Integrand size = 25, antiderivative size = 208 \[ \int x^5 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {8 b d^3 n \sqrt {d+e x^2}}{105 e^3}-\frac {8 b d^2 n \left (d+e x^2\right )^{3/2}}{315 e^3}+\frac {9 b d n \left (d+e x^2\right )^{5/2}}{175 e^3}-\frac {b n \left (d+e x^2\right )^{7/2}}{49 e^3}+\frac {8 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{105 e^3}+\frac {d^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {2 d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3} \]
-8/315*b*d^2*n*(e*x^2+d)^(3/2)/e^3+9/175*b*d*n*(e*x^2+d)^(5/2)/e^3-1/49*b* n*(e*x^2+d)^(7/2)/e^3+8/105*b*d^(7/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e ^3+1/3*d^2*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/e^3-2/5*d*(e*x^2+d)^(5/2)*(a+b* ln(c*x^n))/e^3+1/7*(e*x^2+d)^(7/2)*(a+b*ln(c*x^n))/e^3-8/105*b*d^3*n*(e*x^ 2+d)^(1/2)/e^3
Time = 0.13 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.21 \[ \int x^5 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {8 b d^{7/2} n \log (x)}{105 e^3}+\frac {b n \sqrt {d+e x^2} \left (8 d^3-4 d^2 e x^2+3 d e^2 x^4+15 e^3 x^6\right ) \log (x)}{105 e^3}+\sqrt {d+e x^2} \left (\frac {1}{49} x^6 \left (7 a-b n+7 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+\frac {d x^4 \left (35 a-12 b n+35 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{1225 e}+\frac {2 d^3 \left (420 a-389 b n+420 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{11025 e^3}-\frac {d^2 x^2 \left (420 a-179 b n+420 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{11025 e^2}\right )+\frac {8 b d^{7/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{105 e^3} \]
(-8*b*d^(7/2)*n*Log[x])/(105*e^3) + (b*n*Sqrt[d + e*x^2]*(8*d^3 - 4*d^2*e* x^2 + 3*d*e^2*x^4 + 15*e^3*x^6)*Log[x])/(105*e^3) + Sqrt[d + e*x^2]*((x^6* (7*a - b*n + 7*b*(-(n*Log[x]) + Log[c*x^n])))/49 + (d*x^4*(35*a - 12*b*n + 35*b*(-(n*Log[x]) + Log[c*x^n])))/(1225*e) + (2*d^3*(420*a - 389*b*n + 42 0*b*(-(n*Log[x]) + Log[c*x^n])))/(11025*e^3) - (d^2*x^2*(420*a - 179*b*n + 420*b*(-(n*Log[x]) + Log[c*x^n])))/(11025*e^2)) + (8*b*d^(7/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(105*e^3)
Time = 0.50 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.87, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2792, 27, 1578, 1192, 25, 1584, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int x^5 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int \frac {\left (e x^2+d\right )^{3/2} \left (15 e^2 x^4-12 d e x^2+8 d^2\right )}{105 e^3 x}dx+\frac {d^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {2 d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b n \int \frac {\left (e x^2+d\right )^{3/2} \left (15 e^2 x^4-12 d e x^2+8 d^2\right )}{x}dx}{105 e^3}+\frac {d^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {2 d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle -\frac {b n \int \frac {\left (e x^2+d\right )^{3/2} \left (15 e^2 x^4-12 d e x^2+8 d^2\right )}{x^2}dx^2}{210 e^3}+\frac {d^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {2 d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle -\frac {b n \int -\frac {x^8 \left (15 e^2 x^8-42 d e^2 x^4+35 d^2 e^2\right )}{d-x^4}d\sqrt {e x^2+d}}{105 e^5}+\frac {d^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {2 d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b n \int \frac {x^8 \left (15 e^2 x^8-42 d e^2 x^4+35 d^2 e^2\right )}{d-x^4}d\sqrt {e x^2+d}}{105 e^5}+\frac {d^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {2 d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {b n \int \left (-15 e^2 x^{12}+27 d e^2 x^8-8 d^2 e^2 x^4-8 d^3 e^2+\frac {8 d^4 e^2}{d-x^4}\right )d\sqrt {e x^2+d}}{105 e^5}+\frac {d^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {2 d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {2 d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {b n \left (-8 d^{7/2} e^2 \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )+8 d^3 e^2 \sqrt {d+e x^2}+\frac {8}{3} d^2 e^2 x^6-\frac {27}{5} d e^2 x^{10}+\frac {15 e^2 x^{14}}{7}\right )}{105 e^5}\) |
