3.2.74 \(\int \frac {(a+b \tanh ^{-1}(\frac {c}{x^2}))^2}{x^3} \, dx\) [174]

Optimal. Leaf size=99 \[ -\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{2 c}-\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{2 x^2}+\frac {b \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right ) \log \left (\frac {2}{1-\frac {c}{x^2}}\right )}{c}+\frac {b^2 \text {PolyLog}\left (2,1-\frac {2}{1-\frac {c}{x^2}}\right )}{2 c} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6039, 6021, 6131, 6055, 2449, 2352} \begin {gather*} -\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{2 c}-\frac {\left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )^2}{2 x^2}+\frac {b \log \left (\frac {2}{1-\frac {c}{x^2}}\right ) \left (a+b \coth ^{-1}\left (\frac {x^2}{c}\right )\right )}{c}+\frac {b^2 \text {Li}_2\left (1-\frac {2}{1-\frac {c}{x^2}}\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c/x^2])^2/x^3,x]

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 6021

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x^n])^p, x] - Dist[b
*c*n*p, Int[x^n*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x], x] /; FreeQ[{a, b, c, n}, x] && IGtQ[p
, 0] && (EqQ[n, 1] || EqQ[p, 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 6055

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^p)
*(Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[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 6131

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

Rubi steps

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

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Mathematica [A]
time = 0.12, size = 114, normalized size = 1.15 \begin {gather*} -\frac {a^2}{2 x^2}-\frac {a b \left (\frac {c \tanh ^{-1}\left (\frac {c}{x^2}\right )}{x^2}-\log \left (\frac {1}{\sqrt {1-\frac {c^2}{x^4}}}\right )\right )}{c}-\frac {b^2 \left (\tanh ^{-1}\left (\frac {c}{x^2}\right ) \left (-\tanh ^{-1}\left (\frac {c}{x^2}\right )+\frac {c \tanh ^{-1}\left (\frac {c}{x^2}\right )}{x^2}-2 \log \left (1+e^{-2 \tanh ^{-1}\left (\frac {c}{x^2}\right )}\right )\right )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (\frac {c}{x^2}\right )}\right )\right )}{2 c} \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])/x^2 - Log[1/Sqrt[1 - c^2/x^4]]))/c - (b^2*(ArcTanh[c/x^2]*(-ArcTanh[c/
x^2] + (c*ArcTanh[c/x^2])/x^2 - 2*Log[1 + E^(-2*ArcTanh[c/x^2])]) + PolyLog[2, -E^(-2*ArcTanh[c/x^2])]))/(2*c)

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Maple [A]
time = 0.64, size = 137, normalized size = 1.38

method	result	size
derivativedivides	\(-\frac {\frac {c \,a^{2}}{x^{2}}+\frac {\arctanh \left (\frac {c}{x^{2}}\right )^{2} b^{2} c}{x^{2}}+b^{2} \arctanh \left (\frac {c}{x^{2}}\right )^{2}-2 \arctanh \left (\frac {c}{x^{2}}\right ) \ln \left (1+\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right ) b^{2}-\polylog \left (2, -\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right ) b^{2}+\frac {2 a b c \arctanh \left (\frac {c}{x^{2}}\right )}{x^{2}}+a b \ln \left (1-\frac {c^{2}}{x^{4}}\right )}{2 c}\)	\(137\)
default	\(-\frac {\frac {c \,a^{2}}{x^{2}}+\frac {\arctanh \left (\frac {c}{x^{2}}\right )^{2} b^{2} c}{x^{2}}+b^{2} \arctanh \left (\frac {c}{x^{2}}\right )^{2}-2 \arctanh \left (\frac {c}{x^{2}}\right ) \ln \left (1+\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right ) b^{2}-\polylog \left (2, -\frac {\left (1+\frac {c}{x^{2}}\right )^{2}}{1-\frac {c^{2}}{x^{4}}}\right ) b^{2}+\frac {2 a b c \arctanh \left (\frac {c}{x^{2}}\right )}{x^{2}}+a b \ln \left (1-\frac {c^{2}}{x^{4}}\right )}{2 c}\)	\(137\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^3,x, algorithm="maxima")

[Out]

1/8*(8*c^3*integrate(log(x)^2/(c*x^7 - c^3*x^3), x) + c^2*(log(x^2 + c)/c^3 + log(x^2 - c)/c^3 - 4*log(x)/c^3)
- 8*c^2*integrate(x^2*log(x^2 + c)/(c*x^7 - c^3*x^3), x) + 8*c^2*integrate(x^2*log(x)/(c*x^7 - c^3*x^3), x) +
2*c*(log(x^2 - c)/c^2 - log(x^2)/c^2 + 1/(c*x^2))*log(-c/x^2 + 1) - c*(log(x^2 + c)/c^2 - log(x^2 - c)/c^2) -
8*c*integrate(x^4*log(x)^2/(c*x^7 - c^3*x^3), x) - 4*c*integrate(x^4*log(x^2 + c)/(c*x^7 - c^3*x^3), x) + 16*
c*integrate(x^4*log(x)/(c*x^7 - c^3*x^3), x) - log(-c/x^2 + 1)^2/x^2 - (x^2*log(x^2 - c)^2 + 4*x^2*log(x)^2 -
4*x^2*log(x) - 2*(2*x^2*log(x) - x^2)*log(x^2 - c) + 2*c)/(c*x^2) - (c*log(x^2 + c)^2 - 2*((x^2 + c)*log(x^2 +
c) - 2*(x^2 + c)*log(x) - c)*log(x^2 - c))/(c*x^2) - 4*integrate(x^6*log(x^2 + c)/(c*x^7 - c^3*x^3), x) + 8*i
ntegrate(x^6*log(x)/(c*x^7 - c^3*x^3), x))*b^2 - 1/2*a*b*(2*c*arctanh(c/x^2)/x^2 + log(-c^2/x^4 + 1))/c - 1/2*
a^2/x^2

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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^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)

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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 (\frac {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)

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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^3,x, algorithm="giac")

[Out]

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

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