Integrand size = 23, antiderivative size = 102 \[ \int x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {b d n \sqrt {d+e x^2}}{3 e}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e}+\frac {b d^{3/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e} \]
-1/9*b*n*(e*x^2+d)^(3/2)/e+1/3*b*d^(3/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/2) )/e+1/3*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/e-1/3*b*d*n*(e*x^2+d)^(1/2)/e
Time = 0.07 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.33 \[ \int x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {3 a d \sqrt {d+e x^2}-4 b d n \sqrt {d+e x^2}+3 a e x^2 \sqrt {d+e x^2}-b e n x^2 \sqrt {d+e x^2}-3 b d^{3/2} n \log (x)+3 b \left (d+e x^2\right )^{3/2} \log \left (c x^n\right )+3 b d^{3/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{9 e} \]
(3*a*d*Sqrt[d + e*x^2] - 4*b*d*n*Sqrt[d + e*x^2] + 3*a*e*x^2*Sqrt[d + e*x^ 2] - b*e*n*x^2*Sqrt[d + e*x^2] - 3*b*d^(3/2)*n*Log[x] + 3*b*(d + e*x^2)^(3 /2)*Log[c*x^n] + 3*b*d^(3/2)*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(9*e)
Time = 0.25 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.93, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2776, 243, 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.
\(\displaystyle \int x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2776 |
\(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {b n \int \frac {\left (e x^2+d\right )^{3/2}}{x}dx}{3 e}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {b n \int \frac {\left (e x^2+d\right )^{3/2}}{x^2}dx^2}{6 e}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {b n \left (d \int \frac {\sqrt {e x^2+d}}{x^2}dx^2+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )}{6 e}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {b n \left (d \left (d \int \frac {1}{x^2 \sqrt {e x^2+d}}dx^2+2 \sqrt {d+e x^2}\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )}{6 e}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {b n \left (d \left (\frac {2 d \int \frac {1}{\frac {x^4}{e}-\frac {d}{e}}d\sqrt {e x^2+d}}{e}+2 \sqrt {d+e x^2}\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )}{6 e}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e}-\frac {b n \left (d \left (2 \sqrt {d+e x^2}-2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )\right )+\frac {2}{3} \left (d+e x^2\right )^{3/2}\right )}{6 e}\) |
-1/6*(b*n*((2*(d + e*x^2)^(3/2))/3 + d*(2*Sqrt[d + e*x^2] - 2*Sqrt[d]*ArcT anh[Sqrt[d + e*x^2]/Sqrt[d]])))/e + ((d + e*x^2)^(3/2)*(a + b*Log[c*x^n])) /(3*e)
3.3.53.3.1 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]
\[\int x \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e \,x^{2}+d}d x\]
Time = 0.33 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.98 \[ \int x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {3 \, b d^{\frac {3}{2}} n \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - 2 \, {\left (4 \, b d n + {\left (b e n - 3 \, a e\right )} x^{2} - 3 \, a d - 3 \, {\left (b e x^{2} + b d\right )} \log \left (c\right ) - 3 \, {\left (b e n x^{2} + b d n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{18 \, e}, -\frac {3 \, b \sqrt {-d} d n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (4 \, b d n + {\left (b e n - 3 \, a e\right )} x^{2} - 3 \, a d - 3 \, {\left (b e x^{2} + b d\right )} \log \left (c\right ) - 3 \, {\left (b e n x^{2} + b d n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{9 \, e}\right ] \]
[1/18*(3*b*d^(3/2)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - 2*(4*b*d*n + (b*e*n - 3*a*e)*x^2 - 3*a*d - 3*(b*e*x^2 + b*d)*log(c) - 3*( b*e*n*x^2 + b*d*n)*log(x))*sqrt(e*x^2 + d))/e, -1/9*(3*b*sqrt(-d)*d*n*arct an(sqrt(-d)/sqrt(e*x^2 + d)) + (4*b*d*n + (b*e*n - 3*a*e)*x^2 - 3*a*d - 3* (b*e*x^2 + b*d)*log(c) - 3*(b*e*n*x^2 + b*d*n)*log(x))*sqrt(e*x^2 + d))/e]
Time = 11.60 (sec) , antiderivative size = 218, normalized size of antiderivative = 2.14 \[ \int x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=a \left (\begin {cases} \frac {d \sqrt {d + e x^{2}}}{3 e} + \frac {x^{2} \sqrt {d + e x^{2}}}{3} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{3 e} + \frac {d^{2}}{3 e^{\frac {3}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {d x}{3 \sqrt {e} \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {\begin {cases} \frac {d \sqrt {d + e x^{2}}}{3 e} + \frac {x^{2} \sqrt {d + e x^{2}}}{3} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}}{3} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{2}}{4} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {d \sqrt {d + e x^{2}}}{3 e} + \frac {x^{2} \sqrt {d + e x^{2}}}{3} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*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)) - b*n*Piecewise((-d**(3/2)*asinh(sqrt(d)/(sqrt(e )*x))/(3*e) + d**2/(3*e**(3/2)*x*sqrt(d/(e*x**2) + 1)) + d*x/(3*sqrt(e)*sq rt(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*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)
Exception generated. \[ \int x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Exception raised: ValueError} \]
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 = 0.34 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.38 \[ \int x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{3} \, \sqrt {e x^{2} + d} b x^{2} \log \left (c\right ) + \frac {1}{3} \, \sqrt {e x^{2} + d} a x^{2} + \frac {1}{9} \, {\left (\frac {3 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} \log \left (x\right )}{e} - \frac {\frac {3 \, d^{2} \arctan \left (\frac {\sqrt {e x^{2} + d}}{\sqrt {-d}}\right )}{\sqrt {-d}} + {\left (e x^{2} + d\right )}^{\frac {3}{2}} + 3 \, \sqrt {e x^{2} + d} d}{e}\right )} b n + \frac {\sqrt {e x^{2} + d} b d \log \left (c\right )}{3 \, e} + \frac {\sqrt {e x^{2} + d} a d}{3 \, e} \]
1/3*sqrt(e*x^2 + d)*b*x^2*log(c) + 1/3*sqrt(e*x^2 + d)*a*x^2 + 1/9*(3*(e*x ^2 + d)^(3/2)*log(x)/e - (3*d^2*arctan(sqrt(e*x^2 + d)/sqrt(-d))/sqrt(-d) + (e*x^2 + d)^(3/2) + 3*sqrt(e*x^2 + d)*d)/e)*b*n + 1/3*sqrt(e*x^2 + d)*b* d*log(c)/e + 1/3*sqrt(e*x^2 + d)*a*d/e
Timed out. \[ \int x \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x\,\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]