3.256 \(\int x^4 \sqrt{d+e x^2} (a+b \log (c x^n)) \, dx\)

Optimal. Leaf size=469 \[ -\frac{b d^{5/2} n \sqrt{d+e x^2} \text{PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}\right )}{32 e^{5/2} \sqrt{\frac{e x^2}{d}+1}}-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{\frac{e x^2}{d}+1}}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{7 b d^2 n x \sqrt{d+e x^2}}{192 e^2}+\frac{b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )^2}{32 e^{5/2} \sqrt{\frac{e x^2}{d}+1}}+\frac{5 b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{192 e^{5/2} \sqrt{\frac{e x^2}{d}+1}}-\frac{b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}\right )}{16 e^{5/2} \sqrt{\frac{e x^2}{d}+1}}-\frac{1}{36} b n x^5 \sqrt{d+e x^2}-\frac{5 b d n x^3 \sqrt{d+e x^2}}{288 e} \]

[Out]

(7*b*d^2*n*x*Sqrt[d + e*x^2])/(192*e^2) - (5*b*d*n*x^3*Sqrt[d + e*x^2])/(288*e) - (b*n*x^5*Sqrt[d + e*x^2])/36
+ (5*b*d^(5/2)*n*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])/(192*e^(5/2)*Sqrt[1 + (e*x^2)/d]) + (b*d^(5/2)
*n*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]^2)/(32*e^(5/2)*Sqrt[1 + (e*x^2)/d]) - (b*d^(5/2)*n*Sqrt[d + e*
x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[1 - E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])])/(16*e^(5/2)*Sqrt[1 + (e*x^2)/d]
) - (d^2*x*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/(16*e^2) + (d*x^3*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/(24*e) +
(x^5*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/6 + (d^(5/2)*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[
c*x^n]))/(16*e^(5/2)*Sqrt[1 + (e*x^2)/d]) - (b*d^(5/2)*n*Sqrt[d + e*x^2]*PolyLog[2, E^(2*ArcSinh[(Sqrt[e]*x)/S
qrt[d]])])/(32*e^(5/2)*Sqrt[1 + (e*x^2)/d])

________________________________________________________________________________________

Rubi [A]  time = 0.597734, antiderivative size = 469, normalized size of antiderivative = 1., number of steps used = 19, number of rules used = 13, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.52, Rules used = {2341, 279, 321, 215, 2350, 12, 14, 195, 5659, 3716, 2190, 2279, 2391} \[ -\frac{b d^{5/2} n \sqrt{d+e x^2} \text{PolyLog}\left (2,e^{2 \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}\right )}{32 e^{5/2} \sqrt{\frac{e x^2}{d}+1}}-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{\frac{e x^2}{d}+1}}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{7 b d^2 n x \sqrt{d+e x^2}}{192 e^2}+\frac{b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )^2}{32 e^{5/2} \sqrt{\frac{e x^2}{d}+1}}+\frac{5 b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{192 e^{5/2} \sqrt{\frac{e x^2}{d}+1}}-\frac{b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \log \left (1-e^{2 \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}\right )}{16 e^{5/2} \sqrt{\frac{e x^2}{d}+1}}-\frac{1}{36} b n x^5 \sqrt{d+e x^2}-\frac{5 b d n x^3 \sqrt{d+e x^2}}{288 e} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]),x]

[Out]

(7*b*d^2*n*x*Sqrt[d + e*x^2])/(192*e^2) - (5*b*d*n*x^3*Sqrt[d + e*x^2])/(288*e) - (b*n*x^5*Sqrt[d + e*x^2])/36
+ (5*b*d^(5/2)*n*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])/(192*e^(5/2)*Sqrt[1 + (e*x^2)/d]) + (b*d^(5/2)
*n*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]^2)/(32*e^(5/2)*Sqrt[1 + (e*x^2)/d]) - (b*d^(5/2)*n*Sqrt[d + e*
x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[1 - E^(2*ArcSinh[(Sqrt[e]*x)/Sqrt[d]])])/(16*e^(5/2)*Sqrt[1 + (e*x^2)/d]
) - (d^2*x*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/(16*e^2) + (d*x^3*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/(24*e) +
(x^5*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/6 + (d^(5/2)*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[
c*x^n]))/(16*e^(5/2)*Sqrt[1 + (e*x^2)/d]) - (b*d^(5/2)*n*Sqrt[d + e*x^2]*PolyLog[2, E^(2*ArcSinh[(Sqrt[e]*x)/S
qrt[d]])])/(32*e^(5/2)*Sqrt[1 + (e*x^2)/d])

Rule 2341

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*x^2)/d)^FracPart[q], Int[x^m*(1 + (e*x^2)/d)^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 279

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

Rule 321

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 215

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 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 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 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 5659

