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

Optimal. Leaf size=308 \[ -\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right )}{2 f}+\frac {3 b^2 \text {Li}_3\left (1-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{2 f}-\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right )}{2 f}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3 \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{f}+\frac {3 b \text {Li}_2\left (1-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f}-\frac {\log \left (\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{f}-\frac {3 b^3 \text {Li}_4\left (1-\frac {2 d (e+f x)}{(d e-c f+f) (c+d x+1)}\right )}{4 f}+\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{c+d x+1}\right )}{4 f} \]

[Out]

-(a+b*arctanh(d*x+c))^3*ln(2/(d*x+c+1))/f+(a+b*arctanh(d*x+c))^3*ln(2*d*(f*x+e)/(-c*f+d*e+f)/(d*x+c+1))/f+3/2*
b*(a+b*arctanh(d*x+c))^2*polylog(2,1-2/(d*x+c+1))/f-3/2*b*(a+b*arctanh(d*x+c))^2*polylog(2,1-2*d*(f*x+e)/(-c*f
+d*e+f)/(d*x+c+1))/f+3/2*b^2*(a+b*arctanh(d*x+c))*polylog(3,1-2/(d*x+c+1))/f-3/2*b^2*(a+b*arctanh(d*x+c))*poly
log(3,1-2*d*(f*x+e)/(-c*f+d*e+f)/(d*x+c+1))/f+3/4*b^3*polylog(4,1-2/(d*x+c+1))/f-3/4*b^3*polylog(4,1-2*d*(f*x+
e)/(-c*f+d*e+f)/(d*x+c+1))/f

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Rubi [A]  time = 0.19, antiderivative size = 308, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6111, 5924} \[ -\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {PolyLog}\left (3,1-\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{2 f}+\frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{2 f}-\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{2 f}+\frac {3 b \text {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^2}{2 f}-\frac {3 b^3 \text {PolyLog}\left (4,1-\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{4 f}+\frac {3 b^3 \text {PolyLog}\left (4,1-\frac {2}{c+d x+1}\right )}{4 f}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3 \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{f}-\frac {\log \left (\frac {2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )^3}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 5924

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^3/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^3*Log[
2/(1 + c*x)])/e, x] + (Simp[((a + b*ArcTanh[c*x])^3*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(
3*b*(a + b*ArcTanh[c*x])^2*PolyLog[2, 1 - 2/(1 + c*x)])/(2*e), x] - Simp[(3*b*(a + b*ArcTanh[c*x])^2*PolyLog[2
, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e), x] + Simp[(3*b^2*(a + b*ArcTanh[c*x])*PolyLog[3, 1 - 2/(1
+ c*x)])/(2*e), x] - Simp[(3*b^2*(a + b*ArcTanh[c*x])*PolyLog[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/
(2*e), x] + Simp[(3*b^3*PolyLog[4, 1 - 2/(1 + c*x)])/(4*e), x] - Simp[(3*b^3*PolyLog[4, 1 - (2*c*(d + e*x))/((
c*d + e)*(1 + c*x))])/(4*e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]

Rule 6111

Int[((a_.) + ArcTanh[(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*ArcTanh[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &
& IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{e+f x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(x)\right )^3}{\frac {d e-c f}{d}+\frac {f x}{d}} \, dx,x,c+d x\right )}{d}\\ &=-\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3 \log \left (\frac {2}{1+c+d x}\right )}{f}+\frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3 \log \left (\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{f}+\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2}{1+c+d x}\right )}{2 f}-\frac {3 b \left (a+b \tanh ^{-1}(c+d x)\right )^2 \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 f}+\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2}{1+c+d x}\right )}{2 f}-\frac {3 b^2 \left (a+b \tanh ^{-1}(c+d x)\right ) \text {Li}_3\left (1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{2 f}+\frac {3 b^3 \text {Li}_4\left (1-\frac {2}{1+c+d x}\right )}{4 f}-\frac {3 b^3 \text {Li}_4\left (1-\frac {2 d (e+f x)}{(d e+f-c f) (1+c+d x)}\right )}{4 f}\\ \end {align*}

