∫ (a+b coth−1(c+dx))3 _______________________________

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

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

[Out]

-(a+b*arccoth(d*x+c))^3*ln(2/(d*x+c+1))/f+(a+b*arccoth(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*arccoth(d*x+c))^2*polylog(2,1-2/(d*x+c+1))/f-3/2*b*(a+b*arccoth(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*arccoth(d*x+c))*polylog(3,1-2/(d*x+c+1))/f-3/2*b^2*(a+b*arccoth(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 = {6112, 5925} \[ -\frac {3 b^2 \left (a+b \coth ^{-1}(c+d x)\right ) \text {PolyLog}\left (3,1-\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{2 f}+\frac {3 b^2 \text {PolyLog}\left (3,1-\frac {2}{c+d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )}{2 f}-\frac {3 b \left (a+b \coth ^{-1}(c+d x)\right )^2 \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{2 f}+\frac {3 b \text {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )^2}{2 f}-\frac {3 b^3 \text {PolyLog}\left (4,1-\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{4 f}+\frac {3 b^3 \text {PolyLog}\left (4,1-\frac {2}{c+d x+1}\right )}{4 f}+\frac {\left (a+b \coth ^{-1}(c+d x)\right )^3 \log \left (\frac {2 d (e+f x)}{(c+d x+1) (-c f+d e+f)}\right )}{f}-\frac {\log \left (\frac {2}{c+d x+1}\right ) \left (a+b \coth ^{-1}(c+d x)\right )^3}{f} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

*(a + b*ArcCoth[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*ArcCoth[c + d*x])*PolyLog[3, 1 - 2/(1 + c + d*x)])/(2*f) - (3*b^2*(a + b*ArcCoth[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 5925

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))^3/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcCoth[c*x])^3*Log[

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

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

+ c*x)])/(2*e), x] - Simp[(3*b^2*(a + b*ArcCoth[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 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]

Rubi steps

\begin {align*} \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (a+b \coth ^{-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 \coth ^{-1}(c+d x)\right )^3 \log \left (\frac {2}{1+c+d x}\right )}{f}+\frac {\left (a+b \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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 \coth ^{-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 = 28.57, size = 0, normalized size = 0.00 \[ \int \frac {\left (a+b \coth ^{-1}(c+d x)\right )^3}{e+f x} \, dx \]

Veriﬁcation is Not applicable to the result.

[In]

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

[Out]

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

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fricas [F] time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {arcoth}\left (d x + c\right )^{3} + 3 \, a b^{2} \operatorname {arcoth}\left (d x + c\right )^{2} + 3 \, a^{2} b \operatorname {arcoth}\left (d x + c\right ) + a^{3}}{f x + e}, x\right ) \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C] time = 1.55, size = 3796, normalized size = 12.32 \[ \text {output too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

6*a*b^2/f*polylog(3,-1/((d*x+c-1)/(d*x+c+1))^(1/2))+3/2*a*b^2/(c*f-d*e-f)*polylog(3,(c*f-d*e-f)*(d*x+c+1)/(d*x

+c-1)/(c*f-d*e+f))+3/4*b^3*c/(c*f-d*e-f)*polylog(4,(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))+6*a*b^2/f*poly

log(3,1/((d*x+c-1)/(d*x+c+1))^(1/2))+3/2*b^3/(c*f-d*e-f)*arccoth(d*x+c)*polylog(3,(c*f-d*e-f)*(d*x+c+1)/(d*x+c

-1)/(c*f-d*e+f))-3/2*b^3/(c*f-d*e-f)*arccoth(d*x+c)^2*polylog(2,(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))-3

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

1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)+b^3/f*arccoth(d*x+c)^3*ln((d*x+c+1)/(d*x+c-1

)-1)+6*b^3/f*arccoth(d*x+c)*polylog(3,-1/((d*x+c-1)/(d*x+c+1))^(1/2))-b^3/f*arccoth(d*x+c)^3*ln(1-1/((d*x+c-1)

/(d*x+c+1))^(1/2))-b^3/f*arccoth(d*x+c)^3*ln(1+1/((d*x+c-1)/(d*x+c+1))^(1/2))-3*b^3/f*arccoth(d*x+c)^2*polylog

(2,1/((d*x+c-1)/(d*x+c+1))^(1/2))+6*b^3/f*arccoth(d*x+c)*polylog(3,1/((d*x+c-1)/(d*x+c+1))^(1/2))+b^3*ln((d*x+

c)*f-c*f+d*e)/f*arccoth(d*x+c)^3-b^3/(c*f-d*e-f)*arccoth(d*x+c)^3*ln(1-(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(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))-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))-6*b^3/f*polylog(4,1/((d*x+c-1)/(d*x+c+1))^(1/2))-6*b^3/f*polylog(4,-1/((d*x+c-1)/(d*x+

