3.1.47 \(\int \frac {a+b \tanh ^{-1}(c \sqrt {x})}{x (d+e x)} \, dx\) [47]

Optimal. Leaf size=358 \[ \frac {2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{d}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{d}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{d}+\frac {a \log (x)}{d}-\frac {b \text {PolyLog}\left (2,1-\frac {2}{1+c \sqrt {x}}\right )}{d}+\frac {b \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 d}+\frac {b \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 d}-\frac {b \text {PolyLog}\left (2,-c \sqrt {x}\right )}{d}+\frac {b \text {PolyLog}\left (2,c \sqrt {x}\right )}{d} \]

[Out]

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

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Rubi [A]
time = 0.43, antiderivative size = 358, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {36, 29, 31, 1607, 6139, 6031, 6191, 6057, 2449, 2352, 2497} \begin {gather*} -\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}-\sqrt {e}\right )}\right )}{d}-\frac {\left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {x}+1\right ) \left (c \sqrt {-d}+\sqrt {e}\right )}\right )}{d}+\frac {2 \log \left (\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d}+\frac {a \log (x)}{d}+\frac {b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (\sqrt {x} c+1\right )}\right )}{2 d}+\frac {b \text {Li}_2\left (1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (\sqrt {-d} c+\sqrt {e}\right ) \left (\sqrt {x} c+1\right )}\right )}{2 d}-\frac {b \text {Li}_2\left (1-\frac {2}{\sqrt {x} c+1}\right )}{d}-\frac {b \text {Li}_2\left (-c \sqrt {x}\right )}{d}+\frac {b \text {Li}_2\left (c \sqrt {x}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a + b*ArcTanh[c*Sqrt[x]])*Log[2/(1 + c*Sqrt[x])])/d - ((a + b*ArcTanh[c*Sqrt[x]])*Log[(2*c*(Sqrt[-d] - Sqr
t[e]*Sqrt[x]))/((c*Sqrt[-d] - Sqrt[e])*(1 + c*Sqrt[x]))])/d - ((a + b*ArcTanh[c*Sqrt[x]])*Log[(2*c*(Sqrt[-d] +
Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/d + (a*Log[x])/d - (b*PolyLog[2, 1 - 2/(1 + c*Sq
rt[x])])/d + (b*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] - Sqrt[e])*(1 + c*Sqrt[x]))])/(
2*d) + (b*PolyLog[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/(2*d) -
(b*PolyLog[2, -(c*Sqrt[x])])/d + (b*PolyLog[2, c*Sqrt[x]])/d

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 1607

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 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 6139

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*(x_)^(m_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[a
+ b*ArcTanh[c*x], x^m/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] && IntegerQ[m] &&  !(EqQ[m, 1] && NeQ[
a, 0])

Rule 6191

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a,
b, c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && (GtQ[q, 0] || IntegerQ[m])

Rubi steps

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

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Mathematica [A]
time = 0.83, size = 302, normalized size = 0.84 \begin {gather*} -\frac {-4 b \tanh ^{-1}\left (c \sqrt {x}\right )^2-4 b \tanh ^{-1}\left (c \sqrt {x}\right ) \log \left (1-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )+2 b \tanh ^{-1}\left (c \sqrt {x}\right ) \log \left (1+\frac {\left (c^2 d+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}}{c^2 d-2 c \sqrt {-d} \sqrt {e}-e}\right )+2 b \tanh ^{-1}\left (c \sqrt {x}\right ) \log \left (1+\frac {\left (c^2 d+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}}{c^2 d+2 c \sqrt {-d} \sqrt {e}-e}\right )-2 a \log (x)+2 a \log (d+e x)+2 b \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )+b \text {PolyLog}\left (2,-\frac {\left (c^2 d+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}}{c^2 d-2 c \sqrt {-d} \sqrt {e}-e}\right )+b \text {PolyLog}\left (2,-\frac {\left (c^2 d+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt {x}\right )}}{c^2 d+2 c \sqrt {-d} \sqrt {e}-e}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(-4*b*ArcTanh[c*Sqrt[x]]^2 - 4*b*ArcTanh[c*Sqrt[x]]*Log[1 - E^(-2*ArcTanh[c*Sqrt[x]])] + 2*b*ArcTanh[c*Sq
rt[x]]*Log[1 + ((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))/(c^2*d - 2*c*Sqrt[-d]*Sqrt[e] - e)] + 2*b*ArcTanh[c*Sqrt
[x]]*Log[1 + ((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))/(c^2*d + 2*c*Sqrt[-d]*Sqrt[e] - e)] - 2*a*Log[x] + 2*a*Log
[d + e*x] + 2*b*PolyLog[2, E^(-2*ArcTanh[c*Sqrt[x]])] + b*PolyLog[2, -(((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))/
(c^2*d - 2*c*Sqrt[-d]*Sqrt[e] - e))] + b*PolyLog[2, -(((c^2*d + e)*E^(2*ArcTanh[c*Sqrt[x]]))/(c^2*d + 2*c*Sqrt
[-d]*Sqrt[e] - e))])/d

