3.4.3 \(\int \frac {x^6 (a+b \log (c x^n))}{(d+e x^2)^{5/2}} \, dx\) [303]

Optimal. Leaf size=443 \[ \frac {b d n x}{3 e^3 \sqrt {d+e x^2}}-\frac {b n x \sqrt {d+e x^2}}{4 e^3}-\frac {31 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{12 e^{7/2} \sqrt {d+e x^2}}-\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{4 e^{7/2} \sqrt {d+e x^2}}+\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {d+e x^2}}+\frac {5 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^{3/2} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}+\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \text {Li}_2\left (e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{4 e^{7/2} \sqrt {d+e x^2}} \]

[Out]

-1/3*x^5*(a+b*ln(c*x^n))/e/(e*x^2+d)^(3/2)+5/6*b*d*n*x/e^3/(e*x^2+d)^(1/2)+1/2*b*n*x^3/e^2/(e*x^2+d)^(1/2)-5/3
*x^3*(a+b*ln(c*x^n))/e^2/(e*x^2+d)^(1/2)-3/4*b*n*x*(e*x^2+d)^(1/2)/e^3+5/2*x*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e
^3-31/12*b*d^(3/2)*n*arcsinh(x*e^(1/2)/d^(1/2))*(1+e*x^2/d)^(1/2)/e^(7/2)/(e*x^2+d)^(1/2)-5/4*b*d^(3/2)*n*arcs
inh(x*e^(1/2)/d^(1/2))^2*(1+e*x^2/d)^(1/2)/e^(7/2)/(e*x^2+d)^(1/2)-5*b*d^(3/2)*n*arcsinh(x*e^(1/2)/d^(1/2))*ar
ctanh((x*e^(1/2)/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*(1+e*x^2/d)^(1/2)/e^(7/2)/(e*x^2+d)^(1/2)+5/2*b*d^(3/2)*n*arcsi
nh(x*e^(1/2)/d^(1/2))*ln(1+(x*e^(1/2)/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*(1+e*x^2/d)^(1/2)/e^(7/2)/(e*x^2+d)^(1/2)-
5/2*d^(3/2)*arcsinh(x*e^(1/2)/d^(1/2))*(a+b*ln(c*x^n))*(1+e*x^2/d)^(1/2)/e^(7/2)/(e*x^2+d)^(1/2)+5/4*b*d^(3/2)
*n*polylog(2,(x*e^(1/2)/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*(1+e*x^2/d)^(1/2)/e^(7/2)/(e*x^2+d)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.37, antiderivative size = 443, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 13, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {2386, 294, 327, 221, 2392, 21, 1171, 396, 5775, 3797, 2221, 2317, 2438} \begin {gather*} \frac {5 b d^{3/2} n \sqrt {\frac {e x^2}{d}+1} \text {PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{4 e^{7/2} \sqrt {d+e x^2}}-\frac {5 d^{3/2} \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}+\frac {5 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {d+e x^2}}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {5 b d^{3/2} n \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{4 e^{7/2} \sqrt {d+e x^2}}-\frac {31 b d^{3/2} n \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{12 e^{7/2} \sqrt {d+e x^2}}+\frac {5 b d^{3/2} n \sqrt {\frac {e x^2}{d}+1} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {b n x \sqrt {d+e x^2}}{4 e^3}+\frac {b d n x}{3 e^3 \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^6*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 221

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 396

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

Rule 1171

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, Simp[(-R)*x*((d + e*x^2)^(q + 1)/(2*d*(q + 1))), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1
)*ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2317

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2386

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[d^IntPart[q
]*((d + e*x^2)^FracPart[q]/(1 + (e/d)*x^2)^FracPart[q]), Int[x^m*(1 + (e/d)*x^2)^q*(a + b*Log[c*x^n]), x], x]
/; FreeQ[{a, b, c, d, e, n}, x] && IntegerQ[m/2] && IntegerQ[q - 1/2] &&  !(LtQ[m + 2*q, -2] || GtQ[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 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 3797

