3.4.14 \(\int (e+f x) \tanh ^{-1}(\tan (a+b x)) \, dx\) [314]

Optimal. Leaf size=162 \[ \frac {i (e+f x)^2 \text {ArcTan}\left (e^{2 i (a+b x)}\right )}{2 f}+\frac {(e+f x)^2 \tanh ^{-1}(\tan (a+b x))}{2 f}-\frac {i (e+f x) \text {PolyLog}\left (2,-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \text {PolyLog}\left (2,i e^{2 i (a+b x)}\right )}{4 b}+\frac {f \text {PolyLog}\left (3,-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {f \text {PolyLog}\left (3,i e^{2 i (a+b x)}\right )}{8 b^2} \]

[Out]

1/2*I*(f*x+e)^2*arctan(exp(2*I*(b*x+a)))/f+1/2*(f*x+e)^2*arctanh(tan(b*x+a))/f-1/4*I*(f*x+e)*polylog(2,-I*exp(
2*I*(b*x+a)))/b+1/4*I*(f*x+e)*polylog(2,I*exp(2*I*(b*x+a)))/b+1/8*f*polylog(3,-I*exp(2*I*(b*x+a)))/b^2-1/8*f*p
olylog(3,I*exp(2*I*(b*x+a)))/b^2

________________________________________________________________________________________

Rubi [A]
time = 0.08, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.385, Rules used = {6386, 4266, 2611, 2320, 6724} \begin {gather*} \frac {i (e+f x)^2 \text {ArcTan}\left (e^{2 i (a+b x)}\right )}{2 f}+\frac {f \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {f \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {i (e+f x) \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {(e+f x)^2 \tanh ^{-1}(\tan (a+b x))}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e + f*x)*ArcTanh[Tan[a + b*x]],x]

[Out]

((I/2)*(e + f*x)^2*ArcTan[E^((2*I)*(a + b*x))])/f + ((e + f*x)^2*ArcTanh[Tan[a + b*x]])/(2*f) - ((I/4)*(e + f*
x)*PolyLog[2, (-I)*E^((2*I)*(a + b*x))])/b + ((I/4)*(e + f*x)*PolyLog[2, I*E^((2*I)*(a + b*x))])/b + (f*PolyLo
g[3, (-I)*E^((2*I)*(a + b*x))])/(8*b^2) - (f*PolyLog[3, I*E^((2*I)*(a + b*x))])/(8*b^2)

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{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 2611

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] + Dist[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 4266

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

Rule 6386

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

Rule 6724

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> 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]

