3.12 \(\int \frac{(a+b \tanh ^{-1}(c x))^2}{d+e x} \, dx\)
Optimal. Leaf size=188 \[ -\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}+\frac{b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 e}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}-\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e} \]
[Out]
-(((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/e) + ((a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c
*x))])/e + (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/e - (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - (2*
c*(d + e*x))/((c*d + e)*(1 + c*x))])/e + (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*e) - (b^2*PolyLog[3, 1 - (2*c*(d
+ e*x))/((c*d + e)*(1 + c*x))])/(2*e)
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Rubi [A] time = 0.0499903, antiderivative size = 188, normalized size of antiderivative = 1.,
number of steps used = 1, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used =
{5922} \[ -\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}+\frac{b \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e}-\frac{b^2 \text{PolyLog}\left (3,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e}+\frac{b^2 \text{PolyLog}\left (3,1-\frac{2}{c x+1}\right )}{2 e}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}-\frac{\log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )^2}{e} \]
Antiderivative was successfully verified.
[In]
Int[(a + b*ArcTanh[c*x])^2/(d + e*x),x]
[Out]
-(((a + b*ArcTanh[c*x])^2*Log[2/(1 + c*x)])/e) + ((a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c
*x))])/e + (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/e - (b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - (2*
c*(d + e*x))/((c*d + e)*(1 + c*x))])/e + (b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*e) - (b^2*PolyLog[3, 1 - (2*c*(d
+ e*x))/((c*d + e)*(1 + c*x))])/(2*e)
Rule 5922
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^2/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])^2*Log[
2/(1 + c*x)])/e, x] + (Simp[((a + b*ArcTanh[c*x])^2*Log[(2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(
b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - 2/(1 + c*x)])/e, x] - Simp[(b*(a + b*ArcTanh[c*x])*PolyLog[2, 1 - (2*c*(
d + e*x))/((c*d + e)*(1 + c*x))])/e, x] + Simp[(b^2*PolyLog[3, 1 - 2/(1 + c*x)])/(2*e), x] - Simp[(b^2*PolyLog
[3, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2,
0]
Rubi steps
\begin{align*} \int \frac{\left (a+b \tanh ^{-1}(c x)\right )^2}{d+e x} \, dx &=-\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2}{1+c x}\right )}{e}+\frac{\left (a+b \tanh ^{-1}(c x)\right )^2 \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e}+\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{e}-\frac{b \left (a+b \tanh ^{-1}(c x)\right ) \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e}+\frac{b^2 \text{Li}_3\left (1-\frac{2}{1+c x}\right )}{2 e}-\frac{b^2 \text{Li}_3\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e}\\ \end{align*}
Mathematica [C] time = 12.5105, size = 938, normalized size = 4.99 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + b*ArcTanh[c*x])^2/(d + e*x),x]
[Out]
(6*a^2*Log[d + e*x] + 6*a*b*ArcTanh[c*x]*(Log[1 - c^2*x^2] + 2*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]]) -
(6*I)*a*b*((-I/4)*(Pi - (2*I)*ArcTanh[c*x])^2 + I*(ArcTanh[(c*d)/e] + ArcTanh[c*x])^2 + (Pi - (2*I)*ArcTanh[c
*x])*Log[1 + E^(2*ArcTanh[c*x])] + (2*I)*(ArcTanh[(c*d)/e] + ArcTanh[c*x])*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + A
rcTanh[c*x]))] - (Pi - (2*I)*ArcTanh[c*x])*Log[2/Sqrt[1 - c^2*x^2]] - (2*I)*(ArcTanh[(c*d)/e] + ArcTanh[c*x])*
Log[(2*I)*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] - I*PolyLog[2, -E^(2*ArcTanh[c*x])] - I*PolyLog[2, E^(-2*(Arc
Tanh[(c*d)/e] + ArcTanh[c*x]))]) + (b^2*(-8*c*d*ArcTanh[c*x]^3 + 4*e*ArcTanh[c*x]^3 - (4*Sqrt[1 - (c^2*d^2)/e^
2]*e*ArcTanh[c*x]^3)/E^ArcTanh[(c*d)/e] - 6*c*d*ArcTanh[c*x]^2*Log[1 + E^(-2*ArcTanh[c*x])] - (6*I)*c*d*Pi*Arc
Tanh[c*x]*Log[(E^(-ArcTanh[c*x]) + E^ArcTanh[c*x])/2] - 6*c*d*ArcTanh[c*x]^2*Log[1 + ((c*d + e)*E^(2*ArcTanh[c
*x]))/(c*d - e)] + 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Lo
g[1 + E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]^2*Log[1 - E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x
