Integrand size = 33, antiderivative size = 142 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {b n \left (d^2-e^2 x^2\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {b e n \sqrt {1-\frac {e^2 x^2}{d^2}} \arcsin \left (\frac {e x}{d}\right )}{d \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{d^2 x \sqrt {d-e x} \sqrt {d+e x}} \]
-b*n*(-e^2*x^2+d^2)/d^2/x/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-(-e^2*x^2+d^2)*(a+b *ln(c*x^n))/d^2/x/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-b*e*n*arcsin(e*x/d)*(1-e^2* x^2/d^2)^(1/2)/d/(-e*x+d)^(1/2)/(e*x+d)^(1/2)
Time = 0.16 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.49 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {b e n x \arctan \left (\frac {e x}{\sqrt {d-e x} \sqrt {d+e x}}\right )+\sqrt {d-e x} \sqrt {d+e x} \left (a+b n+b \log \left (c x^n\right )\right )}{d^2 x} \]
-((b*e*n*x*ArcTan[(e*x)/(Sqrt[d - e*x]*Sqrt[d + e*x])] + Sqrt[d - e*x]*Sqr t[d + e*x]*(a + b*n + b*Log[c*x^n]))/(d^2*x))
Time = 0.57 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.77, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {2787, 2773, 247, 223}
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.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx\) |
\(\Big \downarrow \) 2787 |
\(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {a+b \log \left (c x^n\right )}{x^2 \sqrt {1-\frac {e^2 x^2}{d^2}}}dx}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 2773 |
\(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (b n \int \frac {\sqrt {1-\frac {e^2 x^2}{d^2}}}{x^2}dx-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{x}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (b n \left (-\frac {e^2 \int \frac {1}{\sqrt {1-\frac {e^2 x^2}{d^2}}}dx}{d^2}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}}}{x}\right )-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{x}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
\(\Big \downarrow \) 223 |
\(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (b n \left (-\frac {e \arcsin \left (\frac {e x}{d}\right )}{d}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}}}{x}\right )-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{x}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
(Sqrt[1 - (e^2*x^2)/d^2]*(b*n*(-(Sqrt[1 - (e^2*x^2)/d^2]/x) - (e*ArcSin[(e *x)/d])/d) - (Sqrt[1 - (e^2*x^2)/d^2]*(a + b*Log[c*x^n]))/x))/(Sqrt[d - e* x]*Sqrt[d + e*x])
3.4.15.3.1 Defintions of rubi rules used
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[Rt[-b, 2]*(x/Sqrt [a])]/Rt[-b, 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])/(d*f*(m + 1))), x] - Simp[b*(n/(d*(m + 1))) Int[(f*x)^m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && Eq Q[m + r*(q + 1) + 1, 0] && NeQ[m, -1]
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]
\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{2} \sqrt {-e x +d}\, \sqrt {e x +d}}d x\]
Time = 0.30 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.51 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\frac {2 \, b e n x \arctan \left (\frac {\sqrt {e x + d} \sqrt {-e x + d} - d}{e x}\right ) - {\left (b n \log \left (x\right ) + b n + b \log \left (c\right ) + a\right )} \sqrt {e x + d} \sqrt {-e x + d}}{d^{2} x} \]
(2*b*e*n*x*arctan((sqrt(e*x + d)*sqrt(-e*x + d) - d)/(e*x)) - (b*n*log(x) + b*n + b*log(c) + a)*sqrt(e*x + d)*sqrt(-e*x + d))/(d^2*x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int \frac {a + b \log {\left (c x^{n} \right )}}{x^{2} \sqrt {d - e x} \sqrt {d + e x}}\, dx \]
Time = 0.26 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.72 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {{\left (\frac {e^{2} \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}}} + \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{x}\right )} b n}{d^{2}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} b \log \left (c x^{n}\right )}{d^{2} x} - \frac {\sqrt {-e^{2} x^{2} + d^{2}} a}{d^{2} x} \]
-(e^2*arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2) + sqrt(-e^2*x^2 + d^2)/x)*b*n/ d^2 - sqrt(-e^2*x^2 + d^2)*b*log(c*x^n)/(d^2*x) - sqrt(-e^2*x^2 + d^2)*a/( d^2*x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{\sqrt {e x + d} \sqrt {-e x + d} x^{2}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^2\,\sqrt {d+e\,x}\,\sqrt {d-e\,x}} \,d x \]