3.1.13 \(\int \frac {(a+b \tanh ^{-1}(c x))^2}{(d+e x)^2} \, dx\) [13]

Optimal. Leaf size=321 \[ -\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{e (d+e x)}+\frac {b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1-c x}\right )}{e (c d+e)}-\frac {b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{(c d-e) e}+\frac {2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{c^2 d^2-e^2}-\frac {2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{c^2 d^2-e^2}+\frac {b^2 c \text {PolyLog}\left (2,1-\frac {2}{1-c x}\right )}{2 e (c d+e)}+\frac {b^2 c \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 (c d-e) e}-\frac {b^2 c \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{c^2 d^2-e^2}+\frac {b^2 c \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{c^2 d^2-e^2} \]

[Out]

-(a+b*arctanh(c*x))^2/e/(e*x+d)+b*c*(a+b*arctanh(c*x))*ln(2/(-c*x+1))/e/(c*d+e)-b*c*(a+b*arctanh(c*x))*ln(2/(c
*x+1))/(c*d-e)/e+2*b*c*(a+b*arctanh(c*x))*ln(2/(c*x+1))/(c^2*d^2-e^2)-2*b*c*(a+b*arctanh(c*x))*ln(2*c*(e*x+d)/
(c*d+e)/(c*x+1))/(c^2*d^2-e^2)+1/2*b^2*c*polylog(2,1-2/(-c*x+1))/e/(c*d+e)+1/2*b^2*c*polylog(2,1-2/(c*x+1))/(c
*d-e)/e-b^2*c*polylog(2,1-2/(c*x+1))/(c^2*d^2-e^2)+b^2*c*polylog(2,1-2*c*(e*x+d)/(c*d+e)/(c*x+1))/(c^2*d^2-e^2
)

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Rubi [A]
time = 0.23, antiderivative size = 321, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6065, 6055, 2449, 2352, 6057, 2497} \begin {gather*} \frac {2 b c \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2 d^2-e^2}-\frac {2 b c \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{c^2 d^2-e^2}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{e (d+e x)}+\frac {b c \log \left (\frac {2}{1-c x}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e (c d+e)}-\frac {b c \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e (c d-e)}-\frac {b^2 c \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{c^2 d^2-e^2}+\frac {b^2 c \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{c^2 d^2-e^2}+\frac {b^2 c \text {Li}_2\left (1-\frac {2}{1-c x}\right )}{2 e (c d+e)}+\frac {b^2 c \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 e (c d-e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

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 2497

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 6055

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTanh[c*x])^p)
*(Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[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 6057

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 6065

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

Rubi steps

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

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Mathematica [C] Result contains complex when optimal does not.
time = 3.57, size = 317, normalized size = 0.99 \begin {gather*} -\frac {a^2}{e (d+e x)}+\frac {a b c \left (-\frac {2 \tanh ^{-1}(c x)}{c d+c e x}+\frac {(-c d+e) \log (1-c x)+(c d+e) \log (1+c x)-2 e \log (c (d+e x))}{(c d-e) (c d+e)}\right )}{e}+\frac {b^2 \left (-\frac {e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )} \tanh ^{-1}(c x)^2}{\sqrt {1-\frac {c^2 d^2}{e^2}} e}+\frac {x \tanh ^{-1}(c x)^2}{d+e x}+\frac {c d \left (i \pi \log \left (1+e^{2 \tanh ^{-1}(c x)}\right )-2 \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-i \pi \left (\tanh ^{-1}(c x)-\frac {1}{2} \log \left (1-c^2 x^2\right )\right )-2 \tanh ^{-1}\left (\frac {c d}{e}\right ) \left (\tanh ^{-1}(c x)+\log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-\log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )\right )+\text {PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )\right )}{c^2 d^2-e^2}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(648\) vs. \(2(317)=634\).
time = 1.23, size = 649, normalized size = 2.02

