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

Optimal. Leaf size=374 \[ -\frac{b d \text{PolyLog}\left (2,1-\frac{2}{c \sqrt{x}+1}\right )}{e^2}+\frac{b d \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} \sqrt{x}\right )}{\left (c \sqrt{x}+1\right ) \left (c \sqrt{-d}-\sqrt{e}\right )}\right )}{2 e^2}+\frac{b d \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} \sqrt{x}\right )}{\left (c \sqrt{x}+1\right ) \left (c \sqrt{-d}+\sqrt{e}\right )}\right )}{2 e^2}+\frac{2 d \log \left (\frac{2}{c \sqrt{x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} \sqrt{x}\right )}{\left (c \sqrt{x}+1\right ) \left (c \sqrt{-d}-\sqrt{e}\right )}\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} \sqrt{x}\right )}{\left (c \sqrt{x}+1\right ) \left (c \sqrt{-d}+\sqrt{e}\right )}\right )}{e^2}+\frac{x \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )}{e}-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{c^2 e}+\frac{b \sqrt{x}}{c e} \]

[Out]

(b*Sqrt[x])/(c*e) - (b*ArcTanh[c*Sqrt[x]])/(c^2*e) + (x*(a + b*ArcTanh[c*Sqrt[x]]))/e + (2*d*(a + b*ArcTanh[c*
Sqrt[x]])*Log[2/(1 + c*Sqrt[x])])/e^2 - (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]))])/e^2 - (d*(a + b*ArcTanh[c*Sqrt[x]])*Log[(2*c*(Sqrt[-d] + Sqrt[e]*Sqr
t[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/e^2 - (b*d*PolyLog[2, 1 - 2/(1 + c*Sqrt[x])])/e^2 + (b*d*Pol
yLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] - Sqrt[e])*(1 + c*Sqrt[x]))])/(2*e^2) + (b*d*PolyL
og[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/(2*e^2)

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Rubi [A]  time = 0.484596, antiderivative size = 374, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {43, 5980, 5916, 321, 206, 6044, 5920, 2402, 2315, 2447} \[ -\frac{b d \text{PolyLog}\left (2,1-\frac{2}{c \sqrt{x}+1}\right )}{e^2}+\frac{b d \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}-\sqrt{e} \sqrt{x}\right )}{\left (c \sqrt{x}+1\right ) \left (c \sqrt{-d}-\sqrt{e}\right )}\right )}{2 e^2}+\frac{b d \text{PolyLog}\left (2,1-\frac{2 c \left (\sqrt{-d}+\sqrt{e} \sqrt{x}\right )}{\left (c \sqrt{x}+1\right ) \left (c \sqrt{-d}+\sqrt{e}\right )}\right )}{2 e^2}+\frac{2 d \log \left (\frac{2}{c \sqrt{x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \log \left (\frac{2 c \left (\sqrt{-d}-\sqrt{e} \sqrt{x}\right )}{\left (c \sqrt{x}+1\right ) \left (c \sqrt{-d}-\sqrt{e}\right )}\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \log \left (\frac{2 c \left (\sqrt{-d}+\sqrt{e} \sqrt{x}\right )}{\left (c \sqrt{x}+1\right ) \left (c \sqrt{-d}+\sqrt{e}\right )}\right )}{e^2}+\frac{x \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )}{e}-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{c^2 e}+\frac{b \sqrt{x}}{c e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[x])/(c*e) - (b*ArcTanh[c*Sqrt[x]])/(c^2*e) + (x*(a + b*ArcTanh[c*Sqrt[x]]))/e + (2*d*(a + b*ArcTanh[c*
Sqrt[x]])*Log[2/(1 + c*Sqrt[x])])/e^2 - (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]))])/e^2 - (d*(a + b*ArcTanh[c*Sqrt[x]])*Log[(2*c*(Sqrt[-d] + Sqrt[e]*Sqr
t[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/e^2 - (b*d*PolyLog[2, 1 - 2/(1 + c*Sqrt[x])])/e^2 + (b*d*Pol
yLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] - Sqrt[e])*(1 + c*Sqrt[x]))])/(2*e^2) + (b*d*PolyL
og[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/(2*e^2)

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 5980

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f^2)/e, Int[((f*x)^(m - 2)*(a + b*ArcTanh[c*x])
^p)/(d + e*x^2), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]

Rule 5916

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

Rule 321

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

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 6044

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])

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 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 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 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]]