-1/105*(b*n*((8*d^2*e^2*x^6)/3 - (27*d*e^2*x^10)/5 + (15*e^2*x^14)/7 + 8*d ^3*e^2*Sqrt[d + e*x^2] - 8*d^(7/2)*e^2*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]))/ e^5 + (d^2*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*e^3) - (2*d*(d + e*x^2 )^(5/2)*(a + b*Log[c*x^n]))/(5*e^3) + ((d + e*x^2)^(7/2)*(a + b*Log[c*x^n] ))/(7*e^3)
3.3.51.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[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] && Int egerQ[(m - 1)/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^r)^q, x] }, Simp[(a + b*Log[c*x^n]) u, x] - Simp[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])
\[\int x^{5} \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e \,x^{2}+d}d x\]
Time = 0.34 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.99 \[ \int x^5 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {420 \, b d^{\frac {7}{2}} n \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (225 \, {\left (b e^{3} n - 7 \, a e^{3}\right )} x^{6} + 778 \, b d^{3} n + 9 \, {\left (12 \, b d e^{2} n - 35 \, a d e^{2}\right )} x^{4} - 840 \, a d^{3} - {\left (179 \, b d^{2} e n - 420 \, a d^{2} e\right )} x^{2} - 105 \, {\left (15 \, b e^{3} x^{6} + 3 \, b d e^{2} x^{4} - 4 \, b d^{2} e x^{2} + 8 \, b d^{3}\right )} \log \left (c\right ) - 105 \, {\left (15 \, b e^{3} n x^{6} + 3 \, b d e^{2} n x^{4} - 4 \, b d^{2} e n x^{2} + 8 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{11025 \, e^{3}}, -\frac {840 \, b \sqrt {-d} d^{3} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (225 \, {\left (b e^{3} n - 7 \, a e^{3}\right )} x^{6} + 778 \, b d^{3} n + 9 \, {\left (12 \, b d e^{2} n - 35 \, a d e^{2}\right )} x^{4} - 840 \, a d^{3} - {\left (179 \, b d^{2} e n - 420 \, a d^{2} e\right )} x^{2} - 105 \, {\left (15 \, b e^{3} x^{6} + 3 \, b d e^{2} x^{4} - 4 \, b d^{2} e x^{2} + 8 \, b d^{3}\right )} \log \left (c\right ) - 105 \, {\left (15 \, b e^{3} n x^{6} + 3 \, b d e^{2} n x^{4} - 4 \, b d^{2} e n x^{2} + 8 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{11025 \, e^{3}}\right ] \]
[1/11025*(420*b*d^(7/2)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x ^2) - (225*(b*e^3*n - 7*a*e^3)*x^6 + 778*b*d^3*n + 9*(12*b*d*e^2*n - 35*a* d*e^2)*x^4 - 840*a*d^3 - (179*b*d^2*e*n - 420*a*d^2*e)*x^2 - 105*(15*b*e^3 *x^6 + 3*b*d*e^2*x^4 - 4*b*d^2*e*x^2 + 8*b*d^3)*log(c) - 105*(15*b*e^3*n*x ^6 + 3*b*d*e^2*n*x^4 - 4*b*d^2*e*n*x^2 + 8*b*d^3*n)*log(x))*sqrt(e*x^2 + d ))/e^3, -1/11025*(840*b*sqrt(-d)*d^3*n*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (225*(b*e^3*n - 7*a*e^3)*x^6 + 778*b*d^3*n + 9*(12*b*d*e^2*n - 35*a*d*e^2) *x^4 - 840*a*d^3 - (179*b*d^2*e*n - 420*a*d^2*e)*x^2 - 105*(15*b*e^3*x^6 + 3*b*d*e^2*x^4 - 4*b*d^2*e*x^2 + 8*b*d^3)*log(c) - 105*(15*b*e^3*n*x^6 + 3 *b*d*e^2*n*x^4 - 4*b*d^2*e*n*x^2 + 8*b*d^3*n)*log(x))*sqrt(e*x^2 + d))/e^3 ]
Time = 20.89 (sec) , antiderivative size = 490, normalized size of antiderivative = 2.36 \[ \int x^5 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=a \left (\begin {cases} \frac {8 d^{3} \sqrt {d + e x^{2}}}{105 e^{3}} - \frac {4 d^{2} x^{2} \sqrt {d + e x^{2}}}{105 e^{2}} + \frac {d x^{4} \sqrt {d + e x^{2}}}{35 e} + \frac {x^{6} \sqrt {d + e x^{2}}}{7} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{6}}{6} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {8 d^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{105 e^{3}} + \frac {8 d^{4}}{105 e^{\frac {7}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {8 d^{3} x}{105 e^{\frac {5}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} - \frac {4 d^{2} \left (\begin {cases} \frac {d \sqrt {d + e x^{2}}}{3 e} + \frac {x^{2} \sqrt {d + e x^{2}}}{3} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right )}{105 e^{2}} + \frac {d \left (\begin {cases} - \frac {2 d^{2} \sqrt {d + e x^{2}}}{15 