3.1.49 \(\int \frac {a+b \tanh ^{-1}(c \sqrt {x})}{x^3 (d+e x)} \, dx\) [49]

Optimal. Leaf size=506 \[ -\frac {b c}{6 d x^{3/2}}-\frac {b c^3}{2 d \sqrt {x}}+\frac {b c e}{d^2 \sqrt {x}}+\frac {b c^4 \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d}-\frac {b c^2 e \tanh ^{-1}\left (c \sqrt {x}\right )}{d^2}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d x^2}+\frac {e \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d^2 x}+\frac {2 e^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{d^3}-\frac {e^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{d^3}-\frac {e^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{d^3}+\frac {a e^2 \log (x)}{d^3}-\frac {b e^2 \text {PolyLog}\left (2,1-\frac {2}{1+c \sqrt {x}}\right )}{d^3}+\frac {b e^2 \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}-\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}-\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 d^3}+\frac {b e^2 \text {PolyLog}\left (2,1-\frac {2 c \left (\sqrt {-d}+\sqrt {e} \sqrt {x}\right )}{\left (c \sqrt {-d}+\sqrt {e}\right ) \left (1+c \sqrt {x}\right )}\right )}{2 d^3}-\frac {b e^2 \text {PolyLog}\left (2,-c \sqrt {x}\right )}{d^3}+\frac {b e^2 \text {PolyLog}\left (2,c \sqrt {x}\right )}{d^3} \]

[Out]

-1/6*b*c/d/x^(3/2)+1/2*b*c^4*arctanh(c*x^(1/2))/d-b*c^2*e*arctanh(c*x^(1/2))/d^2+1/2*(-a-b*arctanh(c*x^(1/2)))
/d/x^2+e*(a+b*arctanh(c*x^(1/2)))/d^2/x+a*e^2*ln(x)/d^3+2*e^2*(a+b*arctanh(c*x^(1/2)))*ln(2/(1+c*x^(1/2)))/d^3
-e^2*(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^3-e^
2*(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^3-b*e^2
*polylog(2,-c*x^(1/2))/d^3+b*e^2*polylog(2,c*x^(1/2))/d^3-b*e^2*polylog(2,1-2/(1+c*x^(1/2)))/d^3+1/2*b*e^2*pol
ylog(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^3+1/2*b*e^2*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^3-1/2*b*c^3/d/x^(1/2)+b*c*e/d^2/x^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.64, antiderivative size = 506, normalized size of antiderivative = 1.00, number of steps used = 24, number of rules used = 13, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.565, Rules used = {46, 1607, 6129, 6037, 331, 212, 6139, 6031, 6191, 6057, 2449, 2352, 2497} \begin {gather*} \frac {2 e^2 \log \left (\frac {2}{c \sqrt {x}+1}\right ) \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d^3}-\frac {e^2 \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^3}-\frac {e^2 \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^3}+\frac {e \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d^2 x}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d x^2}+\frac {a e^2 \log (x)}{d^3}+\frac {b c^4 \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d}-\frac {b c^3}{2 d \sqrt {x}}-\frac {b c^2 e \tanh ^{-1}\left (c \sqrt {x}\right )}{d^2}-\frac {b e^2 \text {Li}_2\left (1-\frac {2}{\sqrt {x} c+1}\right )}{d^3}+\frac {b e^2 \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^3}+\frac {b e^2 \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^3}-\frac {b e^2 \text {Li}_2\left (-c \sqrt {x}\right )}{d^3}+\frac {b e^2 \text {Li}_2\left (c \sqrt {x}\right )}{d^3}+\frac {b c e}{d^2 \sqrt {x}}-\frac {b c}{6 d x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])/(x^3*(d + e*x)),x]

[Out]

