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

Optimal. Leaf size=224 \[ 3 b^2 \text {Li}_3\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-3 b^2 \text {Li}_3\left (\frac {2}{1-c \sqrt {x}}-1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-3 b \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+3 b \text {Li}_2\left (\frac {2}{1-c \sqrt {x}}-1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-\frac {3}{2} b^3 \text {Li}_4\left (1-\frac {2}{1-c \sqrt {x}}\right )+\frac {3}{2} b^3 \text {Li}_4\left (\frac {2}{1-c \sqrt {x}}-1\right ) \]

[Out]

-4*(a+b*arctanh(c*x^(1/2)))^3*arctanh(-1+2/(1-c*x^(1/2)))-3*b*(a+b*arctanh(c*x^(1/2)))^2*polylog(2,1-2/(1-c*x^
(1/2)))+3*b*(a+b*arctanh(c*x^(1/2)))^2*polylog(2,-1+2/(1-c*x^(1/2)))+3*b^2*(a+b*arctanh(c*x^(1/2)))*polylog(3,
1-2/(1-c*x^(1/2)))-3*b^2*(a+b*arctanh(c*x^(1/2)))*polylog(3,-1+2/(1-c*x^(1/2)))-3/2*b^3*polylog(4,1-2/(1-c*x^(
1/2)))+3/2*b^3*polylog(4,-1+2/(1-c*x^(1/2)))

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Rubi [A]  time = 0.51, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {6095, 5914, 6052, 5948, 6058, 6062, 6610} \[ 3 b^2 \text {PolyLog}\left (3,1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-3 b^2 \text {PolyLog}\left (3,\frac {2}{1-c \sqrt {x}}-1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-3 b \text {PolyLog}\left (2,1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+3 b \text {PolyLog}\left (2,\frac {2}{1-c \sqrt {x}}-1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2-\frac {3}{2} b^3 \text {PolyLog}\left (4,1-\frac {2}{1-c \sqrt {x}}\right )+\frac {3}{2} b^3 \text {PolyLog}\left (4,\frac {2}{1-c \sqrt {x}}-1\right )+4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])^3/x,x]

[Out]

4*ArcTanh[1 - 2/(1 - c*Sqrt[x])]*(a + b*ArcTanh[c*Sqrt[x]])^3 - 3*b*(a + b*ArcTanh[c*Sqrt[x]])^2*PolyLog[2, 1
- 2/(1 - c*Sqrt[x])] + 3*b*(a + b*ArcTanh[c*Sqrt[x]])^2*PolyLog[2, -1 + 2/(1 - c*Sqrt[x])] + 3*b^2*(a + b*ArcT
anh[c*Sqrt[x]])*PolyLog[3, 1 - 2/(1 - c*Sqrt[x])] - 3*b^2*(a + b*ArcTanh[c*Sqrt[x]])*PolyLog[3, -1 + 2/(1 - c*
Sqrt[x])] - (3*b^3*PolyLog[4, 1 - 2/(1 - c*Sqrt[x])])/2 + (3*b^3*PolyLog[4, -1 + 2/(1 - c*Sqrt[x])])/2

Rule 5914

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

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[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 6052

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

Rule 6058

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[((a + b*ArcT
anh[c*x])^p*PolyLog[2, 1 - u])/(2*c*d), x] + Dist[(b*p)/2, Int[((a + b*ArcTanh[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 6062

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

Rule 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Dist[1/n, Subst[Int[(a + b*ArcTanh[c*x])
^p/x, x], x, x^n], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p, 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 \frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3}{x} \, dx &=2 \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{x} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-(12 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3+(6 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )-(6 b c) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )+3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c \sqrt {x}}\right )+\left (6 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )-\left (6 b^2 c\right ) \operatorname {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )+3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c \sqrt {x}}\right )+3 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_3\left (1-\frac {2}{1-c \sqrt {x}}\right )-3 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_3\left (-1+\frac {2}{1-c \sqrt {x}}\right )-\left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )+\left (3 b^3 c\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_3\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )\\ &=4 \tanh ^{-1}\left (1-\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3-3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )+3 b \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2 \text {Li}_2\left (-1+\frac {2}{1-c \sqrt {x}}\right )+3 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_3\left (1-\frac {2}{1-c \sqrt {x}}\right )-3 b^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \text {Li}_3\left (-1+\frac {2}{1-c \sqrt {x}}\right )-\frac {3}{2} b^3 \text {Li}_4\left (1-\frac {2}{1-c \sqrt {x}}\right )+\frac {3}{2} b^3 \text {Li}_4\left (-1+\frac {2}{1-c \sqrt {x}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 248, normalized size = 1.11 \[ \frac {3}{2} b \left (2 \text {Li}_2\left (\frac {\sqrt {x} c+1}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2-2 \text {Li}_2\left (\frac {\sqrt {x} c+1}{c \sqrt {x}-1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2+b \left (-2 \text {Li}_3\left (\frac {\sqrt {x} c+1}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )+2 \text {Li}_3\left (\frac {\sqrt {x} c+1}{c \sqrt {x}-1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )+b \left (\text {Li}_4\left (\frac {\sqrt {x} c+1}{1-c \sqrt {x}}\right )-\text {Li}_4\left (\frac {\sqrt {x} c+1}{c \sqrt {x}-1}\right )\right )\right )\right )+4 \tanh ^{-1}\left (\frac {2}{c \sqrt {x}-1}+1\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.22, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b^{3} \operatorname {artanh}\left (c \sqrt {x}\right )^{3} + 3 \, a b^{2} \operatorname {artanh}\left (c \sqrt {x}\right )^{2} + 3 \, a^{2} b \operatorname {artanh}\left (c \sqrt {x}\right ) + a^{3}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))^3/x,x, algorithm="fricas")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))^3/x,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.18, size = 1542, normalized size = 6.88 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))^3/x,x)

