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

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

[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

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Rubi [A]  time = 0.125208, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {5984, 5918, 2402, 2315} \[ \frac{b \text{PolyLog}\left (2,1-\frac{2}{1-c \sqrt{x}}\right )}{c^2}-\frac{\left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b c^2}+\frac{2 \log \left (\frac{2}{1-c \sqrt{x}}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )}{c^2} \]

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 5984

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]

Rule 5918

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 2402

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 2315

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

Rubi steps

\begin{align*} \int \frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{1-c^2 x} \, dx &=2 \operatorname{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 \operatorname{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) \operatorname{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) \operatorname{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*}

Mathematica [A]  time = 0.0968517, size = 75, normalized size = 0.96 \[ -\frac{b \left (\text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (c \sqrt{x}\right )}\right )-\tanh ^{-1}\left (c \sqrt{x}\right ) \left (\tanh ^{-1}\left (c \sqrt{x}\right )+2 \log \left (e^{-2 \tanh ^{-1}\left (c \sqrt{x}\right )}+1\right )\right )\right )}{c^2}-\frac{a \log \left (1-c^2 x\right )}{c^2} \]

Warning: Unable to verify antiderivative.

[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]  time = 0.046, size = 186, normalized size = 2.4 \begin{align*} -{\frac{a}{{c}^{2}}\ln \left ( c\sqrt{x}-1 \right ) }-{\frac{a}{{c}^{2}}\ln \left ( 1+c\sqrt{x} \right ) }-{\frac{b}{{c}^{2}}{\it Artanh} \left ( c\sqrt{x} \right ) \ln \left ( c\sqrt{x}-1 \right ) }-{\frac{b}{{c}^{2}}{\it Artanh} \left ( c\sqrt{x} \right ) \ln \left ( 1+c\sqrt{x} \right ) }-{\frac{b}{4\,{c}^{2}} \left ( \ln \left ( c\sqrt{x}-1 \right ) \right ) ^{2}}+{\frac{b}{{c}^{2}}{\it dilog} \left ({\frac{1}{2}}+{\frac{c}{2}\sqrt{x}} \right ) }+{\frac{b}{2\,{c}^{2}}\ln \left ( c\sqrt{x}-1 \right ) \ln \left ({\frac{1}{2}}+{\frac{c}{2}\sqrt{x}} \right ) }+{\frac{b}{2\,{c}^{2}}\ln \left ( -{\frac{c}{2}\sqrt{x}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{c}{2}\sqrt{x}} \right ) }-{\frac{b}{2\,{c}^{2}}\ln \left ( -{\frac{c}{2}\sqrt{x}}+{\frac{1}{2}} \right ) \ln \left ( 1+c\sqrt{x} \right ) }+{\frac{b}{4\,{c}^{2}} \left ( \ln \left ( 1+c\sqrt{x} \right ) \right ) ^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 2.07543, size = 136, normalized size = 1.74 \begin{align*} -\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{align*}

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., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{b \operatorname{artanh}\left (c \sqrt{x}\right ) + a}{c^{2} x - 1}, x\right ) \end{align*}

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., size = 0, normalized size = 0. \begin{align*} - \int \frac{a}{c^{2} x - 1}\, dx - \int \frac{b \operatorname{atanh}{\left (c \sqrt{x} \right )}}{c^{2} x - 1}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int -\frac{b \operatorname{artanh}\left (c \sqrt{x}\right ) + a}{c^{2} x - 1}\,{d x} \end{align*}

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)