3.115 \(\int (e+f x) (a+b \coth ^{-1}(c+d x))^3 \, dx\)

Optimal. Leaf size=326 \[ -\frac{3 b^2 (d e-c f) \text{PolyLog}\left (2,1-\frac{2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d^2}+\frac{3 b^3 (d e-c f) \text{PolyLog}\left (3,1-\frac{2}{-c-d x+1}\right )}{2 d^2}-\frac{3 b^3 f \text{PolyLog}\left (2,-\frac{c+d x+1}{-c-d x+1}\right )}{2 d^2}-\frac{3 b^2 f \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d^2}-\frac{\left (\left (c^2+1\right ) f^2-2 c d e f+d^2 e^2\right ) \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d^2}-\frac{3 b (d e-c f) \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d^2}+\frac{3 b f \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{3 b f (c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 f} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.720792, antiderivative size = 325, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 11, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.611, Rules used = {6112, 5929, 5911, 5985, 5919, 2402, 2315, 6049, 5949, 6059, 6610} \[ -\frac{3 b^2 (d e-c f) \text{PolyLog}\left (2,1-\frac{2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d^2}+\frac{3 b^3 (d e-c f) \text{PolyLog}\left (3,1-\frac{2}{-c-d x+1}\right )}{2 d^2}-\frac{3 b^3 f \text{PolyLog}\left (2,-\frac{c+d x+1}{-c-d x+1}\right )}{2 d^2}-\frac{3 b^2 f \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{d^2}+\frac{\left (-\frac{\left (c^2+1\right ) f}{d}+2 c e-\frac{d e^2}{f}\right ) \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 d}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d^2}-\frac{3 b (d e-c f) \log \left (\frac{2}{-c-d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{d^2}+\frac{3 b f \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{3 b f (c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 f} \]

Antiderivative was successfully verified.

[In]

Int[(e + f*x)*(a + b*ArcCoth[c + d*x])^3,x]

[Out]

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

Rule 6112

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

Rule 5929

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[((d + e*x)^(q + 1)*(
a + b*ArcCoth[c*x])^p)/(e*(q + 1)), x] - Dist[(b*c*p)/(e*(q + 1)), Int[ExpandIntegrand[(a + b*ArcCoth[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 5911

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

Rule 5985

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

Rule 5919

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])^p*
Log[2/(1 + (e*x)/d)])/e, x] + Dist[(b*c*p)/e, Int[((a + b*ArcCoth[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]

Rule 6049

Int[(((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :>
Int[ExpandIntegrand[(a + b*ArcCoth[c*x])^p/(d + e*x^2), (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x]
&& IGtQ[p, 0] && EqQ[c^2*d + e, 0] && IGtQ[m, 0]

Rule 5949

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcCoth[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 6059

Int[(Log[u_]*((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcC
oth[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcCoth[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 6610

