3.6.25 \(\int x (a+b \tanh ^{-1}(c x)) (d+e \log (1-c^2 x^2)) \, dx\) [525]

Optimal. Leaf size=140 \[ \frac {b (d-e) x}{2 c}-\frac {b e x}{c}-\frac {b (d-e) \tanh ^{-1}(c x)}{2 c^2}+\frac {b e \tanh ^{-1}(c x)}{c^2}+\frac {1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}-\frac {e \left (1-c^2 x^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2} \]

[Out]

1/2*b*(d-e)*x/c-b*e*x/c-1/2*b*(d-e)*arctanh(c*x)/c^2+b*e*arctanh(c*x)/c^2+1/2*d*x^2*(a+b*arctanh(c*x))-1/2*e*x
^2*(a+b*arctanh(c*x))+1/2*b*e*x*ln(-c^2*x^2+1)/c-1/2*e*(-c^2*x^2+1)*(a+b*arctanh(c*x))*ln(-c^2*x^2+1)/c^2

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Rubi [A]
time = 0.09, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2504, 2436, 2332, 6230, 327, 213, 2498, 212} \begin {gather*} -\frac {e \left (1-c^2 x^2\right ) \log \left (1-c^2 x^2\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 c^2}+\frac {1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {b (d-e) \tanh ^{-1}(c x)}{2 c^2}+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}+\frac {b e \tanh ^{-1}(c x)}{c^2}+\frac {b x (d-e)}{2 c}-\frac {b e x}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]),x]

[Out]

(b*(d - e)*x)/(2*c) - (b*e*x)/c - (b*(d - e)*ArcTanh[c*x])/(2*c^2) + (b*e*ArcTanh[c*x])/c^2 + (d*x^2*(a + b*Ar
cTanh[c*x]))/2 - (e*x^2*(a + b*ArcTanh[c*x]))/2 + (b*e*x*Log[1 - c^2*x^2])/(2*c) - (e*(1 - c^2*x^2)*(a + b*Arc
Tanh[c*x])*Log[1 - c^2*x^2])/(2*c^2)

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 213

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]
, x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 2332

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2498

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)], x_Symbol] :> Simp[x*Log[c*(d + e*x^n)^p], x] - Dist[e*n*p, Int[
x^n/(d + e*x^n), x], x] /; FreeQ[{c, d, e, n, p}, x]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p,
q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) &&  !(EqQ[q, 1] && ILtQ[n, 0] &&
IGtQ[m, 0])

Rule 6230

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Wit
h[{u = IntHide[x^m*(d + e*Log[f + g*x^2]), x]}, Dist[a + b*ArcTanh[c*x], u, x] - Dist[b*c, Int[ExpandIntegrand
[u/(1 - c^2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[(m + 1)/2, 0]

Rubi steps

\begin {align*} \int x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx &=\frac {1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {e \left (1-c^2 x^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2}-(b c) \int \left (-\frac {(d-e) x^2}{2 \left (-1+c^2 x^2\right )}-\frac {e \log \left (1-c^2 x^2\right )}{2 c^2}\right ) \, dx\\ &=\frac {1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {e \left (1-c^2 x^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2}+\frac {1}{2} (b c (d-e)) \int \frac {x^2}{-1+c^2 x^2} \, dx+\frac {(b e) \int \log \left (1-c^2 x^2\right ) \, dx}{2 c}\\ &=\frac {b (d-e) x}{2 c}+\frac {1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}-\frac {e \left (1-c^2 x^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2}+\frac {(b (d-e)) \int \frac {1}{-1+c^2 x^2} \, dx}{2 c}+(b c e) \int \frac {x^2}{1-c^2 x^2} \, dx\\ &=\frac {b (d-e) x}{2 c}-\frac {b e x}{c}-\frac {b (d-e) \tanh ^{-1}(c x)}{2 c^2}+\frac {1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}-\frac {e \left (1-c^2 x^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2}+\frac {(b e) \int \frac {1}{1-c^2 x^2} \, dx}{c}\\ &=\frac {b (d-e) x}{2 c}-\frac {b e x}{c}-\frac {b (d-e) \tanh ^{-1}(c x)}{2 c^2}+\frac {b e \tanh ^{-1}(c x)}{c^2}+\frac {1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \tanh ^{-1}(c x)\right )+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}-\frac {e \left (1-c^2 x^2\right ) \left (a+b \tanh ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{2 c^2}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 129, normalized size = 0.92 \begin {gather*} \frac {2 b c (d-3 e) x+2 a c^2 (d-e) x^2+2 b c^2 (d-e) x^2 \tanh ^{-1}(c x)+(b (d-3 e)-2 a e) \log (1-c x)-(b (d-3 e)+2 a e) \log (1+c x)+2 e \left (c x (b+a c x)+b \left (-1+c^2 x^2\right ) \tanh ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{4 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]),x]

