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{8 b^2}+\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{8 b^3}-\frac {3 i f^2 (e+f x) \operatorname {PolyLog}\left (4,i e^{2 a+2 b x}\right )}{8 b^3}-\frac {3 i f^3 \operatorname {PolyLog}\left (5,-i e^{2 a+2 b x}\right )}{16 b^4}+\frac {3 i f^3 \operatorname {PolyLog}\left (5,i e^{2 a+2 b x}\right )}{16 b^4} \] Output:

-1/4*(f*x+e)^4*arctan(exp(2*b*x+2*a))/f+1/4*(f*x+e)^4*arctan(tanh(b*x+a))/ 
f+1/4*I*(f*x+e)^3*polylog(2,-I*exp(2*b*x+2*a))/b-1/4*I*(f*x+e)^3*polylog(2 
,I*exp(2*b*x+2*a))/b-3/8*I*f*(f*x+e)^2*polylog(3,-I*exp(2*b*x+2*a))/b^2+3/ 
8*I*f*(f*x+e)^2*polylog(3,I*exp(2*b*x+2*a))/b^2+3/8*I*f^2*(f*x+e)*polylog( 
4,-I*exp(2*b*x+2*a))/b^3-3/8*I*f^2*(f*x+e)*polylog(4,I*exp(2*b*x+2*a))/b^3 
-3/16*I*f^3*polylog(5,-I*exp(2*b*x+2*a))/b^4+3/16*I*f^3*polylog(5,I*exp(2* 
b*x+2*a))/b^4
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(600\) vs. \(2(299)=598\).

Time = 0.67 (sec) , antiderivative size = 600, normalized size of antiderivative = 2.01 \[ \int (e+f x)^3 \arctan (\tanh (a+b x)) \, dx=\frac {1}{4} x \left (4 e^3+6 e^2 f x+4 e f^2 x^2+f^3 x^3\right ) \arctan (\tanh (a+b x))-\frac {i \left (8 b^4 e^3 x \log \left (1-i e^{2 (a+b x)}\right )+12 b^4 e^2 f x^2 \log \left (1-i e^{2 (a+b x)}\right )+8 b^4 e f^2 x^3 \log \left (1-i e^{2 (a+b x)}\right )+2 b^4 f^3 x^4 \log \left (1-i e^{2 (a+b x)}\right )-8 b^4 e^3 x \log \left (1+i e^{2 (a+b x)}\right )-12 b^4 e^2 f x^2 \log \left (1+i e^{2 (a+b x)}\right )-8 b^4 e f^2 x^3 \log \left (1+i e^{2 (a+b x)}\right )-2 b^4 f^3 x^4 \log \left (1+i e^{2 (a+b x)}\right )-4 b^3 (e+f x)^3 \operatorname {PolyLog}\left (2,-i e^{2 (a+b x)}\right )+4 b^3 (e+f x)^3 \operatorname {PolyLog}\left (2,i e^{2 (a+b x)}\right )+6 b^2 e^2 f \operatorname {PolyLog}\left (3,-i e^{2 (a+b x)}\right )+12 b^2 e f^2 x \operatorname {PolyLog}\left (3,-i e^{2 (a+b x)}\right )+6 b^2 f^3 x^2 \operatorname {PolyLog}\left (3,-i e^{2 (a+b x)}\right )-6 b^2 e^2 f \operatorname {PolyLog}\left (3,i e^{2 (a+b x)}\right )-12 b^2 e f^2 x \operatorname {PolyLog}\left (3,i e^{2 (a+b x)}\right )-6 b^2 f^3 x^2 \operatorname {PolyLog}\left (3,i e^{2 (a+b x)}\right )-6 b e f^2 \operatorname {PolyLog}\left (4,-i e^{2 (a+b x)}\right )-6 b f^3 x \operatorname {PolyLog}\left (4,-i e^{2 (a+b x)}\right )+6 b e f^2 \operatorname {PolyLog}\left (4,i e^{2 (a+b x)}\right )+6 b f^3 x \operatorname {PolyLog}\left (4,i e^{2 (a+b x)}\right )+3 f^3 \operatorname {PolyLog}\left (5,-i e^{2 (a+b x)}\right )-3 f^3 \operatorname {PolyLog}\left (5,i e^{2 (a+b x)}\right )\right )}{16 b^4} \] Input:

Integrate[(e + f*x)^3*ArcTan[Tanh[a + b*x]],x]
 

Output:

(x*(4*e^3 + 6*e^2*f*x + 4*e*f^2*x^2 + f^3*x^3)*ArcTan[Tanh[a + b*x]])/4 - 
((I/16)*(8*b^4*e^3*x*Log[1 - I*E^(2*(a + b*x))] + 12*b^4*e^2*f*x^2*Log[1 - 
 I*E^(2*(a + b*x))] + 8*b^4*e*f^2*x^3*Log[1 - I*E^(2*(a + b*x))] + 2*b^4*f 
^3*x^4*Log[1 - I*E^(2*(a + b*x))] - 8*b^4*e^3*x*Log[1 + I*E^(2*(a + b*x))] 
 - 12*b^4*e^2*f*x^2*Log[1 + I*E^(2*(a + b*x))] - 8*b^4*e*f^2*x^3*Log[1 + I 
*E^(2*(a + b*x))] - 2*b^4*f^3*x^4*Log[1 + I*E^(2*(a + b*x))] - 4*b^3*(e + 
f*x)^3*PolyLog[2, (-I)*E^(2*(a + b*x))] + 4*b^3*(e + f*x)^3*PolyLog[2, I*E 
^(2*(a + b*x))] + 6*b^2*e^2*f*PolyLog[3, (-I)*E^(2*(a + b*x))] + 12*b^2*e* 
f^2*x*PolyLog[3, (-I)*E^(2*(a + b*x))] + 6*b^2*f^3*x^2*PolyLog[3, (-I)*E^( 
2*(a + b*x))] - 6*b^2*e^2*f*PolyLog[3, I*E^(2*(a + b*x))] - 12*b^2*e*f^2*x 
*PolyLog[3, I*E^(2*(a + b*x))] - 6*b^2*f^3*x^2*PolyLog[3, I*E^(2*(a + b*x) 
)] - 6*b*e*f^2*PolyLog[4, (-I)*E^(2*(a + b*x))] - 6*b*f^3*x*PolyLog[4, (-I 
)*E^(2*(a + b*x))] + 6*b*e*f^2*PolyLog[4, I*E^(2*(a + b*x))] + 6*b*f^3*x*P 
olyLog[4, I*E^(2*(a + b*x))] + 3*f^3*PolyLog[5, (-I)*E^(2*(a + b*x))] - 3* 
f^3*PolyLog[5, I*E^(2*(a + b*x))]))/b^4
 

Rubi [A] (verified)

Time = 1.15 (sec) , antiderivative size = 327, normalized size of antiderivative = 1.09, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.533, Rules used = {5706, 3042, 4668, 3011, 7163, 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.

\(\displaystyle \int (e+f x)^3 \arctan (\tanh (a+b x)) \, dx\)

\(\Big \downarrow \) 5706

\(\displaystyle \frac {(e+f x)^4 \arctan (\tanh (a+b x))}{4 f}-\frac {b \int (e+f x)^4 \text {sech}(2 a+2 b x)dx}{4 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(e+f x)^4 \arctan (\tanh (a+b x))}{4 f}-\frac {b \int (e+f x)^4 \csc \left (2 i a+2 i b x+\frac {\pi }{2}\right )dx}{4 f}\)

\(\Big \downarrow \) 4668

\(\displaystyle \frac {(e+f x)^4 \arctan (\tanh (a+b x))}{4 f}-\frac {b \left (-\frac {2 i f \int (e+f x)^3 \log \left (1-i e^{2 a+2 b x}\right )dx}{b}+\frac {2 i f \int (e+f x)^3 \log \left (1+i e^{2 a+2 b x}\right )dx}{b}+\frac {(e+f x)^4 \arctan \left (e^{2 a+2 b x}\right )}{b}\right )}{4 f}\)

