3.2.52 \(\int \frac {a+b \tanh ^{-1}(c x)}{x^2 (d+e x)} \, dx\) [152]

Optimal. Leaf size=200 \[ -\frac {a+b \tanh ^{-1}(c x)}{d x}+\frac {b c \log (x)}{d}-\frac {a e \log (x)}{d^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d^2}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d}+\frac {b e \text {PolyLog}(2,-c x)}{2 d^2}-\frac {b e \text {PolyLog}(2,c x)}{2 d^2}+\frac {b e \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 d^2}-\frac {b e \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d^2} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 200, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {6087, 6037, 272, 36, 29, 31, 6031, 6057, 2449, 2352, 2497} \begin {gather*} -\frac {e \log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{d^2}-\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {a e \log (x)}{d^2}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d}+\frac {b e \text {Li}_2(-c x)}{2 d^2}-\frac {b e \text {Li}_2(c x)}{2 d^2}+\frac {b e \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 d^2}-\frac {b e \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 d^2}+\frac {b c \log (x)}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 6031

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b/2)*PolyLog[2, (-c)*x]
, x] + Simp[(b/2)*PolyLog[2, c*x], x]) /; FreeQ[{a, b, c}, x]

Rule 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

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 6087

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {a+b \tanh ^{-1}(c x)}{x^2 (d+e x)} \, dx &=\int \left (\frac {a+b \tanh ^{-1}(c x)}{d x^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right )}{d^2 x}+\frac {e^2 \left (a+b \tanh ^{-1}(c x)\right )}{d^2 (d+e x)}\right ) \, dx\\ &=\frac {\int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx}{d}-\frac {e \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx}{d^2}+\frac {e^2 \int \frac {a+b \tanh ^{-1}(c x)}{d+e x} \, dx}{d^2}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {a e \log (x)}{d^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d^2}+\frac {b e \text {Li}_2(-c x)}{2 d^2}-\frac {b e \text {Li}_2(c x)}{2 d^2}+\frac {(b c) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx}{d}+\frac {(b c e) \int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx}{d^2}-\frac {(b c e) \int \frac {\log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{1-c^2 x^2} \, dx}{d^2}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {a e \log (x)}{d^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d^2}+\frac {b e \text {Li}_2(-c x)}{2 d^2}-\frac {b e \text {Li}_2(c x)}{2 d^2}-\frac {b e \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d^2}+\frac {(b c) \text {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )}{2 d}+\frac {(b e) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c x}\right )}{d^2}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d x}-\frac {a e \log (x)}{d^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d^2}+\frac {b e \text {Li}_2(-c x)}{2 d^2}-\frac {b e \text {Li}_2(c x)}{2 d^2}+\frac {b e \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 d^2}-\frac {b e \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d^2}+\frac {(b c) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )}{2 d}+\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )}{2 d}\\ &=-\frac {a+b \tanh ^{-1}(c x)}{d x}+\frac {b c \log (x)}{d}-\frac {a e \log (x)}{d^2}-\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{d^2}+\frac {e \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{d^2}-\frac {b c \log \left (1-c^2 x^2\right )}{2 d}+\frac {b e \text {Li}_2(-c x)}{2 d^2}-\frac {b e \text {Li}_2(c x)}{2 d^2}+\frac {b e \text {Li}_2\left (1-\frac {2}{1+c x}\right )}{2 d^2}-\frac {b e \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 d^2}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 2.23, size = 360, normalized size = 1.80 \begin {gather*} -\frac {\frac {2 a d^2}{x}-i b d e \pi \tanh ^{-1}(c x)+\frac {2 b d^2 \tanh ^{-1}(c x)}{x}-2 b d e \tanh ^{-1}\left (\frac {c d}{e}\right ) \tanh ^{-1}(c x)+b d e \tanh ^{-1}(c x)^2-\frac {b e^2 \tanh ^{-1}(c x)^2}{c}+\frac {b \sqrt {1-\frac {c^2 d^2}{e^2}} e^2 e^{-\tanh ^{-1}\left (\frac {c d}{e}\right )} \tanh ^{-1}(c x)^2}{c}+2 b d e \tanh ^{-1}(c x) \log \left (1-e^{-2 \tanh ^{-1}(c x)}\right )+i b d e \pi \log \left (1+e^{2 \tanh ^{-1}(c x)}\right )-2 b d e \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 b d e \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+2 a d e \log (x)-2 a d e \log (d+e x)-2 b c d^2 \log \left (\frac {c x}{\sqrt {1-c^2 x^2}}\right )+\frac {1}{2} i b d e \pi \log \left (1-c^2 x^2\right )+2 b d e \tanh ^{-1}\left (\frac {c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )-b d e \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}(c x)}\right )+b d e \text {PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )}{2 d^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((2*a*d^2)/x - I*b*d*e*Pi*ArcTanh[c*x] + (2*b*d^2*ArcTanh[c*x])/x - 2*b*d*e*ArcTanh[(c*d)/e]*ArcTanh[c*x]
+ b*d*e*ArcTanh[c*x]^2 - (b*e^2*ArcTanh[c*x]^2)/c + (b*Sqrt[1 - (c^2*d^2)/e^2]*e^2*ArcTanh[c*x]^2)/(c*E^ArcTa
nh[(c*d)/e]) + 2*b*d*e*ArcTanh[c*x]*Log[1 - E^(-2*ArcTanh[c*x])] + I*b*d*e*Pi*Log[1 + E^(2*ArcTanh[c*x])] - 2*
b*d*e*ArcTanh[(c*d)/e]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - 2*b*d*e*ArcTanh[c*x]*Log[1 - E^(-2*
(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] + 2*a*d*e*Log[x] - 2*a*d*e*Log[d + e*x] - 2*b*c*d^2*Log[(c*x)/Sqrt[1 - c^2
*x^2]] + (I/2)*b*d*e*Pi*Log[1 - c^2*x^2] + 2*b*d*e*ArcTanh[(c*d)/e]*Log[I*Sinh[ArcTanh[(c*d)/e] + ArcTanh[c*x]
]] - b*d*e*PolyLog[2, E^(-2*ArcTanh[c*x])] + b*d*e*PolyLog[2, E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))])/d^3