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Mathematica [F]  time = 22.91, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \tanh ^{-1}(c+d x)\right )^3}{e+f x} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {artanh}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname {artanh}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname {artanh}\left (d x + c\right ) + a^{3}}{f x + e}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )}^{3}}{f x + e}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.87, size = 4064, normalized size = 13.19 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*a*b^2/f*arctanh(d*x+c)*polylog(2,-(d*x+c+1)^2/(1-(d*x+c)^2))-3*a*b^2/(c*f-d*e-f)*arctanh(d*x+c)^2*ln(1-(c*f
-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))-3*a*b^2/(c*f-d*e-f)*arctanh(d*x+c)*polylog(2,(c*f-d*e-f)*(d*x+
c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+3*a^2*b*ln((d*x+c)*f-c*f+d*e)/f*arctanh(d*x+c)+3/2*a^2*b/f*ln((d*x+c)*f-c*f
+d*e)*ln(((d*x+c)*f-f)/(c*f-d*e-f))-3/2*a^2*b/f*ln((d*x+c)*f-c*f+d*e)*ln(((d*x+c)*f+f)/(c*f-d*e+f))+b^3*c/(c*f
-d*e-f)*arctanh(d*x+c)^3*ln(1-(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+3/2*b^3*c/(c*f-d*e-f)*arctan
h(d*x+c)^2*polylog(2,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))-3/2*b^3*c/(c*f-d*e-f)*arctanh(d*x+c)*
polylog(3,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))-3/2*a*b^2*c/(c*f-d*e-f)*polylog(3,(c*f-d*e-f)*(d
*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+3*a*b^2*ln((d*x+c)*f-c*f+d*e)/f*arctanh(d*x+c)^2-I*b^3/f*Pi*arctanh(d*x+
c)^3-b^3/(c*f-d*e-f)*arctanh(d*x+c)^3*ln(1-(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+3/2*a*b^2/f*pol
ylog(3,-(d*x+c+1)^2/(1-(d*x+c)^2))+3/2*a*b^2/(c*f-d*e-f)*polylog(3,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f
+d*e-f))-3/4*b^3/f*polylog(4,-(d*x+c+1)^2/(1-(d*x+c)^2))-3/4*b^3/(c*f-d*e-f)*polylog(4,(c*f-d*e-f)*(d*x+c+1)^2
/(1-(d*x+c)^2)/(-c*f+d*e-f))+a^3*ln((d*x+c)*f-c*f+d*e)/f-3*d*a*b^2/f*e/(c*f-d*e-f)*arctanh(d*x+c)^2*ln(1-(c*f-
d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))-3*d*a*b^2/f*e/(c*f-d*e-f)*arctanh(d*x+c)*polylog(2,(c*f-d*e-f)*
(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+1/2*I*b^3/f*arctanh(d*x+c)^3*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csg
n(I*(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f))*cs
gn(I*(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f)/(1
+(d*x+c+1)^2/(1-(d*x+c)^2)))*Pi-3/2*I*a*b^2/f*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*(c
*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f)/(1+(d*x+c
+1)^2/(1-(d*x+c)^2)))^2*Pi-3/2*I*a*b^2/f*arctanh(d*x+c)^2*csgn(I*(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1
)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f))*csgn(I*(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+
1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2*Pi-3*a*b^2/f*arcta
nh(d*x+c)^2*ln(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^
2))*f)+3/4*b^3*c/(c*f-d*e-f)*polylog(4,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+3/2*a^2*b/f*dilog((
(d*x+c)*f-f)/(c*f-d*e-f))-3/2*a^2*b/f*dilog(((d*x+c)*f+f)/(c*f-d*e+f))-3/2*b^3/(c*f-d*e-f)*arctanh(d*x+c)^2*po
