3.1.68 \(\int \frac {(a+b \tanh ^{-1}(c x^2))^2}{x} \, dx\) [68]

Optimal. Leaf size=137 \[ \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {1}{2} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {PolyLog}\left (2,1-\frac {2}{1-c x^2}\right )+\frac {1}{2} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {PolyLog}\left (2,-1+\frac {2}{1-c x^2}\right )+\frac {1}{4} b^2 \text {PolyLog}\left (3,1-\frac {2}{1-c x^2}\right )-\frac {1}{4} b^2 \text {PolyLog}\left (3,-1+\frac {2}{1-c x^2}\right ) \]

[Out]

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

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Rubi [A]
time = 0.23, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6035, 6033, 6199, 6095, 6205, 6745} \begin {gather*} -\frac {1}{2} b \text {Li}_2\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\frac {1}{2} b \text {Li}_2\left (\frac {2}{1-c x^2}-1\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )+\tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right ) \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2+\frac {1}{4} b^2 \text {Li}_3\left (1-\frac {2}{1-c x^2}\right )-\frac {1}{4} b^2 \text {Li}_3\left (\frac {2}{1-c x^2}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])^2/x,x]

[Out]

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

Rule 6033

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 6035

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 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 6199

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 6205

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 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}\left (c x^2\right )\right )^2}{x} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{x} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-(2 b c) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \tanh ^{-1}\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )+(b c) \text {Subst}\left (\int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )-(b c) \text {Subst}\left (\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,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {1}{2} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )+\frac {1}{2} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (-1+\frac {2}{1-c x^2}\right )+\frac {1}{2} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (1-\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )-\frac {1}{2} \left (b^2 c\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (-1+\frac {2}{1-c x}\right )}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-\frac {1}{2} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (1-\frac {2}{1-c x^2}\right )+\frac {1}{2} b \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \text {Li}_2\left (-1+\frac {2}{1-c x^2}\right )+\frac {1}{4} b^2 \text {Li}_3\left (1-\frac {2}{1-c x^2}\right )-\frac {1}{4} b^2 \text {Li}_3\left (-1+\frac {2}{1-c x^2}\right )\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 183, normalized size = 1.34 \begin {gather*} \frac {1}{2} \left (2 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )^2 \tanh ^{-1}\left (1-\frac {2}{1-c x^2}\right )-4 b c \left (\frac {1}{2} \left (\frac {\left (-a-b \tanh ^{-1}\left (c x^2\right )\right ) \text {PolyLog}\left (2,\frac {-1-c x^2}{-1+c x^2}\right )}{2 c}+\frac {b \text {PolyLog}\left (3,\frac {-1-c x^2}{-1+c x^2}\right )}{4 c}\right )+\frac {1}{2} \left (-\frac {\left (-a-b \tanh ^{-1}\left (c x^2\right )\right ) \text {PolyLog}\left (2,\frac {1+c x^2}{-1+c x^2}\right )}{2 c}-\frac {b \text {PolyLog}\left (3,\frac {1+c x^2}{-1+c x^2}\right )}{4 c}\right )\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))^2/x,x, algorithm="maxima")

[Out]

a^2*log(x) + integrate(1/4*b^2*(log(c*x^2 + 1) - log(-c*x^2 + 1))^2/x + a*b*(log(c*x^2 + 1) - log(-c*x^2 + 1))
/x, 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))^2/x,x, algorithm="fricas")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^2))^2/x,x)

[Out]

int((a + b*atanh(c*x^2))^2/x, x)

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