3.150 \(\int \frac{a+b \log (c x^n)}{x^3 \sqrt{d+e x}} \, dx\)

Optimal. Leaf size=304 \[ -\frac{3 b e^2 n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right )}{4 d^{5/2}}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )^2}{4 d^{5/2}}+\frac{7 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{8 d^{5/2}}-\frac{3 b e^2 n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{2 d^{5/2}}+\frac{5 b e n \sqrt{d+e x}}{8 d^2 x}-\frac{b n \sqrt{d+e x}}{4 d x^2} \]

[Out]

-(b*n*Sqrt[d + e*x])/(4*d*x^2) + (5*b*e*n*Sqrt[d + e*x])/(8*d^2*x) + (7*b*e^2*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]
)/(8*d^(5/2)) + (3*b*e^2*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]^2)/(4*d^(5/2)) - (Sqrt[d + e*x]*(a + b*Log[c*x^n]))/
(2*d*x^2) + (3*e*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/(4*d^2*x) - (3*e^2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*(a + b*Lo
g[c*x^n]))/(4*d^(5/2)) - (3*b*e^2*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])
/(2*d^(5/2)) - (3*b*e^2*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])/(4*d^(5/2))

________________________________________________________________________________________

Rubi [A]  time = 0.336228, antiderivative size = 304, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.478, Rules used = {51, 63, 208, 2350, 12, 14, 47, 5984, 5918, 2402, 2315} \[ -\frac{3 b e^2 n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right )}{4 d^{5/2}}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )^2}{4 d^{5/2}}+\frac{7 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{8 d^{5/2}}-\frac{3 b e^2 n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{2 d^{5/2}}+\frac{5 b e n \sqrt{d+e x}}{8 d^2 x}-\frac{b n \sqrt{d+e x}}{4 d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*n*Sqrt[d + e*x])/(4*d*x^2) + (5*b*e*n*Sqrt[d + e*x])/(8*d^2*x) + (7*b*e^2*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]
)/(8*d^(5/2)) + (3*b*e^2*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]^2)/(4*d^(5/2)) - (Sqrt[d + e*x]*(a + b*Log[c*x^n]))/
(2*d*x^2) + (3*e*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/(4*d^2*x) - (3*e^2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*(a + b*Lo
g[c*x^n]))/(4*d^(5/2)) - (3*b*e^2*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])
/(2*d^(5/2)) - (3*b*e^2*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])/(4*d^(5/2))

Rule 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 2350

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[SimplifyIntegrand[u/x,
x], x], x] /; ((EqQ[r, 1] || EqQ[r, 2]) && IntegerQ[m] && IntegerQ[q - 1/2]) || InverseFunctionFreeQ[u, x]] /
; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && IntegerQ[2*q] && ((IntegerQ[m] && IntegerQ[r]) || IGtQ[q, 0])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 5984

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

Rule 5918

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 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]

