Integrand size = 23, antiderivative size = 213 \[ \int x^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {32 b d^4 n \sqrt {d+e x}}{315 e^3}-\frac {32 b d^3 n (d+e x)^{3/2}}{945 e^3}-\frac {32 b d^2 n (d+e x)^{5/2}}{1575 e^3}+\frac {44 b d n (d+e x)^{7/2}}{441 e^3}-\frac {4 b n (d+e x)^{9/2}}{81 e^3}+\frac {32 b d^{9/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{315 e^3}+\frac {2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3} \]
-32/945*b*d^3*n*(e*x+d)^(3/2)/e^3-32/1575*b*d^2*n*(e*x+d)^(5/2)/e^3+44/441 *b*d*n*(e*x+d)^(7/2)/e^3-4/81*b*n*(e*x+d)^(9/2)/e^3+32/315*b*d^(9/2)*n*arc tanh((e*x+d)^(1/2)/d^(1/2))/e^3+2/5*d^2*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^3- 4/7*d*(e*x+d)^(7/2)*(a+b*ln(c*x^n))/e^3+2/9*(e*x+d)^(9/2)*(a+b*ln(c*x^n))/ e^3-32/315*b*d^4*n*(e*x+d)^(1/2)/e^3
Time = 0.15 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.72 \[ \int x^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 \left (5040 b d^{9/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+\sqrt {d+e x} \left (315 a (d+e x)^2 \left (8 d^2-20 d e x+35 e^2 x^2\right )-2 b n \left (2614 d^4-677 d^3 e x+429 d^2 e^2 x^2+2425 d e^3 x^3+1225 e^4 x^4\right )+315 b (d+e x)^2 \left (8 d^2-20 d e x+35 e^2 x^2\right ) \log \left (c x^n\right )\right )\right )}{99225 e^3} \]
(2*(5040*b*d^(9/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(315*a *(d + e*x)^2*(8*d^2 - 20*d*e*x + 35*e^2*x^2) - 2*b*n*(2614*d^4 - 677*d^3*e *x + 429*d^2*e^2*x^2 + 2425*d*e^3*x^3 + 1225*e^4*x^4) + 315*b*(d + e*x)^2* (8*d^2 - 20*d*e*x + 35*e^2*x^2)*Log[c*x^n])))/(99225*e^3)
Time = 0.44 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.97, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2792, 27, 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.
| \(\displaystyle \int x^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int \frac {2 (d+e x)^{5/2} \left (8 d^2-20 e x d+35 e^2 x^2\right )}{315 e^3 x}dx+\frac {2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 b n \int \frac {(d+e x)^{5/2} \left (8 d^2-20 e x d+35 e^2 x^2\right )}{x}dx}{315 e^3}+\frac {2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle -\frac {4 b n \int \frac {(d+e x)^3 \left (63 d^2 e^2+35 (d+e x)^2 e^2-90 d (d+e x) e^2\right )}{e x}d\sqrt {d+e x}}{315 e^5}+\frac {2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4 b n \int -\frac {(d+e x)^3 \left (63 d^2 e^2+35 (d+e x)^2 e^2-90 d (d+e x) e^2\right )}{e x}d\sqrt {d+e x}}{315 e^5}+\frac {2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {4 b n \int \left (-\frac {8 e d^5}{x}-8 e^2 d^4-8 e^2 (d+e x) d^3-8 e^2 (d+e x)^2 d^2+55 e^2 (d+e x)^3 d-35 e^2 (d+e x)^4\right )d\sqrt {d+e x}}{315 e^5}+\frac {2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {4 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {4 b n \left (-8 d^{9/2} e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+8 d^4 e^2 \sqrt {d+e x}+\frac {8}{3} d^3 e^2 (d+e x)^{3/2}+\frac {8}{5} d^2 e^2 (d+e x)^{5/2}-\frac {55}{7} d e^2 (d+e x)^{7/2}+\frac {35}{9} e^2 (d+e x)^{9/2}\right )}{315 e^5}\) |