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 (e+f x) \left (a+b \coth ^{-1}(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \left (\frac{d e-c f}{d}+\frac{f x}{d}\right ) \left (a+b \coth ^{-1}(x)\right )^3 \, dx,x,c+d x\right )}{d}\\ &=\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 f}-\frac{(3 b) \operatorname{Subst}\left (\int \left (-\frac{f^2 \left (a+b \coth ^{-1}(x)\right )^2}{d^2}+\frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2+2 f (d e-c f) x\right ) \left (a+b \coth ^{-1}(x)\right )^2}{d^2 \left (1-x^2\right )}\right ) \, dx,x,c+d x\right )}{2 f}\\ &=\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 f}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2+2 f (d e-c f) x\right ) \left (a+b \coth ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{2 d^2 f}+\frac{(3 b f) \operatorname{Subst}\left (\int \left (a+b \coth ^{-1}(x)\right )^2 \, dx,x,c+d x\right )}{2 d^2}\\ &=\frac{3 b f (c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 f}-\frac{(3 b) \operatorname{Subst}\left (\int \left (\frac{d^2 e^2 \left (1+\frac{f \left (-2 c d e+f+c^2 f\right )}{d^2 e^2}\right ) \left (a+b \coth ^{-1}(x)\right )^2}{1-x^2}-\frac{2 f (-d e+c f) x \left (a+b \coth ^{-1}(x)\right )^2}{1-x^2}\right ) \, dx,x,c+d x\right )}{2 d^2 f}-\frac{\left (3 b^2 f\right ) \operatorname{Subst}\left (\int \frac{x \left (a+b \coth ^{-1}(x)\right )}{1-x^2} \, dx,x,c+d x\right )}{d^2}\\ &=\frac{3 b f \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{3 b f (c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 f}-\frac{\left (3 b^2 f\right ) \operatorname{Subst}\left (\int \frac{a+b \coth ^{-1}(x)}{1-x} \, dx,x,c+d x\right )}{d^2}-\frac{(3 b (d e-c f)) \operatorname{Subst}\left (\int \frac{x \left (a+b \coth ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{d^2}-\frac{\left (3 b \left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right )\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \coth ^{-1}(x)\right )^2}{1-x^2} \, dx,x,c+d x\right )}{2 d^2 f}\\ &=\frac{3 b f \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{3 b f (c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d^2}-\frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 f}-\frac{3 b^2 f \left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac{2}{1-c-d x}\right )}{d^2}+\frac{\left (3 b^3 f\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d^2}-\frac{(3 b (d e-c f)) \operatorname{Subst}\left (\int \frac{\left (a+b \coth ^{-1}(x)\right )^2}{1-x} \, dx,x,c+d x\right )}{d^2}\\ &=\frac{3 b f \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{3 b f (c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d^2}-\frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 f}-\frac{3 b^2 f \left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac{2}{1-c-d x}\right )}{d^2}-\frac{3 b (d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-c-d x}\right )}{d^2}-\frac{\left (3 b^3 f\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-c-d x}\right )}{d^2}+\frac{\left (6 b^2 (d e-c f)\right ) \operatorname{Subst}\left (\int \frac{\left (a+b \coth ^{-1}(x)\right ) \log \left (\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d^2}\\ &=\frac{3 b f \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{3 b f (c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d^2}-\frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 f}-\frac{3 b^2 f \left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac{2}{1-c-d x}\right )}{d^2}-\frac{3 b (d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-c-d x}\right )}{d^2}-\frac{3 b^3 f \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{2 d^2}-\frac{3 b^2 (d e-c f) \left (a+b \coth ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{d^2}+\frac{\left (3 b^3 (d e-c f)\right ) \operatorname{Subst}\left (\int \frac{\text{Li}_2\left (1-\frac{2}{1-x}\right )}{1-x^2} \, dx,x,c+d x\right )}{d^2}\\ &=\frac{3 b f \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{3 b f (c+d x) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 d^2}+\frac{(d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^3}{d^2}-\frac{\left (d^2 e^2-2 c d e f+\left (1+c^2\right ) f^2\right ) \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 d^2 f}+\frac{(e+f x)^2 \left (a+b \coth ^{-1}(c+d x)\right )^3}{2 f}-\frac{3 b^2 f \left (a+b \coth ^{-1}(c+d x)\right ) \log \left (\frac{2}{1-c-d x}\right )}{d^2}-\frac{3 b (d e-c f) \left (a+b \coth ^{-1}(c+d x)\right )^2 \log \left (\frac{2}{1-c-d x}\right )}{d^2}-\frac{3 b^3 f \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{2 d^2}-\frac{3 b^2 (d e-c f) \left (a+b \coth ^{-1}(c+d x)\right ) \text{Li}_2\left (1-\frac{2}{1-c-d x}\right )}{d^2}+\frac{3 b^3 (d e-c f) \text{Li}_3\left (1-\frac{2}{1-c-d x}\right )}{2 d^2}\\ \end{align*}

