3.4.2 \(\int \frac {a+b \log (c x^n)}{x^3 (d+e x^2)^{5/2}} \, dx\) [302]

Optimal. Leaf size=337 \[ \frac {b e n}{3 d^3 \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{4 d^3 x^2}-\frac {31 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{12 d^{7/2}}-\frac {5 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{7/2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{6 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )^{3/2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{2 d^3 \sqrt {d+e x^2}}+\frac {5 e \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{7/2}}+\frac {5 b e n \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{7/2}}+\frac {5 b e n \text {Li}_2\left (1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{4 d^{7/2}} \]

[Out]

-31/12*b*e*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(7/2)-5/4*b*e*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))^2/d^(7/2)-5/6
*e*(a+b*ln(c*x^n))/d^2/(e*x^2+d)^(3/2)+1/2*(-a-b*ln(c*x^n))/d/x^2/(e*x^2+d)^(3/2)+5/2*e*arctanh((e*x^2+d)^(1/2
)/d^(1/2))*(a+b*ln(c*x^n))/d^(7/2)+5/2*b*e*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)-(e*x^2+d)^
(1/2)))/d^(7/2)+5/4*b*e*n*polylog(2,1-2*d^(1/2)/(d^(1/2)-(e*x^2+d)^(1/2)))/d^(7/2)+1/3*b*e*n/d^3/(e*x^2+d)^(1/
2)-5/2*e*(a+b*ln(c*x^n))/d^3/(e*x^2+d)^(1/2)-1/4*b*n*(e*x^2+d)^(1/2)/d^3/x^2

________________________________________________________________________________________

Rubi [A]
time = 0.35, antiderivative size = 337, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 14, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {272, 44, 53, 65, 214, 2392, 1265, 911, 1273, 464, 6131, 6055, 2449, 2352} \begin {gather*} \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 \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 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 {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )^{3/2}}-\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}}-\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}} \end {gather*}

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)) - (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))

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 53

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 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 464

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 911

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 1265

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 1273

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/(d + e*x^2))*(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))], 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 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*} \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) \text {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) \text {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) \text {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) \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 )}{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) \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 )}{2 d^4}-\frac {\left (b e^3 n\right ) \text {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) \text {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) \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 )}{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) \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 )}{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*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.18, size = 227, normalized size = 0.67 \begin {gather*} \frac {b n \sqrt {1+\frac {d}{e x^2}} \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 \, _2F_1\left (\frac {5}{2},\frac {7}{2};\frac {9}{2};-\frac {d}{e x^2}\right ) (1+2 \log (x))\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 n \log (x)+b \log \left (c x^n\right )\right )}{6 d^3 x^2 \left (d+e x^2\right )^{3/2}}-\frac {5 e \log (x) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{2 d^{7/2}}+\frac {5 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 )}{2 d^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[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 = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{3} \left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx\]

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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]

1/6*a*(15*arcsinh(sqrt(d)*e^(-1/2)/abs(x))*e/d^(7/2) - 15*e/(sqrt(x^2*e + d)*d^3) - 5*e/((x^2*e + d)^(3/2)*d^2
) - 3/((x^2*e + d)^(3/2)*d*x^2)) + b*integrate((log(c) + log(x^n))/((x^7*e^2 + 2*d*x^5*e + d^2*x^3)*sqrt(x^2*e
+ d)), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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(x^2*e + d)*b*log(c*x^n) + sqrt(x^2*e + d)*a)/(x^9*e^3 + 3*d*x^7*e^2 + 3*d^2*x^5*e + d^3*x^3), x
)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)/((x^2*e + d)^(5/2)*x^3), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,\ln \left (c\,x^n\right )}{x^3\,{\left (e\,x^2+d\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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