-1/6*(b*c)/(d*x^(3/2)) - (b*c^3)/(2*d*Sqrt[x]) + (b*c*e)/(d^2*Sqrt[x]) + (b*c^4*ArcTanh[c*Sqrt[x]])/(2*d) - (b
*c^2*e*ArcTanh[c*Sqrt[x]])/d^2 - (a + b*ArcTanh[c*Sqrt[x]])/(2*d*x^2) + (e*(a + b*ArcTanh[c*Sqrt[x]]))/(d^2*x)
+ (2*e^2*(a + b*ArcTanh[c*Sqrt[x]])*Log[2/(1 + c*Sqrt[x])])/d^3 - (e^2*(a + b*ArcTanh[c*Sqrt[x]])*Log[(2*c*(S
qrt[-d] - Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] - Sqrt[e])*(1 + c*Sqrt[x]))])/d^3 - (e^2*(a + b*ArcTanh[c*Sqrt[x]])*L
og[(2*c*(Sqrt[-d] + Sqrt[e]*Sqrt[x]))/((c*Sqrt[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/d^3 + (a*e^2*Log[x])/d^3 - (b
*e^2*PolyLog[2, 1 - 2/(1 + c*Sqrt[x])])/d^3 + (b*e^2*PolyLog[2, 1 - (2*c*(Sqrt[-d] - Sqrt[e]*Sqrt[x]))/((c*Sqr
t[-d] - Sqrt[e])*(1 + c*Sqrt[x]))])/(2*d^3) + (b*e^2*PolyLog[2, 1 - (2*c*(Sqrt[-d] + Sqrt[e]*Sqrt[x]))/((c*Sqr
t[-d] + Sqrt[e])*(1 + c*Sqrt[x]))])/(2*d^3) - (b*e^2*PolyLog[2, -(c*Sqrt[x])])/d^3 + (b*e^2*PolyLog[2, c*Sqrt[
x]])/d^3

Rule 46

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && Lt
Q[m + n + 2, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 331

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

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 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 6129

