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

Optimal. Leaf size=279 \[ -\frac {b d \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^2}+\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{e^2}+\frac {d \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c e}-\frac {b^2 d \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 e^2}+\frac {b^2 d \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 e^2}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c e} \]

[Out]

(a+b*arctanh(c*x))^2/c/e+x*(a+b*arctanh(c*x))^2/e-2*b*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/c/e+d*(a+b*arctanh(c*x
))^2*ln(2/(c*x+1))/e^2-d*(a+b*arctanh(c*x))^2*ln(2*c*(e*x+d)/(c*d+e)/(c*x+1))/e^2-b^2*polylog(2,1-2/(-c*x+1))/
c/e-b*d*(a+b*arctanh(c*x))*polylog(2,1-2/(c*x+1))/e^2+b*d*(a+b*arctanh(c*x))*polylog(2,1-2*c*(e*x+d)/(c*d+e)/(
c*x+1))/e^2-1/2*b^2*d*polylog(3,1-2/(c*x+1))/e^2+1/2*b^2*d*polylog(3,1-2*c*(e*x+d)/(c*d+e)/(c*x+1))/e^2

________________________________________________________________________________________

Rubi [A]  time = 0.26, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {5940, 5910, 5984, 5918, 2402, 2315, 5922} \[ -\frac {b d \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^2}+\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}-\frac {b^2 d \text {PolyLog}\left (3,1-\frac {2}{c x+1}\right )}{2 e^2}+\frac {b^2 d \text {PolyLog}\left (3,1-\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e^2}-\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{c e}+\frac {d \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}-\frac {2 b \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 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 5918

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

Rule 5922

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^2*Log[
2/(1 + c*x)])/e, x] + (Simp[((a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(
b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/e, x] - Simp[(b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - (2*c*(
d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*e), x] - Simp[(b^2*PolyLog
[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2,
0]

