\(\int \frac {(d+e x)^{3/2} (a+b \log (c x^n))}{x^3} \, dx\) [143]

Optimal result

Integrand size = 23, antiderivative size = 293 \[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {b d n \sqrt {d+e x}}{4 x^2}-\frac {11 b e n \sqrt {d+e x}}{8 x}-\frac {9 b e^2 n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{8 \sqrt {d}}+\frac {3 b e^2 n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 \sqrt {d}}-\frac {3 b e^2 n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{2 \sqrt {d}}-\frac {3 b e^2 n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{4 \sqrt {d}} \]

[Out]

-1/2*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/x^2-9/8*b*e^2*n*arctanh((e*x+d)^(1/2)/d^(1/2))/d^(1/2)+3/4*b*e^2*n*arctanh(
(e*x+d)^(1/2)/d^(1/2))^2/d^(1/2)-3/4*e^2*arctanh((e*x+d)^(1/2)/d^(1/2))*(a+b*ln(c*x^n))/d^(1/2)-3/2*b*e^2*n*ar
ctanh((e*x+d)^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)-(e*x+d)^(1/2)))/d^(1/2)-3/4*b*e^2*n*polylog(2,1-2*d^(1/2)/(
d^(1/2)-(e*x+d)^(1/2)))/d^(1/2)-1/4*b*d*n*(e*x+d)^(1/2)/x^2-11/8*b*e*n*(e*x+d)^(1/2)/x-3/4*e*(a+b*ln(c*x^n))*(
e*x+d)^(1/2)/x

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {43, 65, 214, 2392, 12, 14, 44, 6131, 6055, 2449, 2352} \[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {3 b e^2 n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{4 \sqrt {d}}-\frac {9 b e^2 n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{8 \sqrt {d}}-\frac {3 b e^2 n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{2 \sqrt {d}}-\frac {3 b e^2 n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{4 \sqrt {d}}-\frac {b d n \sqrt {d+e x}}{4 x^2}-\frac {11 b e n \sqrt {d+e x}}{8 x} \]

[In]

Int[((d + e*x)^(3/2)*(a + b*Log[c*x^n]))/x^3,x]

[Out]

-1/4*(b*d*n*Sqrt[d + e*x])/x^2 - (11*b*e*n*Sqrt[d + e*x])/(8*x) - (9*b*e^2*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(
8*Sqrt[d]) + (3*b*e^2*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]^2)/(4*Sqrt[d]) - (3*e*Sqrt[d + e*x]*(a + b*Log[c*x^n]))
/(4*x) - ((d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(2*x^2) - (3*e^2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*(a + b*Log[c*x^n
]))/(4*Sqrt[d]) - (3*b*e^2*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])/(2*Sqr
t[d]) - (3*b*e^2*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])/(4*Sqrt[d])

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 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

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

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 2392

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 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 6055