Rubi steps

\begin {align*} \int (e+f x) \tanh ^{-1}(\tan (a+b x)) \, dx &=\frac {(e+f x)^2 \tanh ^{-1}(\tan (a+b x))}{2 f}-\frac {b \int (e+f x)^2 \sec (2 a+2 b x) \, dx}{2 f}\\ &=\frac {i (e+f x)^2 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{2 f}+\frac {(e+f x)^2 \tanh ^{-1}(\tan (a+b x))}{2 f}+\frac {1}{2} \int (e+f x) \log \left (1-i e^{i (2 a+2 b x)}\right ) \, dx-\frac {1}{2} \int (e+f x) \log \left (1+i e^{i (2 a+2 b x)}\right ) \, dx\\ &=\frac {i (e+f x)^2 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{2 f}+\frac {(e+f x)^2 \tanh ^{-1}(\tan (a+b x))}{2 f}-\frac {i (e+f x) \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {(i f) \int \text {Li}_2\left (-i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}-\frac {(i f) \int \text {Li}_2\left (i e^{i (2 a+2 b x)}\right ) \, dx}{4 b}\\ &=\frac {i (e+f x)^2 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{2 f}+\frac {(e+f x)^2 \tanh ^{-1}(\tan (a+b x))}{2 f}-\frac {i (e+f x) \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {f \text {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{8 b^2}-\frac {f \text {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i (2 a+2 b x)}\right )}{8 b^2}\\ &=\frac {i (e+f x)^2 \tan ^{-1}\left (e^{2 i (a+b x)}\right )}{2 f}+\frac {(e+f x)^2 \tanh ^{-1}(\tan (a+b x))}{2 f}-\frac {i (e+f x) \text {Li}_2\left (-i e^{2 i (a+b x)}\right )}{4 b}+\frac {i (e+f x) \text {Li}_2\left (i e^{2 i (a+b x)}\right )}{4 b}+\frac {f \text {Li}_3\left (-i e^{2 i (a+b x)}\right )}{8 b^2}-\frac {f \text {Li}_3\left (i e^{2 i (a+b x)}\right )}{8 b^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.32, size = 295, normalized size = 1.82 \begin {gather*} e x \tanh ^{-1}(\tan (a+b x))+\frac {1}{2} f x^2 \tanh ^{-1}(\tan (a+b x))-\frac {e \left ((-4 a+\pi -4 b x) \left (\log \left (1-i e^{-2 i (a+b x)}\right )-\log \left (1+i e^{-2 i (a+b x)}\right )\right )-(-4 a+\pi ) \log \left (\cot \left (a+\frac {\pi }{4}+b x\right )\right )+2 i \left (\text {PolyLog}\left (2,-i e^{-2 i (a+b x)}\right )-\text {PolyLog}\left (2,i e^{-2 i (a+b x)}\right )\right )\right )}{8 b}+\frac {f \left (4 i b^2 x^2 \text {ArcTan}(\cos (2 (a+b x))+i \sin (2 (a+b x)))+2 i b x \text {PolyLog}(2,i \cos (2 (a+b x))-\sin (2 (a+b x)))-2 i b x \text {PolyLog}(2,-i \cos (2 (a+b x))+\sin (2 (a+b x)))-\text {PolyLog}(3,i \cos (2 (a+b x))-\sin (2 (a+b x)))+\text {PolyLog}(3,-i \cos (2 (a+b x))+\sin (2 (a+b x)))\right )}{8 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(e + f*x)*ArcTanh[Tan[a + b*x]],x]

[Out]

e*x*ArcTanh[Tan[a + b*x]] + (f*x^2*ArcTanh[Tan[a + b*x]])/2 - (e*((-4*a + Pi - 4*b*x)*(Log[1 - I/E^((2*I)*(a +
b*x))] - Log[1 + I/E^((2*I)*(a + b*x))]) - (-4*a + Pi)*Log[Cot[a + Pi/4 + b*x]] + (2*I)*(PolyLog[2, (-I)/E^((
2*I)*(a + b*x))] - PolyLog[2, I/E^((2*I)*(a + b*x))])))/(8*b) + (f*((4*I)*b^2*x^2*ArcTan[Cos[2*(a + b*x)] + I*
Sin[2*(a + b*x)]] + (2*I)*b*x*PolyLog[2, I*Cos[2*(a + b*x)] - Sin[2*(a + b*x)]] - (2*I)*b*x*PolyLog[2, (-I)*Co
s[2*(a + b*x)] + Sin[2*(a + b*x)]] - PolyLog[3, I*Cos[2*(a + b*x)] - Sin[2*(a + b*x)]] + PolyLog[3, (-I)*Cos[2
*(a + b*x)] + Sin[2*(a + b*x)]]))/(8*b^2)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.08, size = 2543, normalized size = 15.70

method	result	size
risch	\(\text {Expression too large to display}\)	\(2543\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*arctanh(tan(b*x+a)),x,method=_RETURNVERBOSE)

[Out]