\(\Big \downarrow \) 3011

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 7163

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

\(\Big \downarrow \) 2720

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

\(\Big \downarrow \) 7143

\(\displaystyle \frac {(e+f x)^4 \arctan (\tanh (a+b x))}{4 f}-\frac {b \left (\frac {(e+f x)^4 \arctan \left (e^{2 a+2 b x}\right )}{b}+\frac {2 i f \left (\frac {3 f \left (\frac {(e+f x)^2 \operatorname {PolyLog}\left (3,-i e^{2 a+2 b x}\right )}{2 b}-\frac {f \left (\frac {(e+f x) \operatorname {PolyLog}\left (4,-i e^{2 a+2 b x}\right )}{2 b}-\frac {f \operatorname {PolyLog}\left (5,-i e^{2 a+2 b x}\right )}{4 b^2}\right )}{b}\right )}{2 b}-\frac {(e+f x)^3 \operatorname {PolyLog}\left (2,-i e^{2 a+2 b x}\right )}{2 b}\right )}{b}-\frac {2 i f \left (\frac {3 f \left (\frac {(e+f x)^2 \operatorname {PolyLog}\left (3,i e^{2 a+2 b x}\right )}{2 b}-\frac {f \left (\frac {(e+f x) \operatorname {PolyLog}\left (4,i e^{2 a+2 b x}\right )}{2 b}-\frac {f \operatorname {PolyLog}\left (5,i e^{2 a+2 b x}\right )}{4 b^2}\right )}{b}\right )}{2 b}-\frac {(e+f x)^3 \operatorname {PolyLog}\left (2,i e^{2 a+2 b x}\right )}{2 b}\right )}{b}\right )}{4 f}\)

Input:

Int[(e + f*x)^3*ArcTan[Tanh[a + b*x]],x]
 

Output:

((e + f*x)^4*ArcTan[Tanh[a + b*x]])/(4*f) - (b*(((e + f*x)^4*ArcTan[E^(2*a 
 + 2*b*x)])/b + ((2*I)*f*(-1/2*((e + f*x)^3*PolyLog[2, (-I)*E^(2*a + 2*b*x 
)])/b + (3*f*(((e + f*x)^2*PolyLog[3, (-I)*E^(2*a + 2*b*x)])/(2*b) - (f*(( 
(e + f*x)*PolyLog[4, (-I)*E^(2*a + 2*b*x)])/(2*b) - (f*PolyLog[5, (-I)*E^( 
2*a + 2*b*x)])/(4*b^2)))/b))/(2*b)))/b - ((2*I)*f*(-1/2*((e + f*x)^3*PolyL 
og[2, I*E^(2*a + 2*b*x)])/b + (3*f*(((e + f*x)^2*PolyLog[3, I*E^(2*a + 2*b 
*x)])/(2*b) - (f*(((e + f*x)*PolyLog[4, I*E^(2*a + 2*b*x)])/(2*b) - (f*Pol 
yLog[5, I*E^(2*a + 2*b*x)])/(4*b^2)))/b))/(2*b)))/b))/(4*f)
 

Defintions of rubi rules used

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 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4668

Int[csc[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)]*((c_.) + (d_.)*(x_ 
))^(m_.), x_Symbol] :> Simp[-2*(c + d*x)^m*(ArcTanh[E^((-I)*e + f*fz*x)/E^( 
I*k*Pi)]/(f*fz*I)), x] + (-Simp[d*(m/(f*fz*I))   Int[(c + d*x)^(m - 1)*Log[ 
1 - E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x] + Simp[d*(m/(f*fz*I))   Int[(c 
+ d*x)^(m - 1)*Log[1 + E^((-I)*e + f*fz*x)/E^(I*k*Pi)], x], x]) /; FreeQ[{c 
, d, e, f, fz}, x] && IntegerQ[2*k] && IGtQ[m, 0]
 

rule 5706

Int[ArcTan[Tanh[(a_.) + (b_.)*(x_)]]*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] 
:> Simp[(e + f*x)^(m + 1)*(ArcTan[Tanh[a + b*x]]/(f*(m + 1))), x] - Simp[b/ 
(f*(m + 1))   Int[(e + f*x)^(m + 1)*Sech[2*a + 2*b*x], x], x] /; FreeQ[{a, 
b, e, f}, x] && IGtQ[m, 0]
 

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]
 

rule 7163

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]
 

Maple [C] (warning: unable to verify)

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

Time = 38.76 (sec) , antiderivative size = 3570, normalized size of antiderivative = 11.94

method	result	size
risch	\(\text {Expression too large to display}\)	\(3570\)

Input:

int((f*x+e)^3*arctan(tanh(b*x+a)),x,method=_RETURNVERBOSE)
 