Rule 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 5984

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

Rubi steps

\begin {align*} \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{d+e x} \, dx &=\int \left (\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{e}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2}{e (d+e x)}\right ) \, dx\\ &=\frac {\int \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx}{e}-\frac {d \int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{d+e x} \, dx}{e}\\ &=\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{e^2}+\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b^2 d \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 e^2}+\frac {b^2 d \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}-\frac {(2 b c) \int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{e}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{e^2}+\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b^2 d \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 e^2}+\frac {b^2 d \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}-\frac {(2 b) \int \frac {a+b \tanh ^{-1}(c x)}{1-c x} \, dx}{e}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c e}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{e^2}+\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b^2 d \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 e^2}+\frac {b^2 d \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}+\frac {\left (2 b^2\right ) \int \frac {\log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx}{e}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c e}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{e^2}+\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b^2 d \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 e^2}+\frac {b^2 d \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-c x}\right )}{c e}\\ &=\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c e}+\frac {x \left (a+b \tanh ^{-1}(c x)\right )^2}{e}-\frac {2 b \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{c e}+\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{e^2}-\frac {d \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{c e}-\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{e^2}+\frac {b d \left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac {b^2 d \text {Li}_3\left (1-\frac {2}{1+c x}\right )}{2 e^2}+\frac {b^2 d \text {Li}_3\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 13.55, size = 861, normalized size = 3.09 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*a^2*e*x - 3*a^2*d*Log[d + e*x] + (3*a*b*((-I)*c*d*Pi*ArcTanh[c*x] + 2*c*e*x*ArcTanh[c*x] - 2*c*d*ArcTanh[(c
*d)/e]*ArcTanh[c*x] + c*d*ArcTanh[c*x]^2 - e*ArcTanh[c*x]^2 + (Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^2)/E^Arc
Tanh[(c*d)/e] + 2*c*d*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] + I*c*d*Pi*Log[1 + E^(2*ArcTanh[c*x])] - 2*c*d
*ArcTanh[(c*d)/e]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - 2*c*d*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTan
h[(c*d)/e] + ArcTanh[c*x]))] + e*Log[1 - c^2*x^2] + (I/2)*c*d*Pi*Log[1 - c^2*x^2] + 2*c*d*ArcTanh[(c*d)/e]*Log
[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] - c*d*PolyLog[2, -E^(-2*ArcTanh[c*x])] + c*d*PolyLog[2, E^(-2*(ArcTa
nh[(c*d)/e] + ArcTanh[c*x]))]))/c + (b^2*(-3*e*ArcTanh[c*x]^2 + 3*c*e*x*ArcTanh[c*x]^2 + 4*c*d*ArcTanh[c*x]^3
- 2*e*ArcTanh[c*x]^3 + (2*Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^3)/E^ArcTanh[(c*d)/e] - 6*e*ArcTanh[c*x]*Log[
1 + E^(-2*ArcTanh[c*x])] + 3*c*d*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + (3*I)*c*d*Pi*ArcTanh[c*x]*Log[(
E^(-ArcTanh[c*x]) + E^ArcTanh[c*x])/2] - 3*c*d*ArcTanh[c*x]^2*Log[1 - E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 3
*c*d*ArcTanh[c*x]^2*Log[1 + E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 6*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[(I/
2)*E^(-ArcTanh[(c*d)/e] - ArcTanh[c*x])*(-1 + E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x])))] - 3*c*d*ArcTanh[c*x]^2
*Log[(e*(-1 + E^(2*ArcTanh[c*x])) + c*d*(1 + E^(2*ArcTanh[c*x])))/(2*E^ArcTanh[c*x])] + 3*c*d*ArcTanh[c*x]^2*L
og[(c*(d + e*x))/Sqrt[1 - c^2*x^2]] + ((3*I)/2)*c*d*Pi*ArcTanh[c*x]*Log[1 - c^2*x^2] + 6*c*d*ArcTanh[(c*d)/e]*
ArcTanh[c*x]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] + 3*(e - c*d*ArcTanh[c*x])*PolyLog[2, -E^(-2*ArcTanh
[c*x])] - 6*c*d*ArcTanh[c*x]*PolyLog[2, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 6*c*d*ArcTanh[c*x]*PolyLog[2,
E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - (3*c*d*PolyLog[3, -E^(-2*ArcTanh[c*x])])/2 + 6*c*d*PolyLog[3, -E^(ArcTa
nh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*PolyLog[3, E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])]))/c)/(3*e^2)

________________________________________________________________________________________

fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{2} x \operatorname {artanh}\left (c x\right )^{2} + 2 \, a b x \operatorname {artanh}\left (c x\right ) + a^{2} x}{e x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*x*arctanh(c*x)^2 + 2*a*b*x*arctanh(c*x) + a^2*x)/(e*x + d), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )}^{2} x}{e x + d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctanh(c*x) + a)^2*x/(e*x + d), x)

________________________________________________________________________________________

maple [C]  time = 1.63, size = 13923, normalized size = 49.90 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctanh(c*x))^2/(e*x+d),x)

[Out]

result too large to display

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {b^{2} x \log \left (-c x + 1\right )^{2}}{4 \, e} + a^{2} {\left (\frac {x}{e} - \frac {d \log \left (e x + d\right )}{e^{2}}\right )} - \int -\frac {{\left (b^{2} c e x^{2} - b^{2} e x\right )} \log \left (c x + 1\right )^{2} + 4 \, {\left (a b c e x^{2} - a b e x\right )} \log \left (c x + 1\right ) - 2 \, {\left ({\left (2 \, a b c e + b^{2} c e\right )} x^{2} + {\left (b^{2} c d - 2 \, a b e\right )} x + {\left (b^{2} c e x^{2} - b^{2} e x\right )} \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{4 \, {\left (c e^{2} x^{2} - d e + {\left (c d e - e^{2}\right )} x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{d+e\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*atanh(c*x))^2)/(d + e*x),x)

[Out]

int((x*(a + b*atanh(c*x))^2)/(d + e*x), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{2}}{d + e x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*atanh(c*x))**2/(e*x+d),x)

[Out]

Integral(x*(a + b*atanh(c*x))**2/(d + e*x), x)

________________________________________________________________________________________