Integrand size = 25, antiderivative size = 154 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 b d^2 n \sqrt {d+e x^2}}{15 e^2}+\frac {2 b d n \left (d+e x^2\right )^{3/2}}{45 e^2}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^2}-\frac {2 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{15 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2} \]
2/45*b*d*n*(e*x^2+d)^(3/2)/e^2-1/25*b*n*(e*x^2+d)^(5/2)/e^2-2/15*b*d^(5/2) *n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e^2-1/3*d*(e*x^2+d)^(3/2)*(a+b*ln(c*x^ n))/e^2+1/5*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/e^2+2/15*b*d^2*n*(e*x^2+d)^(1/ 2)/e^2
Time = 0.10 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.32 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {2 b d^{5/2} n \log (x)}{15 e^2}-\frac {b n \sqrt {d+e x^2} \left (2 d^2-d e x^2-3 e^2 x^4\right ) \log (x)}{15 e^2}+\sqrt {d+e x^2} \left (\frac {1}{25} x^4 \left (5 a-b n+5 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+\frac {d x^2 \left (15 a-8 b n+15 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{225 e}-\frac {d^2 \left (30 a-31 b n+30 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{225 e^2}\right )-\frac {2 b d^{5/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{15 e^2} \]
(2*b*d^(5/2)*n*Log[x])/(15*e^2) - (b*n*Sqrt[d + e*x^2]*(2*d^2 - d*e*x^2 - 3*e^2*x^4)*Log[x])/(15*e^2) + Sqrt[d + e*x^2]*((x^4*(5*a - b*n + 5*b*(-(n* Log[x]) + Log[c*x^n])))/25 + (d*x^2*(15*a - 8*b*n + 15*b*(-(n*Log[x]) + Lo g[c*x^n])))/(225*e) - (d^2*(30*a - 31*b*n + 30*b*(-(n*Log[x]) + Log[c*x^n] )))/(225*e^2)) - (2*b*d^(5/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(15*e^2)
Time = 0.34 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.93, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {2792, 27, 354, 90, 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.
| \(\displaystyle \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {\left (2 d-3 e x^2\right ) \left (e x^2+d\right )^{3/2}}{15 e^2 x}dx+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b n \int \frac {\left (2 d-3 e x^2\right ) \left (e x^2+d\right )^{3/2}}{x}dx}{15 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {b n \int \frac {\left (2 d-3 e x^2\right ) \left (e x^2+d\right )^{3/2}}{x^2}dx^2}{30 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {b n \left (2 d \int \frac {\left (e x^2+d\right )^{3/2}}{x^2}dx^2-\frac {6}{5} \left (d+e x^2\right )^{5/2}\right )}{30 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {b n \left (2 d \left (d \int \frac {\sqrt {e x^2+d}}{x^2}dx^2+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )-\frac {6}{5} \left (d+e x^2\right )^{5/2}\right )}{30 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {b n \left (2 d \left (d \left (d \int \frac {1}{x^2 \sqrt {e x^2+d}}dx^2+2 \sqrt {d+e x^2}\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )-\frac {6}{5} \left (d+e x^2\right )^{5/2}\right )}{30 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {b n \left (2 d \left (d \left (\frac {2 d \int \frac {1}{\frac {x^4}{e}-\frac {d}{e}}d\sqrt {e x^2+d}}{e}+2 \sqrt {d+e x^2}\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )-\frac {6}{5} \left (d+e x^2\right )^{5/2}\right )}{30 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {b n \left (2 d \left (d \left (2 \sqrt {d+e x^2}-2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )-\frac {6}{5} \left (d+e x^2\right )^{5/2}\right )}{30 e^2}\) |
(b*n*((-6*(d + e*x^2)^(5/2))/5 + 2*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]]))))/(30*e^2) - (d* (d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(3*e^2) + ((d + e*x^2)^(5/2)*(a + b* Log[c*x^n]))/(5*e^2)
3.3.52.3.1 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])
\[\int x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e \,x^{2}+d}d x\]