Output:

-3/16*I*f^3*polylog(5,-I*exp(2*b*x+2*a))/b^4+3/16*I*f^3*polylog(5,I*exp(2* 
b*x+2*a))/b^4+1/16*Pi*(csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))*csgn( 
(1+I)*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))-csgn((1+I)*(exp(2*b*x+2*a)+I) 
/(exp(2*b*x+2*a)+1))^2-csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))*csgn( 
(1-I)*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))-csgn((1-I)*(exp(2*b*x+2*a)-I) 
/(exp(2*b*x+2*a)+1))^2-csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)-I 
))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))+csgn(I/(exp(2*b*x+2*a)+1) 
)*csgn(I*(exp(2*b*x+2*a)+I))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1)) 
+csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^ 
2-csgn(I/(exp(2*b*x+2*a)+1))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1)) 
^2+csgn(I*(exp(2*b*x+2*a)-I))*csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1) 
)^2-csgn(I*(exp(2*b*x+2*a)+I))*csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1 
))^2-csgn(I*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1))^3+csgn(I*(exp(2*b*x+2*a 
)-I)/(exp(2*b*x+2*a)+1))*csgn((1-I)*(exp(2*b*x+2*a)-I)/(exp(2*b*x+2*a)+1)) 
^2+csgn(I*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^3-csgn(I*(exp(2*b*x+2*a)+ 
I)/(exp(2*b*x+2*a)+1))*csgn((1+I)*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^2 
+csgn((1+I)*(exp(2*b*x+2*a)+I)/(exp(2*b*x+2*a)+1))^3+csgn((1-I)*(exp(2*b*x 
+2*a)-I)/(exp(2*b*x+2*a)+1))^3+1)*(f*x+e)^4/f-1/8*I*f^3*ln(exp(2*b*x+2*a)- 
I)*x^4-1/2*I*ln(exp(2*b*x+2*a)-I)*x*e^3-1/8*I/f*ln(exp(2*b*x+2*a)-I)*e^4+3 
/2*I*f/b*a*e^2*ln(((-I)^(1/2)-exp(b*x+a))/(-I)^(1/2))*x+3/2*I*f/b*a*e^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. 1460 vs. \(2 (236) = 472\).

Time = 0.21 (sec) , antiderivative size = 1460, normalized size of antiderivative = 4.88 \[ \int (e+f x)^3 \arctan (\tanh (a+b x)) \, dx=\text {Too large to display} \] Input:

integrate((f*x+e)^3*arctan(tanh(b*x+a)),x, algorithm="fricas")
 

Output:

1/8*(24*I*f^3*polylog(5, 1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 
24*I*f^3*polylog(5, -1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 24*I 
*f^3*polylog(5, 1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 24*I*f^3 
*polylog(5, -1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + 2*(b^4*f^3* 
x^4 + 4*b^4*e*f^2*x^3 + 6*b^4*e^2*f*x^2 + 4*b^4*e^3*x)*arctan(sinh(b*x + a 
)/cosh(b*x + a)) - 4*(I*b^3*f^3*x^3 + 3*I*b^3*e*f^2*x^2 + 3*I*b^3*e^2*f*x 
+ I*b^3*e^3)*dilog(1/2*sqrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 4*(I*b 
^3*f^3*x^3 + 3*I*b^3*e*f^2*x^2 + 3*I*b^3*e^2*f*x + I*b^3*e^3)*dilog(-1/2*s 
qrt(4*I)*(cosh(b*x + a) + sinh(b*x + a))) - 4*(-I*b^3*f^3*x^3 - 3*I*b^3*e* 
f^2*x^2 - 3*I*b^3*e^2*f*x - I*b^3*e^3)*dilog(1/2*sqrt(-4*I)*(cosh(b*x + a) 
 + sinh(b*x + a))) - 4*(-I*b^3*f^3*x^3 - 3*I*b^3*e*f^2*x^2 - 3*I*b^3*e^2*f 
*x - I*b^3*e^3)*dilog(-1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a))) + ( 
-I*b^4*f^3*x^4 - 4*I*b^4*e*f^2*x^3 - 6*I*b^4*e^2*f*x^2 - 4*I*b^4*e^3*x - 4 
*I*a*b^3*e^3 + 6*I*a^2*b^2*e^2*f - 4*I*a^3*b*e*f^2 + I*a^4*f^3)*log(1/2*sq 
rt(4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (-I*b^4*f^3*x^4 - 4*I*b^4*e 
*f^2*x^3 - 6*I*b^4*e^2*f*x^2 - 4*I*b^4*e^3*x - 4*I*a*b^3*e^3 + 6*I*a^2*b^2 
*e^2*f - 4*I*a^3*b*e*f^2 + I*a^4*f^3)*log(-1/2*sqrt(4*I)*(cosh(b*x + a) + 
sinh(b*x + a)) + 1) + (I*b^4*f^3*x^4 + 4*I*b^4*e*f^2*x^3 + 6*I*b^4*e^2*f*x 
^2 + 4*I*b^4*e^3*x + 4*I*a*b^3*e^3 - 6*I*a^2*b^2*e^2*f + 4*I*a^3*b*e*f^2 - 
 I*a^4*f^3)*log(1/2*sqrt(-4*I)*(cosh(b*x + a) + sinh(b*x + a)) + 1) + (...
 

