3.1.34 \(\int \frac {(a+b \tanh ^{-1}(c x))^3}{x^5} \, dx\) [34]

Optimal. Leaf size=187 \[ -\frac {b^3 c^3}{4 x}+\frac {1}{4} b^3 c^4 \tanh ^{-1}(c x)-\frac {b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^2}+b c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^3}-\frac {3 b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x}+\frac {1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^3-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}+2 b^2 c^4 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-b^3 c^4 \text {PolyLog}\left (2,-1+\frac {2}{1+c x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.44, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6037, 6129, 331, 212, 6135, 6079, 2497, 6095} \begin {gather*} 2 b^2 c^4 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {b^2 c^2 \left (a+b \tanh ^{-1}(c x)\right )}{4 x^2}+\frac {1}{4} c^4 \left (a+b \tanh ^{-1}(c x)\right )^3+b c^4 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {3 b c^3 \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^3}{4 x^4}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )^2}{4 x^3}-b^3 c^4 \text {Li}_2\left (\frac {2}{c x+1}-1\right )+\frac {1}{4} b^3 c^4 \tanh ^{-1}(c x)-\frac {b^3 c^3}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^3/x^5,x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 6037

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

Rule 6079

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Simp[(a + b*ArcTanh[c*x
])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Dist[b*c*(p/d), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 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 6095

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 6129

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), Int[(f*x)^(m + 2)*((a + b*ArcTanh[c*x])^p/(d +
e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rule 6135

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

Rubi steps

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

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Mathematica [A]
time = 0.47, size = 295, normalized size = 1.58 \begin {gather*} -\frac {2 a^3+2 a^2 b c x+2 a b^2 c^2 x^2+6 a^2 b c^3 x^3+2 b^3 c^3 x^3-2 a b^2 c^4 x^4+2 b^2 \left (b c x \left (1+3 c^2 x^2-4 c^3 x^3\right )+a \left (3-3 c^4 x^4\right )\right ) \tanh ^{-1}(c x)^2-2 b^3 \left (-1+c^4 x^4\right ) \tanh ^{-1}(c x)^3+2 b \tanh ^{-1}(c x) \left (3 a^2+b^2 c^2 x^2 \left (1-c^2 x^2\right )+2 a b c x \left (1+3 c^2 x^2\right )-8 b^2 c^4 x^4 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )+3 a^2 b c^4 x^4 \log (1-c x)-3 a^2 b c^4 x^4 \log (1+c x)-16 a b^2 c^4 x^4 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+8 b^3 c^4 x^4 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )}{8 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])^3/x^5,x]

[Out]

-1/8*(2*a^3 + 2*a^2*b*c*x + 2*a*b^2*c^2*x^2 + 6*a^2*b*c^3*x^3 + 2*b^3*c^3*x^3 - 2*a*b^2*c^4*x^4 + 2*b^2*(b*c*x
*(1 + 3*c^2*x^2 - 4*c^3*x^3) + a*(3 - 3*c^4*x^4))*ArcTanh[c*x]^2 - 2*b^3*(-1 + c^4*x^4)*ArcTanh[c*x]^3 + 2*b*A
rcTanh[c*x]*(3*a^2 + b^2*c^2*x^2*(1 - c^2*x^2) + 2*a*b*c*x*(1 + 3*c^2*x^2) - 8*b^2*c^4*x^4*Log[1 - E^(-2*ArcTa
nh[c*x])]) + 3*a^2*b*c^4*x^4*Log[1 - c*x] - 3*a^2*b*c^4*x^4*Log[1 + c*x] - 16*a*b^2*c^4*x^4*Log[(c*x)/Sqrt[1 -
c^2*x^2]] + 8*b^3*c^4*x^4*PolyLog[2, E^(-2*ArcTanh[c*x])])/x^4

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.86, size = 1204, normalized size = 6.44

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(1204\)
default	\(\text {Expression too large to display}\)	\(1204\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^3/x^5,x,method=_RETURNVERBOSE)