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

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^3 (d+e x)} \, dx &=2 \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{d x^5+e x^7} \, dx,x,\sqrt {x}\right )\\ &=2 \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^5 \left (d+e x^2\right )} \, dx,x,\sqrt {x}\right )\\ &=\frac {2 \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^5} \, dx,x,\sqrt {x}\right )}{d}-\frac {(2 e) \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^3 \left (d+e x^2\right )} \, dx,x,\sqrt {x}\right )}{d}\\ &=-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d x^2}+\frac {(b c) \text {Subst}\left (\int \frac {1}{x^4 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )}{2 d}-\frac {(2 e) \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x^3} \, dx,x,\sqrt {x}\right )}{d^2}+\frac {\left (2 e^2\right ) \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x \left (d+e x^2\right )} \, dx,x,\sqrt {x}\right )}{d^2}\\ &=-\frac {b c}{6 d x^{3/2}}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d x^2}+\frac {e \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d^2 x}+\frac {\left (b c^3\right ) \text {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )}{2 d}-\frac {(b c e) \text {Subst}\left (\int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx,x,\sqrt {x}\right )}{d^2}+\frac {\left (2 e^2\right ) \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 )}{d^2}\\ &=-\frac {b c}{6 d x^{3/2}}-\frac {b c^3}{2 d \sqrt {x}}+\frac {b c e}{d^2 \sqrt {x}}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d x^2}+\frac {e \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d^2 x}+\frac {\left (b c^5\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{2 d}-\frac {\left (b c^3 e\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{d^2}+\frac {\left (2 e^2\right ) \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx,x,\sqrt {x}\right )}{d^3}-\frac {\left (2 e^3\right ) \text {Subst}\left (\int \frac {x \left (a+b \tanh ^{-1}(c x)\right )}{d+e x^2} \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {b c}{6 d x^{3/2}}-\frac {b c^3}{2 d \sqrt {x}}+\frac {b c e}{d^2 \sqrt {x}}+\frac {b c^4 \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d}-\frac {b c^2 e \tanh ^{-1}\left (c \sqrt {x}\right )}{d^2}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d x^2}+\frac {e \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d^2 x}+\frac {a e^2 \log (x)}{d^3}-\frac {b e^2 \text {Li}_2\left (-c \sqrt {x}\right )}{d^3}+\frac {b e^2 \text {Li}_2\left (c \sqrt {x}\right )}{d^3}-\frac {\left (2 e^3\right ) \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^3}\\ &=-\frac {b c}{6 d x^{3/2}}-\frac {b c^3}{2 d \sqrt {x}}+\frac {b c e}{d^2 \sqrt {x}}+\frac {b c^4 \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d}-\frac {b c^2 e \tanh ^{-1}\left (c \sqrt {x}\right )}{d^2}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d x^2}+\frac {e \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d^2 x}+\frac {a e^2 \log (x)}{d^3}-\frac {b e^2 \text {Li}_2\left (-c \sqrt {x}\right )}{d^3}+\frac {b e^2 \text {Li}_2\left (c \sqrt {x}\right )}{d^3}+\frac {e^{5/2} \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{\sqrt {-d}-\sqrt {e} x} \, dx,x,\sqrt {x}\right )}{d^3}-\frac {e^{5/2} \text {Subst}\left (\int \frac {a+b \tanh ^{-1}(c x)}{\sqrt {-d}+\sqrt {e} x} \, dx,x,\sqrt {x}\right )}{d^3}\\ &=-\frac {b c}{6 d x^{3/2}}-\frac {b c^3}{2 d \sqrt {x}}+\frac {b c e}{d^2 \sqrt {x}}+\frac {b c^4 \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d}-\frac {b c^2 e \tanh ^{-1}\left (c \sqrt {x}\right )}{d^2}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d x^2}+\frac {e \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d^2 x}+\frac {2 e^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{d^3}-\frac {e^2 \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^3}-\frac {e^2 \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^3}+\frac {a e^2 \log (x)}{d^3}-\frac {b e^2 \text {Li}_2\left (-c \sqrt {x}\right )}{d^3}+\frac {b e^2 \text {Li}_2\left (c \sqrt {x}\right )}{d^3}-2 \frac {\left (b c e^2\right ) \text {Subst}\left (\int \frac {\log \left (\frac {2}{1+c x}\right )}{1-c^2 x^2} \, dx,x,\sqrt {x}\right )}{d^3}+\frac {\left (b c e^2\right ) \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^3}+\frac {\left (b c e^2\right ) \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^3}\\ &=-\frac {b c}{6 d x^{3/2}}-\frac {b c^3}{2 d \sqrt {x}}+\frac {b c e}{d^2 \sqrt {x}}+\frac {b c^4 \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d}-\frac {b c^2 e \tanh ^{-1}\left (c \sqrt {x}\right )}{d^2}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d x^2}+\frac {e \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d^2 x}+\frac {2 e^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{d^3}-\frac {e^2 \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^3}-\frac {e^2 \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^3}+\frac {a e^2 \log (x)}{d^3}+\frac {b e^2 \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^3}+\frac {b e^2 \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^3}-\frac {b e^2 \text {Li}_2\left (-c \sqrt {x}\right )}{d^3}+\frac {b e^2 \text {Li}_2\left (c \sqrt {x}\right )}{d^3}-2 \frac {\left (b e^2\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1+c \sqrt {x}}\right )}{d^3}\\ &=-\frac {b c}{6 d x^{3/2}}-\frac {b c^3}{2 d \sqrt {x}}+\frac {b c e}{d^2 \sqrt {x}}+\frac {b c^4 \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d}-\frac {b c^2 e \tanh ^{-1}\left (c \sqrt {x}\right )}{d^2}-\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 d x^2}+\frac {e \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )}{d^2 x}+\frac {2 e^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \log \left (\frac {2}{1+c \sqrt {x}}\right )}{d^3}-\frac {e^2 \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^3}-\frac {e^2 \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^3}+\frac {a e^2 \log (x)}{d^3}-\frac {b e^2 \text {Li}_2\left (1-\frac {2}{1+c \sqrt {x}}\right )}{d^3}+\frac {b e^2 \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^3}+\frac {b e^2 \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^3}-\frac {b e^2 \text {Li}_2\left (-c \sqrt {x}\right )}{d^3}+\frac {b e^2 \text {Li}_2\left (c \sqrt {x}\right )}{d^3}\\ \end {align*}

