Integrand size = 33, antiderivative size = 489 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx=-\frac {b n \left (d^2-e^2 x^2\right )}{4 d^2 x^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {b e^2 n \sqrt {1-\frac {e^2 x^2}{d^2}} \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{4 d^2 \sqrt {d-e x} \sqrt {d+e x}}+\frac {b e^2 n \sqrt {1-\frac {e^2 x^2}{d^2}} \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )^2}{4 d^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {\left (d^2-e^2 x^2\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^2 x^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {e^2 \sqrt {1-\frac {e^2 x^2}{d^2}} \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b e^2 n \sqrt {1-\frac {e^2 x^2}{d^2}} \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \log \left (\frac {2}{1-\sqrt {1-\frac {e^2 x^2}{d^2}}}\right )}{2 d^2 \sqrt {d-e x} \sqrt {d+e x}}-\frac {b e^2 n \sqrt {1-\frac {e^2 x^2}{d^2}} \operatorname {PolyLog}\left (2,-\frac {1+\sqrt {1-\frac {e^2 x^2}{d^2}}}{1-\sqrt {1-\frac {e^2 x^2}{d^2}}}\right )}{4 d^2 \sqrt {d-e x} \sqrt {d+e x}} \]
-1/4*b*n*(-e^2*x^2+d^2)/d^2/x^2/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-1/2*(-e^2*x^2 +d^2)*(a+b*ln(c*x^n))/d^2/x^2/(-e*x+d)^(1/2)/(e*x+d)^(1/2)+1/4*b*e^2*n*arc tanh((1-e^2*x^2/d^2)^(1/2))*(1-e^2*x^2/d^2)^(1/2)/d^2/(-e*x+d)^(1/2)/(e*x+ d)^(1/2)+1/4*b*e^2*n*arctanh((1-e^2*x^2/d^2)^(1/2))^2*(1-e^2*x^2/d^2)^(1/2 )/d^2/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-1/2*e^2*arctanh((1-e^2*x^2/d^2)^(1/2))* (a+b*ln(c*x^n))*(1-e^2*x^2/d^2)^(1/2)/d^2/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-1/2 *b*e^2*n*arctanh((1-e^2*x^2/d^2)^(1/2))*ln(2/(1-(1-e^2*x^2/d^2)^(1/2)))*(1 -e^2*x^2/d^2)^(1/2)/d^2/(-e*x+d)^(1/2)/(e*x+d)^(1/2)-1/4*b*e^2*n*polylog(2 ,(-1-(1-e^2*x^2/d^2)^(1/2))/(1-(1-e^2*x^2/d^2)^(1/2)))*(1-e^2*x^2/d^2)^(1/ 2)/d^2/(-e*x+d)^(1/2)/(e*x+d)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.64 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.52 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\frac {\frac {b n \left (-d^2+e^2 x^2\right ) \left (2 d^3 \, _3F_2\left (\frac {3}{2},\frac {3}{2},\frac {3}{2};\frac {5}{2},\frac {5}{2};\frac {d^2}{e^2 x^2}\right )+9 e^2 x^2 \left (d \sqrt {1-\frac {d^2}{e^2 x^2}}-e x \arcsin \left (\frac {d}{e x}\right )\right ) (1+2 \log (x))\right )}{e^2 \sqrt {1-\frac {d^2}{e^2 x^2}} x^4 \sqrt {d-e x} \sqrt {d+e x}}-\frac {18 d \sqrt {d-e x} \sqrt {d+e x} \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{x^2}+18 e^2 \log (x) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )-18 e^2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+\sqrt {d-e x} \sqrt {d+e x}\right )}{36 d^3} \]
((b*n*(-d^2 + e^2*x^2)*(2*d^3*HypergeometricPFQ[{3/2, 3/2, 3/2}, {5/2, 5/2 }, d^2/(e^2*x^2)] + 9*e^2*x^2*(d*Sqrt[1 - d^2/(e^2*x^2)] - e*x*ArcSin[d/(e *x)])*(1 + 2*Log[x])))/(e^2*Sqrt[1 - d^2/(e^2*x^2)]*x^4*Sqrt[d - e*x]*Sqrt [d + e*x]) - (18*d*Sqrt[d - e*x]*Sqrt[d + e*x]*(a - b*n*Log[x] + b*Log[c*x ^n]))/x^2 + 18*e^2*Log[x]*(a - b*n*Log[x] + b*Log[c*x^n]) - 18*e^2*(a - b* n*Log[x] + b*Log[c*x^n])*Log[d + Sqrt[d - e*x]*Sqrt[d + e*x]])/(36*d^3)
Time = 1.00 (sec) , antiderivative size = 307, normalized size of antiderivative = 0.63, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {2787, 2792, 2009}
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.
