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

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

[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 = 0.12, 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 = {2454, 2389, 2295, 6083, 321, 207, 2448, 206} \[ -\frac {e \left (1-c^2 x^2\right ) \log \left (1-c^2 x^2\right ) \left (a+b \tanh ^{-1}(c x)\right )}{2 c^2}+\frac {1}{2} d x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{2} e x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {b (d-e) \tanh ^{-1}(c x)}{2 c^2}+\frac {b e x \log \left (1-c^2 x^2\right )}{2 c}+\frac {b e \tanh ^{-1}(c x)}{c^2}+\frac {b x (d-e)}{2 c}-\frac {b e x}{c} \]

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 206

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

Rule 207

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

Rule 321

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 2295

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

Rule 2389

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 2448

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 2454

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 6083

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 = 0.11, size = 129, normalized size = 0.92 \[ \frac {2 e \log \left (1-c^2 x^2\right ) \left (c x (a c x+b)+b \left (c^2 x^2-1\right ) \tanh ^{-1}(c x)\right )+\log (1-c x) (b (d-3 e)-2 a e)-\log (c x+1) (2 a e+b (d-3 e))+2 a c^2 x^2 (d-e)+2 b c^2 x^2 (d-e) \tanh ^{-1}(c x)+2 b c x (d-3 e)}{4 c^2} \]

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)

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fricas [A]  time = 0.63, size = 139, normalized size = 0.99 \[ \frac {2 \, {\left (a c^{2} d - a c^{2} e\right )} x^{2} + 2 \, {\left (b c d - 3 \, b c e\right )} x + 2 \, {\left (a c^{2} e x^{2} + b c e x - a e\right )} \log \left (-c^{2} x^{2} + 1\right ) + {\left ({\left (b c^{2} d - b c^{2} e\right )} x^{2} - b d + 3 \, b e + {\left (b c^{2} e x^{2} - b e\right )} \log \left (-c^{2} x^{2} + 1\right )\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{4 \, c^{2}} \]

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

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giac [B]  time = 0.35, size = 305, normalized size = 2.18 \[ \frac {b c^{2} x^{2} e \log \left (c x + 1\right )^{2} - b c^{2} x^{2} e \log \left (-c x + 1\right )^{2} + 2 \, a c^{2} x^{2} e \log \left (c x + 1\right ) - b c^{2} x^{2} e \log \left (c x + 1\right ) + 2 \, a c^{2} x^{2} e \log \left (-c x + 1\right ) + b c^{2} x^{2} e \log \left (-c x + 1\right ) + b c^{2} d x^{2} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a c^{2} d x^{2} - 2 \, a c^{2} x^{2} e + 2 \, b c x e \log \left (c x + 1\right ) + 2 \, b c x e \log \left (-c x + 1\right ) + 2 \, b c d x - 6 \, b c x e - b e \log \left (c x + 1\right )^{2} - b e \log \left (c x - 1\right )^{2} + 2 \, b e \log \left (c x - 1\right ) \log \left (-c x + 1\right ) - b d \log \left (c x + 1\right ) - 2 \, a e \log \left (c x + 1\right ) + 3 \, b e \log \left (c x + 1\right ) + b d \log \left (c x - 1\right ) - 2 \, a e \log \left (c x - 1\right ) - 3 \, b e \log \left (c x - 1\right )}{4 \, c^{2}} \]

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*c^2*x^2*e*log(c*x + 1)^2 - b*c^2*x^2*e*log(-c*x + 1)^2 + 2*a*c^2*x^2*e*log(c*x + 1) - b*c^2*x^2*e*log(c
*x + 1) + 2*a*c^2*x^2*e*log(-c*x + 1) + b*c^2*x^2*e*log(-c*x + 1) + b*c^2*d*x^2*log(-(c*x + 1)/(c*x - 1)) + 2*
a*c^2*d*x^2 - 2*a*c^2*x^2*e + 2*b*c*x*e*log(c*x + 1) + 2*b*c*x*e*log(-c*x + 1) + 2*b*c*d*x - 6*b*c*x*e - b*e*l
og(c*x + 1)^2 - b*e*log(c*x - 1)^2 + 2*b*e*log(c*x - 1)*log(-c*x + 1) - b*d*log(c*x + 1) - 2*a*e*log(c*x + 1)
+ 3*b*e*log(c*x + 1) + b*d*log(c*x - 1) - 2*a*e*log(c*x - 1) - 3*b*e*log(c*x - 1))/c^2

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maple [C]  time = 3.85, size = 2951, normalized size = 21.08 \[ \text {output too large to display} \]

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)

[Out]

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

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maxima [A]  time = 0.32, size = 171, normalized size = 1.22 \[ \frac {1}{2} \, a d x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b d - \frac {{\left (c^{2} x^{2} - {\left (c^{2} x^{2} - 1\right )} \log \left (-c^{2} x^{2} + 1\right ) - 1\right )} b e \operatorname {artanh}\left (c x\right )}{2 \, c^{2}} - \frac {{\left (c^{2} x^{2} - {\left (c^{2} x^{2} - 1\right )} \log \left (-c^{2} x^{2} + 1\right ) - 1\right )} a e}{2 \, c^{2}} - \frac {{\left (3 \, c x - {\left (c x + 1\right )} \log \left (c x + 1\right ) - {\left (c x - 1\right )} \log \left (-c x + 1\right )\right )} b e}{2 \, c^{2}} \]

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

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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sympy [A]  time = 3.78, size = 202, normalized size = 1.44 \[ \begin {cases} \frac {a d x^{2}}{2} + \frac {a e x^{2} \log {\left (- c^{2} x^{2} + 1 \right )}}{2} - \frac {a e x^{2}}{2} - \frac {a e \log {\left (- c^{2} x^{2} + 1 \right )}}{2 c^{2}} + \frac {b d x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b e x^{2} \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )}}{2} - \frac {b e x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b d x}{2 c} + \frac {b e x \log {\left (- c^{2} x^{2} + 1 \right )}}{2 c} - \frac {3 b e x}{2 c} - \frac {b d \operatorname {atanh}{\left (c x \right )}}{2 c^{2}} - \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )}}{2 c^{2}} + \frac {3 b e \operatorname {atanh}{\left (c x \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\\frac {a d x^{2}}{2} & \text {otherwise} \end {cases} \]

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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