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Mathematica [A]
time = 1.83, size = 394, normalized size = 0.78 \begin {gather*} -\frac {3 a d^2-6 a d e x-6 a e^2 x^2 \log (x)+6 a e^2 x^2 \log (d+e x)+b \left (c d \sqrt {x} \left (d+3 c^2 d x-6 e x\right )-3 \tanh ^{-1}\left (c \sqrt {x}\right ) \left (d \left (-1+c^2 x\right ) \left (d+c^2 d x-2 e x\right )+2 e^2 x^2 \tanh ^{-1}\left (c \sqrt {x}\right )+4 e^2 x^2 \log \left (1-e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )\right )+6 e^2 x^2 \text {PolyLog}\left (2,e^{-2 \tanh ^{-1}\left (c \sqrt {x}\right )}\right )+3 e^2 x^2 \left (2 \tanh ^{-1}\left (c \sqrt {x}\right ) \left (-\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 )+\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 )\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 )+\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 )\right )\right )}{6 d^3 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*Sqrt[x]])/(x^3*(d + e*x)),x]

[Out]

-1/6*(3*a*d^2 - 6*a*d*e*x - 6*a*e^2*x^2*Log[x] + 6*a*e^2*x^2*Log[d + e*x] + b*(c*d*Sqrt[x]*(d + 3*c^2*d*x - 6*
e*x) - 3*ArcTanh[c*Sqrt[x]]*(d*(-1 + c^2*x)*(d + c^2*d*x - 2*e*x) + 2*e^2*x^2*ArcTanh[c*Sqrt[x]] + 4*e^2*x^2*L
og[1 - E^(-2*ArcTanh[c*Sqrt[x]])]) + 6*e^2*x^2*PolyLog[2, E^(-2*ArcTanh[c*Sqrt[x]])] + 3*e^2*x^2*(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)]) + PolyLog[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*ArcTa
nh[c*Sqrt[x]]))/(c^2*d + 2*c*Sqrt[-d]*Sqrt[e] - e))])))/(d^3*x^2)

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Maple [A]
time = 0.65, size = 808, normalized size = 1.60