(-4*b*n*(8*d^4*e^2*Sqrt[d + e*x] + (8*d^3*e^2*(d + e*x)^(3/2))/3 + (8*d^2* e^2*(d + e*x)^(5/2))/5 - (55*d*e^2*(d + e*x)^(7/2))/7 + (35*e^2*(d + e*x)^ (9/2))/9 - 8*d^(9/2)*e^2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]))/(315*e^5) + (2*d ^2*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3) - (4*d*(d + e*x)^(7/2)*(a + b*Log[c*x^n]))/(7*e^3) + (2*(d + e*x)^(9/2)*(a + b*Log[c*x^n]))/(9*e^3)
3.2.38.3.1 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[((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])
\[\int x^{2} \left (e x +d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]
Time = 0.34 (sec) , antiderivative size = 496, normalized size of antiderivative = 2.33 \[ \int x^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {2 \, {\left (2520 \, b d^{\frac {9}{2}} n \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (5228 \, b d^{4} n - 2520 \, a d^{4} + 1225 \, {\left (2 \, b e^{4} n - 9 \, a e^{4}\right )} x^{4} + 50 \, {\left (97 \, b d e^{3} n - 315 \, a d e^{3}\right )} x^{3} + 3 \, {\left (286 \, b d^{2} e^{2} n - 315 \, a d^{2} e^{2}\right )} x^{2} - 2 \, {\left (677 \, b d^{3} e n - 630 \, a d^{3} e\right )} x - 315 \, {\left (35 \, b e^{4} x^{4} + 50 \, b d e^{3} x^{3} + 3 \, b d^{2} e^{2} x^{2} - 4 \, b d^{3} e x + 8 \, b d^{4}\right )} \log \left (c\right ) - 315 \, {\left (35 \, b e^{4} n x^{4} + 50 \, b d e^{3} n x^{3} + 3 \, b d^{2} e^{2} n x^{2} - 4 \, b d^{3} e n x + 8 \, b d^{4} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{99225 \, e^{3}}, -\frac {2 \, {\left (5040 \, b \sqrt {-d} d^{4} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (5228 \, b d^{4} n - 2520 \, a d^{4} + 1225 \, {\left (2 \, b e^{4} n - 9 \, a e^{4}\right )} x^{4} + 50 \, {\left (97 \, b d e^{3} n - 315 \, a d e^{3}\right )} x^{3} + 3 \, {\left (286 \, b d^{2} e^{2} n - 315 \, a d^{2} e^{2}\right )} x^{2} - 2 \, {\left (677 \, b d^{3} e n - 630 \, a d^{3} e\right )} x - 315 \, {\left (35 \, b e^{4} x^{4} + 50 \, b d e^{3} x^{3} + 3 \, b d^{2} e^{2} x^{2} - 4 \, b d^{3} e x + 8 \, b d^{4}\right )} \log \left (c\right ) - 315 \, {\left (35 \, b e^{4} n x^{4} + 50 \, b d e^{3} n x^{3} + 3 \, b d^{2} e^{2} n x^{2} - 4 \, b d^{3} e n x + 8 \, b d^{4} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{99225 \, e^{3}}\right ] \]
[2/99225*(2520*b*d^(9/2)*n*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - (5228*b*d^4*n - 2520*a*d^4 + 1225*(2*b*e^4*n - 9*a*e^4)*x^4 + 50*(97*b*d*e ^3*n - 315*a*d*e^3)*x^3 + 3*(286*b*d^2*e^2*n - 315*a*d^2*e^2)*x^2 - 2*(677 *b*d^3*e*n - 630*a*d^3*e)*x - 315*(35*b*e^4*x^4 + 50*b*d*e^3*x^3 + 3*b*d^2 *e^2*x^2 - 4*b*d^3*e*x + 8*b*d^4)*log(c) - 315*(35*b*e^4*n*x^4 + 50*b*d*e^ 3*n*x^3 + 3*b*d^2*e^2*n*x^2 - 4*b*d^3*e*n*x + 8*b*d^4*n)*log(x))*sqrt(e*x + d))/e^3, -2/99225*(5040*b*sqrt(-d)*d^4*n*arctan(sqrt(e*x + d)*sqrt(-d)/d ) + (5228*b*d^4*n - 2520*a*d^4 + 1225*(2*b*e^4*n - 9*a*e^4)*x^4 + 50*(97*b *d*e^3*n - 315*a*d*e^3)*x^3 + 3*(286*b*d^2*e^2*n - 315*a*d^2*e^2)*x^2 - 2* (677*b*d^3*e*n - 630*a*d^3*e)*x - 315*(35*b*e^4*x^4 + 50*b*d*e^3*x^3 + 3*b *d^2*e^2*x^2 - 4*b*d^3*e*x + 8*b*d^4)*log(c) - 315*(35*b*e^4*n*x^4 + 50*b* d*e^3*n*x^3 + 3*b*d^2*e^2*n*x^2 - 4*b*d^3*e*n*x + 8*b*d^4*n)*log(x))*sqrt( e*x + d))/e^3]
