3.302 \(\int \frac{a+b \log (c x^n)}{x^3 (d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=337 \[ \frac{5 b e n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right )}{4 d^{7/2}}-\frac{5 e \left (a+b \log \left (c x^n\right )\right )}{2 d^3 \sqrt{d+e x^2}}-\frac{5 e \left (a+b \log \left (c x^n\right )\right )}{6 d^2 \left (d+e x^2\right )^{3/2}}+\frac{5 e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac{a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )^{3/2}}-\frac{b n \sqrt{d+e x^2}}{4 d^3 x^2}+\frac{b e n}{3 d^3 \sqrt{d+e x^2}}-\frac{5 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{4 d^{7/2}}-\frac{31 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{12 d^{7/2}}+\frac{5 b e n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 d^{7/2}} \]

[Out]

(b*e*n)/(3*d^3*Sqrt[d + e*x^2]) - (b*n*Sqrt[d + e*x^2])/(4*d^3*x^2) - (31*b*e*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d
]])/(12*d^(7/2)) - (5*b*e*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]^2)/(4*d^(7/2)) - (5*e*(a + b*Log[c*x^n]))/(6*d^2*
(d + e*x^2)^(3/2)) - (a + b*Log[c*x^n])/(2*d*x^2*(d + e*x^2)^(3/2)) - (5*e*(a + b*Log[c*x^n]))/(2*d^3*Sqrt[d +
 e*x^2]) + (5*e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*(a + b*Log[c*x^n]))/(2*d^(7/2)) + (5*b*e*n*ArcTanh[Sqrt[d + e
*x^2]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*d^(7/2)) + (5*b*e*n*PolyLog[2, 1 - (2*Sqrt[d])
/(Sqrt[d] - Sqrt[d + e*x^2])])/(4*d^(7/2))

________________________________________________________________________________________

Rubi [A]  time = 0.487747, antiderivative size = 341, normalized size of antiderivative = 1.01, number of steps used = 13, number of rules used = 13, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.52, Rules used = {266, 51, 63, 208, 2350, 1251, 897, 1259, 453, 5984, 5918, 2402, 2315} \[ \frac{5 b e n \text{PolyLog}\left (2,1-\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right )}{4 d^{7/2}}-\frac{5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^3 x^2}+\frac{5 \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^2 \sqrt{d+e x^2}}+\frac{5 e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{7/2}}+\frac{a+b \log \left (c x^n\right )}{3 d x^2 \left (d+e x^2\right )^{3/2}}-\frac{b n \sqrt{d+e x^2}}{4 d^3 x^2}+\frac{b e n}{3 d^3 \sqrt{d+e x^2}}-\frac{5 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{4 d^{7/2}}-\frac{31 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{12 d^{7/2}}+\frac{5 b e n \log \left (\frac{2 \sqrt{d}}{\sqrt{d}-\sqrt{d+e x^2}}\right ) \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 d^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*x^n])/(x^3*(d + e*x^2)^(5/2)),x]

[Out]

(b*e*n)/(3*d^3*Sqrt[d + e*x^2]) - (b*n*Sqrt[d + e*x^2])/(4*d^3*x^2) - (31*b*e*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d
]])/(12*d^(7/2)) - (5*b*e*n*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]^2)/(4*d^(7/2)) + (a + b*Log[c*x^n])/(3*d*x^2*(d +
 e*x^2)^(3/2)) + (5*(a + b*Log[c*x^n]))/(3*d^2*x^2*Sqrt[d + e*x^2]) - (5*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/(
2*d^3*x^2) + (5*e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*(a + b*Log[c*x^n]))/(2*d^(7/2)) + (5*b*e*n*ArcTanh[Sqrt[d +
 e*x^2]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*d^(7/2)) + (5*b*e*n*PolyLog[2, 1 - (2*Sqrt[d
])/(Sqrt[d] - Sqrt[d + e*x^2])])/(4*d^(7/2))

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 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
 IntegerQ[(m - 1)/2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
 p] && FractionQ[m]

