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

Optimal. Leaf size=104 \[ x \left (a+b \tanh ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )+\frac {e \left (a+b \tanh ^{-1}(c x)\right )^2}{b c}-2 a e x+\frac {b \left (e \log \left (1-c^2 x^2\right )+d\right )^2}{4 c e}-\frac {b e \log \left (1-c^2 x^2\right )}{c}-2 b e x \tanh ^{-1}(c x) \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6073, 2475, 2390, 2301, 5980, 5910, 260, 5948} \[ x \left (a+b \tanh ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )+\frac {e \left (a+b \tanh ^{-1}(c x)\right )^2}{b c}-2 a e x+\frac {b \left (e \log \left (1-c^2 x^2\right )+d\right )^2}{4 c e}-\frac {b e \log \left (1-c^2 x^2\right )}{c}-2 b e x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

Rule 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2475

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.)*((f_) + (g_.)*(x_)^(s_))^(r_.),
x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q,
x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && IntegerQ[r] && IntegerQ[s/n] && Intege
rQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0])

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 5980

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTanh[c*x])
^p)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 6073

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.)), x_Symbol] :> Simp[x*(d + e*
Log[f + g*x^2])*(a + b*ArcTanh[c*x]), x] + (-Dist[b*c, Int[(x*(d + e*Log[f + g*x^2]))/(1 - c^2*x^2), x], x] -
Dist[2*e*g, Int[(x^2*(a + b*ArcTanh[c*x]))/(f + g*x^2), x], x]) /; FreeQ[{a, b, c, d, e, f, g}, x]