1/8*f*polylog(3,-I*exp(2*I*(b*x+a)))/b^2-1/8*f*polylog(3,I*exp(2*I*(b*x+a)))/b^2+1/4*I*Pi*x*e*csgn(I/(exp(2*I*
(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2+1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))+I))*csg
n(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2+1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+a))+I))*csgn(I*(exp(2*I*(b*x+
a))+I)/(exp(2*I*(b*x+a))+1))^2*x^2+1/8*I*Pi*f*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*
I*(b*x+a))+1))^2*x^2+1/2*I*f/b^2*a*(I*b*x+I*a)*ln(1+exp(I*(b*x+a))*(-1)^(3/4))+1/2*I*f/b^2*a*(I*b*x+I*a)*ln(1-
exp(I*(b*x+a))*(-1)^(3/4))-1/2*I*f/b^2*a*(I*b*x+I*a)*ln(((-I)^(1/2)-exp(I*(b*x+a)))/(-I)^(1/2))-1/4*I*Pi*e*x-1
/8*I*Pi*f*x^2-1/2*I/b^2*f*a*dilog(((-I)^(1/2)-exp(I*(b*x+a)))/(-I)^(1/2))-1/2*I*f/b^2*a*(I*b*x+I*a)*ln(((-I)^(
1/2)+exp(I*(b*x+a)))/(-I)^(1/2))-1/8*I*Pi*f*csgn((1+I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^3*x^2+1/4*I*
Pi*x*e*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^3-1/4*I*Pi*x*e*csgn((1+I)*(exp(2*I*(b*x+a))+I)/(exp(2
*I*(b*x+a))+1))^3-1/4*f/b^2*(I*b*x+I*a)^2*ln(1+I*exp(2*I*(b*x+a)))-1/4*f/b^2*(I*b*x+I*a)*polylog(2,-I*exp(2*I*
(b*x+a)))-1/4*I*Pi*x*e*csgn((1-I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^3+1/2*I/b*e*dilog(((-I)^(1/2)-exp
(I*(b*x+a)))/(-I)^(1/2))+1/2*I/b*e*dilog(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))-1/4*ln(exp(2*I*(b*x+a))-I)*f*
x^2+1/2*I*f/b^2*a*dilog(1+exp(I*(b*x+a))*(-1)^(3/4))+1/2*I*f/b^2*a*dilog(1-exp(I*(b*x+a))*(-1)^(3/4))-1/2*I*e/
b*(I*b*x+I*a)*ln(1+exp(I*(b*x+a))*(-1)^(3/4))-1/2*I*e/b*(I*b*x+I*a)*ln(1-exp(I*(b*x+a))*(-1)^(3/4))-1/8*I*Pi*f
*csgn(I*(exp(2*I*(b*x+a))-I))*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^2*x^2-1/4*I*Pi*x*e*csgn(I/(exp
(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))+I))*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))-1/8*I*Pi*f*cs
gn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))+I))*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))*x^2-
1/8*I*Pi*f*csgn((1-I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^3*x^2-1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+a))-I)/
(exp(2*I*(b*x+a))+1))*csgn((1-I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^2*x^2-1/4*I*Pi*x*e*csgn(I*(exp(2*I
*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))*csgn((1-I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^2+1/8*I*Pi*f*csgn(I*(
exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^3*x^2-1/2*ln(exp(2*I*(b*x+a))-I)*e*x-1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b
