Integrand size = 25, antiderivative size = 231 \[ \int x^5 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {8 b d^4 n \sqrt {d+e x^2}}{315 e^3}-\frac {8 b d^3 n \left (d+e x^2\right )^{3/2}}{945 e^3}-\frac {8 b d^2 n \left (d+e x^2\right )^{5/2}}{1575 e^3}+\frac {11 b d n \left (d+e x^2\right )^{7/2}}{441 e^3}-\frac {b n \left (d+e x^2\right )^{9/2}}{81 e^3}+\frac {8 b d^{9/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{315 e^3}+\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3} \]
-8/945*b*d^3*n*(e*x^2+d)^(3/2)/e^3-8/1575*b*d^2*n*(e*x^2+d)^(5/2)/e^3+11/4 41*b*d*n*(e*x^2+d)^(7/2)/e^3-1/81*b*n*(e*x^2+d)^(9/2)/e^3+8/315*b*d^(9/2)* n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e^3+1/5*d^2*(e*x^2+d)^(5/2)*(a+b*ln(c*x ^n))/e^3-2/7*d*(e*x^2+d)^(7/2)*(a+b*ln(c*x^n))/e^3+1/9*(e*x^2+d)^(9/2)*(a+ b*ln(c*x^n))/e^3-8/315*b*d^4*n*(e*x^2+d)^(1/2)/e^3
Time = 0.22 (sec) , antiderivative size = 256, normalized size of antiderivative = 1.11 \[ \int x^5 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {-2520 b d^{9/2} n \log (x)+315 b n \left (d+e x^2\right )^{5/2} \left (8 d^2-20 d e x^2+35 e^2 x^4\right ) \log (x)+\sqrt {d+e x^2} \left (1225 e^4 x^8 \left (9 a-b n-9 b n \log (x)+9 b \log \left (c x^n\right )\right )+3 d^2 e^2 x^4 \left (315 a-143 b n+315 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+25 d e^3 x^6 \left (630 a-97 b n+630 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+2 d^4 \left (1260 a-1307 b n+1260 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )-d^3 e x^2 \left (1260 a-677 b n+1260 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+2520 b d^{9/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{99225 e^3} \]
(-2520*b*d^(9/2)*n*Log[x] + 315*b*n*(d + e*x^2)^(5/2)*(8*d^2 - 20*d*e*x^2 + 35*e^2*x^4)*Log[x] + Sqrt[d + e*x^2]*(1225*e^4*x^8*(9*a - b*n - 9*b*n*Lo g[x] + 9*b*Log[c*x^n]) + 3*d^2*e^2*x^4*(315*a - 143*b*n + 315*b*(-(n*Log[x ]) + Log[c*x^n])) + 25*d*e^3*x^6*(630*a - 97*b*n + 630*b*(-(n*Log[x]) + Lo g[c*x^n])) + 2*d^4*(1260*a - 1307*b*n + 1260*b*(-(n*Log[x]) + Log[c*x^n])) - d^3*e*x^2*(1260*a - 677*b*n + 1260*b*(-(n*Log[x]) + Log[c*x^n]))) + 252 0*b*d^(9/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(99225*e^3)
Time = 0.52 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.84, 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.
| \(\displaystyle \int x^5 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int \frac {\left (e x^2+d\right )^{5/2} \left (35 e^2 x^4-20 d e x^2+8 d^2\right )}{315 e^3 x}dx+\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b n \int \frac {\left (e x^2+d\right )^{5/2} \left (35 e^2 x^4-20 d e x^2+8 d^2\right )}{x}dx}{315 e^3}+\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle -\frac {b n \int \frac {\left (e x^2+d\right )^{5/2} \left (35 e^2 x^4-20 d e x^2+8 d^2\right )}{x^2}dx^2}{630 e^3}+\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle -\frac {b n \int -\frac {x^{12} \left (35 e^2 x^8-90 d e^2 x^4+63 d^2 e^2\right )}{d-x^4}d\sqrt {e x^2+d}}{315 e^5}+\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b n \int \frac {x^{12} \left (35 e^2 x^8-90 d e^2 x^4+63 d^2 e^2\right )}{d-x^4}d\sqrt {e x^2+d}}{315 e^5}+\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {b n \int \left (-35 e^2 x^{16}+55 d e^2 x^{12}-8 d^2 e^2 x^8-8 d^3 e^2 x^4-8 d^4 e^2+\frac {8 d^5 e^2}{d-x^4}\right )d\sqrt {e x^2+d}}{315 e^5}+\frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {\left (d+e x^2\right )^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^3}-\frac {2 d \left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {b n \left (-8 d^{9/2} e^2 \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )+8 d^4 e^2 \sqrt {d+e x^2}+\frac {8}{3} d^3 e^2 x^6+\frac {8}{5} d^2 e^2 x^{10}-\frac {55}{7} d e^2 x^{14}+\frac {35 e^2 x^{18}}{9}\right )}{315 e^5}\) |
-1/315*(b*n*((8*d^3*e^2*x^6)/3 + (8*d^2*e^2*x^10)/5 - (55*d*e^2*x^14)/7 + (35*e^2*x^18)/9 + 8*d^4*e^2*Sqrt[d + e*x^2] - 8*d^(9/2)*e^2*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]))/e^5 + (d^2*(d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*e ^3) - (2*d*(d + e*x^2)^(7/2)*(a + b*Log[c*x^n]))/(7*e^3) + ((d + e*x^2)^(9 /2)*(a + b*Log[c*x^n]))/(9*e^3)