Rubi steps

\begin {align*} \int \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right ) \, dx &=x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )-(b c) \int \frac {x \left (d+e \log \left (1-c^2 x^2\right )\right )}{1-c^2 x^2} \, dx+\left (2 c^2 e\right ) \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {1}{2} (b c) \operatorname {Subst}\left (\int \frac {d+e \log \left (1-c^2 x\right )}{1-c^2 x} \, dx,x,x^2\right )-(2 e) \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx+(2 e) \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx\\ &=-2 a e x+\frac {e \left (a+b \tanh ^{-1}(c x)\right )^2}{b c}+x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \operatorname {Subst}\left (\int \frac {d+e \log (x)}{x} \, dx,x,1-c^2 x^2\right )}{2 c}-(2 b e) \int \tanh ^{-1}(c x) \, dx\\ &=-2 a e x-2 b e x \tanh ^{-1}(c x)+\frac {e \left (a+b \tanh ^{-1}(c x)\right )^2}{b c}+x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 c e}+(2 b c e) \int \frac {x}{1-c^2 x^2} \, dx\\ &=-2 a e x-2 b e x \tanh ^{-1}(c x)+\frac {e \left (a+b \tanh ^{-1}(c x)\right )^2}{b c}-\frac {b e \log \left (1-c^2 x^2\right )}{c}+x \left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )+\frac {b \left (d+e \log \left (1-c^2 x^2\right )\right )^2}{4 c e}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 144, normalized size = 1.38 \[ a e x \log \left (1-c^2 x^2\right )+\frac {2 a e \tanh ^{-1}(c x)}{c}+a d x-2 a e x+\frac {b d \log \left (1-c^2 x^2\right )}{2 c}+\frac {b e \log ^2\left (1-c^2 x^2\right )}{4 c}-\frac {b e \log \left (1-c^2 x^2\right )}{c}+b e x \log \left (1-c^2 x^2\right ) \tanh ^{-1}(c x)+b d x \tanh ^{-1}(c x)+\frac {b e \tanh ^{-1}(c x)^2}{c}-2 b e x \tanh ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.53, size = 132, normalized size = 1.27 \[ \frac {b e \log \left (-c^{2} x^{2} + 1\right )^{2} + b e \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} + 4 \, {\left (a c d - 2 \, a c e\right )} x + 2 \, {\left (2 \, a c e x + b d - 2 \, b e\right )} \log \left (-c^{2} x^{2} + 1\right ) + 2 \, {\left (b c e x \log \left (-c^{2} x^{2} + 1\right ) + 2 \, a e + {\left (b c d - 2 \, b c e\right )} x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{4 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.18, size = 223, normalized size = 2.14 \[ \frac {b c x e \log \left (c x + 1\right )^{2} - b c x e \log \left (-c x + 1\right )^{2} + 2 \, a c x e \log \left (c x + 1\right ) - 2 \, b c x e \log \left (c x + 1\right ) + 2 \, a c x e \log \left (-c x + 1\right ) + 2 \, b c x e \log \left (-c x + 1\right ) + b c d x \log \left (-\frac {c x + 1}{c x - 1}\right ) + 2 \, a c d x - 4 \, a c x e + b e \log \left (c x + 1\right )^{2} - b e \log \left (c x - 1\right )^{2} + 2 \, b e \log \left (c x - 1\right ) \log \left (-c x + 1\right ) + b d \log \left (c^{2} x^{2} - 1\right ) + 2 \, a e \log \left (c x + 1\right ) - 2 \, b e \log \left (c x + 1\right ) - 2 \, a e \log \left (c x - 1\right ) - 2 \, b e \log \left (c x - 1\right )}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.48, size = 2529, normalized size = 24.32 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [C]  time = 0.32, size = 178, normalized size = 1.71 \[ -{\left (c^{2} {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )} - x \log \left (-c^{2} x^{2} + 1\right )\right )} b e \operatorname {artanh}\left (c x\right ) - {\left (c^{2} {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )} - x \log \left (-c^{2} x^{2} + 1\right )\right )} a e + a d x + \frac {{\left (2 \, c x \operatorname {artanh}\left (c x\right ) + \log \left (-c^{2} x^{2} + 1\right )\right )} b d}{2 \, c} + \frac {{\left ({\left (i \, \pi + 2 \, \log \left (c x - 1\right ) - 2\right )} \log \left (c x + 1\right ) + {\left (i \, \pi - 2\right )} \log \left (c x - 1\right )\right )} b e}{2 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.79, size = 385, normalized size = 3.70 \[ a\,d\,x-2\,a\,e\,x+\frac {b\,e\,{\ln \left (c\,x+1\right )}^2}{2\,c}+\frac {b\,e\,{\ln \left (1-c\,x\right )}^2}{2\,c}+a\,e\,x\,\ln \left (1-c^2\,x^2\right )+\frac {b\,d\,x\,\ln \left (c\,x+1\right )}{2}-\frac {b\,d\,x\,\ln \left (1-c\,x\right )}{2}-b\,e\,x\,\ln \left (c\,x+1\right )+b\,e\,x\,\ln \left (1-c\,x\right )-\frac {a\,e\,\ln \left (c\,x-1\right )}{c}+\frac {a\,e\,\ln \left (c\,x+1\right )}{c}+\frac {b\,d\,\ln \left (c\,x-1\right )}{2\,c}+\frac {b\,d\,\ln \left (c\,x+1\right )}{2\,c}-\frac {b\,e\,\ln \left (c\,x-1\right )}{c}-\frac {b\,e\,\ln \left (c\,x+1\right )}{c}+\frac {b\,e\,x\,\ln \left (c\,x+1\right )\,\ln \left (1-c^2\,x^2\right )}{2}-\frac {b\,e\,x\,\ln \left (1-c\,x\right )\,\ln \left (1-c^2\,x^2\right )}{2}+\frac {b\,e\,\ln \left (1-c^2\,x^2\right )\,\ln \left (-2\,a\,e-2\,a\,c\,e\,x\right )}{2\,c}+\frac {b\,e\,\ln \left (1-c^2\,x^2\right )\,\ln \left (2\,a\,c\,e\,x-2\,a\,e\right )}{2\,c}-\frac {b\,e\,\ln \left (c\,x+1\right )\,\ln \left (-2\,a\,e-2\,a\,c\,e\,x\right )}{2\,c}-\frac {b\,e\,\ln \left (c\,x+1\right )\,\ln \left (2\,a\,c\,e\,x-2\,a\,e\right )}{2\,c}-\frac {b\,e\,\ln \left (1-c\,x\right )\,\ln \left (-2\,a\,e-2\,a\,c\,e\,x\right )}{2\,c}-\frac {b\,e\,\ln \left (1-c\,x\right )\,\ln \left (2\,a\,c\,e\,x-2\,a\,e\right )}{2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.26, size = 148, normalized size = 1.42 \[ \begin {cases} a d x + a e x \log {\left (- c^{2} x^{2} + 1 \right )} - 2 a e x + \frac {2 a e \operatorname {atanh}{\left (c x \right )}}{c} + b d x \operatorname {atanh}{\left (c x \right )} + b e x \log {\left (- c^{2} x^{2} + 1 \right )} \operatorname {atanh}{\left (c x \right )} - 2 b e x \operatorname {atanh}{\left (c x \right )} + \frac {b d \log {\left (- c^{2} x^{2} + 1 \right )}}{2 c} + \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}^{2}}{4 c} - \frac {b e \log {\left (- c^{2} x^{2} + 1 \right )}}{c} + \frac {b e \operatorname {atanh}^{2}{\left (c x \right )}}{c} & \text {for}\: c \neq 0 \\a d x & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*d*x + a*e*x*log(-c**2*x**2 + 1) - 2*a*e*x + 2*a*e*atanh(c*x)/c + b*d*x*atanh(c*x) + b*e*x*log(-c*
*2*x**2 + 1)*atanh(c*x) - 2*b*e*x*atanh(c*x) + b*d*log(-c**2*x**2 + 1)/(2*c) + b*e*log(-c**2*x**2 + 1)**2/(4*c
) - b*e*log(-c**2*x**2 + 1)/c + b*e*atanh(c*x)**2/c, Ne(c, 0)), (a*d*x, True))

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