3.2.6 \(\int \frac {x (a+b \tanh ^{-1}(c x))^2}{(d+c d x)^2} \, dx\) [106]

Optimal. Leaf size=188 \[ \frac {b^2}{2 c^2 d^2 (1+c x)}-\frac {b^2 \tanh ^{-1}(c x)}{2 c^2 d^2}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{c^2 d^2 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d^2}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2 (1+c x)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1+c x}\right )}{c^2 d^2}+\frac {b \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c^2 d^2}+\frac {b^2 \text {PolyLog}\left (3,1-\frac {2}{1+c x}\right )}{2 c^2 d^2} \]

[Out]

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

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Rubi [A]
time = 0.25, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6087, 6065, 6063, 641, 46, 213, 6095, 6055, 6203, 6745} \begin {gather*} \frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2 d^2}+\frac {b \left (a+b \tanh ^{-1}(c x)\right )}{c^2 d^2 (c x+1)}+\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2 (c x+1)}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{2 c^2 d^2}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{c^2 d^2}+\frac {b^2 \text {Li}_3\left (1-\frac {2}{c x+1}\right )}{2 c^2 d^2}+\frac {b^2}{2 c^2 d^2 (c x+1)}-\frac {b^2 \tanh ^{-1}(c x)}{2 c^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 46

Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x
)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 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 641

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c/e)*x)^
p, x] /; FreeQ[{a, c, d, e, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && I
ntegerQ[m + p]))

Rule 6055

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 6063

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

Rule 6065

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

Rule 6087

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 6095

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 6203

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

Rule 6745

Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /
;  !FalseQ[w]] /; FreeQ[n, x]

Rubi steps

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

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Mathematica [A]
time = 0.33, size = 233, normalized size = 1.24 \begin {gather*} \frac {\frac {4 a^2}{1+c x}+4 a^2 \log (1+c x)+2 a b \left (\cosh \left (2 \tanh ^{-1}(c x)\right )+2 \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+2 \tanh ^{-1}(c x) \left (\cosh \left (2 \tanh ^{-1}(c x)\right )-2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )\right )+b^2 \left (\cosh \left (2 \tanh ^{-1}(c x)\right )+2 \tanh ^{-1}(c x) \cosh \left (2 \tanh ^{-1}(c x)\right )+2 \tanh ^{-1}(c x)^2 \cosh \left (2 \tanh ^{-1}(c x)\right )-4 \tanh ^{-1}(c x)^2 \log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )+4 \tanh ^{-1}(c x) \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+2 \text {PolyLog}\left (3,-e^{-2 \tanh ^{-1}(c x)}\right )-\sinh \left (2 \tanh ^{-1}(c x)\right )-2 \tanh ^{-1}(c x) \sinh \left (2 \tanh ^{-1}(c x)\right )-2 \tanh ^{-1}(c x)^2 \sinh \left (2 \tanh ^{-1}(c x)\right )\right )}{4 c^2 d^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^2)/(1 + c*x) + 4*a^2*Log[1 + c*x] + 2*a*b*(Cosh[2*ArcTanh[c*x]] + 2*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 2
*ArcTanh[c*x]*(Cosh[2*ArcTanh[c*x]] - 2*Log[1 + E^(-2*ArcTanh[c*x])] - Sinh[2*ArcTanh[c*x]]) - Sinh[2*ArcTanh[
c*x]]) + b^2*(Cosh[2*ArcTanh[c*x]] + 2*ArcTanh[c*x]*Cosh[2*ArcTanh[c*x]] + 2*ArcTanh[c*x]^2*Cosh[2*ArcTanh[c*x
]] - 4*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] + 4*ArcTanh[c*x]*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 2*PolyL
og[3, -E^(-2*ArcTanh[c*x])] - Sinh[2*ArcTanh[c*x]] - 2*ArcTanh[c*x]*Sinh[2*ArcTanh[c*x]] - 2*ArcTanh[c*x]^2*Si
nh[2*ArcTanh[c*x]]))/(4*c^2*d^2)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 6.67, size = 946, normalized size = 5.03