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 6131

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^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 \sqrt {d}}-(b n) \int \frac {-\sqrt {d+e x} (2 d+5 e x)-\frac {3 e^2 x^2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}}{4 x^3} \, dx \\ & = -\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^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 \sqrt {d}}-\frac {1}{4} (b n) \int \frac {-\sqrt {d+e x} (2 d+5 e x)-\frac {3 e^2 x^2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}}{x^3} \, dx \\ & = -\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^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 \sqrt {d}}-\frac {1}{4} (b n) \int \left (-\frac {2 d \sqrt {d+e x}}{x^3}-\frac {5 e \sqrt {d+e x}}{x^2}-\frac {3 e^2 \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d} x}\right ) \, dx \\ & = -\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^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 \sqrt {d}}+\frac {1}{2} (b d n) \int \frac {\sqrt {d+e x}}{x^3} \, dx+\frac {1}{4} (5 b e n) \int \frac {\sqrt {d+e x}}{x^2} \, dx+\frac {\left (3 b e^2 n\right ) \int \frac {\tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{x} \, dx}{4 \sqrt {d}} \\ & = -\frac {b d n \sqrt {d+e x}}{4 x^2}-\frac {5 b e n \sqrt {d+e x}}{4 x}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^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 \sqrt {d}}+\frac {1}{8} (b d e n) \int \frac {1}{x^2 \sqrt {d+e x}} \, dx+\frac {1}{8} \left (5 b e^2 n\right ) \int \frac {1}{x \sqrt {d+e x}} \, dx+\frac {\left (3 b e^2 n\right ) \text {Subst}\left (\int \frac {x \tanh ^{-1}\left (\frac {x}{\sqrt {d}}\right )}{-d+x^2} \, dx,x,\sqrt {d+e x}\right )}{2 \sqrt {d}} \\ & = -\frac {b d n \sqrt {d+e x}}{4 x^2}-\frac {11 b e n \sqrt {d+e x}}{8 x}+\frac {3 b e^2 n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^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 \sqrt {d}}+\frac {1}{4} (5 b e n) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )-\frac {1}{16} \left (b e^2 n\right ) \int \frac {1}{x \sqrt {d+e x}} \, dx-\frac {\left (3 b e^2 n\right ) \text {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} \\ & = -\frac {b d n \sqrt {d+e x}}{4 x^2}-\frac {11 b e n \sqrt {d+e x}}{8 x}-\frac {5 b e^2 n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{4 \sqrt {d}}+\frac {3 b e^2 n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^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 \sqrt {d}}-\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 \sqrt {d}}-\frac {1}{8} (b e n) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x}\right )+\frac {\left (3 b e^2 n\right ) \text {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} \\ & = -\frac {b d n \sqrt {d+e x}}{4 x^2}-\frac {11 b e n \sqrt {d+e x}}{8 x}-\frac {9 b e^2 n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{8 \sqrt {d}}+\frac {3 b e^2 n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^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 \sqrt {d}}-\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 \sqrt {d}}-\frac {\left (3 b e^2 n\right ) \text {Subst}\left (\int \frac {\log (2 x)}{1-2 x} \, dx,x,\frac {1}{1-\frac {\sqrt {d+e x}}{\sqrt {d}}}\right )}{2 \sqrt {d}} \\ & = -\frac {b d n \sqrt {d+e x}}{4 x^2}-\frac {11 b e n \sqrt {d+e x}}{8 x}-\frac {9 b e^2 n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{8 \sqrt {d}}+\frac {3 b e^2 n \tanh ^{-1}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^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 \sqrt {d}}-\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 \sqrt {d}}-\frac {3 b e^2 n \text {Li}_2\left (1-\frac {2}{1-\frac {\sqrt {d+e x}}{\sqrt {d}}}\right )}{4 \sqrt {d}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 501, normalized size of antiderivative = 1.71 \[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {8 a d^{3/2} \sqrt {d+e x}+4 b d^{3/2} n \sqrt {d+e x}+20 a \sqrt {d} e x \sqrt {d+e x}+22 b \sqrt {d} e n x \sqrt {d+e x}+18 b e^2 n x^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+8 b d^{3/2} \sqrt {d+e x} \log \left (c x^n\right )+20 b \sqrt {d} e x \sqrt {d+e x} \log \left (c x^n\right )-6 a e^2 x^2 \log \left (\sqrt {d}-\sqrt {d+e x}\right )-6 b e^2 x^2 \log \left (c x^n\right ) \log \left (\sqrt {d}-\sqrt {d+e x}\right )+3 b e^2 n x^2 \log ^2\left (\sqrt {d}-\sqrt {d+e x}\right )+6 a e^2 x^2 \log \left (\sqrt {d}+\sqrt {d+e x}\right )+6 b e^2 x^2 \log \left (c x^n\right ) \log \left (\sqrt {d}+\sqrt {d+e x}\right )-3 b e^2 n x^2 \log ^2\left (\sqrt {d}+\sqrt {d+e x}\right )-6 b e^2 n x^2 \log \left (\sqrt {d}+\sqrt {d+e x}\right ) \log \left (\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )+6 b e^2 n x^2 \log \left (\sqrt {d}-\sqrt {d+e x}\right ) \log \left (\frac {1}{2} \left (1+\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )+6 b e^2 n x^2 \operatorname {PolyLog}\left (2,\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )-6 b e^2 n x^2 \operatorname {PolyLog}\left (2,\frac {1}{2} \left (1+\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )}{16 \sqrt {d} x^2} \]

[In]

Integrate[((d + e*x)^(3/2)*(a + b*Log[c*x^n]))/x^3,x]

[Out]

-1/16*(8*a*d^(3/2)*Sqrt[d + e*x] + 4*b*d^(3/2)*n*Sqrt[d + e*x] + 20*a*Sqrt[d]*e*x*Sqrt[d + e*x] + 22*b*Sqrt[d]
*e*n*x*Sqrt[d + e*x] + 18*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] +
20*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[Sqrt[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]
- Sqrt[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])/(Sqrt[d]*x^2)

Maple [F]

\[\int \frac {\left (e x +d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )}{x^{3}}d x\]

[In]

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

[Out]

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

Fricas [F]

\[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{3}} \,d x } \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \left (d + e x\right )^{\frac {3}{2}}}{x^{3}}\, dx \]

[In]

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

[Out]

Integral((a + b*log(c*x**n))*(d + e*x)**(3/2)/x**3, x)

Maxima [F]

\[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{3}} \,d x } \]

[In]

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

[Out]

1/8*(3*e^2*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/sqrt(d) - 2*(5*(e*x + d)^(3/2)*e^2 - 3*sqr
t(e*x + d)*d*e^2)/((e*x + d)^2 - 2*(e*x + d)*d + d^2))*a + b*integrate((e*x*log(c) + d*log(c) + (e*x + d)*log(
x^n))*sqrt(e*x + d)/x^3, x)

Giac [F]

\[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{3}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {\left (a+b\,\ln \left (c\,x^n\right )\right )\,{\left (d+e\,x\right )}^{3/2}}{x^3} \,d x \]

[In]

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

[Out]

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