method	result	size
derivativedivides	\(2 c^{4} \left (-\frac {b}{12 d \,c^{3} x^{\frac {3}{2}}}-\frac {a}{4 d \,c^{4} x^{2}}+\frac {b e}{2 c^{3} d^{2} \sqrt {x}}-\frac {b \ln \left (1+c \sqrt {x}\right ) e}{4 c^{2} d^{2}}+\frac {b \ln \left (c \sqrt {x}-1\right ) e}{4 c^{2} d^{2}}-\frac {a \,e^{2} \ln \left (c^{2} e x +c^{2} d \right )}{2 c^{4} d^{3}}+\frac {a \,e^{2} \ln \left (c \sqrt {x}\right )}{c^{4} d^{3}}+\frac {a e}{2 c^{4} d^{2} x}-\frac {b \,e^{2} \dilog \left (1+c \sqrt {x}\right )}{2 c^{4} d^{3}}-\frac {b \,e^{2} \dilog \left (c \sqrt {x}\right )}{2 c^{4} d^{3}}-\frac {b \,e^{2} \dilog \left (\frac {c \sqrt {-d e}-e \left (1+c \sqrt {x}\right )+e}{c \sqrt {-d e}+e}\right )}{4 c^{4} d^{3}}-\frac {b \,e^{2} \dilog \left (\frac {c \sqrt {-d e}+e \left (1+c \sqrt {x}\right )-e}{c \sqrt {-d e}-e}\right )}{4 c^{4} d^{3}}+\frac {b \,e^{2} \dilog \left (\frac {c \sqrt {-d e}-e \left (c \sqrt {x}-1\right )-e}{c \sqrt {-d e}-e}\right )}{4 c^{4} d^{3}}+\frac {b \,e^{2} \dilog \left (\frac {c \sqrt {-d e}+e \left (c \sqrt {x}-1\right )+e}{c \sqrt {-d e}+e}\right )}{4 c^{4} d^{3}}+\frac {b \arctanh \left (c \sqrt {x}\right ) e}{2 c^{4} d^{2} x}-\frac {b \arctanh \left (c \sqrt {x}\right ) e^{2} \ln \left (c^{2} e x +c^{2} d \right )}{2 c^{4} d^{3}}+\frac {b \arctanh \left (c \sqrt {x}\right ) e^{2} \ln \left (c \sqrt {x}\right )}{c^{4} d^{3}}-\frac {b \arctanh \left (c \sqrt {x}\right )}{4 d \,c^{4} x^{2}}+\frac {b \ln \left (1+c \sqrt {x}\right )}{8 d}-\frac {b \ln \left (c \sqrt {x}-1\right )}{8 d}-\frac {b}{4 d c \sqrt {x}}-\frac {b \,e^{2} \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2 c^{4} d^{3}}+\frac {b \,e^{2} \ln \left (1+c \sqrt {x}\right ) \ln \left (c^{2} e x +c^{2} d \right )}{4 c^{4} d^{3}}-\frac {b \,e^{2} \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 )}{4 c^{4} d^{3}}-\frac {b \,e^{2} \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 )}{4 c^{4} d^{3}}-\frac {b \,e^{2} \ln \left (c \sqrt {x}-1\right ) \ln \left (c^{2} e x +c^{2} d \right )}{4 c^{4} d^{3}}+\frac {b \,e^{2} \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 )}{4 c^{4} d^{3}}+\frac {b \,e^{2} \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 )}{4 c^{4} d^{3}}\right )\)	\(808\)
default	\(2 c^{4} \left (-\frac {b}{12 d \,c^{3} x^{\frac {3}{2}}}-\frac {a}{4 d \,c^{4} x^{2}}+\frac {b e}{2 c^{3} d^{2} \sqrt {x}}-\frac {b \ln \left (1+c \sqrt {x}\right ) e}{4 c^{2} d^{2}}+\frac {b \ln \left (c \sqrt {x}-1\right ) e}{4 c^{2} d^{2}}-\frac {a \,e^{2} \ln \left (c^{2} e x +c^{2} d \right )}{2 c^{4} d^{3}}+\frac {a \,e^{2} \ln \left (c \sqrt {x}\right )}{c^{4} d^{3}}+\frac {a e}{2 c^{4} d^{2} x}-\frac {b \,e^{2} \dilog \left (1+c \sqrt {x}\right )}{2 c^{4} d^{3}}-\frac {b \,e^{2} \dilog \left (c \sqrt {x}\right )}{2 c^{4} d^{3}}-\frac {b \,e^{2} \dilog \left (\frac {c \sqrt {-d e}-e \left (1+c \sqrt {x}\right )+e}{c \sqrt {-d e}+e}\right )}{4 c^{4} d^{3}}-\frac {b \,e^{2} \dilog \left (\frac {c \sqrt {-d e}+e \left (1+c \sqrt {x}\right )-e}{c \sqrt {-d e}-e}\right )}{4 c^{4} d^{3}}+\frac {b \,e^{2} \dilog \left (\frac {c \sqrt {-d e}-e \left (c \sqrt {x}-1\right )-e}{c \sqrt {-d e}-e}\right )}{4 c^{4} d^{3}}+\frac {b \,e^{2} \dilog \left (\frac {c \sqrt {-d e}+e \left (c \sqrt {x}-1\right )+e}{c \sqrt {-d e}+e}\right )}{4 c^{4} d^{3}}+\frac {b \arctanh \left (c \sqrt {x}\right ) e}{2 c^{4} d^{2} x}-\frac {b \arctanh \left (c \sqrt {x}\right ) e^{2} \ln \left (c^{2} e x +c^{2} d \right )}{2 c^{4} d^{3}}+\frac {b \arctanh \left (c \sqrt {x}\right ) e^{2} \ln \left (c \sqrt {x}\right )}{c^{4} d^{3}}-\frac {b \arctanh \left (c \sqrt {x}\right )}{4 d \,c^{4} x^{2}}+\frac {b \ln \left (1+c \sqrt {x}\right )}{8 d}-\frac {b \ln \left (c \sqrt {x}-1\right )}{8 d}-\frac {b}{4 d c \sqrt {x}}-\frac {b \,e^{2} \ln \left (c \sqrt {x}\right ) \ln \left (1+c \sqrt {x}\right )}{2 c^{4} d^{3}}+\frac {b \,e^{2} \ln \left (1+c \sqrt {x}\right ) \ln \left (c^{2} e x +c^{2} d \right )}{4 c^{4} d^{3}}-\frac {b \,e^{2} \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 )}{4 c^{4} d^{3}}-\frac {b \,e^{2} \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 )}{4 c^{4} d^{3}}-\frac {b \,e^{2} \ln \left (c \sqrt {x}-1\right ) \ln \left (c^{2} e x +c^{2} d \right )}{4 c^{4} d^{3}}+\frac {b \,e^{2} \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 )}{4 c^{4} d^{3}}+\frac {b \,e^{2} \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 )}{4 c^{4} d^{3}}\right )\)	\(808\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(1/2)))/x^3/(e*x+d),x,method=_RETURNVERBOSE)