Mathematica [C]  time = 1.31426, size = 600, normalized size = 1.84 \[ \frac{12 a b^2 d e \left (\text{PolyLog}\left (2,e^{-2 \coth ^{-1}(c+d x)}\right )+\coth ^{-1}(c+d x) \left ((c+d x-1) \coth ^{-1}(c+d x)-2 \log \left (1-e^{-2 \coth ^{-1}(c+d x)}\right )\right )\right )-12 a b^2 c f \left (\text{PolyLog}\left (2,e^{-2 \coth ^{-1}(c+d x)}\right )+\coth ^{-1}(c+d x) \left ((c+d x-1) \coth ^{-1}(c+d x)-2 \log \left (1-e^{-2 \coth ^{-1}(c+d x)}\right )\right )\right )+2 b^3 f \left (3 \text{PolyLog}\left (2,e^{-2 \coth ^{-1}(c+d x)}\right )+\coth ^{-1}(c+d x) \left (\left (c^2+2 c d x+d^2 x^2-1\right ) \coth ^{-1}(c+d x)^2+3 (c+d x-1) \coth ^{-1}(c+d x)-6 \log \left (1-e^{-2 \coth ^{-1}(c+d x)}\right )\right )\right )+4 b^3 d e \left (-3 \coth ^{-1}(c+d x) \text{PolyLog}\left (2,e^{2 \coth ^{-1}(c+d x)}\right )+\frac{3}{2} \text{PolyLog}\left (3,e^{2 \coth ^{-1}(c+d x)}\right )+(c+d x) \coth ^{-1}(c+d x)^3+\coth ^{-1}(c+d x)^3-3 \coth ^{-1}(c+d x)^2 \log \left (1-e^{2 \coth ^{-1}(c+d x)}\right )-\frac{i \pi ^3}{8}\right )-4 b^3 c f \left (-3 \coth ^{-1}(c+d x) \text{PolyLog}\left (2,e^{2 \coth ^{-1}(c+d x)}\right )+\frac{3}{2} \text{PolyLog}\left (3,e^{2 \coth ^{-1}(c+d x)}\right )+(c+d x) \coth ^{-1}(c+d x)^3+\coth ^{-1}(c+d x)^3-3 \coth ^{-1}(c+d x)^2 \log \left (1-e^{2 \coth ^{-1}(c+d x)}\right )-\frac{i \pi ^3}{8}\right )+2 a^2 (c+d x) (-2 a c f+2 a d e+3 b f)+3 a^2 b (-2 c f+2 d e+f) \log (-c-d x+1)+3 a^2 b (2 d e-(2 c+1) f) \log (c+d x+1)-6 a^2 b (c+d x) \coth ^{-1}(c+d x) (c f-d (2 e+f x))+2 a^3 f (c+d x)^2+12 a b^2 f \left (-\log \left (\frac{1}{(c+d x) \sqrt{1-\frac{1}{(c+d x)^2}}}\right )+\frac{1}{2} \left ((c+d x)^2-1\right ) \coth ^{-1}(c+d x)^2+(c+d x) \coth ^{-1}(c+d x)\right )}{4 d^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(e + f*x)*(a + b*ArcCoth[c + d*x])^3,x]

[Out]

