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

Optimal. Leaf size=258 \[ -\frac {\text {Li}_2\left (1-\frac {2 \left (b+2 c x-\sqrt {b^2-4 a c}\right )}{\left (b+2 c-\sqrt {b^2-4 a c}\right ) (x+1)}\right )}{2 \sqrt {b^2-4 a c}}+\frac {\text {Li}_2\left (1-\frac {2 \left (b+2 c x+\sqrt {b^2-4 a c}\right )}{\left (b+2 c+\sqrt {b^2-4 a c}\right ) (x+1)}\right )}{2 \sqrt {b^2-4 a c}}+\frac {\tanh ^{-1}(x) \log \left (\frac {2 \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{(x+1) \left (-\sqrt {b^2-4 a c}+b+2 c\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {\tanh ^{-1}(x) \log \left (\frac {2 \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{(x+1) \left (\sqrt {b^2-4 a c}+b+2 c\right )}\right )}{\sqrt {b^2-4 a c}} \]

[Out]

arctanh(x)*ln(2*(b+2*c*x-(-4*a*c+b^2)^(1/2))/(1+x)/(b+2*c-(-4*a*c+b^2)^(1/2)))/(-4*a*c+b^2)^(1/2)-arctanh(x)*l
n(2*(b+2*c*x+(-4*a*c+b^2)^(1/2))/(1+x)/(b+2*c+(-4*a*c+b^2)^(1/2)))/(-4*a*c+b^2)^(1/2)-1/2*polylog(2,1-2*(b+2*c
*x-(-4*a*c+b^2)^(1/2))/(1+x)/(b+2*c-(-4*a*c+b^2)^(1/2)))/(-4*a*c+b^2)^(1/2)+1/2*polylog(2,1-2*(b+2*c*x+(-4*a*c
+b^2)^(1/2))/(1+x)/(b+2*c+(-4*a*c+b^2)^(1/2)))/(-4*a*c+b^2)^(1/2)

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Rubi [A]  time = 0.33, antiderivative size = 258, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {618, 206, 6728, 5920, 2402, 2315, 2447} \[ -\frac {\text {PolyLog}\left (2,1-\frac {2 \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{(x+1) \left (-\sqrt {b^2-4 a c}+b+2 c\right )}\right )}{2 \sqrt {b^2-4 a c}}+\frac {\text {PolyLog}\left (2,1-\frac {2 \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{(x+1) \left (\sqrt {b^2-4 a c}+b+2 c\right )}\right )}{2 \sqrt {b^2-4 a c}}+\frac {\tanh ^{-1}(x) \log \left (\frac {2 \left (-\sqrt {b^2-4 a c}+b+2 c x\right )}{(x+1) \left (-\sqrt {b^2-4 a c}+b+2 c\right )}\right )}{\sqrt {b^2-4 a c}}-\frac {\tanh ^{-1}(x) \log \left (\frac {2 \left (\sqrt {b^2-4 a c}+b+2 c x\right )}{(x+1) \left (\sqrt {b^2-4 a c}+b+2 c\right )}\right )}{\sqrt {b^2-4 a c}} \]

Antiderivative was successfully verified.

[In]

Int[ArcTanh[x]/(a + b*x + c*x^2),x]

[Out]

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

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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]

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 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]]

Rule 5920

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])*Log[2/(1
+ c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d +
e*x))/((c*d + e)*(1 + c*x))]/(1 - c^2*x^2), x], x] + Simp[((a + b*ArcTanh[c*x])*Log[(2*c*(d + e*x))/((c*d + e)
*(1 + c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]