Time = 129.07 (sec) , antiderivative size = 643, normalized size of antiderivative = 3.02 \[ \int x^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=a d \left (\begin {cases} \frac {2 d^{2} \left (d + e x\right )^{\frac {3}{2}}}{3 e^{3}} - \frac {4 d \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} + \frac {2 \left (d + e x\right )^{\frac {7}{2}}}{7 e^{3}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{3}}{3} & \text {otherwise} \end {cases}\right ) + a e \left (\begin {cases} - \frac {2 d^{3} \left (d + e x\right )^{\frac {3}{2}}}{3 e^{4}} + \frac {6 d^{2} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} - \frac {6 d \left (d + e x\right )^{\frac {7}{2}}}{7 e^{4}} + \frac {2 \left (d + e x\right )^{\frac {9}{2}}}{9 e^{4}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}\right ) - b d n \left (\begin {cases} \frac {3112 d^{\frac {7}{2}} \sqrt {1 + \frac {e x}{d}}}{11025 e^{3}} + \frac {16 d^{\frac {7}{2}} \log {\left (\frac {e x}{d} \right )}}{105 e^{3}} - \frac {32 d^{\frac {7}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{105 e^{3}} - \frac {716 d^{\frac {5}{2}} x \sqrt {1 + \frac {e x}{d}}}{11025 e^{2}} + \frac {48 d^{\frac {3}{2}} x^{2} \sqrt {1 + \frac {e x}{d}}}{1225 e} + \frac {4 \sqrt {d} x^{3} \sqrt {1 + \frac {e x}{d}}}{49} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{3}}{9} & \text {otherwise} \end {cases}\right ) + b d \left (\begin {cases} \frac {2 d^{2} \left (d + e x\right )^{\frac {3}{2}}}{3 e^{3}} - \frac {4 d \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} + \frac {2 \left (d + e x\right )^{\frac {7}{2}}}{7 e^{3}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} - b e n \left (\begin {cases} - \frac {17552 d^{\frac {9}{2}} \sqrt {1 + \frac {e x}{d}}}{99225 e^{4}} - \frac {32 d^{\frac {9}{2}} \log {\left (\frac {e x}{d} \right )}}{315 e^{4}} + \frac {64 d^{\frac {9}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{315 e^{4}} + \frac {3736 d^{\frac {7}{2}} x \sqrt {1 + \frac {e x}{d}}}{99225 e^{3}} - \frac {724 d^{\frac {5}{2}} x^{2} \sqrt {1 + \frac {e x}{d}}}{33075 e^{2}} + \frac {64 d^{\frac {3}{2}} x^{3} \sqrt {1 + \frac {e x}{d}}}{3969 e} + \frac {4 \sqrt {d} x^{4} \sqrt {1 + \frac {e x}{d}}}{81} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{4}}{16} & \text {otherwise} \end {cases}\right ) + b e \left (\begin {cases} - \frac {2 d^{3} \left (d + e x\right )^{\frac {3}{2}}}{3 e^{4}} + \frac {6 d^{2} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} - \frac {6 d \left (d + e x\right )^{\frac {7}{2}}}{7 e^{4}} + \frac {2 \left (d + e x\right )^{\frac {9}{2}}}{9 e^{4}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*d*Piecewise((2*d**2*(d + e*x)**(3/2)/(3*e**3) - 4*d*(d + e*x)**(5/2)/(5* e**3) + 2*(d + e*x)**(7/2)/(7*e**3), Ne(e, 