(2*a^2*(2*a*d*e + 3*b*f - 2*a*c*f)*(c + d*x) + 2*a^3*f*(c + d*x)^2 - 6*a^2*b*(c + d*x)*(c*f - d*(2*e + f*x))*A
rcCoth[c + d*x] + 3*a^2*b*(2*d*e + f - 2*c*f)*Log[1 - c - d*x] + 3*a^2*b*(2*d*e - (1 + 2*c)*f)*Log[1 + c + d*x
] + 12*a*b^2*f*((c + d*x)*ArcCoth[c + d*x] + ((-1 + (c + d*x)^2)*ArcCoth[c + d*x]^2)/2 - Log[1/((c + d*x)*Sqrt
[1 - (c + d*x)^(-2)])]) + 12*a*b^2*d*e*(ArcCoth[c + d*x]*((-1 + c + d*x)*ArcCoth[c + d*x] - 2*Log[1 - E^(-2*Ar
cCoth[c + d*x])]) + PolyLog[2, E^(-2*ArcCoth[c + d*x])]) - 12*a*b^2*c*f*(ArcCoth[c + d*x]*((-1 + c + d*x)*ArcC
oth[c + d*x] - 2*Log[1 - E^(-2*ArcCoth[c + d*x])]) + PolyLog[2, E^(-2*ArcCoth[c + d*x])]) + 2*b^3*f*(ArcCoth[c
+ d*x]*(3*(-1 + c + d*x)*ArcCoth[c + d*x] + (-1 + c^2 + 2*c*d*x + d^2*x^2)*ArcCoth[c + d*x]^2 - 6*Log[1 - E^(
-2*ArcCoth[c + d*x])]) + 3*PolyLog[2, E^(-2*ArcCoth[c + d*x])]) + 4*b^3*d*e*((-I/8)*Pi^3 + ArcCoth[c + d*x]^3
+ (c + d*x)*ArcCoth[c + d*x]^3 - 3*ArcCoth[c + d*x]^2*Log[1 - E^(2*ArcCoth[c + d*x])] - 3*ArcCoth[c + d*x]*Pol
yLog[2, E^(2*ArcCoth[c + d*x])] + (3*PolyLog[3, E^(2*ArcCoth[c + d*x])])/2) - 4*b^3*c*f*((-I/8)*Pi^3 + ArcCoth
[c + d*x]^3 + (c + d*x)*ArcCoth[c + d*x]^3 - 3*ArcCoth[c + d*x]^2*Log[1 - E^(2*ArcCoth[c + d*x])] - 3*ArcCoth[
c + d*x]*PolyLog[2, E^(2*ArcCoth[c + d*x])] + (3*PolyLog[3, E^(2*ArcCoth[c + d*x])])/2))/(4*d^2)

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Maple [C]  time = 1.107, size = 12285, normalized size = 37.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x+e)*(a+b*arccoth(d*x+c))^3,x)