[Out]

(2*b*c*(d - 3*e)*x + 2*a*c^2*(d - e)*x^2 + 2*b*c^2*(d - e)*x^2*ArcTanh[c*x] + (b*(d - 3*e) - 2*a*e)*Log[1 - c*
x] - (b*(d - 3*e) + 2*a*e)*Log[1 + c*x] + 2*e*(c*x*(b + a*c*x) + b*(-1 + c^2*x^2)*ArcTanh[c*x])*Log[1 - c^2*x^
2])/(4*c^2)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 11.23, size = 2951, normalized size = 21.08

method	result	size
default	\(\text {Expression too large to display}\)	\(2951\)
risch	\(\text {Expression too large to display}\)	\(8433\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1)),x,method=_RETURNVERBOSE)

[Out]

-3/2*b*e*x/c+5/2*b*e*arctanh(c*x)/c^2+1/2*b*d*x/c-1/2*a*e*x^2+1/2*a*d*x^2-1/2*b*arctanh(c*x)*x^2*e+1/2*a*e*x^2
*ln(-c^2*x^2+1)+1/2*a*e/c^2+1/2*I*b*Pi*arctanh(c*x)*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1
)^(1/2))*x^2*e-1/4*I*b*Pi*arctanh(c*x)*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2
/(-c^2*x^2+1))^2)^2*x^2*e+1/4*I*b*Pi*arctanh(c*x)*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1
/2))^2*x^2*e+1/4*I*b*Pi*arctanh(c*x)*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x
+1)^2/(-c^2*x^2+1))^2)^2*x^2*e-1/2*I*b*Pi*arctanh(c*x)*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*csgn(I*(1+(c*x+1
)^2/(-c^2*x^2+1)))*x^2*e+1/4*I*b*Pi*arctanh(c*x)*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(1+(c*x+1)^2/(-c^
2*x^2+1)))^2*x^2*e+1/2*I*b/c*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*x*e-1/4*I*b
/c*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*x*e+1/4*I*b/c
*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*x*e+1/4*I*b/c*Pi*csgn(I/(1+(c*x+1)^2/(-
c^2*x^2+1))^2)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*x*e-1/2*I*b/c*Pi*csgn(I*(1+(c*x+1)
^2/(-c^2*x^2+1))^2)^2*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))*x*e+1/4*I*b/c*Pi*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)
*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))^2*x*e-1/2*I*b/c^2*arctanh(c*x)*Pi*e*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csg
n(I*(c*x+1)^2/(c^2*x^2-1))^2+1/4*I*b/c^2*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*csgn(I*(
c*x+1)^2/(c^2*x^2-1))*Pi*e*arctanh(c*x)-1/4*I*b/c^2*arctanh(c*x)*Pi*e*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*csg
n(I*(c*x+1)^2/(c^2*x^2-1))-1/4*I*b/c^2*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2*csgn(I/(1+
(c*x+1)^2/(-c^2*x^2+1))^2)*Pi*e*arctanh(c*x)+1/2*I*b/c^2*arctanh(c*x)*Pi*e*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))*
csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2-1/4*I*b/c^2*arctanh(c*x)*Pi*e*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))^2*csgn
(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)+1/4*I*b/c^2*e*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+
1))^2)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)+1/4*I*b/c*Pi*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1
))^2)^3*x*e-1/4*I*b/c^2*arctanh(c*x)*Pi*e*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3-1/4*I*b/c^2*arctanh(c*x)*Pi*e*csgn(I
*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3-1/4*I*b/c^2*arctanh(c*x)*Pi*e*csgn(I*(1+(c*x+1)^2/(-c^2
*x^2+1))^2)^3-1/2*I*b/c^2*e*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*csgn(I*(c*x+1)^2/(c^2*x^2-1))^2+1/4*I*b/c^2*
e*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2-1/4*I*b/c^2*cs
gn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))^2*Pi*e-1/4*I*b/c^2*e*Pi*csgn(I/(1+(c*x+1)^2/(-c
^2*x^2+1))^2)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2+1/2*I*b/c^2*Pi*e*csgn(I*(1+(c*x+1)^
2/(-c^2*x^2+1)))*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)^2-1/4*I*b/c^2*Pi*e*csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1)))^2*
csgn(I*(1+(c*x+1)^2/(-c^2*x^2+1))^2)+1/4*I*b*Pi*arctanh(c*x)*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*x^2*e+1/4*I*b*Pi*
arctanh(c*x)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3*x^2*e+1/4*I*b*Pi*arctanh(c*x)*csgn(I
*(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3*x^2*e+1/4*I*b/c*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*x*e+1/4*I*b/c*Pi*csgn(I*(c
*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3*x*e-1/4*I*b/c^2*Pi*e*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3-1/4*I
*b/c^2*e*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)^3-1/4*I*b/c^2*e*Pi*csgn(I*(1+(c*x+1)^2/
(-c^2*x^2+1))^2)^3+b/c^2*e*(arctanh(c*x)*x*c+arctanh(c*x)+1)*(c*x-1)*ln((c*x+1)/(-c^2*x^2+1)^(1/2))+b*ln(2)*ar
ctanh(c*x)*x^2*e-b*arctanh(c*x)*ln(1+(c*x+1)^2/(-c^2*x^2+1))*x^2*e+b/c*ln(2)*x*e-b/c*ln(1+(c*x+1)^2/(-c^2*x^2+
1))*x*e-b/c^2*arctanh(c*x)*ln(2)*e+b/c^2*arctanh(c*x)*e*ln(1+(c*x+1)^2/(-c^2*x^2+1))-1/2*a*e/c^2*ln(-c^2*x^2+1
)+1/2*b*arctanh(c*x)*x^2*d-1/2*b/c^2*arctanh(c*x)*d-b/c^2*ln(2)*e-1/2*b/c^2*d-1/4*I*b*Pi*arctanh(c*x)*csgn(I*(
c*x+1)^2/(c^2*x^2-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2
+1))^2)*x^2*e-1/4*I*b/c*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(c*x+1)^2
/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1))^2)*x*e+1/4*I*b/c^2*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2
+1))^2)*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1))^2)*csgn(I*(c*x+1)^2/(c^2*x^2-1))*Pi*e*arctanh(c*x)+3/2*e*b/c^2