3.3.63.3.1 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])
\[\int x^{5} \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]
Time = 0.37 (sec) , antiderivative size = 514, normalized size of antiderivative = 2.23 \[ \int x^5 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {1260 \, b d^{\frac {9}{2}} n \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (1225 \, {\left (b e^{4} n - 9 \, a e^{4}\right )} x^{8} + 25 \, {\left (97 \, b d e^{3} n - 630 \, a d e^{3}\right )} x^{6} + 2614 \, b d^{4} n - 2520 \, a d^{4} + 3 \, {\left (143 \, b d^{2} e^{2} n - 315 \, a d^{2} e^{2}\right )} x^{4} - {\left (677 \, b d^{3} e n - 1260 \, a d^{3} e\right )} x^{2} - 315 \, {\left (35 \, b e^{4} x^{8} + 50 \, b d e^{3} x^{6} + 3 \, b d^{2} e^{2} x^{4} - 4 \, b d^{3} e x^{2} + 8 \, b d^{4}\right )} \log \left (c\right ) - 315 \, {\left (35 \, b e^{4} n x^{8} + 50 \, b d e^{3} n x^{6} + 3 \, b d^{2} e^{2} n x^{4} - 4 \, b d^{3} e n x^{2} + 8 \, b d^{4} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{99225 \, e^{3}}, -\frac {2520 \, b \sqrt {-d} d^{4} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (1225 \, {\left (b e^{4} n - 9 \, a e^{4}\right )} x^{8} + 25 \, {\left (97 \, b d e^{3} n - 630 \, a d e^{3}\right )} x^{6} + 2614 \, b d^{4} n - 2520 \, a d^{4} + 3 \, {\left (143 \, b d^{2} e^{2} n - 315 \, a d^{2} e^{2}\right )} x^{4} - {\left (677 \, b d^{3} e n - 1260 \, a d^{3} e\right )} x^{2} - 315 \, {\left (35 \, b e^{4} x^{8} + 50 \, b d e^{3} x^{6} + 3 \, b d^{2} e^{2} x^{4} - 4 \, b d^{3} e x^{2} + 8 \, b d^{4}\right )} \log \left (c\right ) - 315 \, {\left (35 \, b e^{4} n x^{8} + 50 \, b d e^{3} n x^{6} + 3 \, b d^{2} e^{2} n x^{4} - 4 \, b d^{3} e n x^{2} + 8 \, b d^{4} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{99225 \, e^{3}}\right ] \]
[1/99225*(1260*b*d^(9/2)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/ x^2) - (1225*(b*e^4*n - 9*a*e^4)*x^8 + 25*(97*b*d*e^3*n - 630*a*d*e^3)*x^6 + 2614*b*d^4*n - 2520*a*d^4 + 3*(143*b*d^2*e^2*n - 315*a*d^2*e^2)*x^4 - ( 677*b*d^3*e*n - 1260*a*d^3*e)*x^2 - 315*(35*b*e^4*x^8 + 50*b*d*e^3*x^6 + 3 *b*d^2*e^2*x^4 - 4*b*d^3*e*x^2 + 8*b*d^4)*log(c) - 315*(35*b*e^4*n*x^8 + 5 0*b*d*e^3*n*x^6 + 3*b*d^2*e^2*n*x^4 - 4*b*d^3*e*n*x^2 + 8*b*d^4*n)*log(x)) *sqrt(e*x^2 + d))/e^3, -1/99225*(2520*b*sqrt(-d)*d^4*n*arctan(sqrt(-d)/sqr t(e*x^2 + d)) + (1225*(b*e^4*n - 9*a*e^4)*x^8 + 25*(97*b*d*e^3*n - 630*a*d *e^3)*x^6 + 2614*b*d^4*n - 2520*a*d^4 + 3*(143*b*d^2*e^2*n - 315*a*d^2*e^2 )*x^4 - (677*b*d^3*e*n - 1260*a*d^3*e)*x^2 - 315*(35*b*e^4*x^8 + 50*b*d*e^ 3*x^6 + 3*b*d^2*e^2*x^4 - 4*b*d^3*e*x^2 + 8*b*d^4)*log(c) - 315*(35*b*e^4* n*x^8 + 50*b*d*e^3*n*x^6 + 3*b*d^2*e^2*n*x^4 - 4*b*d^3*e*n*x^2 + 8*b*d^4*n )*log(x))*sqrt(e*x^2 + d))/e^3]
Time = 95.55 (sec) , antiderivative size = 1161, normalized size of antiderivative = 5.03 \[ \int x^5 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
a*d*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)) + a*e*Piecewise((-16*d**4*sqrt(d + e*x**2)/(315*e**4) + 8*d**3*x**2*sqrt(d + e*x**2)/(315*e**3) - 2*d**2*x**4 *sqrt(d + e*x**2)/(105*e**2) + d*x**6*sqrt(d + e*x**2)/(63*e) + x**8*sqrt( d + e*x**2)/9, Ne(e, 0)), (sqrt(d)*x**8/8, True)) - b*d*n*Piecewise((-8*d* *(7/2)*asinh(sqrt(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*sqrt(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*d*Piec ewise((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))*log(c*x**n) - b*e*n*Piecewise((16*d**(9/2)* asinh(sqrt(d)/(sqrt(e)*x))/(315*e**4) - 16*d**5/(315*e**(9/2)*x*sqrt(d/(e* x**2) + 1)) - 16*d**4*x/(315*e**(7/2)*sqrt(d/(e*x**2) + 1)) + 8*d**3*Pi...
Exception generated. \[ \int x^5 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]
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
\[ \int x^5 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{5} \,d x } \]
Timed out. \[ \int x^5 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^5\,{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]