[Out]

2*c^4*(1/2*b/c^4*arctanh(c*x^(1/2))/d^2*e/x-1/12*b/d/c^3/x^(3/2)-1/4*a/d/c^4/x^2+1/2*b/c^3/d^2/x^(1/2)*e-1/4*b
*arctanh(c*x^(1/2))/d/c^4/x^2-1/4*b/c^2/d^2*ln(1+c*x^(1/2))*e+1/4*b/c^2/d^2*ln(c*x^(1/2)-1)*e-1/2*b/c^4/d^3*e^
2*dilog(1+c*x^(1/2))-1/2*b/c^4/d^3*e^2*dilog(c*x^(1/2))-1/4*b/c^4/d^3*e^2*dilog((c*(-d*e)^(1/2)-e*(1+c*x^(1/2)
)+e)/(c*(-d*e)^(1/2)+e))-1/4*b/c^4/d^3*e^2*dilog((c*(-d*e)^(1/2)+e*(1+c*x^(1/2))-e)/(c*(-d*e)^(1/2)-e))+1/4*b/
c^4/d^3*e^2*dilog((c*(-d*e)^(1/2)-e*(c*x^(1/2)-1)-e)/(c*(-d*e)^(1/2)-e))+1/4*b/c^4/d^3*e^2*dilog((c*(-d*e)^(1/
2)+e*(c*x^(1/2)-1)+e)/(c*(-d*e)^(1/2)+e))-1/2*a/c^4*e^2/d^3*ln(c^2*e*x+c^2*d)+a/c^4/d^3*e^2*ln(c*x^(1/2))+1/2*
a/c^4/d^2*e/x+1/8*b/d*ln(1+c*x^(1/2))-1/8*b/d*ln(c*x^(1/2)-1)-1/4*b/d/c/x^(1/2)-1/2*b/c^4*arctanh(c*x^(1/2))*e
^2/d^3*ln(c^2*e*x+c^2*d)+b/c^4*arctanh(c*x^(1/2))/d^3*e^2*ln(c*x^(1/2))-1/2*b/c^4/d^3*e^2*ln(c*x^(1/2))*ln(1+c
*x^(1/2))+1/4*b/c^4/d^3*e^2*ln(1+c*x^(1/2))*ln(c^2*e*x+c^2*d)-1/4*b/c^4/d^3*e^2*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))-1/4*b/c^4/d^3*e^2*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))-1/4*b/c^4/d^3*e^2*ln(c*x^(1/2)-1)*ln(c^2*e*x+c^2*d)+1/4*b/c^4/d^3*e^2*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))+1/4*b/c^4/d^3*e^2*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)))

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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^3/(e*x+d),x, algorithm="maxima")

[Out]

-1/2*a*(2*e^2*log(x*e + d)/d^3 - 2*e^2*log(x)/d^3 - (2*x*e - d)/(d^2*x^2)) + b*integrate(1/2*log(c*sqrt(x) + 1
)/((x^(7/2)*e + d*x^(5/2))*sqrt(x)), x) - b*integrate(1/2*log(-c*sqrt(x) + 1)/((x^(7/2)*e + d*x^(5/2))*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^3/(e*x+d),x, algorithm="fricas")

[Out]

integral((b*arctanh(c*sqrt(x)) + a)/(x^4*e + d*x^3), 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**3/(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^3/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*sqrt(x)) + a)/((e*x + d)*x^3), 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^3\,\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^3*(d + e*x)),x)

[Out]

int((a + b*atanh(c*x^(1/2)))/(x^3*(d + e*x)), x)

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