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Maxima [A]
time = 0.26, size = 174, normalized size = 1.24 \begin {gather*} \frac {1}{2} \, a d x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d - \frac {{\left (c^{2} x^{2} - {\left (c^{2} x^{2} - 1\right )} \log \left (-c^{2} x^{2} + 1\right ) - 1\right )} b \operatorname {artanh}\left (c x\right ) e}{2 \, c^{2}} - \frac {{\left (c^{2} x^{2} - {\left (c^{2} x^{2} - 1\right )} \log \left (-c^{2} x^{2} + 1\right ) - 1\right )} a e}{2 \, c^{2}} - \frac {{\left (3 \, c x - {\left (c x + 1\right )} \log \left (c x + 1\right ) - {\left (c x - 1\right )} \log \left (-c x + 1\right )\right )} b e}{2 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="maxima")

[Out]

1/2*a*d*x^2 + 1/4*(2*x^2*arctanh(c*x) + c*(2*x/c^2 - log(c*x + 1)/c^3 + log(c*x - 1)/c^3))*b*d - 1/2*(c^2*x^2
- (c^2*x^2 - 1)*log(-c^2*x^2 + 1) - 1)*b*arctanh(c*x)*e/c^2 - 1/2*(c^2*x^2 - (c^2*x^2 - 1)*log(-c^2*x^2 + 1) -
1)*a*e/c^2 - 1/2*(3*c*x - (c*x + 1)*log(c*x + 1) - (c*x - 1)*log(-c*x + 1))*b*e/c^2