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

Rule 3716

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)))/(E^(2*I*k*Pi)*(1 + E^(2*
(-(I*e) + f*fz*x))/E^(2*I*k*Pi))), x], x] /; FreeQ[{c, d, e, f, fz}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rule 2190

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*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), 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 2279

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 2391

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]

Rubi steps

\begin{align*} \int x^4 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx &=\frac{\sqrt{d+e x^2} \int x^4 \sqrt{1+\frac{e x^2}{d}} \left (a+b \log \left (c x^n\right )\right ) \, dx}{\sqrt{1+\frac{e x^2}{d}}}\\ &=-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{\left (b n \sqrt{d+e x^2}\right ) \int \frac{\sqrt{e} x \sqrt{1+\frac{e x^2}{d}} \left (-3 d^2+2 d e x^2+8 e^2 x^4\right )+3 d^{5/2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{48 e^{5/2} x} \, dx}{\sqrt{1+\frac{e x^2}{d}}}\\ &=-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{\left (b n \sqrt{d+e x^2}\right ) \int \frac{\sqrt{e} x \sqrt{1+\frac{e x^2}{d}} \left (-3 d^2+2 d e x^2+8 e^2 x^4\right )+3 d^{5/2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{x} \, dx}{48 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}\\ &=-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{\left (b n \sqrt{d+e x^2}\right ) \int \left (-3 d^2 \sqrt{e} \sqrt{1+\frac{e x^2}{d}}+2 d e^{3/2} x^2 \sqrt{1+\frac{e x^2}{d}}+8 e^{5/2} x^4 \sqrt{1+\frac{e x^2}{d}}+\frac{3 d^{5/2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{x}\right ) \, dx}{48 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}\\ &=-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{\left (b n \sqrt{d+e x^2}\right ) \int x^4 \sqrt{1+\frac{e x^2}{d}} \, dx}{6 \sqrt{1+\frac{e x^2}{d}}}-\frac{\left (b d^{5/2} n \sqrt{d+e x^2}\right ) \int \frac{\sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{x} \, dx}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}+\frac{\left (b d^2 n \sqrt{d+e x^2}\right ) \int \sqrt{1+\frac{e x^2}{d}} \, dx}{16 e^2 \sqrt{1+\frac{e x^2}{d}}}-\frac{\left (b d n \sqrt{d+e x^2}\right ) \int x^2 \sqrt{1+\frac{e x^2}{d}} \, dx}{24 e \sqrt{1+\frac{e x^2}{d}}}\\ &=\frac{b d^2 n x \sqrt{d+e x^2}}{32 e^2}-\frac{b d n x^3 \sqrt{d+e x^2}}{96 e}-\frac{1}{36} b n x^5 \sqrt{d+e x^2}-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{\left (b n \sqrt{d+e x^2}\right ) \int \frac{x^4}{\sqrt{1+\frac{e x^2}{d}}} \, dx}{36 \sqrt{1+\frac{e x^2}{d}}}-\frac{\left (b d^{5/2} n \sqrt{d+e x^2}\right ) \operatorname{Subst}\left (\int x \coth (x) \, dx,x,\sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}+\frac{\left (b d^2 n \sqrt{d+e x^2}\right ) \int \frac{1}{\sqrt{1+\frac{e x^2}{d}}} \, dx}{32 e^2 \sqrt{1+\frac{e x^2}{d}}}-\frac{\left (b d n \sqrt{d+e x^2}\right ) \int \frac{x^2}{\sqrt{1+\frac{e x^2}{d}}} \, dx}{96 e \sqrt{1+\frac{e x^2}{d}}}\\ &=\frac{5 b d^2 n x \sqrt{d+e x^2}}{192 e^2}-\frac{5 b d n x^3 \sqrt{d+e x^2}}{288 e}-\frac{1}{36} b n x^5 \sqrt{d+e x^2}+\frac{b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{32 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}+\frac{b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )^2}{32 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}+\frac{\left (b d^{5/2} n \sqrt{d+e x^2}\right ) \operatorname{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 )}{8 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}+\frac{\left (b d^2 n \sqrt{d+e x^2}\right ) \int \frac{1}{\sqrt{1+\frac{e x^2}{d}}} \, dx}{192 e^2 \sqrt{1+\frac{e x^2}{d}}}+\frac{\left (b d n \sqrt{d+e x^2}\right ) \int \frac{x^2}{\sqrt{1+\frac{e x^2}{d}}} \, dx}{48 e \sqrt{1+\frac{e x^2}{d}}}\\ &=\frac{7 b d^2 n x \sqrt{d+e x^2}}{192 e^2}-\frac{5 b d n x^3 \sqrt{d+e x^2}}{288 e}-\frac{1}{36} b n x^5 \sqrt{d+e x^2}+\frac{7 b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{192 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}+\frac{b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )^2}{32 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{b d^{5/2} n \sqrt{d+e x^2} \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 )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}+\frac{\left (b d^{5/2} n \sqrt{d+e x^2}\right ) \operatorname{Subst}\left (\int \log \left (1-e^{2 x}\right ) \, dx,x,\sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{\left (b d^2 n \sqrt{d+e x^2}\right ) \int \frac{1}{\sqrt{1+\frac{e x^2}{d}}} \, dx}{96 e^2 \sqrt{1+\frac{e x^2}{d}}}\\ &=\frac{7 b d^2 n x \sqrt{d+e x^2}}{192 e^2}-\frac{5 b d n x^3 \sqrt{d+e x^2}}{288 e}-\frac{1}{36} b n x^5 \sqrt{d+e x^2}+\frac{5 b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{192 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}+\frac{b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )^2}{32 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{b d^{5/2} n \sqrt{d+e x^2} \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 )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}+\frac{\left (b d^{5/2} n \sqrt{d+e x^2}\right ) \operatorname{Subst}\left (\int \frac{\log (1-x)}{x} \, dx,x,e^{2 \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}\right )}{32 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}\\ &=\frac{7 b d^2 n x \sqrt{d+e x^2}}{192 e^2}-\frac{5 b d n x^3 \sqrt{d+e x^2}}{288 e}-\frac{1}{36} b n x^5 \sqrt{d+e x^2}+\frac{5 b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{192 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}+\frac{b d^{5/2} n \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )^2}{32 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{b d^{5/2} n \sqrt{d+e x^2} \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 )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{d^2 x \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac{d x^3 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac{1}{6} x^5 \sqrt{d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac{d^{5/2} \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}-\frac{b d^{5/2} n \sqrt{d+e x^2} \text{Li}_2\left (e^{2 \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}\right )}{32 e^{5/2} \sqrt{1+\frac{e x^2}{d}}}\\ \end{align*}