*x+a))+I)/(exp(2*I*(b*x+a))+1))*csgn((1+I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))-1/8*I*Pi*f*csgn(I*(exp(2
*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))*csgn((1+I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))*x^2+1/8*I*Pi*f*csgn
(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))*csgn((1-I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))*x^2+1/4*I*
Pi*x*e*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))*csgn((1-I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))
+1/4*f/b^2*(I*b*x+I*a)^2*ln(1-I*exp(2*I*(b*x+a)))+1/4*f/b^2*(I*b*x+I*a)*polylog(2,I*exp(2*I*(b*x+a)))-1/4*I*Pi
*x*e*csgn(I*(exp(2*I*(b*x+a))-I))*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^2+1/8*I*Pi*f*csgn((1-I)*(e
xp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^2*x^2+1/8*I*Pi*f*csgn((1+I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1)
)^2*x^2+1/4*I*Pi*x*e*csgn((1+I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2+1/2/b*e*a*ln(exp(2*I*(b*x+a))+I)-
1/2*e/b*a*ln(-exp(2*I*(b*x+a))+I)+1/4*f/b^2*a^2*ln(-exp(2*I*(b*x+a))+I)-1/4/b^2*f*a^2*ln(exp(2*I*(b*x+a))+I)-1
/2*I*e/b*dilog(1+exp(I*(b*x+a))*(-1)^(3/4))-1/2*I*e/b*dilog(1-exp(I*(b*x+a))*(-1)^(3/4))-1/4*I*Pi*x*e*csgn(I/(
exp(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))^2+1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+a))+I
)/(exp(2*I*(b*x+a))+1))*csgn((1+I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2*x^2+1/4*I*Pi*x*e*csgn(I*(exp(2
*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))*csgn((1+I)*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^2-1/2*I/b^2*f*a*dil
og(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))-1/8*I*Pi*f*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))-I)
/(exp(2*I*(b*x+a))+1))^2*x^2+1/2*I*e/b*(I*b*x+I*a)*ln(((-I)^(1/2)-exp(I*(b*x+a)))/(-I)^(1/2))+1/2*I*e/b*(I*b*x
+I*a)*ln(((-I)^(1/2)+exp(I*(b*x+a)))/(-I)^(1/2))+1/4*I*Pi*x*e*csgn((1-I)*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a)
)+1))^2-1/4*I*Pi*x*e*csgn(I*(exp(2*I*(b*x+a))+I)/(exp(2*I*(b*x+a))+1))^3-1/8*I*Pi*f*csgn(I*(exp(2*I*(b*x+a))+I
)/(exp(2*I*(b*x+a))+1))^3*x^2+1/8*I*Pi*f*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))-I))*csgn(I*(exp
(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))*x^2+1/4*I*Pi*x*e*csgn(I/(exp(2*I*(b*x+a))+1))*csgn(I*(exp(2*I*(b*x+a))-
I))*csgn(I*(exp(2*I*(b*x+a))-I)/(exp(2*I*(b*x+a))+1))+1/2*(1/2*f*x^2+e*x)*ln(exp(2*I*(b*x+a))+I)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*arctanh(tan(b*x+a)),x, algorithm="maxima")

