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

Optimal. Leaf size=117 \[ -b c^2 \text{PolyLog}\left (2,\frac{2}{c \sqrt{x}+1}-1\right )+\frac{c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b}+2 c^2 \log \left (2-\frac{2}{c \sqrt{x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{x}+b c^2 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{b c}{\sqrt{x}} \]

[Out]

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

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Rubi [A]  time = 0.363359, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {44, 1593, 5982, 5916, 325, 206, 5988, 5932, 2447} \[ -b c^2 \text{PolyLog}\left (2,\frac{2}{c \sqrt{x}+1}-1\right )+\frac{c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b}+2 c^2 \log \left (2-\frac{2}{c \sqrt{x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{x}+b c^2 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{b c}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 1593

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 5982

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

Rule 5916

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 5988

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]

Rule 5932

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 2447

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]]

Rubi steps

\begin{align*} \int \frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{x^2 \left (1-c^2 x\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(c x)}{x^3-c^2 x^5} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(c x)}{x^3 \left (1-c^2 x^2\right )} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(c x)}{x^3} \, dx,x,\sqrt{x}\right )+\left (2 c^2\right ) \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(c x)}{x \left (1-c^2 x^2\right )} \, dx,x,\sqrt{x}\right )\\ &=-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{x}+\frac{c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b}+(b c) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx,x,\sqrt{x}\right )+\left (2 c^2\right ) \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx,x,\sqrt{x}\right )\\ &=-\frac{b c}{\sqrt{x}}-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{x}+\frac{c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b}+2 c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \log \left (2-\frac{2}{1+c \sqrt{x}}\right )+\left (b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )-\left (2 b c^3\right ) \operatorname{Subst}\left (\int \frac{\log \left (2-\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{b c}{\sqrt{x}}+b c^2 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{x}+\frac{c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b}+2 c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \log \left (2-\frac{2}{1+c \sqrt{x}}\right )-b c^2 \text{Li}_2\left (-1+\frac{2}{1+c \sqrt{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.324301, size = 118, normalized size = 1.01 \[ -b c^2 \left (\text{PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c \sqrt{x}\right )}\right )-\tanh ^{-1}\left (c \sqrt{x}\right ) \left (-\frac{1-c^2 x}{c^2 x}+\tanh ^{-1}\left (c \sqrt{x}\right )+2 \log \left (1-e^{-2 \tanh ^{-1}\left (c \sqrt{x}\right )}\right )\right )+\frac{1}{c \sqrt{x}}\right )+2 a c^2 \log \left (\sqrt{x}\right )-a c^2 \log \left (1-c^2 x\right )-\frac{a}{x} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [B]  time = 0.057, size = 315, normalized size = 2.7 \begin{align*} -{c}^{2}a\ln \left ( c\sqrt{x}-1 \right ) -{\frac{a}{x}}+2\,{c}^{2}a\ln \left ( c\sqrt{x} \right ) -{c}^{2}a\ln \left ( 1+c\sqrt{x} \right ) -{c}^{2}b{\it Artanh} \left ( c\sqrt{x} \right ) \ln \left ( c\sqrt{x}-1 \right ) -{\frac{b}{x}{\it Artanh} \left ( c\sqrt{x} \right ) }+2\,{c}^{2}b{\it Artanh} \left ( c\sqrt{x} \right ) \ln \left ( c\sqrt{x} \right ) -{c}^{2}b{\it Artanh} \left ( c\sqrt{x} \right ) \ln \left ( 1+c\sqrt{x} \right ) -{bc{\frac{1}{\sqrt{x}}}}-{\frac{{c}^{2}b}{2}\ln \left ( c\sqrt{x}-1 \right ) }+{\frac{{c}^{2}b}{2}\ln \left ( 1+c\sqrt{x} \right ) }-{c}^{2}b{\it dilog} \left ( c\sqrt{x} \right ) -{c}^{2}b{\it dilog} \left ( 1+c\sqrt{x} \right ) -{c}^{2}b\ln \left ( c\sqrt{x} \right ) \ln \left ( 1+c\sqrt{x} \right ) -{\frac{{c}^{2}b}{4} \left ( \ln \left ( c\sqrt{x}-1 \right ) \right ) ^{2}}+{c}^{2}b{\it dilog} \left ({\frac{1}{2}}+{\frac{c}{2}\sqrt{x}} \right ) +{\frac{{c}^{2}b}{2}\ln \left ( c\sqrt{x}-1 \right ) \ln \left ({\frac{1}{2}}+{\frac{c}{2}\sqrt{x}} \right ) }+{\frac{{c}^{2}b}{2}\ln \left ( -{\frac{c}{2}\sqrt{x}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{c}{2}\sqrt{x}} \right ) }-{\frac{{c}^{2}b}{2}\ln \left ( -{\frac{c}{2}\sqrt{x}}+{\frac{1}{2}} \right ) \ln \left ( 1+c\sqrt{x} \right ) }+{\frac{{c}^{2}b}{4} \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)))/x^2/(-c^2*x+1),x)