0)), (sqrt(d)*x**3/3, True)) + a*e*Piecewise((-2*d**3*(d + e*x)**(3/2)/(3*e**4) + 6*d**2*(d + e*x)**(5/2) /(5*e**4) - 6*d*(d + e*x)**(7/2)/(7*e**4) + 2*(d + e*x)**(9/2)/(9*e**4), N e(e, 0)), (sqrt(d)*x**4/4, True)) - b*d*n*Piecewise((3112*d**(7/2)*sqrt(1 + e*x/d)/(11025*e**3) + 16*d**(7/2)*log(e*x/d)/(105*e**3) - 32*d**(7/2)*lo g(sqrt(1 + e*x/d) + 1)/(105*e**3) - 716*d**(5/2)*x*sqrt(1 + e*x/d)/(11025* e**2) + 48*d**(3/2)*x**2*sqrt(1 + e*x/d)/(1225*e) + 4*sqrt(d)*x**3*sqrt(1 + e*x/d)/49, (e > -oo) & (e < oo) & Ne(e, 0)), (sqrt(d)*x**3/9, True)) + b *d*Piecewise((2*d**2*(d + e*x)**(3/2)/(3*e**3) - 4*d*(d + e*x)**(5/2)/(5*e **3) + 2*(d + e*x)**(7/2)/(7*e**3), Ne(e, 0)), (sqrt(d)*x**3/3, True))*log (c*x**n) - b*e*n*Piecewise((-17552*d**(9/2)*sqrt(1 + e*x/d)/(99225*e**4) - 32*d**(9/2)*log(e*x/d)/(315*e**4) + 64*d**(9/2)*log(sqrt(1 + e*x/d) + 1)/ (315*e**4) + 3736*d**(7/2)*x*sqrt(1 + e*x/d)/(99225*e**3) - 724*d**(5/2)*x **2*sqrt(1 + e*x/d)/(33075*e**2) + 64*d**(3/2)*x**3*sqrt(1 + e*x/d)/(3969* e) + 4*sqrt(d)*x**4*sqrt(1 + e*x/d)/81, (e > -oo) & (e < oo) & Ne(e, 0)), (sqrt(d)*x**4/16, True)) + b*e*Piecewise((-2*d**3*(d + e*x)**(3/2)/(3*e**4 ) + 6*d**2*(d + e*x)**(5/2)/(5*e**4) - 6*d*(d + e*x)**(7/2)/(7*e**4) + 2*( d + e*x)**(9/2)/(9*e**4), Ne(e, 0)), (sqrt(d)*x**4/4, True))*log(c*x**n)
Time = 0.27 (sec) , antiderivative size = 196, normalized size of antiderivative = 0.92 \[ \int x^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {4}{99225} \, {\left (\frac {1260 \, d^{\frac {9}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{3}} + \frac {1225 \, {\left (e x + d\right )}^{\frac {9}{2}} - 2475 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 504 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} + 840 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} + 2520 \, \sqrt {e x + d} d^{4}}{e^{3}}\right )} b n + \frac {2}{315} \, {\left (\frac {35 \, {\left (e x + d\right )}^{\frac {9}{2}}}{e^{3}} - \frac {90 \, {\left (e x + d\right )}^{\frac {7}{2}} d}{e^{3}} + \frac {63 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2}}{e^{3}}\right )} b \log \left (c x^{n}\right ) + \frac {2}{315} \, {\left (\frac {35 \, {\left (e x + d\right )}^{\frac {9}{2}}}{e^{3}} - \frac {90 \, {\left (e x + d\right )}^{\frac {7}{2}} d}{e^{3}} + \frac {63 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2}}{e^{3}}\right )} a \]
-4/99225*(1260*d^(9/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt (d)))/e^3 + (1225*(e*x + d)^(9/2) - 2475*(e*x + d)^(7/2)*d + 504*(e*x + d) ^(5/2)*d^2 + 840*(e*x + d)^(3/2)*d^3 + 2520*sqrt(e*x + d)*d^4)/e^3)*b*n + 2/315*(35*(e*x + d)^(9/2)/e^3 - 90*(e*x + d)^(7/2)*d/e^3 + 63*(e*x + d)^(5 /2)*d^2/e^3)*b*log(c*x^n) + 2/315*(35*(e*x + d)^(9/2)/e^3 - 90*(e*x + d)^( 7/2)*d/e^3 + 63*(e*x + d)^(5/2)*d^2/e^3)*a
\[ \int x^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (e x + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \,d x } \]
Timed out. \[ \int x^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )\,{\left (d+e\,x\right )}^{3/2} \,d x \]