3.58 \(\int \frac{x^4 (a+b \tanh ^{-1}(c x))}{(d+c d x)^3} \, dx\)

Optimal. Leaf size=227 \[ \frac{3 b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{c^5 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}+\frac{4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (c x+1)}-\frac{a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (c x+1)^2}-\frac{6 \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3}-\frac{3 a x}{c^4 d^3}-\frac{3 b \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac{b x}{2 c^4 d^3}+\frac{15 b}{8 c^5 d^3 (c x+1)}-\frac{b}{8 c^5 d^3 (c x+1)^2}-\frac{3 b x \tanh ^{-1}(c x)}{c^4 d^3}-\frac{19 b \tanh ^{-1}(c x)}{8 c^5 d^3} \]

[Out]

(-3*a*x)/(c^4*d^3) + (b*x)/(2*c^4*d^3) - b/(8*c^5*d^3*(1 + c*x)^2) + (15*b)/(8*c^5*d^3*(1 + c*x)) - (19*b*ArcT
anh[c*x])/(8*c^5*d^3) - (3*b*x*ArcTanh[c*x])/(c^4*d^3) + (x^2*(a + b*ArcTanh[c*x]))/(2*c^3*d^3) - (a + b*ArcTa
nh[c*x])/(2*c^5*d^3*(1 + c*x)^2) + (4*(a + b*ArcTanh[c*x]))/(c^5*d^3*(1 + c*x)) - (6*(a + b*ArcTanh[c*x])*Log[
2/(1 + c*x)])/(c^5*d^3) - (3*b*Log[1 - c^2*x^2])/(2*c^5*d^3) + (3*b*PolyLog[2, 1 - 2/(1 + c*x)])/(c^5*d^3)

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Rubi [A]  time = 0.287584, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 13, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.65, Rules used = {5940, 5910, 260, 5916, 321, 206, 5926, 627, 44, 207, 5918, 2402, 2315} \[ \frac{3 b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{c^5 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}+\frac{4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (c x+1)}-\frac{a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (c x+1)^2}-\frac{6 \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3}-\frac{3 a x}{c^4 d^3}-\frac{3 b \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac{b x}{2 c^4 d^3}+\frac{15 b}{8 c^5 d^3 (c x+1)}-\frac{b}{8 c^5 d^3 (c x+1)^2}-\frac{3 b x \tanh ^{-1}(c x)}{c^4 d^3}-\frac{19 b \tanh ^{-1}(c x)}{8 c^5 d^3} \]

Antiderivative was successfully verified.

[In]

Int[(x^4*(a + b*ArcTanh[c*x]))/(d + c*d*x)^3,x]

[Out]

(-3*a*x)/(c^4*d^3) + (b*x)/(2*c^4*d^3) - b/(8*c^5*d^3*(1 + c*x)^2) + (15*b)/(8*c^5*d^3*(1 + c*x)) - (19*b*ArcT
anh[c*x])/(8*c^5*d^3) - (3*b*x*ArcTanh[c*x])/(c^4*d^3) + (x^2*(a + b*ArcTanh[c*x]))/(2*c^3*d^3) - (a + b*ArcTa
nh[c*x])/(2*c^5*d^3*(1 + c*x)^2) + (4*(a + b*ArcTanh[c*x]))/(c^5*d^3*(1 + c*x)) - (6*(a + b*ArcTanh[c*x])*Log[
2/(1 + c*x)])/(c^5*d^3) - (3*b*Log[1 - c^2*x^2])/(2*c^5*d^3) + (3*b*PolyLog[2, 1 - 2/(1 + c*x)])/(c^5*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 5916