[Out]

1/8*(f*x^2 + 2*x*e)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x + 2*a)^2 + 4*sin(2*b*x + 2*a) + 2) - 1/8*(f*x^2 + 2
*x*e)*log(2*cos(2*b*x + 2*a)^2 + 2*sin(2*b*x + 2*a)^2 - 4*sin(2*b*x + 2*a) + 2) - integrate(((b*f*x^2 + 2*b*x*
e)*cos(4*b*x + 4*a)*cos(2*b*x + 2*a) + (b*f*x^2 + 2*b*x*e)*sin(4*b*x + 4*a)*sin(2*b*x + 2*a) + (b*f*x^2 + 2*b*
x*e)*cos(2*b*x + 2*a))/(cos(4*b*x + 4*a)^2 + sin(4*b*x + 4*a)^2 + 2*cos(4*b*x + 4*a) + 1), x)

________________________________________________________________________________________

Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 948 vs. \(2 (134) = 268\).
time = 0.38, size = 948, normalized size = 5.85 \begin {gather*} -\frac {2 \, {\left (-i \, b f x - i \, b \cosh \left (1\right ) - i \, b \sinh \left (1\right )\right )} {\rm Li}_2\left (-\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} + 2 \, \tan \left (b x + a\right ) - i + 1}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 2 \, {\left (-i \, b f x - i \, b \cosh \left (1\right ) - i \, b \sinh \left (1\right )\right )} {\rm Li}_2\left (-\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} - 2 \, \tan \left (b x + a\right ) - i + 1}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 2 \, {\left (i \, b f x + i \, b \cosh \left (1\right ) + i \, b \sinh \left (1\right )\right )} {\rm Li}_2\left (-\frac {-\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} + 2 \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 2 \, {\left (i \, b f x + i \, b \cosh \left (1\right ) + i \, b \sinh \left (1\right )\right )} {\rm Li}_2\left (-\frac {-\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} - 2 \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1} + 1\right ) + 2 \, {\left (b^{2} f x^{2} - a^{2} f + 2 \, {\left (b^{2} x + a b\right )} \cosh \left (1\right ) + 2 \, {\left (b^{2} x + a b\right )} \sinh \left (1\right )\right )} \log \left (\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} + 2 \, \tan \left (b x + a\right ) - i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, {\left (a^{2} f - 2 \, a b \cosh \left (1\right ) - 2 \, a b \sinh \left (1\right )\right )} \log \left (\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) + i - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 \, {\left (a^{2} f - 2 \, a b \cosh \left (1\right ) - 2 \, a b \sinh \left (1\right )\right )} \log \left (\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) + i - 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 \, {\left (b^{2} f x^{2} - a^{2} f + 2 \, {\left (b^{2} x + a b\right )} \cosh \left (1\right ) + 2 \, {\left (b^{2} x + a b\right )} \sinh \left (1\right )\right )} \log \left (\frac {\left (i + 1\right ) \, \tan \left (b x + a\right )^{2} - 2 \, \tan \left (b x + a\right ) - i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, {\left (b^{2} f x^{2} - a^{2} f + 2 \, {\left (b^{2} x + a b\right )} \cosh \left (1\right ) + 2 \, {\left (b^{2} x + a b\right )} \sinh \left (1\right )\right )} \log \left (\frac {-\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} + 2 \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 \, {\left (b^{2} f x^{2} - a^{2} f + 2 \, {\left (b^{2} x + a b\right )} \cosh \left (1\right ) + 2 \, {\left (b^{2} x + a b\right )} \sinh \left (1\right )\right )} \log \left (\frac {-\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} - 2 \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) + 2 \, {\left (a^{2} f - 2 \, a b \cosh \left (1\right ) - 2 \, a b \sinh \left (1\right )\right )} \log \left (\frac {\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} + 2 i \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 2 \, {\left (a^{2} f - 2 \, a b \cosh \left (1\right ) - 2 \, a b \sinh \left (1\right )\right )} \log \left (\frac {\left (i - 1\right ) \, \tan \left (b x + a\right )^{2} - 2 i \, \tan \left (b x + a\right ) + i + 1}{\tan \left (b x + a\right )^{2} + 1}\right ) - 4 \, {\left (b^{2} f x^{2} + 2 \, b^{2} x \cosh \left (1\right ) + 2 \, b^{2} x \sinh \left (1\right )\right )} \log \left (-\frac {\tan \left (b x + a\right ) + 1}{\tan \left (b x + a\right ) - 1}\right ) - f {\rm polylog}\left (3, \frac {i \, \tan \left (b x + a\right )^{2} + 2 \, \tan \left (b x + a\right ) - i}{\tan \left (b x + a\right )^{2} + 1}\right ) + f {\rm polylog}\left (3, \frac {i \, \tan \left (b x + a\right )^{2} - 2 \, \tan \left (b x + a\right ) - i}{\tan \left (b x + a\right )^{2} + 1}\right ) - f {\rm polylog}\left (3, \frac {-i \, \tan \left (b x + a\right )^{2} + 2 \, \tan \left (b x + a\right ) + i}{\tan \left (b x + a\right )^{2} + 1}\right ) + f {\rm polylog}\left (3, \frac {-i \, \tan \left (b x + a\right )^{2} - 2 \, \tan \left (b x + a\right ) + i}{\tan \left (b x + a\right )^{2} + 1}\right )}{16 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*arctanh(tan(b*x+a)),x, algorithm="fricas")

[Out]