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Maple [A]
time = 0.64, size = 540, normalized size = 1.51

method	result	size
derivativedivides	\(-\frac {a \ln \left (c^{2} e x +c^{2} d \right )}{d}+\frac {2 a \ln \left (c \sqrt {x}\right )}{d}-\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (c^{2} e x +c^{2} d \right )}{d}+\frac {2 b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}\right )}{d}+\frac {b \ln \left (1+c \sqrt {x}\right ) \ln \left (c^{2} e x +c^{2} d \right )}{2 d}-\frac {b \ln \left (1+c \sqrt {x}\right ) \ln \left (\frac {c \sqrt {-d e}-e \left (1+c \sqrt {x}\right )+e}{c \sqrt {-d e}+e}\right )}{2 d}-\frac {b \ln \left (1+c \sqrt {x}\right ) \ln \left (\frac {c \sqrt {-d e}+e \left (1+c \sqrt {x}\right )-e}{c \sqrt {-d e}-e}\right )}{2 d}-\frac {b \dilog \left (\frac {c \sqrt {-d e}-e \left (1+c \sqrt {x}\right )+e}{c \sqrt {-d e}+e}\right )}{2 d}-\frac {b \dilog \left (\frac {c \sqrt {-d e}+e \left (1+c \sqrt {x}\right )-e}{c \sqrt {-d e}-e}\right )}{2 d}-\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (c^{2} e x +c^{2} d \right )}{2 d}+\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {-d e}-e \left (c \sqrt {x}-1\right )-e}{c \sqrt {-d e}-e}\right )}{2 d}+\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {-d e}+e \left (c \sqrt {x}-1\right )+e}{c \sqrt {-d e}+e}\right )}{2 d}+\frac {b \dilog \left (\frac {c \sqrt {-d e}-e \left (c \sqrt {x}-1\right )-e}{c \sqrt {-d e}-e}\right )}{2 d}+\frac {b \dilog \left (\frac {c \sqrt {-d e}+e \left (c \sqrt {x}-1\right )+e}{c \sqrt {-d e}+e}\right )}{2 d}-\frac {b \dilog \left (1+c \sqrt {x}\right )}{d}-\frac {b \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{d}-\frac {b \dilog \left (c \sqrt {x}\right )}{d}\)	\(540\)
default	\(-\frac {a \ln \left (c^{2} e x +c^{2} d \right )}{d}+\frac {2 a \ln \left (c \sqrt {x}\right )}{d}-\frac {b \arctanh \left (c \sqrt {x}\right ) \ln \left (c^{2} e x +c^{2} d \right )}{d}+\frac {2 b \arctanh \left (c \sqrt {x}\right ) \ln \left (c \sqrt {x}\right )}{d}+\frac {b \ln \left (1+c \sqrt {x}\right ) \ln \left (c^{2} e x +c^{2} d \right )}{2 d}-\frac {b \ln \left (1+c \sqrt {x}\right ) \ln \left (\frac {c \sqrt {-d e}-e \left (1+c \sqrt {x}\right )+e}{c \sqrt {-d e}+e}\right )}{2 d}-\frac {b \ln \left (1+c \sqrt {x}\right ) \ln \left (\frac {c \sqrt {-d e}+e \left (1+c \sqrt {x}\right )-e}{c \sqrt {-d e}-e}\right )}{2 d}-\frac {b \dilog \left (\frac {c \sqrt {-d e}-e \left (1+c \sqrt {x}\right )+e}{c \sqrt {-d e}+e}\right )}{2 d}-\frac {b \dilog \left (\frac {c \sqrt {-d e}+e \left (1+c \sqrt {x}\right )-e}{c \sqrt {-d e}-e}\right )}{2 d}-\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (c^{2} e x +c^{2} d \right )}{2 d}+\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {-d e}-e \left (c \sqrt {x}-1\right )-e}{c \sqrt {-d e}-e}\right )}{2 d}+\frac {b \ln \left (c \sqrt {x}-1\right ) \ln \left (\frac {c \sqrt {-d e}+e \left (c \sqrt {x}-1\right )+e}{c \sqrt {-d e}+e}\right )}{2 d}+\frac {b \dilog \left (\frac {c \sqrt {-d e}-e \left (c \sqrt {x}-1\right )-e}{c \sqrt {-d e}-e}\right )}{2 d}+\frac {b \dilog \left (\frac {c \sqrt {-d e}+e \left (c \sqrt {x}-1\right )+e}{c \sqrt {-d e}+e}\right )}{2 d}-\frac {b \dilog \left (1+c \sqrt {x}\right )}{d}-\frac {b \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{d}-\frac {b \dilog \left (c \sqrt {x}\right )}{d}\)	\(540\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))/x/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

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

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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^(1/2)))/x/(e*x+d),x, algorithm="maxima")

[Out]

-a*(log(x*e + d)/d - log(x)/d) + b*integrate(1/2*log(c*sqrt(x) + 1)/((x^(3/2)*e + d*sqrt(x))*sqrt(x)), x) - b*
integrate(1/2*log(-c*sqrt(x) + 1)/((x^(3/2)*e + d*sqrt(x))*sqrt(x)), 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^(1/2)))/x/(e*x+d),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**(1/2)))/x/(e*x+d),x)

[Out]

Timed out

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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^(1/2)))/x/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*sqrt(x)) + a)/((e*x + d)*x), 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\,\sqrt {x}\right )}{x\,\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^(1/2)))/(x*(d + e*x)),x)

[Out]

int((a + b*atanh(c*x^(1/2)))/(x*(d + e*x)), x)

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