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (Complex[0, fz_])*(f_.)*(x_)], x_Symbol] :> Simp[(-I)*((
c + d*x)^(m + 1)/(d*(m + 1))), x] + Dist[2*I, Int[((c + d*x)^m*(E^(2*((-I)*e + f*fz*x))/(1 + E^(2*((-I)*e + f*
fz*x))/E^(2*I*k*Pi))))/E^(2*I*k*Pi), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 5775

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/(x_), x_Symbol] :> Dist[1/b, Subst[Int[x^n*Coth[-a/b + x/b], x],
x, a + b*ArcSinh[c*x]], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac {\sqrt {1+\frac {e x^2}{d}} \int \frac {x^6 \left (a+b \log \left (c x^n\right )\right )}{\left (1+\frac {e x^2}{d}\right )^{5/2}} \, dx}{d^2 \sqrt {d+e x^2}}\\ &=-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {d+e x^2}}+\frac {5 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^{3/2} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {\left (b n \sqrt {1+\frac {e x^2}{d}}\right ) \int \left (\frac {d^3 \sqrt {1+\frac {e x^2}{d}} \left (15 d^2+20 d e x^2+3 e^2 x^4\right )}{6 e^3 \left (d+e x^2\right )^2}-\frac {5 d^{7/2} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{2 e^{7/2} x}\right ) \, dx}{d^2 \sqrt {d+e x^2}}\\ &=-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {d+e x^2}}+\frac {5 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^{3/2} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}+\frac {\left (5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {\sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{x} \, dx}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {\left (b d n \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {\sqrt {1+\frac {e x^2}{d}} \left (15 d^2+20 d e x^2+3 e^2 x^4\right )}{\left (d+e x^2\right )^2} \, dx}{6 e^3 \sqrt {d+e x^2}}\\ &=-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {d+e x^2}}+\frac {5 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^{3/2} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}+\frac {\left (5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}}\right ) \text {Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {\left (b n \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {15 d^2+20 d e x^2+3 e^2 x^4}{\left (1+\frac {e x^2}{d}\right )^{3/2}} \, dx}{6 d e^3 \sqrt {d+e x^2}}\\ &=\frac {b d n x}{3 e^3 \sqrt {d+e x^2}}-\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{4 e^{7/2} \sqrt {d+e x^2}}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {d+e x^2}}+\frac {5 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^{3/2} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {\left (5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}}\right ) \text {Subst}\left (\int \frac {e^{2 x} x}{1-e^{2 x}} \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )\right )}{e^{7/2} \sqrt {d+e x^2}}+\frac {\left (b n \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {-17 d^2-3 d e x^2}{\sqrt {1+\frac {e x^2}{d}}} \, dx}{6 d e^3 \sqrt {d+e x^2}}\\ &=\frac {b d n x}{3 e^3 \sqrt {d+e x^2}}-\frac {b n x \sqrt {d+e x^2}}{4 e^3}-\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{4 e^{7/2} \sqrt {d+e x^2}}+\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {d+e x^2}}+\frac {5 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^{3/2} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {\left (5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}}\right ) \text {Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {\left (31 b d n \sqrt {1+\frac {e x^2}{d}}\right ) \int \frac {1}{\sqrt {1+\frac {e x^2}{d}}} \, dx}{12 e^3 \sqrt {d+e x^2}}\\ &=\frac {b d n x}{3 e^3 \sqrt {d+e x^2}}-\frac {b n x \sqrt {d+e x^2}}{4 e^3}-\frac {31 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{12 e^{7/2} \sqrt {d+e x^2}}-\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{4 e^{7/2} \sqrt {d+e x^2}}+\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {d+e x^2}}+\frac {5 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^{3/2} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {\left (5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}}\right ) \text {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{4 e^{7/2} \sqrt {d+e x^2}}\\ &=\frac {b d n x}{3 e^3 \sqrt {d+e x^2}}-\frac {b n x \sqrt {d+e x^2}}{4 e^3}-\frac {31 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{12 e^{7/2} \sqrt {d+e x^2}}-\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{4 e^{7/2} \sqrt {d+e x^2}}+\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{2 e^{7/2} \sqrt {d+e x^2}}-\frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{3 e \left (d+e x^2\right )^{3/2}}-\frac {5 x^3 \left (a+b \log \left (c x^n\right )\right )}{3 e^2 \sqrt {d+e x^2}}+\frac {5 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 e^3}-\frac {5 d^{3/2} \sqrt {1+\frac {e x^2}{d}} \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^{7/2} \sqrt {d+e x^2}}+\frac {5 b d^{3/2} n \sqrt {1+\frac {e x^2}{d}} \text {Li}_2\left (e^{2 \sinh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{4 e^{7/2} \sqrt {d+e x^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 = 0.16, size = 199, normalized size = 0.45 \begin {gather*} \frac {b n x^7 \sqrt {1+\frac {e x^2}{d}} \left (5 \, _3F_2\left (\frac {7}{2},\frac {7}{2},\frac {7}{2};\frac {9}{2},\frac {9}{2};-\frac {e x^2}{d}\right )+7 \, _2F_1\left (\frac {5}{2},\frac {7}{2};\frac {9}{2};-\frac {e x^2}{d}\right ) (-1+2 \log (x))\right )}{98 d^2 \sqrt {d+e x^2}}+\frac {x \left (15 d^2+20 d e x^2+3 e^2 x^4\right ) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{6 e^3 \left (d+e x^2\right )^{3/2}}-\frac {5 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{2 e^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(a + b*Log[c*x^n]))/(d + e*x^2)^(5/2),x]