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Fricas [A]
time = 0.37, size = 215, normalized size = 1.54 \begin {gather*} \frac {2 \, a c^{2} d x^{2} + 2 \, b c d x - 2 \, {\left (a c^{2} x^{2} + 3 \, b c x\right )} \cosh \left (1\right ) + 2 \, {\left ({\left (a c^{2} x^{2} + b c x - a\right )} \cosh \left (1\right ) + {\left (a c^{2} x^{2} + b c x - a\right )} \sinh \left (1\right )\right )} \log \left (-c^{2} x^{2} + 1\right ) + {\left (b c^{2} d x^{2} - b d - {\left (b c^{2} x^{2} - 3 \, b\right )} \cosh \left (1\right ) + {\left ({\left (b c^{2} x^{2} - b\right )} \cosh \left (1\right ) + {\left (b c^{2} x^{2} - b\right )} \sinh \left (1\right )\right )} \log \left (-c^{2} x^{2} + 1\right ) - {\left (b c^{2} x^{2} - 3 \, b\right )} \sinh \left (1\right )\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) - 2 \, {\left (a c^{2} x^{2} + 3 \, b c x\right )} \sinh \left (1\right )}{4 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="fricas")

[Out]

1/4*(2*a*c^2*d*x^2 + 2*b*c*d*x - 2*(a*c^2*x^2 + 3*b*c*x)*cosh(1) + 2*((a*c^2*x^2 + b*c*x - a)*cosh(1) + (a*c^2
*x^2 + b*c*x - a)*sinh(1))*log(-c^2*x^2 + 1) + (b*c^2*d*x^2 - b*d - (b*c^2*x^2 - 3*b)*cosh(1) + ((b*c^2*x^2 -
b)*cosh(1) + (b*c^2*x^2 - b)*sinh(1))*log(-c^2*x^2 + 1) - (b*c^2*x^2 - 3*b)*sinh(1))*log(-(c*x + 1)/(c*x - 1))
- 2*(a*c^2*x^2 + 3*b*c*x)*sinh(1))/c^2

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Sympy [A]
time = 1.58, size = 202, normalized size = 1.44 \begin {gather*} \begin {cases} \frac {a d x^{2}}{2} + \frac {a e x^{2} \log {\left (- c^{2} x^{2} + 1 \right )}}{2} - \frac {a e x^{2}}{2} - \frac {a e \log {\left (- c^{2} x^{2} + 1 \right )}}{2 c^{2}} + \frac {b d x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b e x^{2} \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )}}{2} - \frac {b e x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b d x}{2 c} + \frac {b e x \log {\left (- c^{2} x^{2} + 1 \right )}}{2 c} - \frac {3 b e x}{2 c} - \frac {b d \operatorname {atanh}{\left (c x \right )}}{2 c^{2}} - \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )}}{2 c^{2}} + \frac {3 b e \operatorname {atanh}{\left (c x \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\\frac {a d x^{2}}{2} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*atanh(c*x))*(d+e*ln(-c**2*x**2+1)),x)

[Out]

Piecewise((a*d*x**2/2 + a*e*x**2*log(-c**2*x**2 + 1)/2 - a*e*x**2/2 - a*e*log(-c**2*x**2 + 1)/(2*c**2) + b*d*x
**2*atanh(c*x)/2 + b*e*x**2*log(-c**2*x**2 + 1)*atanh(c*x)/2 - b*e*x**2*atanh(c*x)/2 + b*d*x/(2*c) + b*e*x*log
(-c**2*x**2 + 1)/(2*c) - 3*b*e*x/(2*c) - b*d*atanh(c*x)/(2*c**2) - b*e*log(-c**2*x**2 + 1)*atanh(c*x)/(2*c**2)
+ 3*b*e*atanh(c*x)/(2*c**2), Ne(c, 0)), (a*d*x**2/2, True))

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Giac [A]
time = 0.63, size = 209, normalized size = 1.49 \begin {gather*} -\frac {1}{4} \, b e x^{2} \log \left (-c x + 1\right )^{2} + \frac {1}{2} \, {\left (a d - a e\right )} x^{2} + \frac {1}{4} \, {\left (b e x^{2} - \frac {b e}{c^{2}}\right )} \log \left (c x + 1\right )^{2} + \frac {1}{4} \, {\left ({\left (b d + 2 \, a e - b e\right )} x^{2} + \frac {2 \, b e x}{c}\right )} \log \left (c x + 1\right ) - \frac {b e \log \left (c x - 1\right )^{2}}{4 \, c^{2}} - \frac {1}{4} \, {\left ({\left (b d - 2 \, a e - b e\right )} x^{2} - \frac {2 \, b e x}{c} - \frac {2 \, b e \log \left (c x - 1\right )}{c^{2}}\right )} \log \left (-c x + 1\right ) + \frac {{\left (b d - 3 \, b e\right )} x}{2 \, c} - \frac {{\left (b d + 2 \, a e - 3 \, b e\right )} \log \left (c x + 1\right )}{4 \, c^{2}} + \frac {{\left (b d - 2 \, a e - 3 \, b e\right )} \log \left (c x - 1\right )}{4 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1)),x, algorithm="giac")

