\(\int x^4 \sqrt {d+e x^2} (a+b \log (c x^n)) \, dx\) [256]

Optimal result

Integrand size = 25, antiderivative size = 469 \[ \int x^4 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {7 b d^2 n x \sqrt {d+e x^2}}{192 e^2}-\frac {5 b d n x^3 \sqrt {d+e x^2}}{288 e}-\frac {1}{36} b n x^5 \sqrt {d+e x^2}+\frac {5 b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{192 e^{5/2} \sqrt {1+\frac {e x^2}{d}}}+\frac {b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{32 e^{5/2} \sqrt {1+\frac {e x^2}{d}}}-\frac {b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{16 e^{5/2} \sqrt {1+\frac {e x^2}{d}}}-\frac {d^2 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac {d x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}+\frac {1}{6} x^5 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac {d^{5/2} \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt {1+\frac {e x^2}{d}}}-\frac {b d^{5/2} n \sqrt {d+e x^2} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{32 e^{5/2} \sqrt {1+\frac {e x^2}{d}}} \]

[Out]

7/192*b*d^2*n*x*(e*x^2+d)^(1/2)/e^2-5/288*b*d*n*x^3*(e*x^2+d)^(1/2)/e-1/36*b*n*x^5*(e*x^2+d)^(1/2)-1/16*d^2*x*
(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e^2+1/24*d*x^3*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e+1/6*x^5*(a+b*ln(c*x^n))*(e*x^
2+d)^(1/2)+5/192*b*d^(5/2)*n*arcsinh(x*e^(1/2)/d^(1/2))*(e*x^2+d)^(1/2)/e^(5/2)/(1+e*x^2/d)^(1/2)+1/32*b*d^(5/
2)*n*arcsinh(x*e^(1/2)/d^(1/2))^2*(e*x^2+d)^(1/2)/e^(5/2)/(1+e*x^2/d)^(1/2)-1/16*b*d^(5/2)*n*arcsinh(x*e^(1/2)
/d^(1/2))*ln(1-(x*e^(1/2)/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*(e*x^2+d)^(1/2)/e^(5/2)/(1+e*x^2/d)^(1/2)+1/16*d^(5/2)
*arcsinh(x*e^(1/2)/d^(1/2))*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e^(5/2)/(1+e*x^2/d)^(1/2)-1/32*b*d^(5/2)*n*polylog
(2,(x*e^(1/2)/d^(1/2)+(1+e*x^2/d)^(1/2))^2)*(e*x^2+d)^(1/2)/e^(5/2)/(1+e*x^2/d)^(1/2)

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 469, normalized size of antiderivative = 1.00, number of steps used = 19, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.520, Rules used = {2386, 285, 327, 221, 2392, 12, 14, 201, 5775, 3797, 2221, 2317, 2438} \[ \int x^4 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {d^{5/2} \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{16 e^{5/2} \sqrt {\frac {e x^2}{d}+1}}-\frac {d^2 x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{16 e^2}+\frac {1}{6} x^5 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )+\frac {d x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{24 e}-\frac {b d^{5/2} n \sqrt {d+e x^2} \operatorname {PolyLog}\left (2,e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{32 e^{5/2} \sqrt {\frac {e x^2}{d}+1}}+\frac {b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )^2}{32 e^{5/2} \sqrt {\frac {e x^2}{d}+1}}+\frac {5 b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{192 e^{5/2} \sqrt {\frac {e x^2}{d}+1}}-\frac {b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log \left (1-e^{2 \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}\right )}{16 e^{5/2} \sqrt {\frac {e x^2}{d}+1}}+\frac {7 b d^2 n x \sqrt {d+e x^2}}{192 e^2}-\frac {1}{36} b n x^5 \sqrt {d+e x^2}-\frac {5 b d n x^3 \sqrt {d+e x^2}}{288 e} \]

[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 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 201

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 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 285

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 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 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*} \text {integral}& = \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 ) \text {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 ) \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 )}{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 ) \text {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 ) \text {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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.38 (sec) , antiderivative size = 276, normalized size of antiderivative = 0.59 \[ \int x^4 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {-48 b e^{5/2} n x^5 \sqrt {d+e x^2} \, _3F_2\left (-\frac {1}{2},\frac {5}{2},\frac {5}{2};\frac {7}{2},\frac {7}{2};-\frac {e x^2}{d}\right )+75 b d^{5/2} n \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right ) \log (x)+25 \sqrt {1+\frac {e x^2}{d}} \left (a \sqrt {e} x \sqrt {d+e x^2} \left (-3 d^2+2 d e x^2+8 e^2 x^4\right )+3 d^3 (a-b n \log (x)) \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )+b \log \left (c x^n\right ) \left (\sqrt {e} x \sqrt {d+e x^2} \left (-3 d^2+2 d e x^2+8 e^2 x^4\right )+3 d^3 \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )\right )\right )}{1200 e^{5/2} \sqrt {1+\frac {e x^2}{d}}} \]

[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])

Maple [F]

\[\int x^{4} \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e \,x^{2}+d}d x\]

[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)

Fricas [F]

\[ \int x^4 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sqrt {e x^{2} + d} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{4} \,d x } \]

[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)

Sympy [F]

\[ \int x^4 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^{4} \left (a + b \log {\left (c x^{n} \right )}\right ) \sqrt {d + e x^{2}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int x^4 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [F]

\[ \int x^4 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sqrt {e x^{2} + d} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{4} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int x^4 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^4\,\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]

[In]

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

[Out]

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