Rubi steps

\begin{align*} \int \frac{a+b \log \left (c x^n\right )}{x^3 \sqrt{d+e x}} \, dx &=-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}-(b n) \int \frac{\sqrt{d} \sqrt{d+e x} (-2 d+3 e x)-3 e^2 x^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 d^{5/2} x^3} \, dx\\ &=-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}-\frac{(b n) \int \frac{\sqrt{d} \sqrt{d+e x} (-2 d+3 e x)-3 e^2 x^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{x^3} \, dx}{4 d^{5/2}}\\ &=-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}-\frac{(b n) \int \left (-\frac{2 d^{3/2} \sqrt{d+e x}}{x^3}+\frac{3 \sqrt{d} e \sqrt{d+e x}}{x^2}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{x}\right ) \, dx}{4 d^{5/2}}\\ &=-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}+\frac{(b n) \int \frac{\sqrt{d+e x}}{x^3} \, dx}{2 d}-\frac{(3 b e n) \int \frac{\sqrt{d+e x}}{x^2} \, dx}{4 d^2}+\frac{\left (3 b e^2 n\right ) \int \frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{x} \, dx}{4 d^{5/2}}\\ &=-\frac{b n \sqrt{d+e x}}{4 d x^2}+\frac{3 b e n \sqrt{d+e x}}{4 d^2 x}-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}+\frac{(b e n) \int \frac{1}{x^2 \sqrt{d+e x}} \, dx}{8 d}+\frac{\left (3 b e^2 n\right ) \operatorname{Subst}\left (\int \frac{x \tanh ^{-1}\left (\frac{x}{\sqrt{d}}\right )}{-d+x^2} \, dx,x,\sqrt{d+e x}\right )}{2 d^{5/2}}-\frac{\left (3 b e^2 n\right ) \int \frac{1}{x \sqrt{d+e x}} \, dx}{8 d^2}\\ &=-\frac{b n \sqrt{d+e x}}{4 d x^2}+\frac{5 b e n \sqrt{d+e x}}{8 d^2 x}+\frac{3 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )^2}{4 d^{5/2}}-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}-\frac{(3 b e n) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 d^2}-\frac{\left (3 b e^2 n\right ) \operatorname{Subst}\left (\int \frac{\tanh ^{-1}\left (\frac{x}{\sqrt{d}}\right )}{1-\frac{x}{\sqrt{d}}} \, dx,x,\sqrt{d+e x}\right )}{2 d^3}-\frac{\left (b e^2 n\right ) \int \frac{1}{x \sqrt{d+e x}} \, dx}{16 d^2}\\ &=-\frac{b n \sqrt{d+e x}}{4 d x^2}+\frac{5 b e n \sqrt{d+e x}}{8 d^2 x}+\frac{3 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{4 d^{5/2}}+\frac{3 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )^2}{4 d^{5/2}}-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}-\frac{3 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right )}{2 d^{5/2}}-\frac{(b e n) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{8 d^2}+\frac{\left (3 b e^2 n\right ) \operatorname{Subst}\left (\int \frac{\log \left (\frac{2}{1-\frac{x}{\sqrt{d}}}\right )}{1-\frac{x^2}{d}} \, dx,x,\sqrt{d+e x}\right )}{2 d^3}\\ &=-\frac{b n \sqrt{d+e x}}{4 d x^2}+\frac{5 b e n \sqrt{d+e x}}{8 d^2 x}+\frac{7 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{8 d^{5/2}}+\frac{3 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )^2}{4 d^{5/2}}-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}-\frac{3 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right )}{2 d^{5/2}}-\frac{\left (3 b e^2 n\right ) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1-\frac{\sqrt{d+e x}}{\sqrt{d}}}\right )}{2 d^{5/2}}\\ &=-\frac{b n \sqrt{d+e x}}{4 d x^2}+\frac{5 b e n \sqrt{d+e x}}{8 d^2 x}+\frac{7 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )}{8 d^{5/2}}+\frac{3 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )^2}{4 d^{5/2}}-\frac{\sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{2 d x^2}+\frac{3 e \sqrt{d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 d^2 x}-\frac{3 e^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^{5/2}}-\frac{3 b e^2 n \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right ) \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x}}\right )}{2 d^{5/2}}-\frac{3 b e^2 n \text{Li}_2\left (1-\frac{2}{1-\frac{\sqrt{d+e x}}{\sqrt{d}}}\right )}{4 d^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.341853, size = 501, normalized size = 1.65 \[ \frac{-6 b e^2 n x^2 \text{PolyLog}\left (2,\frac{1}{2}-\frac{\sqrt{d+e x}}{2 \sqrt{d}}\right )+6 b e^2 n x^2 \text{PolyLog}\left (2,\frac{1}{2} \left (\frac{\sqrt{d+e x}}{\sqrt{d}}+1\right )\right )-8 a d^{3/2} \sqrt{d+e x}+6 a e^2 x^2 \log \left (\sqrt{d}-\sqrt{d+e x}\right )-6 a e^2 x^2 \log \left (\sqrt{d+e x}+\sqrt{d}\right )+12 a \sqrt{d} e x \sqrt{d+e x}-8 b d^{3/2} \sqrt{d+e x} \log \left (c x^n\right )+6 b e^2 x^2 \log \left (c x^n\right ) \log \left (\sqrt{d}-\sqrt{d+e x}\right )-6 b e^2 x^2 \log \left (c x^n\right ) \log \left (\sqrt{d+e x}+\sqrt{d}\right )+12 b \sqrt{d} e x \sqrt{d+e x} \log \left (c x^n\right )-4 b d^{3/2} n \sqrt{d+e x}-3 b e^2 n x^2 \log ^2\left (\sqrt{d}-\sqrt{d+e x}\right )+3 b e^2 n x^2 \log ^2\left (\sqrt{d+e x}+\sqrt{d}\right )+6 b e^2 n x^2 \log \left (\sqrt{d+e x}+\sqrt{d}\right ) \log \left (\frac{1}{2}-\frac{\sqrt{d+e x}}{2 \sqrt{d}}\right )-6 b e^2 n x^2 \log \left (\sqrt{d}-\sqrt{d+e x}\right ) \log \left (\frac{1}{2} \left (\frac{\sqrt{d+e x}}{\sqrt{d}}+1\right )\right )+14 b e^2 n x^2 \tanh ^{-1}\left (\frac{\sqrt{d+e x}}{\sqrt{d}}\right )+10 b \sqrt{d} e n x \sqrt{d+e x}}{16 d^{5/2} x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*d^(3/2)*Sqrt[d + e*x] - 4*b*d^(3/2)*n*Sqrt[d + e*x] + 12*a*Sqrt[d]*e*x*Sqrt[d + e*x] + 10*b*Sqrt[d]*e*n*
x*Sqrt[d + e*x] + 14*b*e^2*n*x^2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] - 8*b*d^(3/2)*Sqrt[d + e*x]*Log[c*x^n] + 12*b*
Sqrt[d]*e*x*Sqrt[d + e*x]*Log[c*x^n] + 6*a*e^2*x^2*Log[Sqrt[d] - Sqrt[d + e*x]] + 6*b*e^2*x^2*Log[c*x^n]*Log[S
qrt[d] - Sqrt[d + e*x]] - 3*b*e^2*n*x^2*Log[Sqrt[d] - Sqrt[d + e*x]]^2 - 6*a*e^2*x^2*Log[Sqrt[d] + Sqrt[d + e*
x]] - 6*b*e^2*x^2*Log[c*x^n]*Log[Sqrt[d] + Sqrt[d + e*x]] + 3*b*e^2*n*x^2*Log[Sqrt[d] + Sqrt[d + e*x]]^2 + 6*b
*e^2*n*x^2*Log[Sqrt[d] + Sqrt[d + e*x]]*Log[1/2 - Sqrt[d + e*x]/(2*Sqrt[d])] - 6*b*e^2*n*x^2*Log[Sqrt[d] - Sqr
t[d + e*x]]*Log[(1 + Sqrt[d + e*x]/Sqrt[d])/2] - 6*b*e^2*n*x^2*PolyLog[2, 1/2 - Sqrt[d + e*x]/(2*Sqrt[d])] + 6
*b*e^2*n*x^2*PolyLog[2, (1 + Sqrt[d + e*x]/Sqrt[d])/2])/(16*d^(5/2)*x^2)

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Maple [F]  time = 0.513, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c{x}^{n} \right ) }{{x}^{3}}{\frac{1}{\sqrt{ex+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{e x + d} b \log \left (c x^{n}\right ) + \sqrt{e x + d} a}{e x^{4} + d x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((sqrt(e*x + d)*b*log(c*x^n) + sqrt(e*x + d)*a)/(e*x^4 + d*x^3), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b \log{\left (c x^{n} \right )}}{x^{3} \sqrt{d + e x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))/x**3/(e*x+d)**(1/2),x)

[Out]

Integral((a + b*log(c*x**n))/(x**3*sqrt(d + e*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \log \left (c x^{n}\right ) + a}{\sqrt{e x + d} x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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