method	result	size
derivativedivides	\(\frac {-\frac {a^{2} c^{2}}{\left (c e x +d c \right ) e}-\frac {b^{2} c^{2} \arctanh \left (c x \right )^{2}}{\left (c e x +d c \right ) e}-\frac {2 b^{2} c^{2} \arctanh \left (c x \right ) \ln \left (c e x +d c \right )}{\left (d c +e \right ) \left (d c -e \right )}+\frac {2 b^{2} c^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{e \left (2 d c -2 e \right )}-\frac {2 b^{2} c^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{e \left (2 d c +2 e \right )}-\frac {b^{2} c^{2} \ln \left (c e x +d c \right ) \ln \left (\frac {c e x -e}{-d c -e}\right )}{\left (d c +e \right ) \left (d c -e \right )}-\frac {b^{2} c^{2} \dilog \left (\frac {c e x -e}{-d c -e}\right )}{\left (d c +e \right ) \left (d c -e \right )}+\frac {b^{2} c^{2} \ln \left (c e x +d c \right ) \ln \left (\frac {c e x +e}{-d c +e}\right )}{\left (d c +e \right ) \left (d c -e \right )}+\frac {b^{2} c^{2} \dilog \left (\frac {c e x +e}{-d c +e}\right )}{\left (d c +e \right ) \left (d c -e \right )}+\frac {b^{2} c^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 e \left (d c -e \right )}-\frac {b^{2} c^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 e \left (d c -e \right )}-\frac {b^{2} c^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 e \left (d c -e \right )}-\frac {b^{2} c^{2} \ln \left (c x +1\right )^{2}}{4 e \left (d c -e \right )}-\frac {b^{2} c^{2} \ln \left (c x -1\right )^{2}}{4 e \left (d c +e \right )}+\frac {b^{2} c^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 e \left (d c +e \right )}+\frac {b^{2} c^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 e \left (d c +e \right )}-\frac {2 a b \,c^{2} \arctanh \left (c x \right )}{\left (c e x +d c \right ) e}-\frac {2 a b \,c^{2} \ln \left (c e x +d c \right )}{\left (d c +e \right ) \left (d c -e \right )}+\frac {2 a b \,c^{2} \ln \left (c x +1\right )}{e \left (2 d c -2 e \right )}-\frac {2 a b \,c^{2} \ln \left (c x -1\right )}{e \left (2 d c +2 e \right )}}{c}\)	\(649\)
default	\(\frac {-\frac {a^{2} c^{2}}{\left (c e x +d c \right ) e}-\frac {b^{2} c^{2} \arctanh \left (c x \right )^{2}}{\left (c e x +d c \right ) e}-\frac {2 b^{2} c^{2} \arctanh \left (c x \right ) \ln \left (c e x +d c \right )}{\left (d c +e \right ) \left (d c -e \right )}+\frac {2 b^{2} c^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{e \left (2 d c -2 e \right )}-\frac {2 b^{2} c^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{e \left (2 d c +2 e \right )}-\frac {b^{2} c^{2} \ln \left (c e x +d c \right ) \ln \left (\frac {c e x -e}{-d c -e}\right )}{\left (d c +e \right ) \left (d c -e \right )}-\frac {b^{2} c^{2} \dilog \left (\frac {c e x -e}{-d c -e}\right )}{\left (d c +e \right ) \left (d c -e \right )}+\frac {b^{2} c^{2} \ln \left (c e x +d c \right ) \ln \left (\frac {c e x +e}{-d c +e}\right )}{\left (d c +e \right ) \left (d c -e \right )}+\frac {b^{2} c^{2} \dilog \left (\frac {c e x +e}{-d c +e}\right )}{\left (d c +e \right ) \left (d c -e \right )}+\frac {b^{2} c^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 e \left (d c -e \right )}-\frac {b^{2} c^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 e \left (d c -e \right )}-\frac {b^{2} c^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 e \left (d c -e \right )}-\frac {b^{2} c^{2} \ln \left (c x +1\right )^{2}}{4 e \left (d c -e \right )}-\frac {b^{2} c^{2} \ln \left (c x -1\right )^{2}}{4 e \left (d c +e \right )}+\frac {b^{2} c^{2} \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 e \left (d c +e \right )}+\frac {b^{2} c^{2} \ln \left (c x -1\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 e \left (d c +e \right )}-\frac {2 a b \,c^{2} \arctanh \left (c x \right )}{\left (c e x +d c \right ) e}-\frac {2 a b \,c^{2} \ln \left (c e x +d c \right )}{\left (d c +e \right ) \left (d c -e \right )}+\frac {2 a b \,c^{2} \ln \left (c x +1\right )}{e \left (2 d c -2 e \right )}-\frac {2 a b \,c^{2} \ln \left (c x -1\right )}{e \left (2 d c +2 e \right )}}{c}\)	\(649\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right )^{2}}{\left (d + e x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )}^2}{{\left (d+e\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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