3.59 \(\int \frac{x^3 (a+b \tanh ^{-1}(c x))}{(d+c d x)^3} \, dx\)

Optimal. Leaf size=194 \[ -\frac{3 b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{2 c^4 d^3}-\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (c x+1)}+\frac{a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (c x+1)^2}+\frac{3 \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3}+\frac{a x}{c^3 d^3}+\frac{b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}-\frac{11 b}{8 c^4 d^3 (c x+1)}+\frac{b}{8 c^4 d^3 (c x+1)^2}+\frac{b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac{11 b \tanh ^{-1}(c x)}{8 c^4 d^3} \]

[Out]

(a*x)/(c^3*d^3) + b/(8*c^4*d^3*(1 + c*x)^2) - (11*b)/(8*c^4*d^3*(1 + c*x)) + (11*b*ArcTanh[c*x])/(8*c^4*d^3) +
(b*x*ArcTanh[c*x])/(c^3*d^3) + (a + b*ArcTanh[c*x])/(2*c^4*d^3*(1 + c*x)^2) - (3*(a + b*ArcTanh[c*x]))/(c^4*d
^3*(1 + c*x)) + (3*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/(c^4*d^3) + (b*Log[1 - c^2*x^2])/(2*c^4*d^3) - (3*b*
PolyLog[2, 1 - 2/(1 + c*x)])/(2*c^4*d^3)

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Rubi [A]  time = 0.248, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 18, number of rules used = 10, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5940, 5910, 260, 5926, 627, 44, 207, 5918, 2402, 2315} \[ -\frac{3 b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{2 c^4 d^3}-\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (c x+1)}+\frac{a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (c x+1)^2}+\frac{3 \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3}+\frac{a x}{c^3 d^3}+\frac{b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}-\frac{11 b}{8 c^4 d^3 (c x+1)}+\frac{b}{8 c^4 d^3 (c x+1)^2}+\frac{b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac{11 b \tanh ^{-1}(c x)}{8 c^4 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcTanh[c*x]))/(d + c*d*x)^3,x]

[Out]

(a*x)/(c^3*d^3) + b/(8*c^4*d^3*(1 + c*x)^2) - (11*b)/(8*c^4*d^3*(1 + c*x)) + (11*b*ArcTanh[c*x])/(8*c^4*d^3) +
(b*x*ArcTanh[c*x])/(c^3*d^3) + (a + b*ArcTanh[c*x])/(2*c^4*d^3*(1 + c*x)^2) - (3*(a + b*ArcTanh[c*x]))/(c^4*d
^3*(1 + c*x)) + (3*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/(c^4*d^3) + (b*Log[1 - c^2*x^2])/(2*c^4*d^3) - (3*b*
PolyLog[2, 1 - 2/(1 + c*x)])/(2*c^4*d^3)

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 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 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 5926

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 627

Int[((d_) + (e_.)*(x_))^(m_.)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a/d + (c*x)/e)^
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 44

