3.1.22 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{x^4} \, dx\) [22]

Optimal. Leaf size=130 \[ -\frac {b^2 c^2}{3 x}+\frac {1}{3} b^2 c^3 \tanh ^{-1}(c x)-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}+\frac {1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}+\frac {2}{3} b c^3 \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )-\frac {1}{3} b^2 c^3 \text {PolyLog}\left (2,-1+\frac {2}{1+c x}\right ) \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6037, 6129, 331, 212, 6135, 6079, 2497} \begin {gather*} \frac {1}{3} c^3 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {2}{3} b c^3 \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{3 x^3}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right )}{3 x^2}-\frac {1}{3} b^2 c^3 \text {Li}_2\left (\frac {2}{c x+1}-1\right )+\frac {1}{3} b^2 c^3 \tanh ^{-1}(c x)-\frac {b^2 c^2}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/x^4,x]

[Out]

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

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

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Mathematica [A]
time = 0.25, size = 145, normalized size = 1.12 \begin {gather*} -\frac {a^2+a b c x+b^2 c^2 x^2+b^2 \left (1-c^3 x^3\right ) \tanh ^{-1}(c x)^2+b \tanh ^{-1}(c x) \left (2 a+b c x-b c^3 x^3-2 b c^3 x^3 \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )-2 a b c^3 x^3 \log (c x)+a b c^3 x^3 \log \left (1-c^2 x^2\right )+b^2 c^3 x^3 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )}{3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])^2/x^4,x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(304\) vs. \(2(116)=232\).
time = 0.09, size = 305, normalized size = 2.35

method	result	size
derivativedivides	\(c^{3} \left (-\frac {a^{2}}{3 c^{3} x^{3}}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{3 c^{3} x^{3}}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{3}-\frac {b^{2} \arctanh \left (c x \right )}{3 c^{2} x^{2}}+\frac {2 b^{2} \ln \left (c x \right ) \arctanh \left (c x \right )}{3}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{3}-\frac {b^{2} \ln \left (c x -1\right )}{6}+\frac {b^{2} \ln \left (c x +1\right )}{6}-\frac {b^{2}}{3 c x}+\frac {b^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{3}+\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{6}-\frac {b^{2} \ln \left (c x -1\right )^{2}}{12}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{6}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{6}+\frac {b^{2} \ln \left (c x +1\right )^{2}}{12}-\frac {b^{2} \dilog \left (c x \right )}{3}-\frac {b^{2} \dilog \left (c x +1\right )}{3}-\frac {b^{2} \ln \left (c x \right ) \ln \left (c x +1\right )}{3}-\frac {2 a b \arctanh \left (c x \right )}{3 c^{3} x^{3}}-\frac {a b \ln \left (c x -1\right )}{3}-\frac {a b}{3 c^{2} x^{2}}+\frac {2 a b \ln \left (c x \right )}{3}-\frac {a b \ln \left (c x +1\right )}{3}\right )\)	\(305\)
default	\(c^{3} \left (-\frac {a^{2}}{3 c^{3} x^{3}}-\frac {b^{2} \arctanh \left (c x \right )^{2}}{3 c^{3} x^{3}}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{3}-\frac {b^{2} \arctanh \left (c x \right )}{3 c^{2} x^{2}}+\frac {2 b^{2} \ln \left (c x \right ) \arctanh \left (c x \right )}{3}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{3}-\frac {b^{2} \ln \left (c x -1\right )}{6}+\frac {b^{2} \ln \left (c x +1\right )}{6}-\frac {b^{2}}{3 c x}+\frac {b^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{3}+\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{6}-\frac {b^{2} \ln \left (c x -1\right )^{2}}{12}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{6}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{6}+\frac {b^{2} \ln \left (c x +1\right )^{2}}{12}-\frac {b^{2} \dilog \left (c x \right )}{3}-\frac {b^{2} \dilog \left (c x +1\right )}{3}-\frac {b^{2} \ln \left (c x \right ) \ln \left (c x +1\right )}{3}-\frac {2 a b \arctanh \left (c x \right )}{3 c^{3} x^{3}}-\frac {a b \ln \left (c x -1\right )}{3}-\frac {a b}{3 c^{2} x^{2}}+\frac {2 a b \ln \left (c x \right )}{3}-\frac {a b \ln \left (c x +1\right )}{3}\right )\)	\(305\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))^2/x^4,x,method=_RETURNVERBOSE)

[Out]

c^3*(-1/3*a^2/c^3/x^3-1/3*b^2/c^3/x^3*arctanh(c*x)^2-1/3*b^2*arctanh(c*x)*ln(c*x-1)-1/3*b^2*arctanh(c*x)/c^2/x
^2+2/3*b^2*ln(c*x)*arctanh(c*x)-1/3*b^2*arctanh(c*x)*ln(c*x+1)-1/6*b^2*ln(c*x-1)+1/6*b^2*ln(c*x+1)-1/3*b^2/c/x
+1/3*b^2*dilog(1/2*c*x+1/2)+1/6*b^2*ln(c*x-1)*ln(1/2*c*x+1/2)-1/12*b^2*ln(c*x-1)^2-1/6*b^2*ln(-1/2*c*x+1/2)*ln
(c*x+1)+1/6*b^2*ln(-1/2*c*x+1/2)*ln(1/2*c*x+1/2)+1/12*b^2*ln(c*x+1)^2-1/3*b^2*dilog(c*x)-1/3*b^2*dilog(c*x+1)-
1/3*b^2*ln(c*x)*ln(c*x+1)-2/3*a*b/c^3/x^3*arctanh(c*x)-1/3*a*b*ln(c*x-1)-1/3*a*b/c^2/x^2+2/3*a*b*ln(c*x)-1/3*a
*b*ln(c*x+1))

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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))^2/x^4,x, algorithm="maxima")

[Out]

-1/3*((c^2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c + 2*arctanh(c*x)/x^3)*a*b - 1/12*b^2*(log(-c*x + 1)^2/x^
3 + 3*integrate(-1/3*(3*(c*x - 1)*log(c*x + 1)^2 + 2*(c*x - 3*(c*x - 1)*log(c*x + 1))*log(-c*x + 1))/(c*x^5 -
x^4), x)) - 1/3*a^2/x^3

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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))^2/x^4,x, algorithm="fricas")

[Out]

integral((b^2*arctanh(c*x)^2 + 2*a*b*arctanh(c*x) + a^2)/x^4, 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 )^{2}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))**2/x**4,x)

[Out]

Integral((a + b*atanh(c*x))**2/x**4, 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))^2/x^4,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)^2/x^4, 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 )}^2}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^2/x^4,x)

[Out]

int((a + b*atanh(c*x))^2/x^4, x)

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