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a^{3} f x^{2} + \frac{3}{4} \,{\left (2 \, x^{2} \operatorname{arcoth}\left (d x + c\right ) + d{\left (\frac{2 \, x}{d^{2}} - \frac{{\left (c^{2} + 2 \, c + 1\right )} \log \left (d x + c + 1\right )}{d^{3}} + \frac{{\left (c^{2} - 2 \, c + 1\right )} \log \left (d x + c - 1\right )}{d^{3}}\right )}\right )} a^{2} b f + a^{3} e x + \frac{3 \,{\left (2 \,{\left (d x + c\right )} \operatorname{arcoth}\left (d x + c\right ) + \log \left (-{\left (d x + c\right )}^{2} + 1\right )\right )} a^{2} b e}{2 \, d} + \frac{{\left (b^{3} d^{2} f x^{2} + 2 \, b^{3} d^{2} e x -{\left (c^{2} f - 2 \,{\left (d e - f\right )} c - 2 \, d e + f\right )} b^{3}\right )} \log \left (d x + c + 1\right )^{3} + 3 \,{\left (2 \, a b^{2} d^{2} f x^{2} + 2 \,{\left (2 \, a b^{2} d^{2} e + b^{3} d f\right )} x -{\left (b^{3} d^{2} f x^{2} + 2 \, b^{3} d^{2} e x -{\left (c^{2} f - 2 \,{\left (d e + f\right )} c + 2 \, d e + f\right )} b^{3}\right )} \log \left (d x + c - 1\right )\right )} \log \left (d x + c + 1\right )^{2}}{16 \, d^{2}} + \int -\frac{{\left (b^{3} d^{2} f x^{2} +{\left (d^{2} e + c d f + d f\right )} b^{3} x +{\left (c d e + d e\right )} b^{3}\right )} \log \left (d x + c - 1\right )^{3} - 6 \,{\left (a b^{2} d^{2} f x^{2} +{\left (d^{2} e + c d f + d f\right )} a b^{2} x +{\left (c d e + d e\right )} a b^{2}\right )} \log \left (d x + c - 1\right )^{2} + 3 \,{\left (2 \, a b^{2} d^{2} f x^{2} -{\left (b^{3} d^{2} f x^{2} +{\left (d^{2} e + c d f + d f\right )} b^{3} x +{\left (c d e + d e\right )} b^{3}\right )} \log \left (d x + c - 1\right )^{2} + 2 \,{\left (2 \, a b^{2} d^{2} e + b^{3} d f\right )} x +{\left (4 \,{\left (c d e + d e\right )} a b^{2} +{\left (c^{2} f - 2 \,{\left (d e + f\right )} c + 2 \, d e + f\right )} b^{3} +{\left (4 \, a b^{2} d^{2} f - b^{3} d^{2} f\right )} x^{2} - 2 \,{\left (b^{3} d^{2} e - 2 \,{\left (d^{2} e + c d f + d f\right )} a b^{2}\right )} x\right )} \log \left (d x + c - 1\right )\right )} \log \left (d x + c + 1\right )}{8 \,{\left (d^{2} x + c d + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(a+b*arccoth(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*a^3*f*x^2 + 3/4*(2*x^2*arccoth(d*x + c) + d*(2*x/d^2 - (c^2 + 2*c + 1)*log(d*x + c + 1)/d^3 + (c^2 - 2*c +
1)*log(d*x + c - 1)/d^3))*a^2*b*f + a^3*e*x + 3/2*(2*(d*x + c)*arccoth(d*x + c) + log(-(d*x + c)^2 + 1))*a^2*
b*e/d + 1/16*((b^3*d^2*f*x^2 + 2*b^3*d^2*e*x - (c^2*f - 2*(d*e - f)*c - 2*d*e + f)*b^3)*log(d*x + c + 1)^3 + 3
*(2*a*b^2*d^2*f*x^2 + 2*(2*a*b^2*d^2*e + b^3*d*f)*x - (b^3*d^2*f*x^2 + 2*b^3*d^2*e*x - (c^2*f - 2*(d*e + f)*c
+ 2*d*e + f)*b^3)*log(d*x + c - 1))*log(d*x + c + 1)^2)/d^2 + integrate(-1/8*((b^3*d^2*f*x^2 + (d^2*e + c*d*f
+ d*f)*b^3*x + (c*d*e + d*e)*b^3)*log(d*x + c - 1)^3 - 6*(a*b^2*d^2*f*x^2 + (d^2*e + c*d*f + d*f)*a*b^2*x + (c
*d*e + d*e)*a*b^2)*log(d*x + c - 1)^2 + 3*(2*a*b^2*d^2*f*x^2 - (b^3*d^2*f*x^2 + (d^2*e + c*d*f + d*f)*b^3*x +
(c*d*e + d*e)*b^3)*log(d*x + c - 1)^2 + 2*(2*a*b^2*d^2*e + b^3*d*f)*x + (4*(c*d*e + d*e)*a*b^2 + (c^2*f - 2*(d
*e + f)*c + 2*d*e + f)*b^3 + (4*a*b^2*d^2*f - b^3*d^2*f)*x^2 - 2*(b^3*d^2*e - 2*(d^2*e + c*d*f + d*f)*a*b^2)*x
)*log(d*x + c - 1))*log(d*x + c + 1))/(d^2*x + c*d + d), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(a+b*arccoth(d*x+c))^3,x, algorithm="fricas")

[Out]

integral(a^3*f*x + a^3*e + (b^3*f*x + b^3*e)*arccoth(d*x + c)^3 + 3*(a*b^2*f*x + a*b^2*e)*arccoth(d*x + c)^2 +
3*(a^2*b*f*x + a^2*b*e)*arccoth(d*x + c), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(a+b*acoth(d*x+c))**3,x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)*(a+b*arccoth(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((f*x + e)*(b*arccoth(d*x + c) + a)^3, x)