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

(c*f-d*e-f)*arccoth(d*x+c)*polylog(3,(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))-3/2*d*b^3/f*e/(c*f-d*e-f)*ar

ccoth(d*x+c)^2*polylog(2,(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))-d*b^3/f*e/(c*f-d*e-f)*arccoth(d*x+c)^3*l

n(1-(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))+3/2*d*a*b^2/f*e/(c*f-d*e-f)*polylog(3,(c*f-d*e-f)*(d*x+c+1)/(

d*x+c-1)/(c*f-d*e+f))-1/2*I*b^3/f*Pi*arccoth(d*x+c)^3*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-

1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f))*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+

1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-1)-1))^2-1/2*I*b^3/f*Pi*arccoth(d*x+c)^3*csgn(I/((d*x+c+1)/(d*x+c-1)-1))*

csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-

1)-1))^2-3/2*b^3*c/(c*f-d*e-f)*arccoth(d*x+c)*polylog(3,(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))+3/2*b^3*c

/(c*f-d*e-f)*arccoth(d*x+c)^2*polylog(2,(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))+b^3*c/(c*f-d*e-f)*arccoth

(d*x+c)^3*ln(1-(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))+3*a*b^2*ln((d*x+c)*f-c*f+d*e)/f*arccoth(d*x+c)^2-I

*b^3/f*Pi*arccoth(d*x+c)^3-3/2*a*b^2*c/(c*f-d*e-f)*polylog(3,(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))-3*a*

b^2/f*arccoth(d*x+c)^2*ln(1+1/((d*x+c-1)/(d*x+c+1))^(1/2))-6*a*b^2/f*arccoth(d*x+c)*polylog(2,-1/((d*x+c-1)/(d

*x+c+1))^(1/2))-3*a*b^2/f*arccoth(d*x+c)^2*ln(1-1/((d*x+c-1)/(d*x+c+1))^(1/2))-6*a*b^2/f*arccoth(d*x+c)*polylo

g(2,1/((d*x+c-1)/(d*x+c+1))^(1/2))+3*a*b^2/f*arccoth(d*x+c)^2*ln((d*x+c+1)/(d*x+c-1)-1)-3*a*b^2/(c*f-d*e-f)*ar

ccoth(d*x+c)^2*ln(1-(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))-3*a*b^2/(c*f-d*e-f)*arccoth(d*x+c)*polylog(2,

(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))-3*a*b^2/f*arccoth(d*x+c)^2*ln(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x

+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)+3*a^2*b*ln((d*x+c)*f-c*f+d*e)/f*arccoth(d*x+c)-3*d*a*b^2/f*e/

(c*f-d*e-f)*arccoth(d*x+c)^2*ln(1-(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))-3*d*a*b^2/f*e/(c*f-d*e-f)*arcco

th(d*x+c)*polylog(2,(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))+1/2*I*b^3/f*Pi*arccoth(d*x+c)^3*csgn(I/((d*x+

c+1)/(d*x+c-1)-1))*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)

)*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+

c-1)-1))-3/2*I*a*b^2/f*Pi*arccoth(d*x+c)^2*csgn(I/((d*x+c+1)/(d*x+c-1)-1))*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f

+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-1)-1))^2-3/2*I*a*b^2/f*Pi*arccoth(d

*x+c)^2*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f))*csgn(I*((

(d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-1)-1))^2+

3/2*I*a*b^2/f*Pi*arccoth(d*x+c)^2*csgn(I/((d*x+c+1)/(d*x+c-1)-1))*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+

c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f))*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e

*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-1)-1))-3/2*I*a*b^2/f*Pi*arccoth(d*x+c)^2*csgn(I*(((d*x+c+1)/(

d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-1)-1))^3+3*I*a*b^2/f

*Pi*arccoth(d*x+c)^2*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*

f)/((d*x+c+1)/(d*x+c-1)-1))^2+I*b^3/f*Pi*arccoth(d*x+c)^3*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*

x+c-1))*e*d+(-(d*x+c+1)/(d*x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-1)-1))^2+3*a*b^2*c/(c*f-d*e-f)*arccoth(d*x+c)^2*ln(1

-(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))+3*a*b^2*c/(c*f-d*e-f)*arccoth(d*x+c)*polylog(2,(c*f-d*e-f)*(d*x+

c+1)/(d*x+c-1)/(c*f-d*e+f))-3/4*d*b^3/f*e/(c*f-d*e-f)*polylog(4,(c*f-d*e-f)*(d*x+c+1)/(d*x+c-1)/(c*f-d*e+f))-1

/2*I*b^3/f*Pi*arccoth(d*x+c)^3*csgn(I*(((d*x+c+1)/(d*x+c-1)-1)*c*f+(1-(d*x+c+1)/(d*x+c-1))*e*d+(-(d*x+c+1)/(d*

x+c-1)-1)*f)/((d*x+c+1)/(d*x+c-1)-1))^3-3*I*a*b^2/f*Pi*arccoth(d*x+c)^2

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

) + 1))/(f*x + e), x)

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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