Sympy [F]

\[ \int (e+f x)^3 \arctan (\tanh (a+b x)) \, dx=\int \left (e + f x\right )^{3} \operatorname {atan}{\left (\tanh {\left (a + b x \right )} \right )}\, dx \] Input:

integrate((f*x+e)**3*atan(tanh(b*x+a)),x)
 

Output:

Integral((e + f*x)**3*atan(tanh(a + b*x)), x)
 

Maxima [F]

\[ \int (e+f x)^3 \arctan (\tanh (a+b x)) \, dx=\int { {\left (f x + e\right )}^{3} \arctan \left (\tanh \left (b x + a\right )\right ) \,d x } \] Input:

integrate((f*x+e)^3*arctan(tanh(b*x+a)),x, algorithm="maxima")
 

Output:

1/4*(f^3*x^4 + 4*e*f^2*x^3 + 6*e^2*f*x^2 + 4*e^3*x)*arctan((e^(2*b*x + 2*a 
) - 1)/(e^(2*b*x + 2*a) + 1)) - integrate(1/2*(b*f^3*x^4*e^(2*a) + 4*b*e*f 
^2*x^3*e^(2*a) + 6*b*e^2*f*x^2*e^(2*a) + 4*b*e^3*x*e^(2*a))*e^(2*b*x)/(e^( 
4*b*x + 4*a) + 1), x)
 

Giac [F(-1)]

Timed out. \[ \int (e+f x)^3 \arctan (\tanh (a+b x)) \, dx=\text {Timed out} \] Input:

integrate((f*x+e)^3*arctan(tanh(b*x+a)),x, algorithm="giac")
 

Output:

Timed out
 

Mupad [F(-1)]

Timed out. \[ \int (e+f x)^3 \arctan (\tanh (a+b x)) \, dx=\int \mathrm {atan}\left (\mathrm {tanh}\left (a+b\,x\right )\right )\,{\left (e+f\,x\right )}^3 \,d x \] Input:

int(atan(tanh(a + b*x))*(e + f*x)^3,x)
                                                                                    
                                                                                    
 

Output:

int(atan(tanh(a + b*x))*(e + f*x)^3, x)
 

Reduce [F]

\[ \int (e+f x)^3 \arctan (\tanh (a+b x)) \, dx=\left (\int \mathit {atan} \left (\tanh \left (b x +a \right )\right )d x \right ) e^{3}+\left (\int \mathit {atan} \left (\tanh \left (b x +a \right )\right ) x^{3}d x \right ) f^{3}+3 \left (\int \mathit {atan} \left (\tanh \left (b x +a \right )\right ) x^{2}d x \right ) e \,f^{2}+3 \left (\int \mathit {atan} \left (\tanh \left (b x +a \right )\right ) x d x \right ) e^{2} f \] Input:

int((f*x+e)^3*atan(tanh(b*x+a)),x)
 

Output:

int(atan(tanh(a + b*x)),x)*e**3 + int(atan(tanh(a + b*x))*x**3,x)*f**3 + 3 
*int(atan(tanh(a + b*x))*x**2,x)*e*f**2 + 3*int(atan(tanh(a + b*x))*x,x)*e 
**2*f