Mathematica [C]  time = 0.660345, size = 276, normalized size = 0.59 \[ \frac{-48 b e^{5/2} n x^5 \sqrt{d+e x^2} \, _3F_2\left (-\frac{1}{2},\frac{5}{2},\frac{5}{2};\frac{7}{2},\frac{7}{2};-\frac{e x^2}{d}\right )+25 \sqrt{\frac{e x^2}{d}+1} \left (3 d^3 \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right ) (a-b n \log (x))+a \sqrt{e} x \sqrt{d+e x^2} \left (-3 d^2+2 d e x^2+8 e^2 x^4\right )+b \log \left (c x^n\right ) \left (\sqrt{e} x \sqrt{d+e x^2} \left (-3 d^2+2 d e x^2+8 e^2 x^4\right )+3 d^3 \log \left (\sqrt{e} \sqrt{d+e x^2}+e x\right )\right )\right )+75 b d^{5/2} n \log (x) \sqrt{d+e x^2} \sinh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right )}{1200 e^{5/2} \sqrt{\frac{e x^2}{d}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]),x]

[Out]

(-48*b*e^(5/2)*n*x^5*Sqrt[d + e*x^2]*HypergeometricPFQ[{-1/2, 5/2, 5/2}, {7/2, 7/2}, -((e*x^2)/d)] + 75*b*d^(5
/2)*n*Sqrt[d + e*x^2]*ArcSinh[(Sqrt[e]*x)/Sqrt[d]]*Log[x] + 25*Sqrt[1 + (e*x^2)/d]*(a*Sqrt[e]*x*Sqrt[d + e*x^2
]*(-3*d^2 + 2*d*e*x^2 + 8*e^2*x^4) + 3*d^3*(a - b*n*Log[x])*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]] + b*Log[c*x^n]*
(Sqrt[e]*x*Sqrt[d + e*x^2]*(-3*d^2 + 2*d*e*x^2 + 8*e^2*x^4) + 3*d^3*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])))/(120
0*e^(5/2)*Sqrt[1 + (e*x^2)/d])

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Maple [F]  time = 0.448, size = 0, normalized size = 0. \begin{align*} \int{x}^{4} \left ( a+b\ln \left ( c{x}^{n} \right ) \right ) \sqrt{e{x}^{2}+d}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2),x)

[Out]

int(x^4*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2),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(x^4*(a+b*log(c*x^n))*(e*x^2+d)^(1/2),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 (\sqrt{e x^{2} + d} b x^{4} \log \left (c x^{n}\right ) + \sqrt{e x^{2} + d} a x^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*log(c*x^n))*(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(e*x^2 + d)*b*x^4*log(c*x^n) + sqrt(e*x^2 + d)*a*x^4, 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(x**4*(a+b*ln(c*x**n))*(e*x**2+d)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e x^{2} + d}{\left (b \log \left (c x^{n}\right ) + a\right )} x^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*(a+b*log(c*x^n))*(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(e*x^2 + d)*(b*log(c*x^n) + a)*x^4, x)