3.42 \(\int \frac{a+b \tanh ^{-1}(c \sqrt{x})}{x^3 (1-c^2 x)} \, dx\)

Optimal. Leaf size=157 \[ -b c^4 \text{PolyLog}\left (2,\frac{2}{c \sqrt{x}+1}-1\right )+\frac{c^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b}-\frac{c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )}{x}+2 c^4 \log \left (2-\frac{2}{c \sqrt{x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}-\frac{3 b c^3}{2 \sqrt{x}}+\frac{3}{2} b c^4 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{b c}{6 x^{3/2}} \]

[Out]

-(b*c)/(6*x^(3/2)) - (3*b*c^3)/(2*Sqrt[x]) + (3*b*c^4*ArcTanh[c*Sqrt[x]])/2 - (a + b*ArcTanh[c*Sqrt[x]])/(2*x^
2) - (c^2*(a + b*ArcTanh[c*Sqrt[x]]))/x + (c^4*(a + b*ArcTanh[c*Sqrt[x]])^2)/b + 2*c^4*(a + b*ArcTanh[c*Sqrt[x
]])*Log[2 - 2/(1 + c*Sqrt[x])] - b*c^4*PolyLog[2, -1 + 2/(1 + c*Sqrt[x])]

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Rubi [A]  time = 0.455852, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 14, 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} \[ -b c^4 \text{PolyLog}\left (2,\frac{2}{c \sqrt{x}+1}-1\right )+\frac{c^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b}-\frac{c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )}{x}+2 c^4 \log \left (2-\frac{2}{c \sqrt{x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}-\frac{3 b c^3}{2 \sqrt{x}}+\frac{3}{2} b c^4 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{b c}{6 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])/(x^3*(1 - c^2*x)),x]

[Out]

-(b*c)/(6*x^(3/2)) - (3*b*c^3)/(2*Sqrt[x]) + (3*b*c^4*ArcTanh[c*Sqrt[x]])/2 - (a + b*ArcTanh[c*Sqrt[x]])/(2*x^
2) - (c^2*(a + b*ArcTanh[c*Sqrt[x]]))/x + (c^4*(a + b*ArcTanh[c*Sqrt[x]])^2)/b + 2*c^4*(a + b*ArcTanh[c*Sqrt[x
]])*Log[2 - 2/(1 + c*Sqrt[x])] - b*c^4*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^3 \left (1-c^2 x\right )} \, dx &=2 \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(c x)}{x^5-c^2 x^7} \, dx,x,\sqrt{x}\right )\\ &=2 \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(c x)}{x^5 \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^5} \, dx,x,\sqrt{x}\right )+\left (2 c^2\right ) \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 )\\ &=-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}+\frac{1}{2} (b c) \operatorname{Subst}\left (\int \frac{1}{x^4 \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^3} \, dx,x,\sqrt{x}\right )+\left (2 c^4\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{b c}{6 x^{3/2}}-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}-\frac{c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )}{x}+\frac{c^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b}+\frac{1}{2} \left (b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx,x,\sqrt{x}\right )+\left (b c^3\right ) \operatorname{Subst}\left (\int \frac{1}{x^2 \left (1-c^2 x^2\right )} \, dx,x,\sqrt{x}\right )+\left (2 c^4\right ) \operatorname{Subst}\left (\int \frac{a+b \tanh ^{-1}(c x)}{x (1+c x)} \, dx,x,\sqrt{x}\right )\\ &=-\frac{b c}{6 x^{3/2}}-\frac{3 b c^3}{2 \sqrt{x}}-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}-\frac{c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )}{x}+\frac{c^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b}+2 c^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \log \left (2-\frac{2}{1+c \sqrt{x}}\right )+\frac{1}{2} \left (b c^5\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )+\left (b c^5\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )-\left (2 b c^5\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}{6 x^{3/2}}-\frac{3 b c^3}{2 \sqrt{x}}+\frac{3}{2} b c^4 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}-\frac{c^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )}{x}+\frac{c^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )^2}{b}+2 c^4 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \log \left (2-\frac{2}{1+c \sqrt{x}}\right )-b c^4 \text{Li}_2\left (-1+\frac{2}{1+c \sqrt{x}}\right )\\ \end{align*}