Rule 6728

Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a +
b*x^n + c*x^(2*n)), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\tanh ^{-1}(x)}{a+b x+c x^2} \, dx &=\int \left (\frac {2 c \tanh ^{-1}(x)}{\sqrt {b^2-4 a c} \left (b-\sqrt {b^2-4 a c}+2 c x\right )}-\frac {2 c \tanh ^{-1}(x)}{\sqrt {b^2-4 a c} \left (b+\sqrt {b^2-4 a c}+2 c x\right )}\right ) \, dx\\ &=\frac {(2 c) \int \frac {\tanh ^{-1}(x)}{b-\sqrt {b^2-4 a c}+2 c x} \, dx}{\sqrt {b^2-4 a c}}-\frac {(2 c) \int \frac {\tanh ^{-1}(x)}{b+\sqrt {b^2-4 a c}+2 c x} \, dx}{\sqrt {b^2-4 a c}}\\ &=\frac {\tanh ^{-1}(x) \log \left (\frac {2 \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+2 c-\sqrt {b^2-4 a c}\right ) (1+x)}\right )}{\sqrt {b^2-4 a c}}-\frac {\tanh ^{-1}(x) \log \left (\frac {2 \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+2 c+\sqrt {b^2-4 a c}\right ) (1+x)}\right )}{\sqrt {b^2-4 a c}}-\frac {\int \frac {\log \left (\frac {2 \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+2 c-\sqrt {b^2-4 a c}\right ) (1+x)}\right )}{1-x^2} \, dx}{\sqrt {b^2-4 a c}}+\frac {\int \frac {\log \left (\frac {2 \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+2 c+\sqrt {b^2-4 a c}\right ) (1+x)}\right )}{1-x^2} \, dx}{\sqrt {b^2-4 a c}}\\ &=\frac {\tanh ^{-1}(x) \log \left (\frac {2 \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+2 c-\sqrt {b^2-4 a c}\right ) (1+x)}\right )}{\sqrt {b^2-4 a c}}-\frac {\tanh ^{-1}(x) \log \left (\frac {2 \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+2 c+\sqrt {b^2-4 a c}\right ) (1+x)}\right )}{\sqrt {b^2-4 a c}}-\frac {\text {Li}_2\left (1-\frac {2 \left (b-\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+2 c-\sqrt {b^2-4 a c}\right ) (1+x)}\right )}{2 \sqrt {b^2-4 a c}}+\frac {\text {Li}_2\left (1-\frac {2 \left (b+\sqrt {b^2-4 a c}+2 c x\right )}{\left (b+2 c+\sqrt {b^2-4 a c}\right ) (1+x)}\right )}{2 \sqrt {b^2-4 a c}}\\ \end {align*}

