Integrand size = 23, antiderivative size = 221 \[ \int \frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {b n \sqrt {d+e x}}{x}-\frac {b e n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}+\frac {b e n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {2 b e n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}-\frac {b e n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}} \]
-b*e*n*arctanh((e*x+d)^(1/2)/d^(1/2))/d^(1/2)+b*e*n*arctanh((e*x+d)^(1/2)/ d^(1/2))^2/d^(1/2)-e*arctanh((e*x+d)^(1/2)/d^(1/2))*(a+b*ln(c*x^n))/d^(1/2 )-2*b*e*n*arctanh((e*x+d)^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)-(e*x+d)^(1/ 2)))/d^(1/2)-b*e*n*polylog(2,1-2*d^(1/2)/(d^(1/2)-(e*x+d)^(1/2)))/d^(1/2)- b*n*(e*x+d)^(1/2)/x-(a+b*ln(c*x^n))*(e*x+d)^(1/2)/x
Time = 0.20 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.77 \[ \int \frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {4 a \sqrt {d} \sqrt {d+e x}+4 b \sqrt {d} n \sqrt {d+e x}+4 b e n x \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+4 b \sqrt {d} \sqrt {d+e x} \log \left (c x^n\right )-2 a e x \log \left (\sqrt {d}-\sqrt {d+e x}\right )-2 b e x \log \left (c x^n\right ) \log \left (\sqrt {d}-\sqrt {d+e x}\right )+b e n x \log ^2\left (\sqrt {d}-\sqrt {d+e x}\right )+2 a e x \log \left (\sqrt {d}+\sqrt {d+e x}\right )+2 b e x \log \left (c x^n\right ) \log \left (\sqrt {d}+\sqrt {d+e x}\right )-b e n x \log ^2\left (\sqrt {d}+\sqrt {d+e x}\right )-2 b e n x \log \left (\sqrt {d}+\sqrt {d+e x}\right ) \log \left (\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )+2 b e n x \log \left (\sqrt {d}-\sqrt {d+e x}\right ) \log \left (\frac {1}{2} \left (1+\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )+2 b e n x \operatorname {PolyLog}\left (2,\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )-2 b e n x \operatorname {PolyLog}\left (2,\frac {1}{2} \left (1+\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )}{4 \sqrt {d} x} \]
-1/4*(4*a*Sqrt[d]*Sqrt[d + e*x] + 4*b*Sqrt[d]*n*Sqrt[d + e*x] + 4*b*e*n*x* ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 4*b*Sqrt[d]*Sqrt[d + e*x]*Log[c*x^n] - 2* a*e*x*Log[Sqrt[d] - Sqrt[d + e*x]] - 2*b*e*x*Log[c*x^n]*Log[Sqrt[d] - Sqrt [d + e*x]] + b*e*n*x*Log[Sqrt[d] - Sqrt[d + e*x]]^2 + 2*a*e*x*Log[Sqrt[d] + Sqrt[d + e*x]] + 2*b*e*x*Log[c*x^n]*Log[Sqrt[d] + Sqrt[d + e*x]] - b*e*n *x*Log[Sqrt[d] + Sqrt[d + e*x]]^2 - 2*b*e*n*x*Log[Sqrt[d] + Sqrt[d + e*x]] *Log[1/2 - Sqrt[d + e*x]/(2*Sqrt[d])] + 2*b*e*n*x*Log[Sqrt[d] - Sqrt[d + e *x]]*Log[(1 + Sqrt[d + e*x]/Sqrt[d])/2] + 2*b*e*n*x*PolyLog[2, 1/2 - Sqrt[ d + e*x]/(2*Sqrt[d])] - 2*b*e*n*x*PolyLog[2, (1 + Sqrt[d + e*x]/Sqrt[d])/2 ])/(Sqrt[d]*x)
Time = 0.54 (sec) , antiderivative size = 215, normalized size of antiderivative = 0.97, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2792, 25, 2010, 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.
\(\displaystyle \int \frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 2792 |
\(\displaystyle -b n \int -\frac {\frac {e x \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}+\sqrt {d+e x}}{x^2}dx-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle b n \int \frac {\frac {e x \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}+\sqrt {d+e x}}{x^2}dx-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle b n \int \left (\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d} x}+\frac {\sqrt {d+e x}}{x^2}\right )dx-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d}}-\frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x}+b n \left (\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{\sqrt {d}}-\frac {2 e \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}-\frac {e \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}-\frac {\sqrt {d+e x}}{x}\right )\) |
-((Sqrt[d + e*x]*(a + b*Log[c*x^n]))/x) - (e*ArcTanh[Sqrt[d + e*x]/Sqrt[d] ]*(a + b*Log[c*x^n]))/Sqrt[d] + b*n*(-(Sqrt[d + e*x]/x) - (e*ArcTanh[Sqrt[ d + e*x]/Sqrt[d]])/Sqrt[d] + (e*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]^2)/Sqrt[d] - (2*e*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])/Sqrt[d] - (e*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])] )/Sqrt[d])
3.2.35.3.1 Defintions of rubi rules used
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
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])
\[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e x +d}}{x^{2}}d x\]
\[ \int \frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {\sqrt {e x + d} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{2}} \,d x } \]
\[ \int \frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \sqrt {d + e x}}{x^{2}}\, dx \]
\[ \int \frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {\sqrt {e x + d} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{2}} \,d x } \]
1/2*(e*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/sqrt(d) - 2*sqrt(e*x + d)/x)*a + b*integrate(sqrt(e*x + d)*(log(c) + log(x^n))/x^2, x)
\[ \int \frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int { \frac {\sqrt {e x + d} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\int \frac {\left (a+b\,\ln \left (c\,x^n\right )\right )\,\sqrt {d+e\,x}}{x^2} \,d x \]