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] && L
tQ[m + n + 2, 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b \tanh ^{-1}(c x)\right )}{(d+c d x)^3} \, dx &=\int \left (\frac{a+b \tanh ^{-1}(c x)}{c^3 d^3}-\frac{a+b \tanh ^{-1}(c x)}{c^3 d^3 (1+c x)^3}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^3 (1+c x)^2}-\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^3 (1+c x)}\right ) \, dx\\ &=\frac{\int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^3 d^3}-\frac{\int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{c^3 d^3}+\frac{3 \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{c^3 d^3}-\frac{3 \int \frac{a+b \tanh ^{-1}(c x)}{1+c x} \, dx}{c^3 d^3}\\ &=\frac{a x}{c^3 d^3}+\frac{a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{c^4 d^3}-\frac{b \int \frac{1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{2 c^3 d^3}+\frac{b \int \tanh ^{-1}(c x) \, dx}{c^3 d^3}+\frac{(3 b) \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{c^3 d^3}-\frac{(3 b) \int \frac{\log \left (\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^3 d^3}\\ &=\frac{a x}{c^3 d^3}+\frac{b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac{a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{c^4 d^3}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+c x}\right )}{c^4 d^3}-\frac{b \int \frac{1}{(1-c x) (1+c x)^3} \, dx}{2 c^3 d^3}+\frac{(3 b) \int \frac{1}{(1-c x) (1+c x)^2} \, dx}{c^3 d^3}-\frac{b \int \frac{x}{1-c^2 x^2} \, dx}{c^2 d^3}\\ &=\frac{a x}{c^3 d^3}+\frac{b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac{a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{c^4 d^3}+\frac{b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}-\frac{3 b \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{2 c^4 d^3}-\frac{b \int \left (\frac{1}{2 (1+c x)^3}+\frac{1}{4 (1+c x)^2}-\frac{1}{4 \left (-1+c^2 x^2\right )}\right ) \, dx}{2 c^3 d^3}+\frac{(3 b) \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^3 d^3}\\ &=\frac{a x}{c^3 d^3}+\frac{b}{8 c^4 d^3 (1+c x)^2}-\frac{11 b}{8 c^4 d^3 (1+c x)}+\frac{b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac{a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{c^4 d^3}+\frac{b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}-\frac{3 b \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{2 c^4 d^3}+\frac{b \int \frac{1}{-1+c^2 x^2} \, dx}{8 c^3 d^3}-\frac{(3 b) \int \frac{1}{-1+c^2 x^2} \, dx}{2 c^3 d^3}\\ &=\frac{a x}{c^3 d^3}+\frac{b}{8 c^4 d^3 (1+c x)^2}-\frac{11 b}{8 c^4 d^3 (1+c x)}+\frac{11 b \tanh ^{-1}(c x)}{8 c^4 d^3}+\frac{b x \tanh ^{-1}(c x)}{c^3 d^3}+\frac{a+b \tanh ^{-1}(c x)}{2 c^4 d^3 (1+c x)^2}-\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}+\frac{3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{c^4 d^3}+\frac{b \log \left (1-c^2 x^2\right )}{2 c^4 d^3}-\frac{3 b \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{2 c^4 d^3}\\ \end{align*}

Mathematica [A]  time = 0.708057, size = 167, normalized size = 0.86 \[ \frac{b \left (-48 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+16 \log \left (1-c^2 x^2\right )+20 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )-20 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )+4 \tanh ^{-1}(c x) \left (8 c x+24 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+10 \sinh \left (2 \tanh ^{-1}(c x)\right )-\sinh \left (4 \tanh ^{-1}(c x)\right )-10 \cosh \left (2 \tanh ^{-1}(c x)\right )+\cosh \left (4 \tanh ^{-1}(c x)\right )\right )\right )+32 a c x-\frac{96 a}{c x+1}+\frac{16 a}{(c x+1)^2}-96 a \log (c x+1)}{32 c^4 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^3*(a + b*ArcTanh[c*x]))/(d + c*d*x)^3,x]

[Out]

(32*a*c*x + (16*a)/(1 + c*x)^2 - (96*a)/(1 + c*x) - 96*a*Log[1 + c*x] + b*(-20*Cosh[2*ArcTanh[c*x]] + Cosh[4*A
rcTanh[c*x]] + 16*Log[1 - c^2*x^2] - 48*PolyLog[2, -E^(-2*ArcTanh[c*x])] + 20*Sinh[2*ArcTanh[c*x]] + 4*ArcTanh
[c*x]*(8*c*x - 10*Cosh[2*ArcTanh[c*x]] + Cosh[4*ArcTanh[c*x]] + 24*Log[1 + E^(-2*ArcTanh[c*x])] + 10*Sinh[2*Ar
cTanh[c*x]] - Sinh[4*ArcTanh[c*x]]) - Sinh[4*ArcTanh[c*x]]))/(32*c^4*d^3)

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Maple [A]  time = 0.053, size = 270, normalized size = 1.4 \begin{align*}{\frac{ax}{{c}^{3}{d}^{3}}}+{\frac{a}{2\,{c}^{4}{d}^{3} \left ( cx+1 \right ) ^{2}}}-3\,{\frac{a}{{c}^{4}{d}^{3} \left ( cx+1 \right ) }}-3\,{\frac{a\ln \left ( cx+1 \right ) }{{c}^{4}{d}^{3}}}+{\frac{bx{\it Artanh} \left ( cx \right ) }{{c}^{3}{d}^{3}}}+{\frac{b{\it Artanh} \left ( cx \right ) }{2\,{c}^{4}{d}^{3} \left ( cx+1 \right ) ^{2}}}-3\,{\frac{b{\it Artanh} \left ( cx \right ) }{{c}^{4}{d}^{3} \left ( cx+1 \right ) }}-3\,{\frac{b{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{{c}^{4}{d}^{3}}}+{\frac{3\,b}{2\,{c}^{4}{d}^{3}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }-{\frac{3\,b\ln \left ( cx+1 \right ) }{2\,{c}^{4}{d}^{3}}\ln \left ( -{\frac{cx}{2}}+{\frac{1}{2}} \right ) }+{\frac{3\,b}{2\,{c}^{4}{d}^{3}}{\it dilog} \left ({\frac{1}{2}}+{\frac{cx}{2}} \right ) }+{\frac{3\,b \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{4\,{c}^{4}{d}^{3}}}-{\frac{3\,b\ln \left ( cx-1 \right ) }{16\,{c}^{4}{d}^{3}}}+{\frac{b}{8\,{c}^{4}{d}^{3} \left ( cx+1 \right ) ^{2}}}-{\frac{11\,b}{8\,{c}^{4}{d}^{3} \left ( cx+1 \right ) }}+{\frac{19\,b\ln \left ( cx+1 \right ) }{16\,{c}^{4}{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c*x))/(c*d*x+d)^3,x)