-1/16*(2*(-I*b*f*x - I*b*cosh(1) - I*b*sinh(1))*dilog(-((I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I + 1)/(tan(
b*x + a)^2 + 1) + 1) + 2*(-I*b*f*x - I*b*cosh(1) - I*b*sinh(1))*dilog(-((I + 1)*tan(b*x + a)^2 - 2*tan(b*x + a
) - I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2*(I*b*f*x + I*b*cosh(1) + I*b*sinh(1))*dilog(-(-(I - 1)*tan(b*x + a)^2
+ 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2*(I*b*f*x + I*b*cosh(1) + I*b*sinh(1))*dilog(-(-(I - 1
)*tan(b*x + a)^2 - 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1) + 1) + 2*(b^2*f*x^2 - a^2*f + 2*(b^2*x + a*b)*
cosh(1) + 2*(b^2*x + a*b)*sinh(1))*log(((I + 1)*tan(b*x + a)^2 + 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1))
+ 2*(a^2*f - 2*a*b*cosh(1) - 2*a*b*sinh(1))*log(((I + 1)*tan(b*x + a)^2 + 2*I*tan(b*x + a) + I - 1)/(tan(b*x
+ a)^2 + 1)) - 2*(a^2*f - 2*a*b*cosh(1) - 2*a*b*sinh(1))*log(((I + 1)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I -
1)/(tan(b*x + a)^2 + 1)) - 2*(b^2*f*x^2 - a^2*f + 2*(b^2*x + a*b)*cosh(1) + 2*(b^2*x + a*b)*sinh(1))*log(((I +
1)*tan(b*x + a)^2 - 2*tan(b*x + a) - I + 1)/(tan(b*x + a)^2 + 1)) + 2*(b^2*f*x^2 - a^2*f + 2*(b^2*x + a*b)*co
sh(1) + 2*(b^2*x + a*b)*sinh(1))*log((-(I - 1)*tan(b*x + a)^2 + 2*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1))
- 2*(b^2*f*x^2 - a^2*f + 2*(b^2*x + a*b)*cosh(1) + 2*(b^2*x + a*b)*sinh(1))*log((-(I - 1)*tan(b*x + a)^2 - 2*t
an(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) + 2*(a^2*f - 2*a*b*cosh(1) - 2*a*b*sinh(1))*log(((I - 1)*tan(b*x +
a)^2 + 2*I*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) - 2*(a^2*f - 2*a*b*cosh(1) - 2*a*b*sinh(1))*log(((I - 1
)*tan(b*x + a)^2 - 2*I*tan(b*x + a) + I + 1)/(tan(b*x + a)^2 + 1)) - 4*(b^2*f*x^2 + 2*b^2*x*cosh(1) + 2*b^2*x*
sinh(1))*log(-(tan(b*x + a) + 1)/(tan(b*x + a) - 1)) - f*polylog(3, (I*tan(b*x + a)^2 + 2*tan(b*x + a) - I)/(t
an(b*x + a)^2 + 1)) + f*polylog(3, (I*tan(b*x + a)^2 - 2*tan(b*x + a) - I)/(tan(b*x + a)^2 + 1)) - f*polylog(3
, (-I*tan(b*x + a)^2 + 2*tan(b*x + a) + I)/(tan(b*x + a)^2 + 1)) + f*polylog(3, (-I*tan(b*x + a)^2 - 2*tan(b*x
+ a) + I)/(tan(b*x + a)^2 + 1)))/b^2

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (e + f x\right ) \operatorname {atanh}{\left (\tan {\left (a + b x \right )} \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*atanh(tan(b*x+a)),x)

[Out]

Integral((e + f*x)*atanh(tan(a + b*x)), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*arctanh(tan(b*x+a)),x, algorithm="giac")

[Out]

integrate((f*x + e)*arctanh(tan(b*x + a)), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \mathrm {atanh}\left (\mathrm {tan}\left (a+b\,x\right )\right )\,\left (e+f\,x\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(tan(a + b*x))*(e + f*x),x)

[Out]

int(atanh(tan(a + b*x))*(e + f*x), x)

________________________________________________________________________________________