[Out]

-1/4*b*e*x^2*log(-c*x + 1)^2 + 1/2*(a*d - a*e)*x^2 + 1/4*(b*e*x^2 - b*e/c^2)*log(c*x + 1)^2 + 1/4*((b*d + 2*a*
e - b*e)*x^2 + 2*b*e*x/c)*log(c*x + 1) - 1/4*b*e*log(c*x - 1)^2/c^2 - 1/4*((b*d - 2*a*e - b*e)*x^2 - 2*b*e*x/c
- 2*b*e*log(c*x - 1)/c^2)*log(-c*x + 1) + 1/2*(b*d - 3*b*e)*x/c - 1/4*(b*d + 2*a*e - 3*b*e)*log(c*x + 1)/c^2
+ 1/4*(b*d - 2*a*e - 3*b*e)*log(c*x - 1)/c^2

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Mupad [B]
time = 1.42, size = 557, normalized size = 3.98 \begin {gather*} {\ln \left (1-c\,x\right )}^2\,\left (\frac {b\,e}{4\,c^2}-\frac {b\,e\,x^2}{4}\right )-{\ln \left (c\,x+1\right )}^2\,\left (\frac {b\,e}{4\,c^2}-\frac {b\,e\,x^2}{4}\right )+\ln \left (1-c\,x\right )\,\left (\frac {x^2\,\left (a\,e-\frac {b\,d}{2}+\frac {b\,e}{2}+\frac {b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )}{2}\right )}{2}+\frac {b\,e\,x}{2\,c}\right )+c\,\ln \left (c\,x+1\right )\,\left (\frac {x^2\,\left (2\,a\,e+b\,d-b\,e-b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{4\,c}+\frac {b\,e\,x}{2\,c^2}\right )-\frac {a\,x^2\,\left (e-d+e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{2}-\frac {\ln \left (\frac {x\,\left (2\,a\,e+b\,d-3\,b\,e-b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{2}-\frac {3\,b\,e-b\,d+b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )}{2\,c}-a\,e\,x\right )\,\left (2\,a\,e+b\,d-3\,b\,e-b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{4\,c^2}-\frac {\ln \left (\frac {x\,\left (2\,a\,e-b\,d+3\,b\,e+b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{2}-\frac {3\,b\,e-b\,d+b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )}{2\,c}-a\,e\,x\right )\,\left (2\,a\,e-b\,d+3\,b\,e+b\,e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{4\,c^2}-\frac {b\,x\,\left (3\,e-d+e\,\left (\ln \left (c\,x+1\right )+\ln \left (1-c\,x\right )-\ln \left (1-c^2\,x^2\right )\right )\right )}{2\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)),x)

[Out]

log(1 - c*x)^2*((b*e)/(4*c^2) - (b*e*x^2)/4) - log(c*x + 1)^2*((b*e)/(4*c^2) - (b*e*x^2)/4) + log(1 - c*x)*((x
^2*(a*e - (b*d)/2 + (b*e)/2 + (b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2)))/2))/2 + (b*e*x)/(2*c)) +
c*log(c*x + 1)*((x^2*(2*a*e + b*d - b*e - b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/(4*c) + (b*e*
x)/(2*c^2)) - (a*x^2*(e - d + e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/2 - (log((x*(2*a*e + b*d -
3*b*e - b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/2 - (3*b*e - b*d + b*e*(log(c*x + 1) + log(1 -
c*x) - log(1 - c^2*x^2)))/(2*c) - a*e*x)*(2*a*e + b*d - 3*b*e - b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2
*x^2))))/(4*c^2) - (log((x*(2*a*e - b*d + 3*b*e + b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/2 - (
3*b*e - b*d + b*e*(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2)))/(2*c) - a*e*x)*(2*a*e - b*d + 3*b*e + b*e*
(log(c*x + 1) + log(1 - c*x) - log(1 - c^2*x^2))))/(4*c^2) - (b*x*(3*e - d + e*(log(c*x + 1) + log(1 - c*x) -
log(1 - c^2*x^2))))/(2*c)

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