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

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 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 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^4 \left (a+b \tanh ^{-1}(c x)\right )}{(d+c d x)^3} \, dx &=\int \left (-\frac{3 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3}+\frac{x \left (a+b \tanh ^{-1}(c x)\right )}{c^3 d^3}+\frac{a+b \tanh ^{-1}(c x)}{c^4 d^3 (1+c x)^3}-\frac{4 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)^2}+\frac{6 \left (a+b \tanh ^{-1}(c x)\right )}{c^4 d^3 (1+c x)}\right ) \, dx\\ &=\frac{\int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^3} \, dx}{c^4 d^3}-\frac{3 \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^4 d^3}-\frac{4 \int \frac{a+b \tanh ^{-1}(c x)}{(1+c x)^2} \, dx}{c^4 d^3}+\frac{6 \int \frac{a+b \tanh ^{-1}(c x)}{1+c x} \, dx}{c^4 d^3}+\frac{\int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{c^3 d^3}\\ &=-\frac{3 a x}{c^4 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac{a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac{4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac{6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{c^5 d^3}+\frac{b \int \frac{1}{(1+c x)^2 \left (1-c^2 x^2\right )} \, dx}{2 c^4 d^3}-\frac{(3 b) \int \tanh ^{-1}(c x) \, dx}{c^4 d^3}-\frac{(4 b) \int \frac{1}{(1+c x) \left (1-c^2 x^2\right )} \, dx}{c^4 d^3}+\frac{(6 b) \int \frac{\log \left (\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{c^4 d^3}-\frac{b \int \frac{x^2}{1-c^2 x^2} \, dx}{2 c^2 d^3}\\ &=-\frac{3 a x}{c^4 d^3}+\frac{b x}{2 c^4 d^3}-\frac{3 b x \tanh ^{-1}(c x)}{c^4 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac{a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac{4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac{6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{c^5 d^3}+\frac{(6 b) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+c x}\right )}{c^5 d^3}+\frac{b \int \frac{1}{(1-c x) (1+c x)^3} \, dx}{2 c^4 d^3}-\frac{b \int \frac{1}{1-c^2 x^2} \, dx}{2 c^4 d^3}-\frac{(4 b) \int \frac{1}{(1-c x) (1+c x)^2} \, dx}{c^4 d^3}+\frac{(3 b) \int \frac{x}{1-c^2 x^2} \, dx}{c^3 d^3}\\ &=-\frac{3 a x}{c^4 d^3}+\frac{b x}{2 c^4 d^3}-\frac{b \tanh ^{-1}(c x)}{2 c^5 d^3}-\frac{3 b x \tanh ^{-1}(c x)}{c^4 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac{a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac{4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac{6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{c^5 d^3}-\frac{3 b \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac{3 b \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^5 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^4 d^3}-\frac{(4 b) \int \left (\frac{1}{2 (1+c x)^2}-\frac{1}{2 \left (-1+c^2 x^2\right )}\right ) \, dx}{c^4 d^3}\\ &=-\frac{3 a x}{c^4 d^3}+\frac{b x}{2 c^4 d^3}-\frac{b}{8 c^5 d^3 (1+c x)^2}+\frac{15 b}{8 c^5 d^3 (1+c x)}-\frac{b \tanh ^{-1}(c x)}{2 c^5 d^3}-\frac{3 b x \tanh ^{-1}(c x)}{c^4 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac{a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac{4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac{6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{c^5 d^3}-\frac{3 b \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac{3 b \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^5 d^3}-\frac{b \int \frac{1}{-1+c^2 x^2} \, dx}{8 c^4 d^3}+\frac{(2 b) \int \frac{1}{-1+c^2 x^2} \, dx}{c^4 d^3}\\ &=-\frac{3 a x}{c^4 d^3}+\frac{b x}{2 c^4 d^3}-\frac{b}{8 c^5 d^3 (1+c x)^2}+\frac{15 b}{8 c^5 d^3 (1+c x)}-\frac{19 b \tanh ^{-1}(c x)}{8 c^5 d^3}-\frac{3 b x \tanh ^{-1}(c x)}{c^4 d^3}+\frac{x^2 \left (a+b \tanh ^{-1}(c x)\right )}{2 c^3 d^3}-\frac{a+b \tanh ^{-1}(c x)}{2 c^5 d^3 (1+c x)^2}+\frac{4 \left (a+b \tanh ^{-1}(c x)\right )}{c^5 d^3 (1+c x)}-\frac{6 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{c^5 d^3}-\frac{3 b \log \left (1-c^2 x^2\right )}{2 c^5 d^3}+\frac{3 b \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{c^5 d^3}\\ \end{align*}

Mathematica [A]  time = 0.829683, size = 189, normalized size = 0.83 \[ \frac{b \left (96 \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )-48 \log \left (1-c^2 x^2\right )+4 \tanh ^{-1}(c x) \left (4 c^2 x^2-24 c x-48 \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )-14 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+14 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )-4\right )+16 c x-28 \sinh \left (2 \tanh ^{-1}(c x)\right )+\sinh \left (4 \tanh ^{-1}(c x)\right )+28 \cosh \left (2 \tanh ^{-1}(c x)\right )-\cosh \left (4 \tanh ^{-1}(c x)\right )\right )+16 a c^2 x^2-96 a c x+\frac{128 a}{c x+1}-\frac{16 a}{(c x+1)^2}+192 a \log (c x+1)}{32 c^5 d^3} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x^4*(a + b*ArcTanh[c*x]))/(d + c*d*x)^3,x]

