Integrand size = 25, antiderivative size = 252 \[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {b n \sqrt {d+e x^2}}{4 x^2}-\frac {b e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{4 \sqrt {d}}+\frac {b e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 \sqrt {d}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 \sqrt {d}}-\frac {b e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 \sqrt {d}}-\frac {b e n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{4 \sqrt {d}} \]
-1/4*b*e*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(1/2)+1/4*b*e*n*arctanh((e*x ^2+d)^(1/2)/d^(1/2))^2/d^(1/2)-1/2*e*arctanh((e*x^2+d)^(1/2)/d^(1/2))*(a+b *ln(c*x^n))/d^(1/2)-1/2*b*e*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))*ln(2*d^(1/2 )/(d^(1/2)-(e*x^2+d)^(1/2)))/d^(1/2)-1/4*b*e*n*polylog(2,1-2*d^(1/2)/(d^(1 /2)-(e*x^2+d)^(1/2)))/d^(1/2)-1/4*b*n*(e*x^2+d)^(1/2)/x^2-1/2*(a+b*ln(c*x^ n))*(e*x^2+d)^(1/2)/x^2
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.30 (sec) , antiderivative size = 303, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\frac {-2 b \sqrt {d} n \sqrt {d+e x^2} \, _3F_2\left (\frac {1}{2},\frac {1}{2},\frac {1}{2};\frac {3}{2},\frac {3}{2};-\frac {d}{e x^2}\right )-b \sqrt {e} n x \sqrt {d+e x^2} \text {arcsinh}\left (\frac {\sqrt {d}}{\sqrt {e} x}\right ) (1+2 \log (x))+\sqrt {1+\frac {d}{e x^2}} \left (-2 a \sqrt {d} \sqrt {d+e x^2}-b \sqrt {d} n \sqrt {d+e x^2}-2 b e n x^2 \log ^2(x)-2 a e x^2 \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )+2 e x^2 \log (x) \left (a+b \log \left (c x^n\right )+b n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )\right )-2 b \log \left (c x^n\right ) \left (\sqrt {d} \sqrt {d+e x^2}+e x^2 \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )\right )\right )}{4 \sqrt {d} \sqrt {1+\frac {d}{e x^2}} x^2} \]
(-2*b*Sqrt[d]*n*Sqrt[d + e*x^2]*HypergeometricPFQ[{1/2, 1/2, 1/2}, {3/2, 3 /2}, -(d/(e*x^2))] - b*Sqrt[e]*n*x*Sqrt[d + e*x^2]*ArcSinh[Sqrt[d]/(Sqrt[e ]*x)]*(1 + 2*Log[x]) + Sqrt[1 + d/(e*x^2)]*(-2*a*Sqrt[d]*Sqrt[d + e*x^2] - b*Sqrt[d]*n*Sqrt[d + e*x^2] - 2*b*e*n*x^2*Log[x]^2 - 2*a*e*x^2*Log[d + Sq rt[d]*Sqrt[d + e*x^2]] + 2*e*x^2*Log[x]*(a + b*Log[c*x^n] + b*n*Log[d + Sq rt[d]*Sqrt[d + e*x^2]]) - 2*b*Log[c*x^n]*(Sqrt[d]*Sqrt[d + e*x^2] + e*x^2* Log[d + Sqrt[d]*Sqrt[d + e*x^2]])))/(4*Sqrt[d]*Sqrt[1 + d/(e*x^2)]*x^2)
Time = 0.66 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {2792, 27, 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.
\(\displaystyle \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx\) |
\(\Big \downarrow \) 2792 |
\(\displaystyle -b n \int -\frac {\frac {e \text {arctanh}\left (\frac {\sqrt {e x^2+d}}{\sqrt {d}}\right ) x^2}{\sqrt {d}}+\sqrt {e x^2+d}}{2 x^3}dx-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 \sqrt {d}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{2} b n \int \frac {\frac {e \text {arctanh}\left (\frac {\sqrt {e x^2+d}}{\sqrt {d}}\right ) x^2}{\sqrt {d}}+\sqrt {e x^2+d}}{x^3}dx-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 \sqrt {d}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle \frac {1}{2} b n \int \left (\frac {e \text {arctanh}\left (\frac {\sqrt {e x^2+d}}{\sqrt {d}}\right )}{\sqrt {d} x}+\frac {\sqrt {e x^2+d}}{x^3}\right )dx-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 \sqrt {d}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 \sqrt {d}}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {1}{2} b n \left (\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{2 \sqrt {d}}-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{2 \sqrt {d}}-\frac {e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{\sqrt {d}}-\frac {e \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^2+d}}\right )}{2 \sqrt {d}}-\frac {\sqrt {d+e x^2}}{2 x^2}\right )\) |
-1/2*(Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/x^2 - (e*ArcTanh[Sqrt[d + e*x^2] /Sqrt[d]]*(a + b*Log[c*x^n]))/(2*Sqrt[d]) + (b*n*(-1/2*Sqrt[d + e*x^2]/x^2 - (e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(2*Sqrt[d]) + (e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]^2)/(2*Sqrt[d]) - (e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*Log[( 2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/Sqrt[d] - (e*PolyLog[2, 1 - (2*Sq rt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*Sqrt[d])))/2
3.3.55.3.1 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[(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])
\[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e \,x^{2}+d}}{x^{3}}d x\]
\[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {\sqrt {e x^{2} + d} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{3}} \,d x } \]
\[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \sqrt {d + e x^{2}}}{x^{3}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {\sqrt {e x^{2} + d} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^3} \,d x \]