Time = 0.34 (sec) , antiderivative size = 309, normalized size of antiderivative = 2.01 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {15 \, b d^{\frac {5}{2}} n \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (9 \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} - 31 \, b d^{2} n + 30 \, a d^{2} + {\left (8 \, b d e n - 15 \, a d e\right )} x^{2} - 15 \, {\left (3 \, b e^{2} x^{4} + b d e x^{2} - 2 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b e^{2} n x^{4} + b d e n x^{2} - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{225 \, e^{2}}, \frac {30 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) - {\left (9 \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} - 31 \, b d^{2} n + 30 \, a d^{2} + {\left (8 \, b d e n - 15 \, a d e\right )} x^{2} - 15 \, {\left (3 \, b e^{2} x^{4} + b d e x^{2} - 2 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b e^{2} n x^{4} + b d e n x^{2} - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{225 \, e^{2}}\right ] \]
[1/225*(15*b*d^(5/2)*n*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - (9*(b*e^2*n - 5*a*e^2)*x^4 - 31*b*d^2*n + 30*a*d^2 + (8*b*d*e*n - 15*a* d*e)*x^2 - 15*(3*b*e^2*x^4 + b*d*e*x^2 - 2*b*d^2)*log(c) - 15*(3*b*e^2*n*x ^4 + b*d*e*n*x^2 - 2*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/e^2, 1/225*(30*b*sq rt(-d)*d^2*n*arctan(sqrt(-d)/sqrt(e*x^2 + d)) - (9*(b*e^2*n - 5*a*e^2)*x^4 - 31*b*d^2*n + 30*a*d^2 + (8*b*d*e*n - 15*a*d*e)*x^2 - 15*(3*b*e^2*x^4 + b*d*e*x^2 - 2*b*d^2)*log(c) - 15*(3*b*e^2*n*x^4 + b*d*e*n*x^2 - 2*b*d^2*n) *log(x))*sqrt(e*x^2 + d))/e^2]
Time = 14.27 (sec) , antiderivative size = 343, normalized size of antiderivative = 2.23 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=a \left (\begin {cases} - \frac {2 d^{2} \sqrt {d + e x^{2}}}{15 e^{2}} + \frac {d x^{2} \sqrt {d + e x^{2}}}{15 e} + \frac {x^{4} \sqrt {d + e x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {2 d^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{15 e^{2}} - \frac {2 d^{3}}{15 e^{\frac {5}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} - \frac {2 d^{2} x}{15 e^{\frac {3}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {d \left (\begin {cases} \frac {d \sqrt {d + e x^{2}}}{3 e} + \frac {x^{2} \sqrt {d + e x^{2}}}{3} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right )}{15 e} + \frac {\begin {cases} - \frac {2 d^{2} \sqrt {d + e x^{2}}}{15 e^{2}} + \frac {d x^{2} \sqrt {d + e x^{2}}}{15 e} + \frac {x^{4} \sqrt {d + e x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}}{5} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{4}}{16} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {2 d^{2} \sqrt {d + e x^{2}}}{15 e^{2}} + \frac {d x^{2} \sqrt {d + e x^{2}}}{15 e} + \frac {x^{4} \sqrt {d + e x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*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)) - b*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, N e(e, 0)), (sqrt(d)*x**4/4, True))/5, (e > -oo) & (e < oo) & Ne(e, 0)), (sq rt(d)*x**4/16, True)) + b*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))*log(c*x**n)
Exception generated. \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]
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 = 0.35 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.40 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{5} \, \sqrt {e x^{2} + d} b x^{4} \log \left (c\right ) + \frac {1}{5} \, \sqrt {e x^{2} + d} a x^{4} + \frac {\sqrt {e x^{2} + d} b d x^{2} \log \left (c\right )}{15 \, e} + \frac {\sqrt {e x^{2} + d} a d x^{2}}{15 \, e} + \frac {1}{225} \, b n {\left (\frac {15 \, {\left (3 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} - 5 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d\right )} \log \left (x\right )}{e^{2}} + \frac {\frac {30 \, d^{3} \arctan \left (\frac {\sqrt {e x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} - 9 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} + 10 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d + 30 \, \sqrt {e x^{2} + d} d^{2}}{e^{2}}\right )} - \frac {2 \, \sqrt {e x^{2} + d} b d^{2} \log \left (c\right )}{15 \, e^{2}} - \frac {2 \, \sqrt {e x^{2} + d} a d^{2}}{15 \, e^{2}} \]
1/5*sqrt(e*x^2 + d)*b*x^4*log(c) + 1/5*sqrt(e*x^2 + d)*a*x^4 + 1/15*sqrt(e *x^2 + d)*b*d*x^2*log(c)/e + 1/15*sqrt(e*x^2 + d)*a*d*x^2/e + 1/225*b*n*(1 5*(3*(e*x^2 + d)^(5/2) - 5*(e*x^2 + d)^(3/2)*d)*log(x)/e^2 + (30*d^3*arcta n(sqrt(e*x^2 + d)/sqrt(-d))/sqrt(-d) - 9*(e*x^2 + d)^(5/2) + 10*(e*x^2 + d )^(3/2)*d + 30*sqrt(e*x^2 + d)*d^2)/e^2) - 2/15*sqrt(e*x^2 + d)*b*d^2*log( c)/e^2 - 2/15*sqrt(e*x^2 + d)*a*d^2/e^2
Timed out. \[ \int x^3 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^3\,\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]