Integrand size = 23, antiderivative size = 125 \[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b d^2 n \sqrt {d+e x^2}}{5 e}-\frac {b d n \left (d+e x^2\right )^{3/2}}{15 e}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e}+\frac {b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{5 e}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e} \]
-1/15*b*d*n*(e*x^2+d)^(3/2)/e-1/25*b*n*(e*x^2+d)^(5/2)/e+1/5*b*d^(5/2)*n*a rctanh((e*x^2+d)^(1/2)/d^(1/2))/e+1/5*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/e-1/ 5*b*d^2*n*(e*x^2+d)^(1/2)/e
Time = 0.10 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.45 \[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b d^{5/2} n \log (x)}{5 e}+\frac {b n \left (d+e x^2\right )^{5/2} \log (x)}{5 e}+\sqrt {d+e x^2} \left (\frac {1}{25} e x^4 \left (5 a-b n+5 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+\frac {d^2 \left (15 a-23 b n+15 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{75 e}+\frac {1}{75} d x^2 \left (30 a-11 b n+30 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )\right )+\frac {b d^{5/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{5 e} \]
-1/5*(b*d^(5/2)*n*Log[x])/e + (b*n*(d + e*x^2)^(5/2)*Log[x])/(5*e) + Sqrt[ d + e*x^2]*((e*x^4*(5*a - b*n + 5*b*(-(n*Log[x]) + Log[c*x^n])))/25 + (d^2 *(15*a - 23*b*n + 15*b*(-(n*Log[x]) + Log[c*x^n])))/(75*e) + (d*x^2*(30*a - 11*b*n + 30*b*(-(n*Log[x]) + Log[c*x^n])))/75) + (b*d^(5/2)*n*Log[d + Sq rt[d]*Sqrt[d + e*x^2]])/(5*e)
Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2776, 243, 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.
| \(\displaystyle \int x \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2776 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {b n \int \frac {\left (e x^2+d\right )^{5/2}}{x}dx}{5 e}\) |
| \(\Big \downarrow \) 243 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {b n \int \frac {\left (e x^2+d\right )^{5/2}}{x^2}dx^2}{10 e}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {b n \left (d \int \frac {\left (e x^2+d\right )^{3/2}}{x^2}dx^2+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )}{10 e}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {b n \left (d \left (d \int \frac {\sqrt {e x^2+d}}{x^2}dx^2+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )}{10 e}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {b n \left (d \left (d \left (d \int \frac {1}{x^2 \sqrt {e x^2+d}}dx^2+2 \sqrt {d+e x^2}\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )}{10 e}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {b n \left (d \left (d \left (\frac {2 d \int \frac {1}{\frac {x^4}{e}-\frac {d}{e}}d\sqrt {e x^2+d}}{e}+2 \sqrt {d+e x^2}\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )}{10 e}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e}-\frac {b n \left (d \left (d \left (2 \sqrt {d+e x^2}-2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )}{10 e}\) |
-1/10*(b*n*((2*(d + e*x^2)^(5/2))/5 + d*((2*(d + e*x^2)^(3/2))/3 + d*(2*Sq rt[d + e*x^2] - 2*Sqrt[d]*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]))))/e + ((d + e *x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*e)
3.3.65.3.1 Defintions of rubi rules used
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_)^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_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :> Simp[f^m*(d + e*x^r)^(q + 1)*((a + b*L og[c*x^n])^p/(e*r*(q + 1))), x] - Simp[b*f^m*n*(p/(e*r*(q + 1))) Int[(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d , e, f, m, n, q, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || G tQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]
\[\int x \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]
Time = 0.36 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.43 \[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {15 \, b d^{\frac {5}{2}} n \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - 2 \, {\left (3 \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 23 \, b d^{2} n - 15 \, a d^{2} + {\left (11 \, b d e n - 30 \, a d e\right )} x^{2} - 15 \, {\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{150 \, e}, -\frac {15 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (3 \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 23 \, b d^{2} n - 15 \, a d^{2} + {\left (11 \, b d e n - 30 \, a d e\right )} x^{2} - 15 \, {\left (b e^{2} x^{4} + 2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{75 \, e}\right ] \]
[1/150*(15*b*d^(5/2)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - 2*(3*(b*e^2*n - 5*a*e^2)*x^4 + 23*b*d^2*n - 15*a*d^2 + (11*b*d*e*n - 30 *a*d*e)*x^2 - 15*(b*e^2*x^4 + 2*b*d*e*x^2 + b*d^2)*log(c) - 15*(b*e^2*n*x^ 4 + 2*b*d*e*n*x^2 + b*d^2*n)*log(x))*sqrt(e*x^2 + d))/e, -1/75*(15*b*sqrt( -d)*d^2*n*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (3*(b*e^2*n - 5*a*e^2)*x^4 + 23*b*d^2*n - 15*a*d^2 + (11*b*d*e*n - 30*a*d*e)*x^2 - 15*(b*e^2*x^4 + 2*b* d*e*x^2 + b*d^2)*log(c) - 15*(b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*log(x ))*sqrt(e*x^2 + d))/e]
Time = 30.64 (sec) , antiderivative size = 573, normalized size of antiderivative = 4.58 \[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=a 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 ) + a e \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 d n \left (\begin {cases} - \frac {d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{3 e} + \frac {d^{2}}{3 e^{\frac {3}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {d x}{3 \sqrt {e} \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {\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}}{3} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{2}}{4} & \text {otherwise} \end {cases}\right ) + b 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 ) \log {\left (c x^{n} \right )} - b e 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 e \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*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)) + a*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, True)) - b*d*n*Piecewise((-d**(3/2)*asinh(sqrt(d)/(sqr t(e)*x))/(3*e) + d**2/(3*e**(3/2)*x*sqrt(d/(e*x**2) + 1)) + d*x/(3*sqrt(e) *sqrt(d/(e*x**2) + 1)) + 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))/3, (e > -oo) & (e < oo) & Ne(e, 0)), (sqrt(d)*x**2/4, True)) + b*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))*log(c*x* *n) - b*e*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*sqr t(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))/5, (e > -oo) & (e < oo) & Ne( e, 0)), (sqrt(d)*x**4/16, True)) + b*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, True))*log(c*x**n)
Exception generated. \[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]
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
\[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )} x \,d x } \]
Timed out. \[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x\,{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]