Integrand size = 25, antiderivative size = 177 \[ \int x^3 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 b d^3 n \sqrt {d+e x^2}}{35 e^2}+\frac {2 b d^2 n \left (d+e x^2\right )^{3/2}}{105 e^2}+\frac {2 b d n \left (d+e x^2\right )^{5/2}}{175 e^2}-\frac {b n \left (d+e x^2\right )^{7/2}}{49 e^2}-\frac {2 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{35 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2} \]
2/105*b*d^2*n*(e*x^2+d)^(3/2)/e^2+2/175*b*d*n*(e*x^2+d)^(5/2)/e^2-1/49*b*n *(e*x^2+d)^(7/2)/e^2-2/35*b*d^(7/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e^2 -1/5*d*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/e^2+1/7*(e*x^2+d)^(7/2)*(a+b*ln(c*x ^n))/e^2+2/35*b*d^3*n*(e*x^2+d)^(1/2)/e^2
Time = 0.13 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.28 \[ \int x^3 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 b d^{7/2} n \log (x)}{35 e^2}-\frac {b n \left (2 d-5 e x^2\right ) \left (d+e x^2\right )^{5/2} \log (x)}{35 e^2}+\sqrt {d+e x^2} \left (\frac {1}{49} e x^6 \left (7 a-b n+7 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+\frac {d^2 x^2 \left (105 a-71 b n+105 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{3675 e}-\frac {d^3 \left (210 a-247 b n+210 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{3675 e^2}+\frac {d x^4 \left (280 a-61 b n+280 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{1225}\right )-\frac {2 b d^{7/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{35 e^2} \]
(2*b*d^(7/2)*n*Log[x])/(35*e^2) - (b*n*(2*d - 5*e*x^2)*(d + e*x^2)^(5/2)*L og[x])/(35*e^2) + Sqrt[d + e*x^2]*((e*x^6*(7*a - b*n + 7*b*(-(n*Log[x]) + Log[c*x^n])))/49 + (d^2*x^2*(105*a - 71*b*n + 105*b*(-(n*Log[x]) + Log[c*x ^n])))/(3675*e) - (d^3*(210*a - 247*b*n + 210*b*(-(n*Log[x]) + Log[c*x^n]) ))/(3675*e^2) + (d*x^4*(280*a - 61*b*n + 280*b*(-(n*Log[x]) + Log[c*x^n])) )/1225) - (2*b*d^(7/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(35*e^2)
Time = 0.37 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.91, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {2792, 27, 354, 90, 60, 60, 60, 73, 221}
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.
| \(\displaystyle \int x^3 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {\left (2 d-5 e x^2\right ) \left (e x^2+d\right )^{5/2}}{35 e^2 x}dx+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b n \int \frac {\left (2 d-5 e x^2\right ) \left (e x^2+d\right )^{5/2}}{x}dx}{35 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {b n \int \frac {\left (2 d-5 e x^2\right ) \left (e x^2+d\right )^{5/2}}{x^2}dx^2}{70 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {b n \left (2 d \int \frac {\left (e x^2+d\right )^{5/2}}{x^2}dx^2-\frac {10}{7} \left (d+e x^2\right )^{7/2}\right )}{70 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {b n \left (2 d \left (d \int \frac {\left (e x^2+d\right )^{3/2}}{x^2}dx^2+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )-\frac {10}{7} \left (d+e x^2\right )^{7/2}\right )}{70 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {b n \left (2 d \left (d \left (d \int \frac {\sqrt {e x^2+d}}{x^2}dx^2+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )-\frac {10}{7} \left (d+e x^2\right )^{7/2}\right )}{70 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {b n \left (2 d \left (d \left (d \left (d \int \frac {1}{x^2 \sqrt {e x^2+d}}dx^2+2 \sqrt {d+e x^2}\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )-\frac {10}{7} \left (d+e x^2\right )^{7/2}\right )}{70 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {b n \left (2 d \left (d \left (d \left (\frac {2 d \int \frac {1}{\frac {x^4}{e}-\frac {d}{e}}d\sqrt {e x^2+d}}{e}+2 \sqrt {d+e x^2}\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )-\frac {10}{7} \left (d+e x^2\right )^{7/2}\right )}{70 e^2}+\frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {d \left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {b n \left (2 d \left (d \left (d \left (2 \sqrt {d+e x^2}-2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )-\frac {10}{7} \left (d+e x^2\right )^{7/2}\right )}{70 e^2}\) |