Rubi steps

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

Mathematica [A]  time = 1.44733, size = 337, normalized size = 0.9 \[ \frac{\frac{2 b \left (-c^2 d \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}\left (c \sqrt{x}\right )}\right )+\tanh ^{-1}\left (c \sqrt{x}\right ) \left (2 c^2 d \log \left (e^{-2 \tanh ^{-1}\left (c \sqrt{x}\right )}+1\right )+c^2 e x-e\right )+c^2 d \tanh ^{-1}\left (c \sqrt{x}\right )^2+c e \sqrt{x}\right )}{c^2}-b d \left (\text{PolyLog}\left (2,-\frac{\left (c^2 d+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt{x}\right )}}{c^2 d-2 c \sqrt{-d} \sqrt{e}-e}\right )+\text{PolyLog}\left (2,-\frac{\left (c^2 d+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt{x}\right )}}{c^2 d+2 c \sqrt{-d} \sqrt{e}-e}\right )+2 \tanh ^{-1}\left (c \sqrt{x}\right ) \left (\log \left (\frac{\left (c^2 d+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt{x}\right )}}{c^2 d-2 c \sqrt{-d} \sqrt{e}-e}+1\right )+\log \left (\frac{\left (c^2 d+e\right ) e^{2 \tanh ^{-1}\left (c \sqrt{x}\right )}}{c^2 d+2 c \sqrt{-d} \sqrt{e}-e}+1\right )-\tanh ^{-1}\left (c \sqrt{x}\right )\right )\right )-2 a d \log (d+e x)+2 a e x}{2 e^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [A]  time = 0.057, size = 539, normalized size = 1.4 \begin{align*}{\frac{ax}{e}}-{\frac{ad\ln \left ({c}^{2}ex+{c}^{2}d \right ) }{{e}^{2}}}+{\frac{bx}{e}{\it Artanh} \left ( c\sqrt{x} \right ) }-{\frac{bd\ln \left ({c}^{2}ex+{c}^{2}d \right ) }{{e}^{2}}{\it Artanh} \left ( c\sqrt{x} \right ) }-{\frac{bd\ln \left ({c}^{2}ex+{c}^{2}d \right ) }{2\,{e}^{2}}\ln \left ( c\sqrt{x}-1 \right ) }+{\frac{bd}{2\,{e}^{2}}\ln \left ( c\sqrt{x}-1 \right ) \ln \left ({ \left ( c\sqrt{-de}-e \left ( c\sqrt{x}-1 \right ) -e \right ) \left ( c\sqrt{-de}-e \right ) ^{-1}} \right ) }+{\frac{bd}{2\,{e}^{2}}\ln \left ( c\sqrt{x}-1 \right ) \ln \left ({ \left ( c\sqrt{-de}+e \left ( c\sqrt{x}-1 \right ) +e \right ) \left ( c\sqrt{-de}+e \right ) ^{-1}} \right ) }+{\frac{bd}{2\,{e}^{2}}{\it dilog} \left ({ \left ( c\sqrt{-de}-e \left ( c\sqrt{x}-1 \right ) -e \right ) \left ( c\sqrt{-de}-e \right ) ^{-1}} \right ) }+{\frac{bd}{2\,{e}^{2}}{\it dilog} \left ({ \left ( c\sqrt{-de}+e \left ( c\sqrt{x}-1 \right ) +e \right ) \left ( c\sqrt{-de}+e \right ) ^{-1}} \right ) }+{\frac{bd\ln \left ({c}^{2}ex+{c}^{2}d \right ) }{2\,{e}^{2}}\ln \left ( 1+c\sqrt{x} \right ) }-{\frac{bd}{2\,{e}^{2}}\ln \left ( 1+c\sqrt{x} \right ) \ln \left ({ \left ( c\sqrt{-de}-e \left ( 1+c\sqrt{x} \right ) +e \right ) \left ( c\sqrt{-de}+e \right ) ^{-1}} \right ) }-{\frac{bd}{2\,{e}^{2}}\ln \left ( 1+c\sqrt{x} \right ) \ln \left ({ \left ( c\sqrt{-de}+e \left ( 1+c\sqrt{x} \right ) -e \right ) \left ( c\sqrt{-de}-e \right ) ^{-1}} \right ) }-{\frac{bd}{2\,{e}^{2}}{\it dilog} \left ({ \left ( c\sqrt{-de}-e \left ( 1+c\sqrt{x} \right ) +e \right ) \left ( c\sqrt{-de}+e \right ) ^{-1}} \right ) }-{\frac{bd}{2\,{e}^{2}}{\it dilog} \left ({ \left ( c\sqrt{-de}+e \left ( 1+c\sqrt{x} \right ) -e \right ) \left ( c\sqrt{-de}-e \right ) ^{-1}} \right ) }+{\frac{b}{ce}\sqrt{x}}+{\frac{b}{2\,{c}^{2}e}\ln \left ( c\sqrt{x}-1 \right ) }-{\frac{b}{2\,{c}^{2}e}\ln \left ( 1+c\sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a{\left (\frac{x}{e} - \frac{d \log \left (e x + d\right )}{e^{2}}\right )} + b \int \frac{x \log \left (c \sqrt{x} + 1\right )}{2 \,{\left (e x + d\right )}}\,{d x} - b \int \frac{x \log \left (-c \sqrt{x} + 1\right )}{2 \,{\left (e x + d\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x^(1/2)))/(e*x+d),x, algorithm="maxima")

[Out]

a*(x/e - d*log(e*x + d)/e^2) + b*integrate(1/2*x*log(c*sqrt(x) + 1)/(e*x + d), x) - b*integrate(1/2*x*log(-c*s
qrt(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 x \operatorname{artanh}\left (c \sqrt{x}\right ) + a x}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c \sqrt{x}\right ) + a\right )} x}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x^(1/2)))/(e*x+d),x, algorithm="giac")

[Out]

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