lylog(2,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+3/2*b^3/(c*f-d*e-f)*arctanh(d*x+c)*polylog(3,(c*f-
d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+b^3*ln((d*x+c)*f-c*f+d*e)/f*arctanh(d*x+c)^3-b^3/f*arctanh(d*x+
c)^3*ln(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f)-
3/2*b^3/f*arctanh(d*x+c)^2*polylog(2,-(d*x+c+1)^2/(1-(d*x+c)^2))+3/2*b^3/f*arctanh(d*x+c)*polylog(3,-(d*x+c+1)
^2/(1-(d*x+c)^2))+3/2*I*a*b^2/f*arctanh(d*x+c)^2*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*(c*f*(1+(d*x+c+1
)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f))*csgn(I*(c*f*(1+(d*x+c+
1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f)/(1+(d*x+c+1)^2/(1-(d*x
+c)^2)))*Pi-3/4*d*b^3/f*e/(c*f-d*e-f)*polylog(4,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+3*a*b^2*c/
(c*f-d*e-f)*arctanh(d*x+c)^2*ln(1-(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+3*a*b^2*c/(c*f-d*e-f)*ar
ctanh(d*x+c)*polylog(2,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+I*b^3/f*arctanh(d*x+c)^3*csgn(I*(c*
f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f)/(1+(d*x+c+
1)^2/(1-(d*x+c)^2)))^2*Pi-1/2*I*b^3/f*arctanh(d*x+c)^3*csgn(I*(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2
/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^3*Pi-3*I*a*b^2/f*Pi*arct
anh(d*x+c)^2+3/2*d*a*b^2/f*e/(c*f-d*e-f)*polylog(3,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))-1/2*I*b
^3/f*arctanh(d*x+c)^3*csgn(I/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))*csgn(I*(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c
+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2*Pi-1/2*I*b^3/f*ar
ctanh(d*x+c)^3*csgn(I*(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-
(d*x+c)^2))*f))*csgn(I*(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1
-(d*x+c)^2))*f)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^2*Pi+3*I*a*b^2/f*arctanh(d*x+c)^2*csgn(I*(c*f*(1+(d*x+c+1)^2/(1
-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f)/(1+(d*x+c+1)^2/(1-(d*x+c)^2))
)^2*Pi-3/2*I*a*b^2/f*arctanh(d*x+c)^2*csgn(I*(c*f*(1+(d*x+c+1)^2/(1-(d*x+c)^2))+(-(d*x+c+1)^2/(1-(d*x+c)^2)-1)
*e*d+(1-(d*x+c+1)^2/(1-(d*x+c)^2))*f)/(1+(d*x+c+1)^2/(1-(d*x+c)^2)))^3*Pi-d*b^3/f*e/(c*f-d*e-f)*arctanh(d*x+c)
^3*ln(1-(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))-3/2*d*b^3/f*e/(c*f-d*e-f)*arctanh(d*x+c)^2*polylog
(2,(c*f-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))+3/2*d*b^3/f*e/(c*f-d*e-f)*arctanh(d*x+c)*polylog(3,(c*f
-d*e-f)*(d*x+c+1)^2/(1-(d*x+c)^2)/(-c*f+d*e-f))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {a^{3} \log \left (f x + e\right )}{f} + \int \frac {b^{3} {\left (\log \left (d x + c + 1\right ) - \log \left (-d x - c + 1\right )\right )}^{3}}{8 \, {\left (f x + e\right )}} + \frac {3 \, a b^{2} {\left (\log \left (d x + c + 1\right ) - \log \left (-d x - c + 1\right )\right )}^{2}}{4 \, {\left (f x + e\right )}} + \frac {3 \, a^{2} b {\left (\log \left (d x + c + 1\right ) - \log \left (-d x - c + 1\right )\right )}}{2 \, {\left (f x + e\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}^3}{e+f\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c + d x \right )}\right )^{3}}{e + f x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atanh(c + d*x))**3/(e + f*x), x)

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