3.1.54 \(\int \frac {a+b \tanh ^{-1}(c x^2)}{x} \, dx\) [54]

Optimal. Leaf size=30 \[ a \log (x)-\frac {1}{4} b \text {PolyLog}\left (2,-c x^2\right )+\frac {1}{4} b \text {PolyLog}\left (2,c x^2\right ) \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6035, 6031} \begin {gather*} a \log (x)-\frac {1}{4} b \text {Li}_2\left (-c x^2\right )+\frac {1}{4} b \text {Li}_2\left (c x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])/x,x]

[Out]

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

Rule 6031

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b/2)*PolyLog[2, (-c)*x]
, x] + Simp[(b/2)*PolyLog[2, c*x], x]) /; FreeQ[{a, b, c}, x]

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]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 28, normalized size = 0.93 \begin {gather*} a \log (x)+\frac {1}{4} b \left (-\text {PolyLog}\left (2,-c x^2\right )+\text {PolyLog}\left (2,c x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^2])/x,x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(123\) vs. \(2(26)=52\).
time = 0.05, size = 124, normalized size = 4.13

method	result	size
default	\(a \ln \left (x \right )+b \ln \left (x \right ) \arctanh \left (c \,x^{2}\right )+\frac {b \ln \left (x \right ) \ln \left (1-x \sqrt {c}\right )}{2}+\frac {b \ln \left (x \right ) \ln \left (1+x \sqrt {c}\right )}{2}+\frac {b \dilog \left (1-x \sqrt {c}\right )}{2}+\frac {b \dilog \left (1+x \sqrt {c}\right )}{2}-\frac {b \ln \left (x \right ) \ln \left (1+x \sqrt {-c}\right )}{2}-\frac {b \ln \left (x \right ) \ln \left (1-x \sqrt {-c}\right )}{2}-\frac {b \dilog \left (1+x \sqrt {-c}\right )}{2}-\frac {b \dilog \left (1-x \sqrt {-c}\right )}{2}\)	\(124\)
risch	\(a \ln \left (x \right )+\frac {b \ln \left (x \right ) \ln \left (1-x \sqrt {c}\right )}{2}+\frac {b \ln \left (x \right ) \ln \left (1+x \sqrt {c}\right )}{2}-\frac {\ln \left (x \right ) \ln \left (-c \,x^{2}+1\right ) b}{2}+\frac {b \dilog \left (1-x \sqrt {c}\right )}{2}+\frac {b \dilog \left (1+x \sqrt {c}\right )}{2}+\frac {\ln \left (x \right ) \ln \left (c \,x^{2}+1\right ) b}{2}-\frac {b \ln \left (x \right ) \ln \left (1+x \sqrt {-c}\right )}{2}-\frac {b \ln \left (x \right ) \ln \left (1-x \sqrt {-c}\right )}{2}-\frac {b \dilog \left (1+x \sqrt {-c}\right )}{2}-\frac {b \dilog \left (1-x \sqrt {-c}\right )}{2}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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,x, algorithm="maxima")

[Out]

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

[Out]

integral((b*arctanh(c*x^2) + a)/x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {a+b\,\mathrm {atanh}\left (c\,x^2\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^2))/x,x)

[Out]

int((a + b*atanh(c*x^2))/x, x)

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