[Out]

I*b^3*Pi*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*arctanh(c*x^(1/2))^3+6*a*b^2*
ln(c*x^(1/2))*arctanh(c*x^(1/2))^2-6*a*b^2*arctanh(c*x^(1/2))*polylog(2,-(1+c*x^(1/2))^2/(-c^2*x+1))-6*a*b^2*a
rctanh(c*x^(1/2))^2*ln((1+c*x^(1/2))^2/(-c^2*x+1)-1)+6*a^2*b*ln(c*x^(1/2))*arctanh(c*x^(1/2))-3*a^2*b*ln(c*x^(
1/2))*ln(1+c*x^(1/2))+6*a*b^2*arctanh(c*x^(1/2))^2*ln(1-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+12*a*b^2*arctanh(c*x^(
1/2))*polylog(2,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+6*a*b^2*arctanh(c*x^(1/2))^2*ln(1+(1+c*x^(1/2))/(-c^2*x+1)^(1/
2))+12*a*b^2*arctanh(c*x^(1/2))*polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-3/2*b^3*polylog(4,-(1+c*x^(1/2))^2/
(-c^2*x+1))+12*b^3*polylog(4,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+12*b^3*polylog(4,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))
+2*a^3*ln(c*x^(1/2))-I*b^3*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1)/(1+
(1+c*x^(1/2))^2/(-c^2*x+1)))^2*arctanh(c*x^(1/2))^3-I*b^3*Pi*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1))*csgn(I*((1
+c*x^(1/2))^2/(-c^2*x+1)-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*arctanh(c*x^(1/2))^3+3*I*a*b^2*Pi*csgn(I*((1+c*x
^(1/2))^2/(-c^2*x+1)-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^3*arctanh(c*x^(1/2))^2+3*a*b^2*polylog(3,-(1+c*x^(1/2)
)^2/(-c^2*x+1))-12*a*b^2*polylog(3,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-12*a*b^2*polylog(3,-(1+c*x^(1/2))/(-c^2*x+1
)^(1/2))-3*a^2*b*dilog(c*x^(1/2))-3*a^2*b*dilog(1+c*x^(1/2))+2*b^3*arctanh(c*x^(1/2))^3*ln(1+(1+c*x^(1/2))/(-c
^2*x+1)^(1/2))+6*b^3*arctanh(c*x^(1/2))^2*polylog(2,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-12*b^3*arctanh(c*x^(1/2))
*polylog(3,-(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+2*b^3*ln(c*x^(1/2))*arctanh(c*x^(1/2))^3-2*b^3*arctanh(c*x^(1/2))^
3*ln((1+c*x^(1/2))^2/(-c^2*x+1)-1)-3*b^3*arctanh(c*x^(1/2))^2*polylog(2,-(1+c*x^(1/2))^2/(-c^2*x+1))+3*b^3*arc
tanh(c*x^(1/2))*polylog(3,-(1+c*x^(1/2))^2/(-c^2*x+1))+2*b^3*arctanh(c*x^(1/2))^3*ln(1-(1+c*x^(1/2))/(-c^2*x+1
)^(1/2))+6*b^3*arctanh(c*x^(1/2))^2*polylog(2,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))-12*b^3*arctanh(c*x^(1/2))*polylo
g(3,(1+c*x^(1/2))/(-c^2*x+1)^(1/2))+3*I*a*b^2*Pi*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1))*csgn(I/(1+(1+c*x^(1/2)
)^2/(-c^2*x+1)))*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*arctanh(c*x^(1/2))^2-3*
I*a*b^2*Pi*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1)/(1+(1+c*x^(1/2))^2/(-c
^2*x+1)))^2*arctanh(c*x^(1/2))^2-3*I*a*b^2*Pi*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1))*csgn(I*((1+c*x^(1/2))^2/(
-c^2*x+1)-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))^2*arctanh(c*x^(1/2))^2+I*b^3*Pi*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1
)-1))*csgn(I/(1+(1+c*x^(1/2))^2/(-c^2*x+1)))*csgn(I*((1+c*x^(1/2))^2/(-c^2*x+1)-1)/(1+(1+c*x^(1/2))^2/(-c^2*x+
1)))*arctanh(c*x^(1/2))^3

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))^3/x,x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^(1/2)))^3/x,x)

[Out]

int((a + b*atanh(c*x^(1/2)))^3/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*atanh(c*sqrt(x)))**3/x, x)

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