3.1.69 \(\int \frac {(a+b \tanh ^{-1}(c x^2))^2}{x^3} \, dx\) [69]

Optimal. Leaf size=87 \[ \frac {1}{2} c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{2 x^2}+b c \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \log \left (2-\frac {2}{1+c x^2}\right )-\frac {1}{2} b^2 c \text {PolyLog}\left (2,-1+\frac {2}{1+c x^2}\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6039, 6037, 6135, 6079, 2497} \begin {gather*} \frac {1}{2} c \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2-\frac {\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2}{2 x^2}+b c \log \left (2-\frac {2}{c x^2+1}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {1}{2} b^2 c \text {Li}_2\left (\frac {2}{c x^2+1}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])^2/x^3,x]

[Out]

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

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 6039

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m
+ 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[S
implify[(m + 1)/n]]

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 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}\left (c x^2\right )\right )^2}{x^3} \, dx &=\int \left (\frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{4 x^3}-\frac {b \left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{2 x^3}+\frac {b^2 \log ^2\left (1+c x^2\right )}{4 x^3}\right ) \, dx\\ &=\frac {1}{4} \int \frac {\left (2 a-b \log \left (1-c x^2\right )\right )^2}{x^3} \, dx-\frac {1}{2} b \int \frac {\left (-2 a+b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{x^3} \, dx+\frac {1}{4} b^2 \int \frac {\log ^2\left (1+c x^2\right )}{x^3} \, dx\\ &=\frac {1}{8} \text {Subst}\left (\int \frac {(2 a-b \log (1-c x))^2}{x^2} \, dx,x,x^2\right )-\frac {1}{4} b \text {Subst}\left (\int \frac {(-2 a+b \log (1-c x)) \log (1+c x)}{x^2} \, dx,x,x^2\right )+\frac {1}{8} b^2 \text {Subst}\left (\int \frac {\log ^2(1+c x)}{x^2} \, dx,x,x^2\right )\\ &=-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}+\frac {1}{4} (b c) \text {Subst}\left (\int \frac {2 a-b \log (1-c x)}{x} \, dx,x,x^2\right )-\frac {1}{4} (b c) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{x (1+c x)} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x (1-c x)} \, dx,x,x^2\right )\\ &=a b c \log (x)-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}-\frac {1}{4} b^2 c \text {Li}_2\left (-c x^2\right )-\frac {1}{4} (b c) \text {Subst}\left (\int \left (\frac {-2 a+b \log (1-c x)}{x}-\frac {c (-2 a+b \log (1-c x))}{1+c x}\right ) \, dx,x,x^2\right )-\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \left (\frac {\log (1+c x)}{x}-\frac {c \log (1+c x)}{-1+c x}\right ) \, dx,x,x^2\right )\\ &=a b c \log (x)-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}-\frac {1}{4} b^2 c \text {Li}_2\left (-c x^2\right )+\frac {1}{4} b^2 c \text {Li}_2\left (c x^2\right )-\frac {1}{4} (b c) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{x} \, dx,x,x^2\right )+\frac {1}{4} \left (b c^2\right ) \text {Subst}\left (\int \frac {-2 a+b \log (1-c x)}{1+c x} \, dx,x,x^2\right )-\frac {1}{4} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log (1+c x)}{-1+c x} \, dx,x,x^2\right )\\ &=2 a b c \log (x)-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac {1}{4} b c \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )-\frac {1}{4} b^2 c \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}-\frac {1}{2} b^2 c \text {Li}_2\left (-c x^2\right )+\frac {1}{4} b^2 c \text {Li}_2\left (c x^2\right )-\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log (1-c x)}{x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1-c x)\right )}{1+c x} \, dx,x,x^2\right )+\frac {1}{4} \left (b^2 c^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {1}{2} (1+c x)\right )}{1-c x} \, dx,x,x^2\right )\\ &=2 a b c \log (x)-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac {1}{4} b c \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )-\frac {1}{4} b^2 c \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}-\frac {1}{2} b^2 c \text {Li}_2\left (-c x^2\right )+\frac {1}{2} b^2 c \text {Li}_2\left (c x^2\right )-\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1-c x^2\right )+\frac {1}{4} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\log \left (1-\frac {x}{2}\right )}{x} \, dx,x,1+c x^2\right )\\ &=2 a b c \log (x)-\frac {\left (1-c x^2\right ) \left (2 a-b \log \left (1-c x^2\right )\right )^2}{8 x^2}-\frac {1}{4} b c \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (\frac {1}{2} \left (1+c x^2\right )\right )-\frac {1}{4} b^2 c \log \left (\frac {1}{2} \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )-\frac {b \left (2 a-b \log \left (1-c x^2\right )\right ) \log \left (1+c x^2\right )}{4 x^2}-\frac {b^2 \left (1+c x^2\right ) \log ^2\left (1+c x^2\right )}{8 x^2}-\frac {1}{2} b^2 c \text {Li}_2\left (-c x^2\right )+\frac {1}{2} b^2 c \text {Li}_2\left (c x^2\right )+\frac {1}{4} b^2 c \text {Li}_2\left (\frac {1}{2} \left (1-c x^2\right )\right )-\frac {1}{4} b^2 c \text {Li}_2\left (\frac {1}{2} \left (1+c x^2\right )\right )\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.10, size = 119, normalized size = 1.37 \begin {gather*} -\frac {a^2}{2 x^2}+a b c \left (-\frac {\tanh ^{-1}\left (c x^2\right )}{c x^2}+\log \left (c x^2\right )-\frac {1}{2} \log \left (1-c^2 x^4\right )\right )+\frac {1}{2} b^2 c \left (\tanh ^{-1}\left (c x^2\right ) \left (\tanh ^{-1}\left (c x^2\right )-\frac {\tanh ^{-1}\left (c x^2\right )}{c x^2}+2 \log \left (1-e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right )-\text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c x^2\right )}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^2])^2/x^3,x]

[Out]

-1/2*a^2/x^2 + a*b*c*(-(ArcTanh[c*x^2]/(c*x^2)) + Log[c*x^2] - Log[1 - c^2*x^4]/2) + (b^2*c*(ArcTanh[c*x^2]*(A
rcTanh[c*x^2] - ArcTanh[c*x^2]/(c*x^2) + 2*Log[1 - E^(-2*ArcTanh[c*x^2])]) - PolyLog[2, E^(-2*ArcTanh[c*x^2])]
))/2

________________________________________________________________________________________

Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arctanh \left (c \,x^{2}\right )\right )^{2}}{x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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))^2/x^3,x, algorithm="maxima")

[Out]

-1/2*(c*(log(c^2*x^4 - 1) - log(x^4)) + 2*arctanh(c*x^2)/x^2)*a*b - 1/8*b^2*(log(-c*x^2 + 1)^2/x^2 + 2*integra
te(-((c*x^2 - 1)*log(c*x^2 + 1)^2 + 2*(c*x^2 - (c*x^2 - 1)*log(c*x^2 + 1))*log(-c*x^2 + 1))/(c*x^5 - x^3), x))
- 1/2*a^2/x^2

________________________________________________________________________________________

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))^2/x^3,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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))^2/x^3,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^2\right )\right )}^2}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^2))^2/x^3,x)

[Out]

int((a + b*atanh(c*x^2))^2/x^3, x)

________________________________________________________________________________________