]))] + 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[(I/2)*E^(-ArcTanh[(c*d)/e] - ArcTanh[c*x])*(-1 + E^(2*(ArcTanh
[(c*d)/e] + ArcTanh[c*x])))] + 6*c*d*ArcTanh[c*x]^2*Log[(e*(-1 + E^(2*ArcTanh[c*x])) + c*d*(1 + E^(2*ArcTanh[c
*x])))/(2*E^ArcTanh[c*x])] - 6*c*d*ArcTanh[c*x]^2*Log[(c*(d + e*x))/Sqrt[1 - c^2*x^2]] - (3*I)*c*d*Pi*ArcTanh[
c*x]*Log[1 - c^2*x^2] - 12*c*d*ArcTanh[(c*d)/e]*ArcTanh[c*x]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]]] + 6*
c*d*ArcTanh[c*x]*PolyLog[2, -E^(-2*ArcTanh[c*x])] - 6*c*d*ArcTanh[c*x]*PolyLog[2, -(((c*d + e)*E^(2*ArcTanh[c*
x]))/(c*d - e))] + 12*c*d*ArcTanh[c*x]*PolyLog[2, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 12*c*d*ArcTanh[c*x]*
PolyLog[2, E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] + 6*c*d*ArcTanh[c*x]*PolyLog[2, E^(2*(ArcTanh[(c*d)/e] + ArcTa
nh[c*x]))] + 3*c*d*PolyLog[3, -E^(-2*ArcTanh[c*x])] + 3*c*d*PolyLog[3, -(((c*d + e)*E^(2*ArcTanh[c*x]))/(c*d -
e))] - 12*c*d*PolyLog[3, -E^(ArcTanh[(c*d)/e] + ArcTanh[c*x])] - 12*c*d*PolyLog[3, E^(ArcTanh[(c*d)/e] + ArcT
anh[c*x])] - 3*c*d*PolyLog[3, E^(2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))]))/(c*d))/(6*e)
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Maple [C] time = 0.549, size = 1170, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*arctanh(c*x))^2/(e*x+d),x)
[Out]
a^2*ln(c*e*x+c*d)/e+b^2*ln(c*e*x+c*d)/e*arctanh(c*x)^2-b^2/e*arctanh(c*x)^2*ln(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*
d*((c*x+1)^2/(-c^2*x^2+1)+1))+1/2*I*b^2/e*arctanh(c*x)^2*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(((c*x+1
)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(I*(((c*x+1)^2/(-c^2*x^2
+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1)))-1/2*I*b^2/e*arctanh(c*x)^2*Pi*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c
*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))^2*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)
^2/(-c^2*x^2+1)+1)))+1/2*I*b^2/e*arctanh(c*x)^2*Pi*csgn(I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x
^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))^3-1/2*I*b^2/e*arctanh(c*x)^2*Pi*csgn(I/((c*x+1)^2/(-c^2*x^2+1)+1))*csgn(
I*(((c*x+1)^2/(-c^2*x^2+1)-1)*e+c*d*((c*x+1)^2/(-c^2*x^2+1)+1))/((c*x+1)^2/(-c^2*x^2+1)+1))^2-b^2/e*arctanh(c*
x)*polylog(2,-(c*x+1)^2/(-c^2*x^2+1))+1/2*b^2/e*polylog(3,-(c*x+1)^2/(-c^2*x^2+1))+b^2/(c*d+e)*arctanh(c*x)^2*
ln(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+b^2/(c*d+e)*arctanh(c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1
)/(-c*d+e))-1/2*b^2/(c*d+e)*polylog(3,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+c*b^2/e*d/(c*d+e)*arctanh(c*x)^
2*ln(1-(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+c*b^2/e*d/(c*d+e)*arctanh(c*x)*polylog(2,(c*d+e)*(c*x+1)^2/(-c
^2*x^2+1)/(-c*d+e))-1/2*c*b^2/e*d/(c*d+e)*polylog(3,(c*d+e)*(c*x+1)^2/(-c^2*x^2+1)/(-c*d+e))+2*a*b*ln(c*e*x+c*
d)/e*arctanh(c*x)-a*b/e*ln(c*e*x+c*d)*ln((c*e*x+e)/(-c*d+e))-a*b/e*dilog((c*e*x+e)/(-c*d+e))+a*b/e*ln(c*e*x+c*
d)*ln((c*e*x-e)/(-c*d-e))+a*b/e*dilog((c*e*x-e)/(-c*d-e))
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \frac{a^{2} \log \left (e x + d\right )}{e} + \int \frac{b^{2}{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}^{2}}{4 \,{\left (e x + d\right )}} + \frac{a b{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{e x + d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctanh(c*x))^2/(e*x+d),x, algorithm="maxima")
[Out]
a^2*log(e*x + d)/e + integrate(1/4*b^2*(log(c*x + 1) - log(-c*x + 1))^2/(e*x + d) + a*b*(log(c*x + 1) - log(-c
*x + 1))/(e*x + d), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b^{2} \operatorname{artanh}\left (c x\right )^{2} + 2 \, a b \operatorname{artanh}\left (c x\right ) + a^{2}}{e x + d}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctanh(c*x))^2/(e*x+d),x, algorithm="fricas")
[Out]
integral((b^2*arctanh(c*x)^2 + 2*a*b*arctanh(c*x) + a^2)/(e*x + d), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{atanh}{\left (c x \right )}\right )^{2}}{d + e x}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*atanh(c*x))**2/(e*x+d),x)
[Out]
Integral((a + b*atanh(c*x))**2/(d + e*x), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )}^{2}}{e x + d}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*arctanh(c*x))^2/(e*x+d),x, algorithm="giac")
[Out]
integrate((b*arctanh(c*x) + a)^2/(e*x + d), x)