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

Optimal. Leaf size=287 \[ \frac{3 b e n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right )}{4 d^{5/2}}-\frac{3 e \left (a+b \log \left (c x^n\right )\right )}{2 d^2 \sqrt{d+e x^2}}+\frac{3 e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}-\frac{a+b \log \left (c x^n\right )}{2 d x^2 \sqrt{d+e x^2}}-\frac{b n \sqrt{d+e x^2}}{4 d^2 x^2}-\frac{3 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{4 d^{5/2}}-\frac{5 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{4 d^{5/2}}+\frac{3 b e n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 d^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.386355, antiderivative size = 286, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 11, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.44, Rules used = {266, 51, 63, 208, 2350, 446, 78, 5984, 5918, 2402, 2315} \[ \frac{3 b e n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right )}{4 d^{5/2}}-\frac{3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2}+\frac{3 e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{5/2}}+\frac{a+b \log \left (c x^n\right )}{d x^2 \sqrt{d+e x^2}}-\frac{b n \sqrt{d+e x^2}}{4 d^2 x^2}-\frac{3 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{4 d^{5/2}}-\frac{5 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{4 d^{5/2}}+\frac{3 b e n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 d^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 266

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 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 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 78

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

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

Mathematica [C]  time = 0.311647, size = 218, normalized size = 0.76 \[ \frac{3 b d^{5/2} n \sqrt{\frac{d}{e x^2}+1} \, _3F_2\left (\frac{5}{2},\frac{5}{2},\frac{5}{2};\frac{7}{2},\frac{7}{2};-\frac{d}{e x^2}\right )-25 e x^2 \left (\sqrt{d} \left (d+3 e x^2\right )+3 e x^2 \log (x) \sqrt{d+e x^2}-3 e x^2 \sqrt{d+e x^2} \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )-5 b d^{5/2} n (2 \log (x)+1) \sqrt{\frac{d}{e x^2}+1} \, _2F_1\left (\frac{3}{2},\frac{5}{2};\frac{7}{2};-\frac{d}{e x^2}\right )}{50 d^{5/2} e x^4 \sqrt{d+e x^2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(3*b*d^(5/2)*n*Sqrt[1 + d/(e*x^2)]*HypergeometricPFQ[{5/2, 5/2, 5/2}, {7/2, 7/2}, -(d/(e*x^2))] - 5*b*d^(5/2)*
n*Sqrt[1 + d/(e*x^2)]*Hypergeometric2F1[3/2, 5/2, 7/2, -(d/(e*x^2))]*(1 + 2*Log[x]) - 25*e*x^2*(a - b*n*Log[x]
+ b*Log[c*x^n])*(Sqrt[d]*(d + 3*e*x^2) + 3*e*x^2*Sqrt[d + e*x^2]*Log[x] - 3*e*x^2*Sqrt[d + e*x^2]*Log[d + Sqr
t[d]*Sqrt[d + e*x^2]]))/(50*d^(5/2)*e*x^4*Sqrt[d + e*x^2])

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Maple [F]  time = 0.412, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\ln \left ( c{x}^{n} \right ) }{{x}^{3}} \left ( e{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}\, 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^2+d)^(3/2),x)

[Out]

int((a+b*ln(c*x^n))/x^3/(e*x^2+d)^(3/2),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^2+d)^(3/2),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^{2} + d} b \log \left (c x^{n}\right ) + \sqrt{e x^{2} + d} a}{e^{2} x^{7} + 2 \, d e x^{5} + d^{2} 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^2+d)^(3/2),x, algorithm="fricas")

[Out]

integral((sqrt(e*x^2 + d)*b*log(c*x^n) + sqrt(e*x^2 + d)*a)/(e^2*x^7 + 2*d*e*x^5 + d^2*x^3), 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((a+b*ln(c*x**n))/x**3/(e*x**2+d)**(3/2),x)

[Out]

Timed out

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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}{{\left (e x^{2} + d\right )}^{\frac{3}{2}} 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^2+d)^(3/2),x, algorithm="giac")

[Out]

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