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 453

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

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 )^{5/2}} \, dx &=\frac{a+b \log \left (c x^n\right )}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^3 x^2}+\frac{5 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^{7/2}}-(b n) \int \left (-\frac{3 d^2+20 d e x^2+15 e^2 x^4}{6 d^3 x^3 \left (d+e x^2\right )^{3/2}}+\frac{5 e \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{2 d^{7/2} x}\right ) \, dx\\ &=\frac{a+b \log \left (c x^n\right )}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^3 x^2}+\frac{5 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^{7/2}}+\frac{(b n) \int \frac{3 d^2+20 d e x^2+15 e^2 x^4}{x^3 \left (d+e x^2\right )^{3/2}} \, dx}{6 d^3}-\frac{(5 b e n) \int \frac{\tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{x} \, dx}{2 d^{7/2}}\\ &=\frac{a+b \log \left (c x^n\right )}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^3 x^2}+\frac{5 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^{7/2}}+\frac{(b n) \operatorname{Subst}\left (\int \frac{3 d^2+20 d e x+15 e^2 x^2}{x^2 (d+e x)^{3/2}} \, dx,x,x^2\right )}{12 d^3}-\frac{(5 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^{7/2}}\\ &=\frac{a+b \log \left (c x^n\right )}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^3 x^2}+\frac{5 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^{7/2}}+\frac{(b n) \operatorname{Subst}\left (\int \frac{-2 d^2-10 d x^2+15 x^4}{x^2 \left (-\frac{d}{e}+\frac{x^2}{e}\right )^2} \, dx,x,\sqrt{d+e x^2}\right )}{6 d^3 e}-\frac{(5 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^{7/2}}\\ &=-\frac{b n \sqrt{d+e x^2}}{4 d^3 x^2}-\frac{5 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{4 d^{7/2}}+\frac{a+b \log \left (c x^n\right )}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^3 x^2}+\frac{5 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^{7/2}}+\frac{(5 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^4}-\frac{\left (b e^3 n\right ) \operatorname{Subst}\left (\int \frac{-\frac{4 d^3}{e^3}-\frac{27 d^2 x^2}{e^3}}{x^2 \left (-\frac{d}{e}+\frac{x^2}{e}\right )} \, dx,x,\sqrt{d+e x^2}\right )}{12 d^5}\\ &=\frac{b e n}{3 d^3 \sqrt{d+e x^2}}-\frac{b n \sqrt{d+e x^2}}{4 d^3 x^2}-\frac{5 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{4 d^{7/2}}+\frac{a+b \log \left (c x^n\right )}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^3 x^2}+\frac{5 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^{7/2}}+\frac{5 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^{7/2}}+\frac{(31 b n) \operatorname{Subst}\left (\int \frac{1}{-\frac{d}{e}+\frac{x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{12 d^3}-\frac{(5 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^4}\\ &=\frac{b e n}{3 d^3 \sqrt{d+e x^2}}-\frac{b n \sqrt{d+e x^2}}{4 d^3 x^2}-\frac{31 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{12 d^{7/2}}-\frac{5 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{4 d^{7/2}}+\frac{a+b \log \left (c x^n\right )}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^3 x^2}+\frac{5 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^{7/2}}+\frac{5 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^{7/2}}+\frac{(5 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^{7/2}}\\ &=\frac{b e n}{3 d^3 \sqrt{d+e x^2}}-\frac{b n \sqrt{d+e x^2}}{4 d^3 x^2}-\frac{31 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{12 d^{7/2}}-\frac{5 b e n \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )^2}{4 d^{7/2}}+\frac{a+b \log \left (c x^n\right )}{3 d x^2 \left (d+e x^2\right )^{3/2}}+\frac{5 \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^2 \sqrt{d+e x^2}}-\frac{5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 d^3 x^2}+\frac{5 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^{7/2}}+\frac{5 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^{7/2}}+\frac{5 b e n \text{Li}_2\left (1-\frac{2}{1-\frac{\sqrt{d+e x^2}}{\sqrt{d}}}\right )}{4 d^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.308776, size = 227, normalized size = 0.67 \[ \frac{b n \sqrt{\frac{d}{e x^2}+1} \left (5 \, _3F_2\left (\frac{7}{2},\frac{7}{2},\frac{7}{2};\frac{9}{2},\frac{9}{2};-\frac{d}{e x^2}\right )-7 (2 \log (x)+1) \, _2F_1\left (\frac{5}{2},\frac{7}{2};\frac{9}{2};-\frac{d}{e x^2}\right )\right )}{98 e^2 x^6 \sqrt{d+e x^2}}-\frac{\left (3 d^2+20 d e x^2+15 e^2 x^4\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{6 d^3 x^2 \left (d+e x^2\right )^{3/2}}-\frac{5 e \log (x) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{2 d^{7/2}}+\frac{5 e \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{2 d^{7/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + b*Log[c*x^n])/(x^3*(d + e*x^2)^(5/2)),x]

[Out]

(b*n*Sqrt[1 + d/(e*x^2)]*(5*HypergeometricPFQ[{7/2, 7/2, 7/2}, {9/2, 9/2}, -(d/(e*x^2))] - 7*Hypergeometric2F1
[5/2, 7/2, 9/2, -(d/(e*x^2))]*(1 + 2*Log[x])))/(98*e^2*x^6*Sqrt[d + e*x^2]) - ((3*d^2 + 20*d*e*x^2 + 15*e^2*x^
4)*(a - b*n*Log[x] + b*Log[c*x^n]))/(6*d^3*x^2*(d + e*x^2)^(3/2)) - (5*e*Log[x]*(a - b*n*Log[x] + b*Log[c*x^n]
))/(2*d^(7/2)) + (5*e*(a - b*n*Log[x] + b*Log[c*x^n])*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(2*d^(7/2))

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Maple [F]  time = 0.411, 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{5}{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)^(5/2),x)

[Out]

int((a+b*ln(c*x^n))/x^3/(e*x^2+d)^(5/2),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)^(5/2),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^{3} x^{9} + 3 \, d e^{2} x^{7} + 3 \, d^{2} e x^{5} + d^{3} 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)^(5/2),x, algorithm="fricas")

[Out]

integral((sqrt(e*x^2 + d)*b*log(c*x^n) + sqrt(e*x^2 + d)*a)/(e^3*x^9 + 3*d*e^2*x^7 + 3*d^2*e*x^5 + d^3*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)**(5/2),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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)/((e*x^2 + d)^(5/2)*x^3), x)