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Maple [A]
time = 3.70, size = 317, normalized size = 1.58

method	result	size
risch	\(\frac {c b \ln \left (-c x \right )}{2 d}-\frac {c b \ln \left (-c x +1\right )}{2 d}+\frac {b \ln \left (-c x +1\right )}{2 d x}-\frac {b e \dilog \left (\frac {\left (-c x +1\right ) e -d c -e}{-d c -e}\right )}{2 d^{2}}-\frac {b e \ln \left (-c x +1\right ) \ln \left (\frac {\left (-c x +1\right ) e -d c -e}{-d c -e}\right )}{2 d^{2}}-\frac {b \dilog \left (-c x +1\right ) e}{2 d^{2}}+\frac {a e \ln \left (\left (-c x +1\right ) e -d c -e \right )}{d^{2}}-\frac {a}{d x}-\frac {a e \ln \left (-c x \right )}{d^{2}}+\frac {b c \ln \left (c x \right )}{2 d}-\frac {b c \ln \left (c x +1\right )}{2 d}-\frac {b \ln \left (c x +1\right )}{2 d x}+\frac {b e \dilog \left (\frac {\left (c x +1\right ) e +d c -e}{d c -e}\right )}{2 d^{2}}+\frac {b e \ln \left (c x +1\right ) \ln \left (\frac {\left (c x +1\right ) e +d c -e}{d c -e}\right )}{2 d^{2}}+\frac {b \dilog \left (c x +1\right ) e}{2 d^{2}}\)	\(301\)
derivativedivides	\(c \left (\frac {a e \ln \left (c e x +d c \right )}{c \,d^{2}}-\frac {a}{d c x}-\frac {a e \ln \left (c x \right )}{c \,d^{2}}+\frac {b \arctanh \left (c x \right ) e \ln \left (c e x +d c \right )}{c \,d^{2}}-\frac {b \arctanh \left (c x \right )}{d c x}-\frac {b \arctanh \left (c x \right ) e \ln \left (c x \right )}{c \,d^{2}}-\frac {b \ln \left (c x +1\right )}{2 d}-\frac {b \ln \left (c x -1\right )}{2 d}+\frac {b \ln \left (c x \right )}{d}+\frac {b e \dilog \left (c x +1\right )}{2 c \,d^{2}}+\frac {b e \ln \left (c x \right ) \ln \left (c x +1\right )}{2 c \,d^{2}}+\frac {b e \dilog \left (c x \right )}{2 c \,d^{2}}+\frac {b \ln \left (c e x +d c \right ) \ln \left (\frac {c e x -e}{-d c -e}\right ) e}{2 c \,d^{2}}+\frac {b \dilog \left (\frac {c e x -e}{-d c -e}\right ) e}{2 c \,d^{2}}-\frac {b \ln \left (c e x +d c \right ) \ln \left (\frac {c e x +e}{-d c +e}\right ) e}{2 c \,d^{2}}-\frac {b \dilog \left (\frac {c e x +e}{-d c +e}\right ) e}{2 c \,d^{2}}\right )\)	\(317\)
default	\(c \left (\frac {a e \ln \left (c e x +d c \right )}{c \,d^{2}}-\frac {a}{d c x}-\frac {a e \ln \left (c x \right )}{c \,d^{2}}+\frac {b \arctanh \left (c x \right ) e \ln \left (c e x +d c \right )}{c \,d^{2}}-\frac {b \arctanh \left (c x \right )}{d c x}-\frac {b \arctanh \left (c x \right ) e \ln \left (c x \right )}{c \,d^{2}}-\frac {b \ln \left (c x +1\right )}{2 d}-\frac {b \ln \left (c x -1\right )}{2 d}+\frac {b \ln \left (c x \right )}{d}+\frac {b e \dilog \left (c x +1\right )}{2 c \,d^{2}}+\frac {b e \ln \left (c x \right ) \ln \left (c x +1\right )}{2 c \,d^{2}}+\frac {b e \dilog \left (c x \right )}{2 c \,d^{2}}+\frac {b \ln \left (c e x +d c \right ) \ln \left (\frac {c e x -e}{-d c -e}\right ) e}{2 c \,d^{2}}+\frac {b \dilog \left (\frac {c e x -e}{-d c -e}\right ) e}{2 c \,d^{2}}-\frac {b \ln \left (c e x +d c \right ) \ln \left (\frac {c e x +e}{-d c +e}\right ) e}{2 c \,d^{2}}-\frac {b \dilog \left (\frac {c e x +e}{-d c +e}\right ) e}{2 c \,d^{2}}\right )\)	\(317\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))/x^2/(e*x+d),x, algorithm="maxima")

[Out]

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

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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))/x^2/(e*x+d),x, algorithm="fricas")

[Out]

integral((b*arctanh(c*x) + a)/(x^3*e + d*x^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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