Integrand size = 33, antiderivative size = 252 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {2 b e^2 n \left (d^2-e^2 x^2\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}}-\frac {b n \left (d^2-e^2 x^2\right )^2}{9 d^4 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 b e^3 n \sqrt {1-\frac {e^2 x^2}{d^2}} \arcsin \left (\frac {e x}{d}\right )}{3 d^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x^3 \sqrt {d-e x} \sqrt {d+e x}}-\frac {2 e^2 \left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{3 d^4 x \sqrt {d-e x} \sqrt {d+e x}} \]
-2/3*b*e^2*n*(-e^2*x^2+d^2)/d^4/x/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-1/9*b*n*(-e ^2*x^2+d^2)^2/d^4/x^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-1/3*(-e^2*x^2+d^2)*(a+b *ln(c*x^n))/d^2/x^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-2/3*e^2*(-e^2*x^2+d^2)*(a +b*ln(c*x^n))/d^4/x/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-2/3*b*e^3*n*arcsin(e*x/d) *(1-e^2*x^2/d^2)^(1/2)/d^3/(-e*x+d)^(1/2)/(e*x+d)^(1/2)
Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.46 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {6 b e^3 n x^3 \arctan \left (\frac {e x}{\sqrt {d-e x} \sqrt {d+e x}}\right )+\sqrt {d-e x} \sqrt {d+e x} \left (3 a \left (d^2+2 e^2 x^2\right )+b n \left (d^2+5 e^2 x^2\right )+3 b \left (d^2+2 e^2 x^2\right ) \log \left (c x^n\right )\right )}{9 d^4 x^3} \]
-1/9*(6*b*e^3*n*x^3*ArcTan[(e*x)/(Sqrt[d - e*x]*Sqrt[d + e*x])] + Sqrt[d - e*x]*Sqrt[d + e*x]*(3*a*(d^2 + 2*e^2*x^2) + b*n*(d^2 + 5*e^2*x^2) + 3*b*( d^2 + 2*e^2*x^2)*Log[c*x^n]))/(d^4*x^3)
Time = 0.66 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.75, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2787, 2792, 27, 358, 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.
| \(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d-e x} \sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 2787 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {1-\frac {e^2 x^2}{d^2}}}dx}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (-b n \int -\frac {\left (d^2+2 e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}}{3 d^2 x^4}dx-\frac {2 e^2 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{3 x^3}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {b n \int \frac {\left (d^2+2 e^2 x^2\right ) \sqrt {1-\frac {e^2 x^2}{d^2}}}{x^4}dx}{3 d^2}-\frac {2 e^2 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{3 x^3}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 358 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {b n \left (2 e^2 \int \frac {\sqrt {1-\frac {e^2 x^2}{d^2}}}{x^2}dx-\frac {d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2}}{3 x^3}\right )}{3 d^2}-\frac {2 e^2 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{3 x^3}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (\frac {b n \left (2 e^2 \left (-\frac {e^2 \int \frac {1}{\sqrt {1-\frac {e^2 x^2}{d^2}}}dx}{d^2}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}}}{x}\right )-\frac {d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2}}{3 x^3}\right )}{3 d^2}-\frac {2 e^2 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{3 x^3}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 223 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (-\frac {2 e^2 \sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{3 x^3}+\frac {b n \left (2 e^2 \left (-\frac {e \arcsin \left (\frac {e x}{d}\right )}{d}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}}}{x}\right )-\frac {d^2 \left (1-\frac {e^2 x^2}{d^2}\right )^{3/2}}{3 x^3}\right )}{3 d^2}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
(Sqrt[1 - (e^2*x^2)/d^2]*((b*n*(-1/3*(d^2*(1 - (e^2*x^2)/d^2)^(3/2))/x^3 + 2*e^2*(-(Sqrt[1 - (e^2*x^2)/d^2]/x) - (e*ArcSin[(e*x)/d])/d)))/(3*d^2) - (Sqrt[1 - (e^2*x^2)/d^2]*(a + b*Log[c*x^n]))/(3*x^3) - (2*e^2*Sqrt[1 - (e^ 2*x^2)/d^2]*(a + b*Log[c*x^n]))/(3*d^2*x)))/(Sqrt[d - e*x]*Sqrt[d + e*x])
3.4.16.3.1 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[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[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x_ Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + S imp[d/e^2 Int[(e*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e , m, p}, x] && NeQ[b*c - a*d, 0] && EqQ[Simplify[m + 2*p + 3], 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[((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])
\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{4} \sqrt {-e x +d}\, \sqrt {e x +d}}d x\]
Time = 0.32 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.54 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\frac {12 \, b e^{3} n x^{3} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-e x + d} - d}{e x}\right ) - {\left (b d^{2} n + 3 \, a d^{2} + {\left (5 \, b e^{2} n + 6 \, a e^{2}\right )} x^{2} + 3 \, {\left (2 \, b e^{2} x^{2} + b d^{2}\right )} \log \left (c\right ) + 3 \, {\left (2 \, b e^{2} n x^{2} + b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x + d} \sqrt {-e x + d}}{9 \, d^{4} x^{3}} \]
1/9*(12*b*e^3*n*x^3*arctan((sqrt(e*x + d)*sqrt(-e*x + d) - d)/(e*x)) - (b* d^2*n + 3*a*d^2 + (5*b*e^2*n + 6*a*e^2)*x^2 + 3*(2*b*e^2*x^2 + b*d^2)*log( c) + 3*(2*b*e^2*n*x^2 + b*d^2*n)*log(x))*sqrt(e*x + d)*sqrt(-e*x + d))/(d^ 4*x^3)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int \frac {a + b \log {\left (c x^{n} \right )}}{x^{4} \sqrt {d - e x} \sqrt {d + e x}}\, dx \]
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{\sqrt {e x + d} \sqrt {-e x + d} x^{4}} \,d x } \]
-1/3*a*(2*sqrt(-e^2*x^2 + d^2)*e^2/(d^4*x) + sqrt(-e^2*x^2 + d^2)/(d^2*x^3 )) + b*integrate((log(c) + log(x^n))/(sqrt(e*x + d)*sqrt(-e*x + d)*x^4), x )
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{\sqrt {e x + d} \sqrt {-e x + d} x^{4}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^4\,\sqrt {d+e\,x}\,\sqrt {d-e\,x}} \,d x \]