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(946\)
default	\(\text {Expression too large to display}\)	\(946\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctanh(c*x))^2/(c*d*x+d)^2,x,method=_RETURNVERBOSE)

[Out]

1/c^2*(-1/2*I*b^2/d^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3*arctanh(c*x)^2+a^2/d^2/(c*x+1)+a^2/d^2*ln(c*x+1)-1/2*
b^2/d^2*arctanh(c*x)^2+2/3*b^2/d^2*arctanh(c*x)^3+1/2*b^2/d^2*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))-1/2*I*b^2/d^2
*Pi*csgn(I/(1+(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*arctanh(c*x)
^2-I*b^2/d^2*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))*arctanh(c*x)^2+1/2*I*b^2/d^
2*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*arctanh(c*x)^2-1
/2*I*b^2/d^2*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*arctanh(c*x)^2+1/2*I*b^2/d^
2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*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)))*arctanh(c*x)^2+1/4*b^2/d^2/(c*x+1)-1/2*I*b^2/d^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2
/(-c^2*x^2+1)))^3*arctanh(c*x)^2-1/2*b^2/d^2*arctanh(c*x)/(c*x+1)*c*x+b^2/d^2*arctanh(c*x)^2/(c*x+1)+b^2/d^2*a
rctanh(c*x)^2*ln(c*x+1)+1/2*b^2/d^2*arctanh(c*x)/(c*x+1)-b^2/d^2*arctanh(c*x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1
))+a*b/d^2/(c*x+1)-a*b/d^2*dilog(1/2*c*x+1/2)-1/2*a*b/d^2*ln(c*x+1)^2+1/2*a*b/d^2*ln(c*x-1)-1/2*a*b/d^2*ln(c*x
+1)-b^2/d^2*arctanh(c*x)^2*ln(2)-2*b^2/d^2*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-1/4*b^2/d^2/(c*x+1)*c
*x+2*a*b/d^2*arctanh(c*x)/(c*x+1)+2*a*b/d^2*arctanh(c*x)*ln(c*x+1)-a*b/d^2*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)+a*
b/d^2*ln(-1/2*c*x+1/2)*ln(c*x+1))

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

[Out]

a^2*(1/(c^3*d^2*x + c^2*d^2) + log(c*x + 1)/(c^2*d^2)) + 1/4*(b^2 + (b^2*c*x + b^2)*log(c*x + 1))*log(-c*x + 1
)^2/(c^3*d^2*x + c^2*d^2) - integrate(-1/4*((b^2*c^2*x^2 - b^2*c*x)*log(c*x + 1)^2 + 4*(a*b*c^2*x^2 - a*b*c*x)
*log(c*x + 1) - 2*(2*a*b*c^2*x^2 + b^2 - (2*a*b*c - b^2*c)*x + (2*b^2*c^2*x^2 + b^2*c*x + b^2)*log(c*x + 1))*l
og(-c*x + 1))/(c^4*d^2*x^3 + c^3*d^2*x^2 - c^2*d^2*x - c*d^2), x)

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2} x}{c^{2} x^{2} + 2 c x + 1}\, dx + \int \frac {b^{2} x \operatorname {atanh}^{2}{\left (c x \right )}}{c^{2} x^{2} + 2 c x + 1}\, dx + \int \frac {2 a b x \operatorname {atanh}{\left (c x \right )}}{c^{2} x^{2} + 2 c x + 1}\, dx}{d^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2*x/(c**2*x**2 + 2*c*x + 1), x) + Integral(b**2*x*atanh(c*x)**2/(c**2*x**2 + 2*c*x + 1), x) + Int
egral(2*a*b*x*atanh(c*x)/(c**2*x**2 + 2*c*x + 1), x))/d**2

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x\,{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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