Mathematica [A]  time = 0.536917, size = 158, normalized size = 1.01 \[ -b c^4 \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c \sqrt{x}\right )}\right )-\frac{-6 a c^4 x^2 \log (x)+6 a c^4 x^2 \log \left (1-c^2 x\right )+6 a c^2 x+3 a+9 b c^3 x^{3/2}-6 b c^4 x^2 \tanh ^{-1}\left (c \sqrt{x}\right )^2-3 b \tanh ^{-1}\left (c \sqrt{x}\right ) \left (3 c^4 x^2+4 c^4 x^2 \log \left (1-e^{-2 \tanh ^{-1}\left (c \sqrt{x}\right )}\right )-2 c^2 x-1\right )+b c \sqrt{x}}{6 x^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c*Sqrt[x]])/(x^3*(1 - c^2*x)),x]

[Out]

-(3*a + b*c*Sqrt[x] + 6*a*c^2*x + 9*b*c^3*x^(3/2) - 6*b*c^4*x^2*ArcTanh[c*Sqrt[x]]^2 - 3*b*ArcTanh[c*Sqrt[x]]*
(-1 - 2*c^2*x + 3*c^4*x^2 + 4*c^4*x^2*Log[1 - E^(-2*ArcTanh[c*Sqrt[x]])]) - 6*a*c^4*x^2*Log[x] + 6*a*c^4*x^2*L
og[1 - c^2*x])/(6*x^2) - b*c^4*PolyLog[2, E^(-2*ArcTanh[c*Sqrt[x]])]

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Maple [B]  time = 0.066, size = 348, normalized size = 2.2 \begin{align*} -{c}^{4}a\ln \left ( c\sqrt{x}-1 \right ) -{\frac{a}{2\,{x}^{2}}}-{\frac{a{c}^{2}}{x}}+2\,{c}^{4}a\ln \left ( c\sqrt{x} \right ) -{c}^{4}a\ln \left ( 1+c\sqrt{x} \right ) -{c}^{4}b{\it Artanh} \left ( c\sqrt{x} \right ) \ln \left ( c\sqrt{x}-1 \right ) -{\frac{b}{2\,{x}^{2}}{\it Artanh} \left ( c\sqrt{x} \right ) }-{\frac{{c}^{2}b}{x}{\it Artanh} \left ( c\sqrt{x} \right ) }+2\,{c}^{4}b{\it Artanh} \left ( c\sqrt{x} \right ) \ln \left ( c\sqrt{x} \right ) -{c}^{4}b{\it Artanh} \left ( c\sqrt{x} \right ) \ln \left ( 1+c\sqrt{x} \right ) -{c}^{4}b{\it dilog} \left ( c\sqrt{x} \right ) -{c}^{4}b{\it dilog} \left ( 1+c\sqrt{x} \right ) -{c}^{4}b\ln \left ( c\sqrt{x} \right ) \ln \left ( 1+c\sqrt{x} \right ) -{\frac{{c}^{4}b}{4} \left ( \ln \left ( c\sqrt{x}-1 \right ) \right ) ^{2}}+{c}^{4}b{\it dilog} \left ({\frac{1}{2}}+{\frac{c}{2}\sqrt{x}} \right ) +{\frac{{c}^{4}b}{2}\ln \left ( c\sqrt{x}-1 \right ) \ln \left ({\frac{1}{2}}+{\frac{c}{2}\sqrt{x}} \right ) }+{\frac{{c}^{4}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}^{4}b}{2}\ln \left ( -{\frac{c}{2}\sqrt{x}}+{\frac{1}{2}} \right ) \ln \left ( 1+c\sqrt{x} \right ) }+{\frac{{c}^{4}b}{4} \left ( \ln \left ( 1+c\sqrt{x} \right ) \right ) ^{2}}-{\frac{3\,{c}^{4}b}{4}\ln \left ( c\sqrt{x}-1 \right ) }-{\frac{bc}{6}{x}^{-{\frac{3}{2}}}}-{\frac{3\,b{c}^{3}}{2}{\frac{1}{\sqrt{x}}}}+{\frac{3\,{c}^{4}b}{4}\ln \left ( 1+c\sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))/x^3/(-c^2*x+1),x)

[Out]