[Out]

a*x/c^3/d^3+1/2/c^4*a/d^3/(c*x+1)^2-3/c^4*a/d^3/(c*x+1)-3/c^4*a/d^3*ln(c*x+1)+b*x*arctanh(c*x)/c^3/d^3+1/2/c^4
*b/d^3*arctanh(c*x)/(c*x+1)^2-3/c^4*b/d^3*arctanh(c*x)/(c*x+1)-3/c^4*b/d^3*arctanh(c*x)*ln(c*x+1)+3/2/c^4*b/d^
3*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)-3/2/c^4*b/d^3*ln(-1/2*c*x+1/2)*ln(c*x+1)+3/2/c^4*b/d^3*dilog(1/2+1/2*c*x)+3
/4/c^4*b/d^3*ln(c*x+1)^2-3/16/c^4*b/d^3*ln(c*x-1)+1/8*b/c^4/d^3/(c*x+1)^2-11/8*b/c^4/d^3/(c*x+1)+19/16/c^4*b/d
^3*ln(c*x+1)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

-1/32*(2*c^4*(2*(7*c*x + 6)/(c^10*d^3*x^2 + 2*c^9*d^3*x + c^8*d^3) - 8*x/(c^7*d^3) + 17*log(c*x + 1)/(c^8*d^3)
- log(c*x - 1)/(c^8*d^3)) - 32*c^4*integrate(1/2*x^4*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x
- c^3*d^3), x) - 6*c^3*(2*(5*c*x + 4)/(c^9*d^3*x^2 + 2*c^8*d^3*x + c^7*d^3) + 7*log(c*x + 1)/(c^7*d^3) + log(c
*x - 1)/(c^7*d^3)) + 128*c^3*integrate(1/2*x^3*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d
^3), x) + 288*c^2*integrate(1/2*x^2*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d^3), x) + 9
*c*(2*x/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - log(c*x + 1)/(c^5*d^3) + log(c*x - 1)/(c^5*d^3)) + 288*c*integ
rate(1/2*x*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d^3), x) + 8*(2*c^3*x^3 + 4*c^2*x^2 -
4*c*x - 6*(c^2*x^2 + 2*c*x + 1)*log(c*x + 1) - 5)*log(-c*x + 1)/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) + 10*(c
*x + 2)/(c^6*d^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - 5*log(c*x + 1)/(c^4*d^3) + 5*log(c*x - 1)/(c^4*d^3) + 96*integ
rate(1/2*log(c*x + 1)/(c^7*d^3*x^4 + 2*c^6*d^3*x^3 - 2*c^4*d^3*x - c^3*d^3), x))*b - 1/2*a*((6*c*x + 5)/(c^6*d
^3*x^2 + 2*c^5*d^3*x + c^4*d^3) - 2*x/(c^3*d^3) + 6*log(c*x + 1)/(c^4*d^3))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{3} \operatorname{artanh}\left (c x\right ) + a x^{3}}{c^{3} d^{3} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c d^{3} x + d^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

integral((b*x^3*arctanh(c*x) + a*x^3)/(c^3*d^3*x^3 + 3*c^2*d^3*x^2 + 3*c*d^3*x + d^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{a x^{3}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac{b x^{3} \operatorname{atanh}{\left (c x \right )}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx}{d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atanh(c*x))/(c*d*x+d)**3,x)

[Out]

(Integral(a*x**3/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b*x**3*atanh(c*x)/(c**3*x**3 + 3*c**2*x*
*2 + 3*c*x + 1), x))/d**3

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )} x^{3}}{{\left (c d x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)*x^3/(c*d*x + d)^3, x)