3.19 \(\int \frac{(a+b \tanh ^{-1}(c+d x))^2}{(c e+d e x)^2} \, dx\)

Optimal. Leaf size=104 \[ -\frac{b^2 \text{PolyLog}\left (2,\frac{2}{c+d x+1}-1\right )}{d e^2}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2}+\frac{2 b \log \left (2-\frac{2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^2} \]

[Out]

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

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Rubi [A]  time = 0.180662, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {6107, 12, 5916, 5988, 5932, 2447} \[ -\frac{b^2 \text{PolyLog}\left (2,\frac{2}{c+d x+1}-1\right )}{d e^2}-\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2 (c+d x)}+\frac{\left (a+b \tanh ^{-1}(c+d x)\right )^2}{d e^2}+\frac{2 b \log \left (2-\frac{2}{c+d x+1}\right ) \left (a+b \tanh ^{-1}(c+d x)\right )}{d e^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c + d*x])^2/(c*e + d*e*x)^2,x]

[Out]

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

Rule 6107

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 0.25711, size = 126, normalized size = 1.21 \[ \frac{-b^2 (c+d x) \text{PolyLog}\left (2,e^{-2 \tanh ^{-1}(c+d x)}\right )+a \left (2 b (c+d x) \log \left (\frac{c+d x}{\sqrt{1-(c+d x)^2}}\right )-a\right )+2 b \tanh ^{-1}(c+d x) \left (b (c+d x) \log \left (1-e^{-2 \tanh ^{-1}(c+d x)}\right )-a\right )+b^2 (c+d x-1) \tanh ^{-1}(c+d x)^2}{d e^2 (c+d x)} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*ArcTanh[c + d*x])^2/(c*e + d*e*x)^2,x]

[Out]

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

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Maple [B]  time = 0.063, size = 396, normalized size = 3.8 \begin{align*} -{\frac{{a}^{2}}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{{b}^{2} \left ({\it Artanh} \left ( dx+c \right ) \right ) ^{2}}{d{e}^{2} \left ( dx+c \right ) }}-{\frac{{b}^{2}{\it Artanh} \left ( dx+c \right ) \ln \left ( dx+c-1 \right ) }{d{e}^{2}}}+2\,{\frac{{b}^{2}\ln \left ( dx+c \right ){\it Artanh} \left ( dx+c \right ) }{d{e}^{2}}}-{\frac{{b}^{2}{\it Artanh} \left ( dx+c \right ) \ln \left ( dx+c+1 \right ) }{d{e}^{2}}}-{\frac{{b}^{2} \left ( \ln \left ( dx+c-1 \right ) \right ) ^{2}}{4\,d{e}^{2}}}+{\frac{{b}^{2}}{d{e}^{2}}{\it dilog} \left ({\frac{1}{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{{b}^{2}\ln \left ( dx+c-1 \right ) }{2\,d{e}^{2}}\ln \left ({\frac{1}{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{{b}^{2}\ln \left ( dx+c+1 \right ) }{2\,d{e}^{2}}\ln \left ( -{\frac{dx}{2}}-{\frac{c}{2}}+{\frac{1}{2}} \right ) }+{\frac{{b}^{2}}{2\,d{e}^{2}}\ln \left ( -{\frac{dx}{2}}-{\frac{c}{2}}+{\frac{1}{2}} \right ) \ln \left ({\frac{1}{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{{b}^{2} \left ( \ln \left ( dx+c+1 \right ) \right ) ^{2}}{4\,d{e}^{2}}}-{\frac{{b}^{2}{\it dilog} \left ( dx+c \right ) }{d{e}^{2}}}-{\frac{{b}^{2}{\it dilog} \left ( dx+c+1 \right ) }{d{e}^{2}}}-{\frac{{b}^{2}\ln \left ( dx+c \right ) \ln \left ( dx+c+1 \right ) }{d{e}^{2}}}-2\,{\frac{ab{\it Artanh} \left ( dx+c \right ) }{d{e}^{2} \left ( dx+c \right ) }}-{\frac{ab\ln \left ( dx+c-1 \right ) }{d{e}^{2}}}+2\,{\frac{ab\ln \left ( dx+c \right ) }{d{e}^{2}}}-{\frac{ab\ln \left ( dx+c+1 \right ) }{d{e}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(d*x+c))^2/(d*e*x+c*e)^2,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -{\left (d{\left (\frac{\log \left (d x + c + 1\right )}{d^{2} e^{2}} - \frac{2 \, \log \left (d x + c\right )}{d^{2} e^{2}} + \frac{\log \left (d x + c - 1\right )}{d^{2} e^{2}}\right )} + \frac{2 \, \operatorname{artanh}\left (d x + c\right )}{d^{2} e^{2} x + c d e^{2}}\right )} a b - \frac{1}{4} \, b^{2}{\left (\frac{\log \left (-d x - c + 1\right )^{2}}{d^{2} e^{2} x + c d e^{2}} + \int -\frac{{\left (d x + c - 1\right )} \log \left (d x + c + 1\right )^{2} + 2 \,{\left (d x -{\left (d x + c - 1\right )} \log \left (d x + c + 1\right ) + c\right )} \log \left (-d x - c + 1\right )}{d^{3} e^{2} x^{3} + c^{3} e^{2} - c^{2} e^{2} +{\left (3 \, c d^{2} e^{2} - d^{2} e^{2}\right )} x^{2} +{\left (3 \, c^{2} d e^{2} - 2 \, c d e^{2}\right )} x}\,{d x}\right )} - \frac{a^{2}}{d^{2} e^{2} x + c d e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(d*(log(d*x + c + 1)/(d^2*e^2) - 2*log(d*x + c)/(d^2*e^2) + log(d*x + c - 1)/(d^2*e^2)) + 2*arctanh(d*x + c)/
(d^2*e^2*x + c*d*e^2))*a*b - 1/4*b^2*(log(-d*x - c + 1)^2/(d^2*e^2*x + c*d*e^2) + integrate(-((d*x + c - 1)*lo
g(d*x + c + 1)^2 + 2*(d*x - (d*x + c - 1)*log(d*x + c + 1) + c)*log(-d*x - c + 1))/(d^3*e^2*x^3 + c^3*e^2 - c^
2*e^2 + (3*c*d^2*e^2 - d^2*e^2)*x^2 + (3*c^2*d*e^2 - 2*c*d*e^2)*x), x)) - a^2/(d^2*e^2*x + c*d*e^2)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{artanh}\left (d x + c\right )^{2} + 2 \, a b \operatorname{artanh}\left (d x + c\right ) + a^{2}}{d^{2} e^{2} x^{2} + 2 \, c d e^{2} x + c^{2} e^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*arctanh(d*x + c)^2 + 2*a*b*arctanh(d*x + c) + a^2)/(d^2*e^2*x^2 + 2*c*d*e^2*x + c^2*e^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(d*x+c))**2/(d*e*x+c*e)**2,x)

[Out]

(Integral(a**2/(c**2 + 2*c*d*x + d**2*x**2), x) + Integral(b**2*atanh(c + d*x)**2/(c**2 + 2*c*d*x + d**2*x**2)
, x) + Integral(2*a*b*atanh(c + d*x)/(c**2 + 2*c*d*x + d**2*x**2), x))/e**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (d x + c\right ) + a\right )}^{2}}{{\left (d e x + c e\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arctanh(d*x + c) + a)^2/(d*e*x + c*e)^2, x)