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Mathematica [C]  time = 19.32, size = 874, normalized size = 3.39 \[ \frac {\frac {2 \sqrt {4 a c-b^2} \left (b \left (\sqrt {\frac {c (a+b+c)}{4 a c-b^2}} e^{i \tan ^{-1}\left (\frac {-b-2 c}{\sqrt {4 a c-b^2}}\right )}-\sqrt {\frac {c (a-b+c)}{4 a c-b^2}} e^{i \tan ^{-1}\left (\frac {2 c-b}{\sqrt {4 a c-b^2}}\right )}\right )-2 c \left (e^{i \tan ^{-1}\left (\frac {2 c-b}{\sqrt {4 a c-b^2}}\right )} \sqrt {\frac {c (a-b+c)}{4 a c-b^2}}+\sqrt {\frac {c (a+b+c)}{4 a c-b^2}} e^{i \tan ^{-1}\left (\frac {-b-2 c}{\sqrt {4 a c-b^2}}\right )}-1\right )\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )^2}{b^2-4 c^2}+2 \left (-i \tan ^{-1}\left (\frac {-b-2 c}{\sqrt {4 a c-b^2}}\right )+i \tan ^{-1}\left (\frac {2 c-b}{\sqrt {4 a c-b^2}}\right )+2 \tanh ^{-1}(x)+\log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac {-b-2 c}{\sqrt {4 a c-b^2}}\right )+\tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )\right )}\right )-\log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac {2 c-b}{\sqrt {4 a c-b^2}}\right )+\tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )\right )}\right )\right ) \tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )+2 \left (\tan ^{-1}\left (\frac {-b-2 c}{\sqrt {4 a c-b^2}}\right ) \left (\log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac {-b-2 c}{\sqrt {4 a c-b^2}}\right )+\tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )\right )}\right )-\log \left (\sin \left (\tan ^{-1}\left (\frac {-b-2 c}{\sqrt {4 a c-b^2}}\right )+\tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )\right )\right )\right )+\tan ^{-1}\left (\frac {2 c-b}{\sqrt {4 a c-b^2}}\right ) \left (\log \left (\sin \left (\tan ^{-1}\left (\frac {2 c-b}{\sqrt {4 a c-b^2}}\right )+\tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )\right )\right )-\log \left (1-e^{2 i \left (\tan ^{-1}\left (\frac {2 c-b}{\sqrt {4 a c-b^2}}\right )+\tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )\right )}\right )\right )\right )-i \text {Li}_2\left (e^{2 i \left (\tan ^{-1}\left (\frac {-b-2 c}{\sqrt {4 a c-b^2}}\right )+\tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )\right )}\right )+i \text {Li}_2\left (e^{2 i \left (\tan ^{-1}\left (\frac {2 c-b}{\sqrt {4 a c-b^2}}\right )+\tan ^{-1}\left (\frac {b+2 c x}{\sqrt {4 a c-b^2}}\right )\right )}\right )}{2 \sqrt {4 a c-b^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((2*Sqrt[-b^2 + 4*a*c]*(b*(Sqrt[(c*(a + b + c))/(-b^2 + 4*a*c)]*E^(I*ArcTan[(-b - 2*c)/Sqrt[-b^2 + 4*a*c]]) -
Sqrt[(c*(a - b + c))/(-b^2 + 4*a*c)]*E^(I*ArcTan[(-b + 2*c)/Sqrt[-b^2 + 4*a*c]])) - 2*c*(-1 + Sqrt[(c*(a + b +
c))/(-b^2 + 4*a*c)]*E^(I*ArcTan[(-b - 2*c)/Sqrt[-b^2 + 4*a*c]]) + Sqrt[(c*(a - b + c))/(-b^2 + 4*a*c)]*E^(I*A
rcTan[(-b + 2*c)/Sqrt[-b^2 + 4*a*c]])))*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]]^2)/(b^2 - 4*c^2) + 2*ArcTan[(b
+ 2*c*x)/Sqrt[-b^2 + 4*a*c]]*((-I)*ArcTan[(-b - 2*c)/Sqrt[-b^2 + 4*a*c]] + I*ArcTan[(-b + 2*c)/Sqrt[-b^2 + 4*a
*c]] + 2*ArcTanh[x] + Log[1 - E^((2*I)*(ArcTan[(-b - 2*c)/Sqrt[-b^2 + 4*a*c]] + ArcTan[(b + 2*c*x)/Sqrt[-b^2 +
4*a*c]]))] - Log[1 - E^((2*I)*(ArcTan[(-b + 2*c)/Sqrt[-b^2 + 4*a*c]] + ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]]
))]) + 2*(ArcTan[(-b - 2*c)/Sqrt[-b^2 + 4*a*c]]*(Log[1 - E^((2*I)*(ArcTan[(-b - 2*c)/Sqrt[-b^2 + 4*a*c]] + Arc
Tan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]]))] - Log[Sin[ArcTan[(-b - 2*c)/Sqrt[-b^2 + 4*a*c]] + ArcTan[(b + 2*c*x)/Sq
rt[-b^2 + 4*a*c]]]]) + ArcTan[(-b + 2*c)/Sqrt[-b^2 + 4*a*c]]*(-Log[1 - E^((2*I)*(ArcTan[(-b + 2*c)/Sqrt[-b^2 +
4*a*c]] + ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]]))] + Log[Sin[ArcTan[(-b + 2*c)/Sqrt[-b^2 + 4*a*c]] + ArcTan[
(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]]]])) - I*PolyLog[2, E^((2*I)*(ArcTan[(-b - 2*c)/Sqrt[-b^2 + 4*a*c]] + ArcTan[(b
+ 2*c*x)/Sqrt[-b^2 + 4*a*c]]))] + I*PolyLog[2, E^((2*I)*(ArcTan[(-b + 2*c)/Sqrt[-b^2 + 4*a*c]] + ArcTan[(b +
2*c*x)/Sqrt[-b^2 + 4*a*c]]))])/(2*Sqrt[-b^2 + 4*a*c])