[Out]

c^4*(3/16*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x
+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))-3/16*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))*
csgn(I*(c*x+1)^2/(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2-3/16*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)/(-c^2*x
^2+1)^(1/2))^2*csgn(I*(c*x+1)^2/(c^2*x^2-1))-3/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)/(-c^2*x^2+1)^(1/2))*cs
gn(I*(c*x+1)^2/(c^2*x^2-1))^2+3/16*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))*csgn(I*(c*x+1)^2/(c^2
*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^2-3/2*a*b^2*arctanh(c*x)/c/x-1/4*a^3/c^4/x^4-1/4*b^3/(c*x+1-(-c^2*x^2+1)^(
1/2))*(-c^2*x^2+1)^(1/2)+1/4*b^3/((-c^2*x^2+1)^(1/2)+c*x+1)*(-c^2*x^2+1)^(1/2)-3/8*a^2*b*ln(c*x-1)+3/8*a^2*b*l
n(c*x+1)-3/4*b^3*arctanh(c*x)^2*ln((c*x+1)/(-c^2*x^2+1)^(1/2))-3/8*b^3*arctanh(c*x)^2*ln(c*x-1)+3/8*b^3*arctan
h(c*x)^2*ln(c*x+1)-3/16*a*b^2*ln(c*x-1)^2-3/16*a*b^2*ln(c*x+1)^2-a*b^2*ln(c*x-1)-a*b^2*ln(c*x+1)+3/4*a*b^2*arc
tanh(c*x)*ln(c*x+1)+3/8*a*b^2*ln(c*x-1)*ln(1/2*c*x+1/2)+3/8*a*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)-3/8*a*b^2*ln(-1/2
*c*x+1/2)*ln(1/2*c*x+1/2)-3/4*a*b^2*arctanh(c*x)*ln(c*x-1)-2*b^3*dilog((c*x+1)/(-c^2*x^2+1)^(1/2))+2*b^3*dilog
(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-3/4*a^2*b/c/x-3/4*b^3*arctanh(c*x)^2/c/x+1/4*b^3*arctanh(c*x)-b^3*arctanh(c*x)^
2+1/4*b^3*arctanh(c*x)^3+2*a*b^2*ln(c*x)+2*b^3*arctanh(c*x)*ln(1+(c*x+1)/(-c^2*x^2+1)^(1/2))-3/16*I*b^3*arctan
h(c*x)^2*Pi*csgn(I*(c*x+1)^2/(c^2*x^2-1))^3-3/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))^2+3
/8*I*b^3*arctanh(c*x)^2*Pi*csgn(I/(1+(c*x+1)^2/(-c^2*x^2+1)))^3-3/16*I*b^3*arctanh(c*x)^2*Pi*csgn(I*(c*x+1)^2/
(c^2*x^2-1)/(1+(c*x+1)^2/(-c^2*x^2+1)))^3-3/4*a^2*b/c^4/x^4*arctanh(c*x)-3/4*a*b^2/c^4/x^4*arctanh(c*x)^2-1/2*
a*b^2*arctanh(c*x)/c^3/x^3-1/4*b^3/c^3/x^3*arctanh(c*x)^2-1/4*b^3*arctanh(c*x)/c^2/x^2-1/4*a*b^2/c^2/x^2-1/4*a
^2*b/c^3/x^3+3/8*I*b^3*arctanh(c*x)^2*Pi-1/4*b^3/c^4/x^4*arctanh(c*x)^3)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3/x^5,x, algorithm="maxima")

[Out]

1/8*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*a^2*b + 1/16*((
32*c^2*log(x) - (3*c^2*x^2*log(c*x + 1)^2 + 3*c^2*x^2*log(c*x - 1)^2 + 16*c^2*x^2*log(c*x - 1) - 2*(3*c^2*x^2*
log(c*x - 1) - 8*c^2*x^2)*log(c*x + 1) + 4)/x^2)*c^2 + 4*(3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x
^2 + 1)/x^3)*c*arctanh(c*x))*a*b^2 - 1/32*b^3*(((c^4*x^4 - 1)*log(-c*x + 1)^3 + (6*c^3*x^3 + 2*c*x - 3*(c^4*x^
4 - 1)*log(c*x + 1))*log(-c*x + 1)^2)/x^4 + 4*integrate(-1/2*(2*(c*x - 1)*log(c*x + 1)^3 + (6*c^4*x^4 + 2*c^2*
x^2 - 6*(c*x - 1)*log(c*x + 1)^2 - 3*(c^5*x^5 - c*x)*log(c*x + 1))*log(-c*x + 1))/(c*x^6 - x^5), x)) - 3/4*a*b
^2*arctanh(c*x)^2/x^4 - 1/4*a^3/x^4

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3/x^5,x, algorithm="fricas")

[Out]

integral((b^3*arctanh(c*x)^3 + 3*a*b^2*arctanh(c*x)^2 + 3*a^2*b*arctanh(c*x) + a^3)/x^5, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{3}}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**3/x**5,x)

[Out]

Integral((a + b*atanh(c*x))**3/x**5, x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))^3/x^5,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^3/x^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^3}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^3/x^5,x)

[Out]

int((a + b*atanh(c*x))^3/x^5, x)

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