3.1.39 \(\int \frac {a+b \tanh ^{-1}(c \sqrt {x})}{1-c^2 x} \, dx\) [39]

Optimal. Leaf size=78 \[ -\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b c^2}+\frac {2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1-c \sqrt {x}}\right )}{c^2}+\frac {b \text {PolyLog}\left (2,1-\frac {2}{1-c \sqrt {x}}\right )}{c^2} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 78, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6069, 6131, 6055, 2449, 2352} \begin {gather*} -\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )^2}{b c^2}+\frac {2 \log \left (\frac {2}{1-c \sqrt {x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{c^2}+\frac {b \text {Li}_2\left (1-\frac {2}{1-c \sqrt {x}}\right )}{c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 6069

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> With[{k = Denominator[n]}, Dist
[k, Subst[Int[x^(k - 1)*((a + b*ArcTanh[c*x^(k*n)])/(d + e*x^k)), x], x, x^(1/k)], x]] /; FreeQ[{a, b, c, d, e
}, x] && FractionQ[n]

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

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Mathematica [A]
time = 0.06, size = 75, normalized size = 0.96 \begin {gather*} -\frac {a \log \left (1-c^2 x\right )}{c^2}-\frac {b \left (-\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 )+\text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )\right )}{c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(161\) vs. \(2(70)=140\).
time = 0.22, size = 162, normalized size = 2.08

method	result	size
derivativedivides	\(-\frac {2 \left (\frac {a \ln \left (c \sqrt {x}-1\right )}{2}+\frac {a \ln \left (1+c \sqrt {x}\right )}{2}+\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}+\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {b \ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {b \dilog \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}-\frac {b \ln \left (1+c \sqrt {x}\right )^{2}}{8}-\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{4}\right )}{c^{2}}\)	\(162\)
default	\(-\frac {2 \left (\frac {a \ln \left (c \sqrt {x}-1\right )}{2}+\frac {a \ln \left (1+c \sqrt {x}\right )}{2}+\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}-1\right )}{2}+\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2}+\frac {b \ln \left (c \sqrt {x}-1\right )^{2}}{8}-\frac {b \dilog \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{2}-\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}-\frac {b \ln \left (1+c \sqrt {x}\right )^{2}}{8}-\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (\frac {c \sqrt {x}}{2}+\frac {1}{2}\right )}{4}+\frac {b \ln \left (-\frac {c \sqrt {x}}{2}+\frac {1}{2}\right ) \ln \left (1+c \sqrt {x}\right )}{4}\right )}{c^{2}}\)	\(162\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/c^2*(1/2*a*ln(c*x^(1/2)-1)+1/2*a*ln(1+c*x^(1/2))+1/2*b*arctanh(c*x^(1/2))*ln(c*x^(1/2)-1)+1/2*b*arctanh(c*x
^(1/2))*ln(1+c*x^(1/2))+1/8*b*ln(c*x^(1/2)-1)^2-1/2*b*dilog(1/2*c*x^(1/2)+1/2)-1/4*b*ln(c*x^(1/2)-1)*ln(1/2*c*
x^(1/2)+1/2)-1/8*b*ln(1+c*x^(1/2))^2-1/4*b*ln(-1/2*c*x^(1/2)+1/2)*ln(1/2*c*x^(1/2)+1/2)+1/4*b*ln(-1/2*c*x^(1/2
)+1/2)*ln(1+c*x^(1/2)))

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Maxima [A]
time = 0.57, size = 101, normalized size = 1.29 \begin {gather*} -\frac {{\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}{c^{2}} - \frac {a \log \left (c^{2} x - 1\right )}{c^{2}} - \frac {b \log \left (c \sqrt {x} + 1\right )^{2} - 2 \, b \log \left (c \sqrt {x} + 1\right ) \log \left (-c \sqrt {x} + 1\right ) - b \log \left (-c \sqrt {x} + 1\right )^{2}}{4 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - \int \frac {a}{c^{2} x - 1}\, dx - \int \frac {b \operatorname {atanh}{\left (c \sqrt {x} \right )}}{c^{2} x - 1}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(-(b*arctanh(c*sqrt(x)) + a)/(c^2*x - 1), 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 )}{c^2\,x-1} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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