(b*n*((-10*(d + e*x^2)^(7/2))/7 + 2*d*((2*(d + e*x^2)^(5/2))/5 + d*((2*(d + e*x^2)^(3/2))/3 + d*(2*Sqrt[d + e*x^2] - 2*Sqrt[d]*ArcTanh[Sqrt[d + e*x^ 2]/Sqrt[d]])))))/(70*e^2) - (d*(d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*e^ 2) + ((d + e*x^2)^(7/2)*(a + b*Log[c*x^n]))/(7*e^2)
3.3.64.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/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])
\[\int x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]
Time = 0.36 (sec) , antiderivative size = 409, normalized size of antiderivative = 2.31 \[ \int x^3 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {105 \, b d^{\frac {7}{2}} n \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (75 \, {\left (b e^{3} n - 7 \, a e^{3}\right )} x^{6} - 247 \, b d^{3} n + 3 \, {\left (61 \, b d e^{2} n - 280 \, a d e^{2}\right )} x^{4} + 210 \, a d^{3} + {\left (71 \, b d^{2} e n - 105 \, a d^{2} e\right )} x^{2} - 105 \, {\left (5 \, b e^{3} x^{6} + 8 \, b d e^{2} x^{4} + b d^{2} e x^{2} - 2 \, b d^{3}\right )} \log \left (c\right ) - 105 \, {\left (5 \, b e^{3} n x^{6} + 8 \, b d e^{2} n x^{4} + b d^{2} e n x^{2} - 2 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{3675 \, e^{2}}, \frac {210 \, b \sqrt {-d} d^{3} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) - {\left (75 \, {\left (b e^{3} n - 7 \, a e^{3}\right )} x^{6} - 247 \, b d^{3} n + 3 \, {\left (61 \, b d e^{2} n - 280 \, a d e^{2}\right )} x^{4} + 210 \, a d^{3} + {\left (71 \, b d^{2} e n - 105 \, a d^{2} e\right )} x^{2} - 105 \, {\left (5 \, b e^{3} x^{6} + 8 \, b d e^{2} x^{4} + b d^{2} e x^{2} - 2 \, b d^{3}\right )} \log \left (c\right ) - 105 \, {\left (5 \, b e^{3} n x^{6} + 8 \, b d e^{2} n x^{4} + b d^{2} e n x^{2} - 2 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{3675 \, e^{2}}\right ] \]
[1/3675*(105*b*d^(7/2)*n*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^ 2) - (75*(b*e^3*n - 7*a*e^3)*x^6 - 247*b*d^3*n + 3*(61*b*d*e^2*n - 280*a*d *e^2)*x^4 + 210*a*d^3 + (71*b*d^2*e*n - 105*a*d^2*e)*x^2 - 105*(5*b*e^3*x^ 6 + 8*b*d*e^2*x^4 + b*d^2*e*x^2 - 2*b*d^3)*log(c) - 105*(5*b*e^3*n*x^6 + 8 *b*d*e^2*n*x^4 + b*d^2*e*n*x^2 - 2*b*d^3*n)*log(x))*sqrt(e*x^2 + d))/e^2, 1/3675*(210*b*sqrt(-d)*d^3*n*arctan(sqrt(-d)/sqrt(e*x^2 + d)) - (75*(b*e^3 *n - 7*a*e^3)*x^6 - 247*b*d^3*n + 3*(61*b*d*e^2*n - 280*a*d*e^2)*x^4 + 210 *a*d^3 + (71*b*d^2*e*n - 105*a*d^2*e)*x^2 - 105*(5*b*e^3*x^6 + 8*b*d*e^2*x ^4 + b*d^2*e*x^2 - 2*b*d^3)*log(c) - 105*(5*b*e^3*n*x^6 + 8*b*d*e^2*n*x^4 + b*d^2*e*n*x^2 - 2*b*d^3*n)*log(x))*sqrt(e*x^2 + d))/e^2]
Time = 48.94 (sec) , antiderivative size = 845, normalized size of antiderivative = 4.77 \[ \int x^3 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Too large to display} \]
a*d*Piecewise((-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)) + a *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)) - b*d*n*Piecewise((2*d**(5/2)*asinh( sqrt(d)/(sqrt(e)*x))/(15*e**2) - 2*d**3/(15*e**(5/2)*x*sqrt(d/(e*x**2) + 1 )) - 2*d**2*x/(15*e**(3/2)*sqrt(d/(e*x**2) + 1)) + d*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 ))/(15*e) + 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, Tr ue))/5, (e > -oo) & (e < oo) & Ne(e, 0)), (sqrt(d)*x**4/16, True)) + b*d*P iecewise((-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))*log(c*x* *n) - b*e*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)*sqr t(d/(e*x**2) + 1)) - 4*d**2*Piecewise((d*sqrt(d + e*x**2)/(3*e) + x**2*sqr t(d + e*x**2)/3, Ne(e, 0)), (sqrt(d)*x**2/2, True))/(105*e**2) + d*Piecewi se((-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) + Piece wise((8*d**3*sqrt(d + e*x**2)/(105*e**3) - 4*d**2*x**2*sqrt(d + e*x**2)...
Exception generated. \[ \int x^3 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]
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
\[ \int x^3 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \,d x } \]
Timed out. \[ \int x^3 \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^3\,{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]