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\relax (x)}{c x^{2} + b x + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\relax (x)}{c x^{2} + b x + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.61, size = 1599, normalized size = 6.20 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/(4*a*c-b^2)/(a^2+2*a*c-b^2+c^2)*arctanh(x)^2*a^2*(-4*a*c+b^2)^(1/2)-1/(a^2+2*a*c-b^2+c^2)*arctanh(x)^2*(-4*
a*c+b^2)^(1/2)+1/(a^2+2*a*c-b^2+c^2)*arctanh(x)^2*a-1/(a^2+2*a*c-b^2+c^2)*arctanh(x)^2*c+1/2/(a^2+2*a*c-b^2+c^
2)*polylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*(-4*a*c+b^2)^(1/2)-1/2/(a^2+2*a*c-b^2+c^2)*po
lylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*a-1/(4*a*c-b^2)/(a^2+2*a*c-b^2+c^2)*arctanh(x)^2*c
*b^2+1/(4*a*c-b^2)/(a^2+2*a*c-b^2+c^2)*arctanh(x)^2*a*b^2-1/(4*a*c-b^2)/(a^2+2*a*c-b^2+c^2)*arctanh(x)^2*c^2*(
-4*a*c+b^2)^(1/2)+1/2/(a^2+2*a*c-b^2+c^2)*polylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*c-1/2*
(-4*a*c+b^2)^(1/2)/(4*a*c-b^2)*polylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/((-4*a*c+b^2)^(1/2)-a+c))+(-4*a*c+b^2)^(1/2)
/(4*a*c-b^2)*arctanh(x)^2+2/(4*a*c-b^2)/(a^2+2*a*c-b^2+c^2)*arctanh(x)^2*a*c*(-4*a*c+b^2)^(1/2)-(4*a*c-b^2+(-4
*a*c+b^2)^(1/2)*a-(-4*a*c+b^2)^(1/2)*c)/(4*a*c-b^2)/(a^2+2*a*c-b^2+c^2)*ln(1-(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*
c+b^2)^(1/2)-a+c))*arctanh(x)*c-1/(4*a*c-b^2)/(a^2+2*a*c-b^2+c^2)*polylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c
+b^2)^(1/2)-a+c))*a*c*(-4*a*c+b^2)^(1/2)+(4*a*c-b^2+(-4*a*c+b^2)^(1/2)*a-(-4*a*c+b^2)^(1/2)*c)/(4*a*c-b^2)/(a^
2+2*a*c-b^2+c^2)*ln(1-(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*arctanh(x)*a+1/2/(4*a*c-b^2)/(a^2+2*
a*c-b^2+c^2)*polylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*a^2*(-4*a*c+b^2)^(1/2)+2/(4*a*c-b^2
)/(a^2+2*a*c-b^2+c^2)*polylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*a^2*c-1/2/(4*a*c-b^2)/(a^2
+2*a*c-b^2+c^2)*polylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*a*b^2+1/2/(4*a*c-b^2)/(a^2+2*a*c
-b^2+c^2)*polylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*c*b^2+1/2/(4*a*c-b^2)/(a^2+2*a*c-b^2+c
^2)*polylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*c^2*(-4*a*c+b^2)^(1/2)-2/(4*a*c-b^2)/(a^2+2*
a*c-b^2+c^2)*polylog(2,(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*c^2*a-(-4*a*c+b^2)^(1/2)/(4*a*c-b^2
)*arctanh(x)*ln(1-(a+b+c)*(1+x)^2/(-x^2+1)/((-4*a*c+b^2)^(1/2)-a+c))-(-(-4*a*c+b^2)^(1/2)+a-c)/(a^2+2*a*c-b^2+
c^2)*ln(1-(a+b+c)*(1+x)^2/(-x^2+1)/(-(-4*a*c+b^2)^(1/2)-a+c))*arctanh(x)+4/(4*a*c-b^2)/(a^2+2*a*c-b^2+c^2)*arc
tanh(x)^2*c^2*a-4/(4*a*c-b^2)/(a^2+2*a*c-b^2+c^2)*arctanh(x)^2*a^2*c

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\mathrm {atanh}\relax (x)}{c\,x^2+b\,x+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh(x)/(a + b*x + c*x^2),x)

[Out]

int(atanh(x)/(a + b*x + c*x^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {atanh}{\relax (x )}}{a + b x + c x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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