e^{2}} + \frac {d x^{2} \sqrt {d + e x^{2}}}{15 e} + \frac {x^{4} \sqrt {d + e x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}\right )}{35 e} + \frac {\begin {cases} \frac {8 d^{3} \sqrt {d + e x^{2}}}{105 e^{3}} - \frac {4 d^{2} x^{2} \sqrt {d + e x^{2}}}{105 e^{2}} + \frac {d x^{4} \sqrt {d + e x^{2}}}{35 e} + \frac {x^{6} \sqrt {d + e x^{2}}}{7} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{6}}{6} & \text {otherwise} \end {cases}}{7} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{6}}{36} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {8 d^{3} \sqrt {d + e x^{2}}}{105 e^{3}} - \frac {4 d^{2} x^{2} \sqrt {d + e x^{2}}}{105 e^{2}} + \frac {d x^{4} \sqrt {d + e x^{2}}}{35 e} + \frac {x^{6} \sqrt {d + e x^{2}}}{7} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{6}}{6} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((8*d**3*sqrt(d + e*x**2)/(105*e**3) - 4*d**2*x**2*sqrt(d + e*x **2)/(105*e**2) + d*x**4*sqrt(d + e*x**2)/(35*e) + x**6*sqrt(d + e*x**2)/7 , Ne(e, 0)), (sqrt(d)*x**6/6, True)) - b*n*Piecewise((-8*d**(7/2)*asinh(sq rt(d)/(sqrt(e)*x))/(105*e**3) + 8*d**4/(105*e**(7/2)*x*sqrt(d/(e*x**2) + 1 )) + 8*d**3*x/(105*e**(5/2)*sqrt(d/(e*x**2) + 1)) - 4*d**2*Piecewise((d*sq rt(d + e*x**2)/(3*e) + x**2*sqrt(d + e*x**2)/3, Ne(e, 0)), (sqrt(d)*x**2/2 , True))/(105*e**2) + d*Piecewise((-2*d**2*sqrt(d + e*x**2)/(15*e**2) + d* x**2*sqrt(d + e*x**2)/(15*e) + x**4*sqrt(d + e*x**2)/5, Ne(e, 0)), (sqrt(d )*x**4/4, True))/(35*e) + Piecewise((8*d**3*sqrt(d + e*x**2)/(105*e**3) - 4*d**2*x**2*sqrt(d + e*x**2)/(105*e**2) + d*x**4*sqrt(d + e*x**2)/(35*e) + x**6*sqrt(d + e*x**2)/7, Ne(e, 0)), (sqrt(d)*x**6/6, True))/7, (e > -oo) & (e < oo) & Ne(e, 0)), (sqrt(d)*x**6/36, True)) + b*Piecewise((8*d**3*sqr t(d + e*x**2)/(105*e**3) - 4*d**2*x**2*sqrt(d + e*x**2)/(105*e**2) + d*x** 4*sqrt(d + e*x**2)/(35*e) + x**6*sqrt(d + e*x**2)/7, Ne(e, 0)), (sqrt(d)*x **6/6, True))*log(c*x**n)
Exception generated. \[ \int x^5 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
Time = 0.38 (sec) , antiderivative size = 288, normalized size of antiderivative = 1.38 \[ \int x^5 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{7} \, \sqrt {e x^{2} + d} b x^{6} \log \left (c\right ) + \frac {1}{7} \, \sqrt {e x^{2} + d} a x^{6} + \frac {\sqrt {e x^{2} + d} b d x^{4} \log \left (c\right )}{35 \, e} + \frac {\sqrt {e x^{2} + d} a d x^{4}}{35 \, e} - \frac {4 \, \sqrt {e x^{2} + d} b d^{2} x^{2} \log \left (c\right )}{105 \, e^{2}} - \frac {4 \, \sqrt {e x^{2} + d} a d^{2} x^{2}}{105 \, e^{2}} + \frac {1}{11025} \, b n {\left (\frac {105 \, {\left (15 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 42 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2}\right )} \log \left (x\right )}{e^{3}} - \frac {\frac {840 \, d^{4} \arctan \left (\frac {\sqrt {e x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} + 225 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 567 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d + 280 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2} + 840 \, \sqrt {e x^{2} + d} d^{3}}{e^{3}}\right )} + \frac {8 \, \sqrt {e x^{2} + d} b d^{3} \log \left (c\right )}{105 \, e^{3}} + \frac {8 \, \sqrt {e x^{2} + d} a d^{3}}{105 \, e^{3}} \]
1/7*sqrt(e*x^2 + d)*b*x^6*log(c) + 1/7*sqrt(e*x^2 + d)*a*x^6 + 1/35*sqrt(e *x^2 + d)*b*d*x^4*log(c)/e + 1/35*sqrt(e*x^2 + d)*a*d*x^4/e - 4/105*sqrt(e *x^2 + d)*b*d^2*x^2*log(c)/e^2 - 4/105*sqrt(e*x^2 + d)*a*d^2*x^2/e^2 + 1/1 1025*b*n*(105*(15*(e*x^2 + d)^(7/2) - 42*(e*x^2 + d)^(5/2)*d + 35*(e*x^2 + d)^(3/2)*d^2)*log(x)/e^3 - (840*d^4*arctan(sqrt(e*x^2 + d)/sqrt(-d))/sqrt (-d) + 225*(e*x^2 + d)^(7/2) - 567*(e*x^2 + d)^(5/2)*d + 280*(e*x^2 + d)^( 3/2)*d^2 + 840*sqrt(e*x^2 + d)*d^3)/e^3) + 8/105*sqrt(e*x^2 + d)*b*d^3*log (c)/e^3 + 8/105*sqrt(e*x^2 + d)*a*d^3/e^3
Timed out. \[ \int x^5 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^5\,\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]