Integrand size = 33, antiderivative size = 251 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=\frac {2 b d^2 n \left (d^2-e^2 x^2\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 b d^4 n \sqrt {1-\frac {e^2 x^2}{d^2}} \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {d^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d-e x} \sqrt {d+e x}}+\frac {\left (d^2-e^2 x^2\right )^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^4 \sqrt {d-e x} \sqrt {d+e x}} \]
2/3*b*d^2*n*(-e^2*x^2+d^2)/e^4/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-1/9*b*n*(-e^2* x^2+d^2)^2/e^4/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-d^2*(-e^2*x^2+d^2)*(a+b*ln(c*x ^n))/e^4/(-e*x+d)^(1/2)/(e*x+d)^(1/2)+1/3*(-e^2*x^2+d^2)^2*(a+b*ln(c*x^n)) /e^4/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-2/3*b*d^4*n*arctanh((1-e^2*x^2/d^2)^(1/2 ))*(1-e^2*x^2/d^2)^(1/2)/e^4/(-e*x+d)^(1/2)/(e*x+d)^(1/2)
Time = 0.30 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.65 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {-6 b d^3 n \log (x)+3 b n \sqrt {d-e x} \sqrt {d+e x} \left (2 d^2+e^2 x^2\right ) \log (x)+\sqrt {d-e x} \sqrt {d+e x} \left (e^2 x^2 \left (3 a-b n-3 b n \log (x)+3 b \log \left (c x^n\right )\right )+d^2 \left (6 a-5 b n-6 b n \log (x)+6 b \log \left (c x^n\right )\right )\right )+6 b d^3 n \log \left (d+\sqrt {d-e x} \sqrt {d+e x}\right )}{9 e^4} \]
-1/9*(-6*b*d^3*n*Log[x] + 3*b*n*Sqrt[d - e*x]*Sqrt[d + e*x]*(2*d^2 + e^2*x ^2)*Log[x] + Sqrt[d - e*x]*Sqrt[d + e*x]*(e^2*x^2*(3*a - b*n - 3*b*n*Log[x ] + 3*b*Log[c*x^n]) + d^2*(6*a - 5*b*n - 6*b*n*Log[x] + 6*b*Log[c*x^n])) + 6*b*d^3*n*Log[d + Sqrt[d - e*x]*Sqrt[d + e*x]])/e^4
Time = 0.70 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.76, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {2787, 2792, 27, 354, 90, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 2787 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {1-\frac {e^2 x^2}{d^2}}}dx}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (-b n \int -\frac {d^2 \left (2 d^2+e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}}{3 e^4 x}dx+\frac {d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {d^4 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{e^4}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {b d^2 n \int \frac {\left (2 d^2+e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}}{x}dx}{3 e^4}+\frac {d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {d^4 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{e^4}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {b d^2 n \int \frac {\left (2 d^2+e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}}{x^2}dx^2}{6 e^4}+\frac {d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {d^4 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{e^4}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {b d^2 n \left (2 d^2 \int \frac {\sqrt {1-\frac {e^2 x^2}{d^2}}}{x^2}dx^2-\frac {2}{3} d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2}\right )}{6 e^4}+\frac {d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {d^4 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{e^4}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {b d^2 n \left (2 d^2 \left (\int \frac {1}{x^2 \sqrt {1-\frac {e^2 x^2}{d^2}}}dx^2+2 \sqrt {1-\frac {e^2 x^2}{d^2}}\right )-\frac {2}{3} d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2}\right )}{6 e^4}+\frac {d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {d^4 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{e^4}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {b d^2 n \left (2 d^2 \left (2 \sqrt {1-\frac {e^2 x^2}{d^2}}-\frac {2 d^2 \int \frac {1}{\frac {d^2}{e^2}-\frac {d^2 x^4}{e^2}}d\sqrt {1-\frac {e^2 x^2}{d^2}}}{e^2}\right )-\frac {2}{3} d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2}\right )}{6 e^4}+\frac {d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {d^4 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{e^4}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {d^4 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}-\frac {d^4 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {b d^2 n \left (2 d^2 \left (2 \sqrt {1-\frac {e^2 x^2}{d^2}}-2 \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )\right )-\frac {2}{3} d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2}\right )}{6 e^4}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
(Sqrt[1 - (e^2*x^2)/d^2]*((b*d^2*n*((-2*d^2*(1 - (e^2*x^2)/d^2)^(3/2))/3 + 2*d^2*(2*Sqrt[1 - (e^2*x^2)/d^2] - 2*ArcTanh[Sqrt[1 - (e^2*x^2)/d^2]])))/ (6*e^4) - (d^4*Sqrt[1 - (e^2*x^2)/d^2]*(a + b*Log[c*x^n]))/e^4 + (d^4*(1 - (e^2*x^2)/d^2)^(3/2)*(a + b*Log[c*x^n]))/(3*e^4)))/(Sqrt[d - e*x]*Sqrt[d + e*x])
3.4.9.3.1 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_.))*(x_)^(m_.)*((d1_) + (e1_.)*(x_))^ (q_)*((d2_) + (e2_.)*(x_))^(q_), x_Symbol] :> Simp[(d1 + e1*x)^q*((d2 + e2* x)^q/(1 + e1*(e2/(d1*d2))*x^2)^q) Int[x^m*(1 + e1*(e2/(d1*d2))*x^2)^q*(a + b*Log[c*x^n]), x], x] /; FreeQ[{a, b, c, d1, e1, d2, e2, n}, x] && EqQ[d2 *e1 + d1*e2, 0] && IntegerQ[m] && IntegerQ[q - 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])
\[\int \frac {x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {-e x +d}\, \sqrt {e x +d}}d x\]
Time = 0.33 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.50 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=\frac {6 \, b d^{3} n \log \left (\frac {\sqrt {e x + d} \sqrt {-e x + d} - d}{x}\right ) + {\left (5 \, b d^{2} n - 6 \, a d^{2} + {\left (b e^{2} n - 3 \, a e^{2}\right )} x^{2} - 3 \, {\left (b e^{2} x^{2} + 2 \, b d^{2}\right )} \log \left (c\right ) - 3 \, {\left (b e^{2} n x^{2} + 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x + d} \sqrt {-e x + d}}{9 \, e^{4}} \]
1/9*(6*b*d^3*n*log((sqrt(e*x + d)*sqrt(-e*x + d) - d)/x) + (5*b*d^2*n - 6* a*d^2 + (b*e^2*n - 3*a*e^2)*x^2 - 3*(b*e^2*x^2 + 2*b*d^2)*log(c) - 3*(b*e^ 2*n*x^2 + 2*b*d^2*n)*log(x))*sqrt(e*x + d)*sqrt(-e*x + d))/e^4
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=\int \frac {x^{3} \left (a + b \log {\left (c x^{n} \right )}\right )}{\sqrt {d - e x} \sqrt {d + e x}}\, dx \]
Time = 0.26 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {1}{9} \, b n {\left (\frac {3 \, d^{3} \log \left (d + \sqrt {-e^{2} x^{2} + d^{2}}\right )}{e^{4}} - \frac {3 \, d^{3} \log \left (-d + \sqrt {-e^{2} x^{2} + d^{2}}\right )}{e^{4}} - \frac {6 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2} - {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}}{e^{4}}\right )} - \frac {1}{3} \, b {\left (\frac {\sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{e^{2}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{e^{4}}\right )} \log \left (c x^{n}\right ) - \frac {1}{3} \, a {\left (\frac {\sqrt {-e^{2} x^{2} + d^{2}} x^{2}}{e^{2}} + \frac {2 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{2}}{e^{4}}\right )} \]
-1/9*b*n*(3*d^3*log(d + sqrt(-e^2*x^2 + d^2))/e^4 - 3*d^3*log(-d + sqrt(-e ^2*x^2 + d^2))/e^4 - (6*sqrt(-e^2*x^2 + d^2)*d^2 - (-e^2*x^2 + d^2)^(3/2)) /e^4) - 1/3*b*(sqrt(-e^2*x^2 + d^2)*x^2/e^2 + 2*sqrt(-e^2*x^2 + d^2)*d^2/e ^4)*log(c*x^n) - 1/3*a*(sqrt(-e^2*x^2 + d^2)*x^2/e^2 + 2*sqrt(-e^2*x^2 + d ^2)*d^2/e^4)
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{\sqrt {e x + d} \sqrt {-e x + d}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {d+e\,x}\,\sqrt {d-e\,x}} \,d x \]