[Out]

-c^2*a*ln(c*x^(1/2)-1)-a/x+2*c^2*a*ln(c*x^(1/2))-c^2*a*ln(1+c*x^(1/2))-c^2*b*arctanh(c*x^(1/2))*ln(c*x^(1/2)-1
)-b*arctanh(c*x^(1/2))/x+2*c^2*b*arctanh(c*x^(1/2))*ln(c*x^(1/2))-c^2*b*arctanh(c*x^(1/2))*ln(1+c*x^(1/2))-b*c
/x^(1/2)-1/2*c^2*b*ln(c*x^(1/2)-1)+1/2*c^2*b*ln(1+c*x^(1/2))-c^2*b*dilog(c*x^(1/2))-c^2*b*dilog(1+c*x^(1/2))-c
^2*b*ln(c*x^(1/2))*ln(1+c*x^(1/2))-1/4*c^2*b*ln(c*x^(1/2)-1)^2+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 [B]  time = 1.7842, size = 335, normalized size = 2.86 \begin{align*} -{\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} -{\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 c^{2} +{\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 c^{2} + \frac{1}{2} \, b c^{2} \log \left (c \sqrt{x} + 1\right ) - \frac{1}{2} \, b c^{2} \log \left (c \sqrt{x} - 1\right ) -{\left (c^{2} \log \left (c \sqrt{x} + 1\right ) + c^{2} \log \left (c \sqrt{x} - 1\right ) - c^{2} \log \left (x\right ) + \frac{1}{x}\right )} a - \frac{b c^{2} x \log \left (c \sqrt{x} + 1\right )^{2} - b c^{2} x \log \left (-c \sqrt{x} + 1\right )^{2} + 4 \, b c \sqrt{x} + 2 \, b \log \left (c \sqrt{x} + 1\right ) - 2 \,{\left (b c^{2} x \log \left (c \sqrt{x} + 1\right ) + b\right )} \log \left (-c \sqrt{x} + 1\right )}{4 \, x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^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 - (log(c*sqrt(x))*log(-c*sq
rt(x) + 1) + dilog(-c*sqrt(x) + 1))*b*c^2 + (log(c*sqrt(x) + 1)*log(-c*sqrt(x)) + dilog(c*sqrt(x) + 1))*b*c^2
+ 1/2*b*c^2*log(c*sqrt(x) + 1) - 1/2*b*c^2*log(c*sqrt(x) - 1) - (c^2*log(c*sqrt(x) + 1) + c^2*log(c*sqrt(x) -
1) - c^2*log(x) + 1/x)*a - 1/4*(b*c^2*x*log(c*sqrt(x) + 1)^2 - b*c^2*x*log(-c*sqrt(x) + 1)^2 + 4*b*c*sqrt(x) +
2*b*log(c*sqrt(x) + 1) - 2*(b*c^2*x*log(c*sqrt(x) + 1) + b)*log(-c*sqrt(x) + 1))/x

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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^{3} - x^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-(b*arctanh(c*sqrt(x)) + a)/(c^2*x^3 - x^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{a}{c^{2} x^{3} - x^{2}}\, dx - \int \frac{b \operatorname{atanh}{\left (c \sqrt{x} \right )}}{c^{2} x^{3} - x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(a/(c**2*x**3 - x**2), x) - Integral(b*atanh(c*sqrt(x))/(c**2*x**3 - x**2), 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}{{\left (c^{2} x - 1\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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