[Out]

(b*n*x^7*Sqrt[1 + (e*x^2)/d]*(5*HypergeometricPFQ[{7/2, 7/2, 7/2}, {9/2, 9/2}, -((e*x^2)/d)] + 7*Hypergeometri
c2F1[5/2, 7/2, 9/2, -((e*x^2)/d)]*(-1 + 2*Log[x])))/(98*d^2*Sqrt[d + e*x^2]) + (x*(15*d^2 + 20*d*e*x^2 + 3*e^2
*x^4)*(a - b*n*Log[x] + b*Log[c*x^n]))/(6*e^3*(d + e*x^2)^(3/2)) - (5*d*(a - b*n*Log[x] + b*Log[c*x^n])*Log[e*
x + Sqrt[e]*Sqrt[d + e*x^2]])/(2*e^(7/2))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{6} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(a+b*ln(c*x^n))/(e*x^2+d)^(5/2),x)

[Out]

int(x^6*(a+b*ln(c*x^n))/(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(x^6*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/6*(3*x^5*e^(-1)/(x^2*e + d)^(3/2) + 5*(3*x^2*e^(-1)/(x^2*e + d)^(3/2) + 2*d*e^(-2)/(x^2*e + d)^(3/2))*d*x*e^
(-1) - 15*d*arcsinh(x*e^(1/2)/sqrt(d))*e^(-7/2) + 5*d*x*e^(-3)/sqrt(x^2*e + d))*a + b*integrate((x^6*log(c) +
x^6*log(x^n))/((x^4*e^2 + 2*d*x^2*e + d^2)*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(x^6*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

integral((sqrt(x^2*e + d)*b*x^6*log(c*x^n) + sqrt(x^2*e + d)*a*x^6)/(x^6*e^3 + 3*d*x^4*e^2 + 3*d^2*x^2*e + d^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(x**6*(a+b*ln(c*x**n))/(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(x^6*(a+b*log(c*x^n))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^6/(x^2*e + d)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^6\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(a + b*log(c*x^n)))/(d + e*x^2)^(5/2),x)

[Out]

int((x^6*(a + b*log(c*x^n)))/(d + e*x^2)^(5/2), x)

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