[Out]

(-96*a*c*x + 16*a*c^2*x^2 - (16*a)/(1 + c*x)^2 + (128*a)/(1 + c*x) + 192*a*Log[1 + c*x] + b*(16*c*x + 28*Cosh[
2*ArcTanh[c*x]] - Cosh[4*ArcTanh[c*x]] - 48*Log[1 - c^2*x^2] + 96*PolyLog[2, -E^(-2*ArcTanh[c*x])] - 28*Sinh[2
*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]] + 4*ArcTanh[c*x]*(-4 - 24*c*x + 4*c^2*x^2 + 14*Cosh[2*ArcTanh[c*x]] - Co
sh[4*ArcTanh[c*x]] - 48*Log[1 + E^(-2*ArcTanh[c*x])] - 14*Sinh[2*ArcTanh[c*x]] + Sinh[4*ArcTanh[c*x]])))/(32*c
^5*d^3)

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Maple [A]  time = 0.049, size = 319, normalized size = 1.4 \begin{align*}{\frac{a{x}^{2}}{2\,{c}^{3}{d}^{3}}}-3\,{\frac{ax}{{c}^{4}{d}^{3}}}-{\frac{a}{2\,{c}^{5}{d}^{3} \left ( cx+1 \right ) ^{2}}}+4\,{\frac{a}{{c}^{5}{d}^{3} \left ( cx+1 \right ) }}+6\,{\frac{a\ln \left ( cx+1 \right ) }{{c}^{5}{d}^{3}}}+{\frac{b{\it Artanh} \left ( cx \right ){x}^{2}}{2\,{c}^{3}{d}^{3}}}-3\,{\frac{bx{\it Artanh} \left ( cx \right ) }{{c}^{4}{d}^{3}}}-{\frac{b{\it Artanh} \left ( cx \right ) }{2\,{c}^{5}{d}^{3} \left ( cx+1 \right ) ^{2}}}+4\,{\frac{b{\it Artanh} \left ( cx \right ) }{{c}^{5}{d}^{3} \left ( cx+1 \right ) }}+6\,{\frac{b{\it Artanh} \left ( cx \right ) \ln \left ( cx+1 \right ) }{{c}^{5}{d}^{3}}}-3\,{\frac{b\ln \left ( -1/2\,cx+1/2 \right ) \ln \left ( 1/2+1/2\,cx \right ) }{{c}^{5}{d}^{3}}}+3\,{\frac{b\ln \left ( -1/2\,cx+1/2 \right ) \ln \left ( cx+1 \right ) }{{c}^{5}{d}^{3}}}-3\,{\frac{b{\it dilog} \left ( 1/2+1/2\,cx \right ) }{{c}^{5}{d}^{3}}}-{\frac{3\,b \left ( \ln \left ( cx+1 \right ) \right ) ^{2}}{2\,{c}^{5}{d}^{3}}}+{\frac{bx}{2\,{c}^{4}{d}^{3}}}+{\frac{b}{2\,{c}^{5}{d}^{3}}}-{\frac{5\,b\ln \left ( cx-1 \right ) }{16\,{c}^{5}{d}^{3}}}-{\frac{b}{8\,{c}^{5}{d}^{3} \left ( cx+1 \right ) ^{2}}}+{\frac{15\,b}{8\,{c}^{5}{d}^{3} \left ( cx+1 \right ) }}-{\frac{43\,b\ln \left ( cx+1 \right ) }{16\,{c}^{5}{d}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*arctanh(c*x))/(c*d*x+d)^3,x)

[Out]

