3.1.100 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{x^2 (d+c d x)} \, dx\) [100]

Optimal. Leaf size=162 \[ \frac {c \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac {2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {c \left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (2-\frac {2}{1+c x}\right )}{d}-\frac {b^2 c \text {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )}{d}+\frac {b c \left (a+b \tanh ^{-1}(c x)\right ) \text {PolyLog}\left (2,-1+\frac {2}{1+c x}\right )}{d}+\frac {b^2 c \text {PolyLog}\left (3,-1+\frac {2}{1+c x}\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.28, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {6081, 6037, 6135, 6079, 2497, 6095, 6203, 6745} \begin {gather*} \frac {b c \text {Li}_2\left (\frac {2}{c x+1}-1\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}+\frac {c \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{d x}+\frac {2 b c \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d}-\frac {c \log \left (2-\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{d}-\frac {b^2 c \text {Li}_2\left (\frac {2}{c x+1}-1\right )}{d}+\frac {b^2 c \text {Li}_3\left (\frac {2}{c x+1}-1\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])^2/(x^2*(d + c*d*x)),x]

[Out]

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

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 6081

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

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 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]

Rule 6203

Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTan
h[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 6745

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

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Mathematica [C] Result contains complex when optimal does not.
time = 0.37, size = 225, normalized size = 1.39 \begin {gather*} \frac {-\frac {a^2}{x}-a^2 c \log (x)+a^2 c \log (1+c x)+\frac {a b \left (-2 \tanh ^{-1}(c x) \left (1+c x \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )\right )+2 c x \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+c x \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )\right )}{x}+b^2 c \left (-\frac {i \pi ^3}{24}+\tanh ^{-1}(c x)^2-\frac {\tanh ^{-1}(c x)^2}{c x}+\frac {2}{3} \tanh ^{-1}(c x)^3+2 \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )-\tanh ^{-1}(c x)^2 \log \left (1-e^{2 \tanh ^{-1}(c x)}\right )-\text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )-\tanh ^{-1}(c x) \text {PolyLog}\left (2,e^{2 \tanh ^{-1}(c x)}\right )+\frac {1}{2} \text {PolyLog}\left (3,e^{2 \tanh ^{-1}(c x)}\right )\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^2/x) - a^2*c*Log[x] + a^2*c*Log[1 + c*x] + (a*b*(-2*ArcTanh[c*x]*(1 + c*x*Log[1 - E^(-2*ArcTanh[c*x])]) +
2*c*x*Log[(c*x)/Sqrt[1 - c^2*x^2]] + c*x*PolyLog[2, E^(-2*ArcTanh[c*x])]))/x + b^2*c*((-1/24*I)*Pi^3 + ArcTan
h[c*x]^2 - ArcTanh[c*x]^2/(c*x) + (2*ArcTanh[c*x]^3)/3 + 2*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] - ArcTanh
[c*x]^2*Log[1 - E^(2*ArcTanh[c*x])] - PolyLog[2, E^(-2*ArcTanh[c*x])] - ArcTanh[c*x]*PolyLog[2, E^(2*ArcTanh[c
*x])] + PolyLog[3, E^(2*ArcTanh[c*x])]/2))/d

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 10.39, size = 7139, normalized size = 44.07

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(7139\)
default	\(\text {Expression too large to display}\)	\(7139\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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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^2/(c*d*x+d),x, algorithm="maxima")

[Out]

a^2*(c*log(c*x + 1)/d - c*log(x)/d - 1/(d*x)) + 1/4*(b^2*c*x*log(c*x + 1) - b^2)*log(-c*x + 1)^2/(d*x) - integ
rate(-1/4*((b^2*c*x - b^2)*log(c*x + 1)^2 + 4*(a*b*c*x - a*b)*log(c*x + 1) + 2*(b^2*c^2*x^2 + 2*a*b - (2*a*b*c
- b^2*c)*x - (b^2*c^3*x^3 + b^2*c^2*x^2 + b^2*c*x - b^2)*log(c*x + 1))*log(-c*x + 1))/(c^2*d*x^4 - d*x^2), x)

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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^2/(c*d*x+d),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {a^{2}}{c x^{3} + x^{2}}\, dx + \int \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{c x^{3} + x^{2}}\, dx + \int \frac {2 a b \operatorname {atanh}{\left (c x \right )}}{c x^{3} + x^{2}}\, dx}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a**2/(c*x**3 + x**2), x) + Integral(b**2*atanh(c*x)**2/(c*x**3 + x**2), x) + Integral(2*a*b*atanh(c*
x)/(c*x**3 + x**2), x))/d

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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^2/(c*d*x+d),x, algorithm="giac")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))^2/(x^2*(d + c*d*x)),x)

[Out]

int((a + b*atanh(c*x))^2/(x^2*(d + c*d*x)), x)

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