3.1.40 \(\int \frac {a+b \tanh ^{-1}(c \sqrt {x})}{x (1-c^2 x)} \, dx\) [40]

Optimal. Leaf size=69 \[ \frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (2-\frac {2}{1+c \sqrt {x}}\right )-b \text {PolyLog}\left (2,-1+\frac {2}{1+c \sqrt {x}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.18, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {36, 29, 31, 1607, 6135, 6079, 2497} \begin {gather*} \frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b}+2 \log \left (2-\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-b \text {Li}_2\left (\frac {2}{\sqrt {x} c+1}-1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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

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Mathematica [A]
time = 0.08, size = 72, normalized size = 1.04 \begin {gather*} b \tanh ^{-1}\left (c \sqrt {x}\right ) \left (\tanh ^{-1}\left (c \sqrt {x}\right )+2 \log \left (1-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )\right )+a \log (x)-a \log \left (1-c^2 x\right )-b \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(216\) vs. \(2(61)=122\).
time = 0.25, size = 217, normalized size = 3.14

method	result	size
derivativedivides	\(-a \ln \left (1+c \sqrt {x}\right )-a \ln \left (c \sqrt {x}-1\right )+2 a \ln \left (c \sqrt {x}\right )-b \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )-b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )+2 b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}\right )-b \dilog \left (1+c \sqrt {x}\right )-b \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )-b \dilog \left (c \sqrt {x}\right )-\frac {b \ln \left (c \sqrt {x}-1\right )^{2}}{4}+b \dilog \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )+\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}+\frac {b \ln \left (1+c \sqrt {x}\right )^{2}}{4}+\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{2}\)	\(217\)
default	\(-a \ln \left (1+c \sqrt {x}\right )-a \ln \left (c \sqrt {x}-1\right )+2 a \ln \left (c \sqrt {x}\right )-b \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )-b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )+2 b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}\right )-b \dilog \left (1+c \sqrt {x}\right )-b \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )-b \dilog \left (c \sqrt {x}\right )-\frac {b \ln \left (c \sqrt {x}-1\right )^{2}}{4}+b \dilog \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )+\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}+\frac {b \ln \left (1+c \sqrt {x}\right )^{2}}{4}+\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{2}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))/x/(-c^2*x+1),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 159 vs. \(2 (60) = 120\).
time = 0.37, size = 159, normalized size = 2.30 \begin {gather*} -\frac {1}{4} \, b \log \left (c \sqrt {x} + 1\right )^{2} + \frac {1}{2} \, b \log \left (c \sqrt {x} + 1\right ) \log \left (-c \sqrt {x} + 1\right ) + \frac {1}{4} \, b \log \left (-c \sqrt {x} + 1\right )^{2} - {\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right ) + {\rm Li}_2\left (\frac {1}{2} \, c \sqrt {x} + \frac {1}{2}\right )\right )} b - {\left (\log \left (c \sqrt {x}\right ) \log \left (-c \sqrt {x} + 1\right ) + {\rm Li}_2\left (-c \sqrt {x} + 1\right )\right )} b + {\left (\log \left (c \sqrt {x} + 1\right ) \log \left (-c \sqrt {x}\right ) + {\rm Li}_2\left (c \sqrt {x} + 1\right )\right )} b - a {\left (\log \left (c \sqrt {x} + 1\right ) + \log \left (c \sqrt {x} - 1\right ) - \log \left (x\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**(1/2)))/x/(-c**2*x+1),x)

[Out]

-Integral(a/(c**2*x**2 - x), x) - Integral(b*atanh(c*sqrt(x))/(c**2*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^(1/2)))/x/(-c^2*x+1),x, algorithm="giac")

[Out]

integrate(-(b*arctanh(c*sqrt(x)) + a)/((c^2*x - 1)*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(a + b*atanh(c*x^(1/2)))/(x*(c^2*x - 1)),x)

[Out]

-int((a + b*atanh(c*x^(1/2)))/(x*(c^2*x - 1)), x)

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