1/2/c^3*a/d^3*x^2-3*a*x/c^4/d^3-1/2/c^5*a/d^3/(c*x+1)^2+4/c^5*a/d^3/(c*x+1)+6/c^5*a/d^3*ln(c*x+1)+1/2/c^3*b/d^
3*arctanh(c*x)*x^2-3*b*x*arctanh(c*x)/c^4/d^3-1/2/c^5*b/d^3*arctanh(c*x)/(c*x+1)^2+4/c^5*b/d^3*arctanh(c*x)/(c
*x+1)+6/c^5*b/d^3*arctanh(c*x)*ln(c*x+1)-3/c^5*b/d^3*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+3/c^5*b/d^3*ln(-1/2*c*x+
1/2)*ln(c*x+1)-3/c^5*b/d^3*dilog(1/2+1/2*c*x)-3/2/c^5*b/d^3*ln(c*x+1)^2+1/2*b*x/c^4/d^3+1/2/c^5*b/d^3-5/16/c^5
*b/d^3*ln(c*x-1)-1/8*b/c^5/d^3/(c*x+1)^2+15/8*b/c^5/d^3/(c*x+1)-43/16/c^5*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^4*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="maxima")

[Out]

1/32*(c^5*(2*(9*c*x + 8)/(c^12*d^3*x^2 + 2*c^11*d^3*x + c^10*d^3) + 4*(c*x^2 - 4*x)/(c^9*d^3) + 31*log(c*x + 1
)/(c^10*d^3) + log(c*x - 1)/(c^10*d^3)) + 32*c^5*integrate(1/2*x^5*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^7*d^3*x^3 -
2*c^5*d^3*x - c^4*d^3), x) + 3*c^4*(2*(7*c*x + 6)/(c^11*d^3*x^2 + 2*c^10*d^3*x + c^9*d^3) - 8*x/(c^8*d^3) + 1
7*log(c*x + 1)/(c^9*d^3) - log(c*x - 1)/(c^9*d^3)) - 32*c^4*integrate(1/2*x^4*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^
7*d^3*x^3 - 2*c^5*d^3*x - c^4*d^3), x) - 15*c^3*(2*(5*c*x + 4)/(c^10*d^3*x^2 + 2*c^9*d^3*x + c^8*d^3) + 7*log(
c*x + 1)/(c^8*d^3) + log(c*x - 1)/(c^8*d^3)) + 192*c^3*integrate(1/2*x^3*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^7*d^3
*x^3 - 2*c^5*d^3*x - c^4*d^3), x) + 9*c^2*(2*(3*c*x + 2)/(c^9*d^3*x^2 + 2*c^8*d^3*x + c^7*d^3) + log(c*x + 1)/
(c^7*d^3) - log(c*x - 1)/(c^7*d^3)) + 576*c^2*integrate(1/2*x^2*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^7*d^3*x^3 - 2*
c^5*d^3*x - c^4*d^3), x) + 9*c*(2*x/(c^7*d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) - log(c*x + 1)/(c^6*d^3) + log(c*x -
1)/(c^6*d^3)) + 576*c*integrate(1/2*x*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^7*d^3*x^3 - 2*c^5*d^3*x - c^4*d^3), x)
- 8*(c^4*x^4 - 4*c^3*x^3 - 11*c^2*x^2 + 2*c*x + 12*(c^2*x^2 + 2*c*x + 1)*log(c*x + 1) + 7)*log(-c*x + 1)/(c^7*
d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) + 14*(c*x + 2)/(c^7*d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) - 7*log(c*x + 1)/(c^5*d^
3) + 7*log(c*x - 1)/(c^5*d^3) + 192*integrate(1/2*log(c*x + 1)/(c^8*d^3*x^4 + 2*c^7*d^3*x^3 - 2*c^5*d^3*x - c^
4*d^3), x))*b + 1/2*a*((8*c*x + 7)/(c^7*d^3*x^2 + 2*c^6*d^3*x + c^5*d^3) + (c*x^2 - 6*x)/(c^4*d^3) + 12*log(c*
x + 1)/(c^5*d^3))

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{4} \operatorname{artanh}\left (c x\right ) + a x^{4}}{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^4*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="fricas")

[Out]

integral((b*x^4*arctanh(c*x) + a*x^4)/(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^{4}}{c^{3} x^{3} + 3 c^{2} x^{2} + 3 c x + 1}\, dx + \int \frac{b x^{4} \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**4*(a+b*atanh(c*x))/(c*d*x+d)**3,x)

[Out]

(Integral(a*x**4/(c**3*x**3 + 3*c**2*x**2 + 3*c*x + 1), x) + Integral(b*x**4*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^{4}}{{\left (c d x + d\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*arctanh(c*x))/(c*d*x+d)^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)*x^4/(c*d*x + d)^3, x)