| \(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 2787 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \int \frac {a+b \log \left (c x^n\right )}{x^3 \sqrt {1-\frac {e^2 x^2}{d^2}}}dx}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (-b n \int \left (-\frac {\text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right ) e^2}{2 d^2 x}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}}}{2 x^3}\right )dx-\frac {e^2 \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^2}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (-\frac {e^2 \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^2}-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-b n \left (-\frac {e^2 \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )^2}{4 d^2}-\frac {e^2 \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right )}{4 d^2}+\frac {e^2 \text {arctanh}\left (\sqrt {1-\frac {e^2 x^2}{d^2}}\right ) \log \left (\frac {2}{1-\sqrt {1-\frac {e^2 x^2}{d^2}}}\right )}{2 d^2}+\frac {e^2 \operatorname {PolyLog}\left (2,-\frac {\sqrt {1-\frac {e^2 x^2}{d^2}}+1}{1-\sqrt {1-\frac {e^2 x^2}{d^2}}}\right )}{4 d^2}+\frac {\sqrt {1-\frac {e^2 x^2}{d^2}}}{4 x^2}\right )\right )}{\sqrt {d-e x} \sqrt {d+e x}}\) |
(Sqrt[1 - (e^2*x^2)/d^2]*(-1/2*(Sqrt[1 - (e^2*x^2)/d^2]*(a + b*Log[c*x^n]) )/x^2 - (e^2*ArcTanh[Sqrt[1 - (e^2*x^2)/d^2]]*(a + b*Log[c*x^n]))/(2*d^2) - b*n*(Sqrt[1 - (e^2*x^2)/d^2]/(4*x^2) - (e^2*ArcTanh[Sqrt[1 - (e^2*x^2)/d ^2]])/(4*d^2) - (e^2*ArcTanh[Sqrt[1 - (e^2*x^2)/d^2]]^2)/(4*d^2) + (e^2*Ar cTanh[Sqrt[1 - (e^2*x^2)/d^2]]*Log[2/(1 - Sqrt[1 - (e^2*x^2)/d^2])])/(2*d^ 2) + (e^2*PolyLog[2, -((1 + Sqrt[1 - (e^2*x^2)/d^2])/(1 - Sqrt[1 - (e^2*x^ 2)/d^2]))])/(4*d^2))))/(Sqrt[d - e*x]*Sqrt[d + e*x])
3.4.12.3.1 Defintions of rubi rules used
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])
\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{3} \sqrt {-e x +d}\, \sqrt {e x +d}}d x\]
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{\sqrt {e x + d} \sqrt {-e x + d} x^{3}} \,d x } \]
integral(-(sqrt(e*x + d)*sqrt(-e*x + d)*b*log(c*x^n) + sqrt(e*x + d)*sqrt( -e*x + d)*a)/(e^2*x^5 - d^2*x^3), x)
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{\sqrt {e x + d} \sqrt {-e x + d} x^{3}} \,d x } \]
-1/2*a*(e^2*log(2*d^2/abs(x) + 2*sqrt(-e^2*x^2 + d^2)*d/abs(x))/d^3 + sqrt (-e^2*x^2 + d^2)/(d^2*x^2)) + b*integrate((log(c) + log(x^n))/(sqrt(e*x + d)*sqrt(-e*x + d)*x^3), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{\sqrt {e x + d} \sqrt {-e x + d} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \sqrt {d-e x} \sqrt {d+e x}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^3\,\sqrt {d+e\,x}\,\sqrt {d-e\,x}} \,d x \]