\(\int \frac {(d+e x^2)^{3/2} (a+b \log (c x^n))}{x^3} \, dx\) [267]

Optimal result

Integrand size = 25, antiderivative size = 295 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-b e n \sqrt {d+e x^2}-\frac {b d n \sqrt {d+e x^2}}{4 x^2}+\frac {3}{4} b \sqrt {d} e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )+\frac {3}{4} b \sqrt {d} e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2+\frac {3}{2} e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {3}{2} \sqrt {d} e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {3}{2} b \sqrt {d} e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )-\frac {3}{4} b \sqrt {d} e n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 295, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {272, 43, 52, 65, 214, 2392, 12, 14, 6131, 6055, 2449, 2352} \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {3}{2} \sqrt {d} e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {3}{2} e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac {3}{4} b \sqrt {d} e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2+\frac {3}{4} b \sqrt {d} e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )-\frac {3}{2} b \sqrt {d} e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )-\frac {3}{4} b \sqrt {d} e n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^2+d}}\right )-b e n \sqrt {d+e x^2}-\frac {b d n \sqrt {d+e x^2}}{4 x^2} \]

[In]

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

[Out]

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

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 52

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

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.59 (sec) , antiderivative size = 349, normalized size of antiderivative = 1.18 \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {b e n \sqrt {d+e x^2} \left (-\, _3F_2\left (-\frac {1}{2},-\frac {1}{2},-\frac {1}{2};\frac {1}{2},\frac {1}{2};-\frac {d}{e x^2}\right )+\sqrt {1+\frac {d}{e x^2}} \log (x)-\frac {\sqrt {d} \text {arcsinh}\left (\frac {\sqrt {d}}{\sqrt {e} x}\right ) \log (x)}{\sqrt {e} x}\right )}{\sqrt {1+\frac {d}{e x^2}}}-\frac {b \sqrt {d} n \sqrt {d+e x^2} \left (2 \sqrt {d} \, _3F_2\left (\frac {1}{2},\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {d}{e x^2}\right )+\left (\sqrt {d} \sqrt {1+\frac {d}{e x^2}}+\sqrt {e} x \text {arcsinh}\left (\frac {\sqrt {d}}{\sqrt {e} x}\right )\right ) (1+2 \log (x))\right )}{4 \sqrt {1+\frac {d}{e x^2}} x^2}-\frac {\left (d-2 e x^2\right ) \sqrt {d+e x^2} \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {3}{2} \sqrt {d} e \log (x) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )-\frac {3}{2} \sqrt {d} e \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right ) \]

[In]

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

[Out]

(b*e*n*Sqrt[d + e*x^2]*(-HypergeometricPFQ[{-1/2, -1/2, -1/2}, {1/2, 1/2}, -(d/(e*x^2))] + Sqrt[1 + d/(e*x^2)]
*Log[x] - (Sqrt[d]*ArcSinh[Sqrt[d]/(Sqrt[e]*x)]*Log[x])/(Sqrt[e]*x)))/Sqrt[1 + d/(e*x^2)] - (b*Sqrt[d]*n*Sqrt[
d + e*x^2]*(2*Sqrt[d]*HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3/2}, -(d/(e*x^2))] + (Sqrt[d]*Sqrt[1 + d/(e*x^
2)] + Sqrt[e]*x*ArcSinh[Sqrt[d]/(Sqrt[e]*x)])*(1 + 2*Log[x])))/(4*Sqrt[1 + d/(e*x^2)]*x^2) - ((d - 2*e*x^2)*Sq
rt[d + e*x^2]*(a - b*n*Log[x] + b*Log[c*x^n]))/(2*x^2) + (3*Sqrt[d]*e*Log[x]*(a - b*n*Log[x] + b*Log[c*x^n]))/
2 - (3*Sqrt[d]*e*(a - b*n*Log[x] + b*Log[c*x^n])*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/2

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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