-c^4*a*ln(c*x^(1/2)-1)-1/2*a/x^2-c^2*a/x+2*c^4*a*ln(c*x^(1/2))-c^4*a*ln(1+c*x^(1/2))-c^4*b*arctanh(c*x^(1/2))*
ln(c*x^(1/2)-1)-1/2*b*arctanh(c*x^(1/2))/x^2-c^2*b*arctanh(c*x^(1/2))/x+2*c^4*b*arctanh(c*x^(1/2))*ln(c*x^(1/2
))-c^4*b*arctanh(c*x^(1/2))*ln(1+c*x^(1/2))-c^4*b*dilog(c*x^(1/2))-c^4*b*dilog(1+c*x^(1/2))-c^4*b*ln(c*x^(1/2)
)*ln(1+c*x^(1/2))-1/4*c^4*b*ln(c*x^(1/2)-1)^2+c^4*b*dilog(1/2+1/2*c*x^(1/2))+1/2*c^4*b*ln(c*x^(1/2)-1)*ln(1/2+
1/2*c*x^(1/2))+1/2*c^4*b*ln(-1/2*c*x^(1/2)+1/2)*ln(1/2+1/2*c*x^(1/2))-1/2*c^4*b*ln(-1/2*c*x^(1/2)+1/2)*ln(1+c*
x^(1/2))+1/4*c^4*b*ln(1+c*x^(1/2))^2-3/4*c^4*b*ln(c*x^(1/2)-1)-1/6*b*c/x^(3/2)-3/2*b*c^3/x^(1/2)+3/4*c^4*b*ln(
1+c*x^(1/2))

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Maxima [B]  time = 1.78481, size = 393, normalized size = 2.5 \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^{4} -{\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^{4} +{\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^{4} + \frac{3}{4} \, b c^{4} \log \left (c \sqrt{x} + 1\right ) - \frac{3}{4} \, b c^{4} \log \left (c \sqrt{x} - 1\right ) - \frac{1}{2} \,{\left (2 \, c^{4} \log \left (c \sqrt{x} + 1\right ) + 2 \, c^{4} \log \left (c \sqrt{x} - 1\right ) - 2 \, c^{4} \log \left (x\right ) + \frac{2 \, c^{2} x + 1}{x^{2}}\right )} a - \frac{3 \, b c^{4} x^{2} \log \left (c \sqrt{x} + 1\right )^{2} - 3 \, b c^{4} x^{2} \log \left (-c \sqrt{x} + 1\right )^{2} + 18 \, b c^{3} x^{\frac{3}{2}} + 2 \, b c \sqrt{x} + 3 \,{\left (2 \, b c^{2} x + b\right )} \log \left (c \sqrt{x} + 1\right ) - 3 \,{\left (2 \, b c^{4} x^{2} \log \left (c \sqrt{x} + 1\right ) + 2 \, b c^{2} x + b\right )} \log \left (-c \sqrt{x} + 1\right )}{12 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^3/(-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^4 - (log(c*sqrt(x))*log(-c*sq
rt(x) + 1) + dilog(-c*sqrt(x) + 1))*b*c^4 + (log(c*sqrt(x) + 1)*log(-c*sqrt(x)) + dilog(c*sqrt(x) + 1))*b*c^4
+ 3/4*b*c^4*log(c*sqrt(x) + 1) - 3/4*b*c^4*log(c*sqrt(x) - 1) - 1/2*(2*c^4*log(c*sqrt(x) + 1) + 2*c^4*log(c*sq
rt(x) - 1) - 2*c^4*log(x) + (2*c^2*x + 1)/x^2)*a - 1/12*(3*b*c^4*x^2*log(c*sqrt(x) + 1)^2 - 3*b*c^4*x^2*log(-c
*sqrt(x) + 1)^2 + 18*b*c^3*x^(3/2) + 2*b*c*sqrt(x) + 3*(2*b*c^2*x + b)*log(c*sqrt(x) + 1) - 3*(2*b*c^4*x^2*log
(c*sqrt(x) + 1) + 2*b*c^2*x + b)*log(-c*sqrt(x) + 1))/x^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^{4} - x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^3/(-c^2*x+1),x, algorithm="fricas")

[Out]

integral(-(b*arctanh(c*sqrt(x)) + a)/(c^2*x^4 - x^3), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**(1/2)))/x**3/(-c**2*x+1),x)

[Out]

Timed out

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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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^3/(-c^2*x+1),x, algorithm="giac")

[Out]

integrate(-(b*arctanh(c*sqrt(x)) + a)/((c^2*x - 1)*x^3), x)