Integrand size = 25, antiderivative size = 337 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\frac {b e n}{3 d^3 \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{4 d^3 x^2}-\frac {31 b e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{12 d^{7/2}}-\frac {5 b e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{7/2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{6 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )^{3/2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{2 d^3 \sqrt {d+e x^2}}+\frac {5 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{7/2}}+\frac {5 b e n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{7/2}}+\frac {5 b e n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{4 d^{7/2}} \]
-31/12*b*e*n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(7/2)-5/4*b*e*n*arctanh((e *x^2+d)^(1/2)/d^(1/2))^2/d^(7/2)-5/6*e*(a+b*ln(c*x^n))/d^2/(e*x^2+d)^(3/2) +1/2*(-a-b*ln(c*x^n))/d/x^2/(e*x^2+d)^(3/2)+5/2*e*arctanh((e*x^2+d)^(1/2)/ d^(1/2))*(a+b*ln(c*x^n))/d^(7/2)+5/2*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^(7/2)+5/4*b*e*n*polylog(2,1-2* d^(1/2)/(d^(1/2)-(e*x^2+d)^(1/2)))/d^(7/2)+1/3*b*e*n/d^3/(e*x^2+d)^(1/2)-5 /2*e*(a+b*ln(c*x^n))/d^3/(e*x^2+d)^(1/2)-1/4*b*n*(e*x^2+d)^(1/2)/d^3/x^2
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.19 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.67 \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\frac {b n \sqrt {1+\frac {d}{e x^2}} \left (5 \, _3F_2\left (\frac {7}{2},\frac {7}{2},\frac {7}{2};\frac {9}{2},\frac {9}{2};-\frac {d}{e x^2}\right )-7 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {7}{2},\frac {9}{2},-\frac {d}{e x^2}\right ) (1+2 \log (x))\right )}{98 e^2 x^6 \sqrt {d+e x^2}}-\frac {\left (3 d^2+20 d e x^2+15 e^2 x^4\right ) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{6 d^3 x^2 \left (d+e x^2\right )^{3/2}}-\frac {5 e \log (x) \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{2 d^{7/2}}+\frac {5 e \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{2 d^{7/2}} \]
(b*n*Sqrt[1 + d/(e*x^2)]*(5*HypergeometricPFQ[{7/2, 7/2, 7/2}, {9/2, 9/2}, -(d/(e*x^2))] - 7*Hypergeometric2F1[5/2, 7/2, 9/2, -(d/(e*x^2))]*(1 + 2*L og[x])))/(98*e^2*x^6*Sqrt[d + e*x^2]) - ((3*d^2 + 20*d*e*x^2 + 15*e^2*x^4) *(a - b*n*Log[x] + b*Log[c*x^n]))/(6*d^3*x^2*(d + e*x^2)^(3/2)) - (5*e*Log [x]*(a - b*n*Log[x] + b*Log[c*x^n]))/(2*d^(7/2)) + (5*e*(a - b*n*Log[x] + b*Log[c*x^n])*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(2*d^(7/2))
Time = 0.75 (sec) , antiderivative size = 330, normalized size of antiderivative = 0.98, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {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.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 2792 |
\(\displaystyle -b n \int \left (\frac {5 e \text {arctanh}\left (\frac {\sqrt {e x^2+d}}{\sqrt {d}}\right )}{2 d^{7/2} x}-\frac {15 e^2 x^4+20 d e x^2+3 d^2}{6 d^3 x^3 \left (e x^2+d\right )^{3/2}}\right )dx+\frac {5 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{2 d^3 \sqrt {d+e x^2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{6 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {5 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^{7/2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{2 d^3 \sqrt {d+e x^2}}-\frac {5 e \left (a+b \log \left (c x^n\right )\right )}{6 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{2 d x^2 \left (d+e x^2\right )^{3/2}}-b n \left (\frac {5 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )^2}{4 d^{7/2}}+\frac {31 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{12 d^{7/2}}-\frac {5 e \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^2}}\right )}{2 d^{7/2}}-\frac {5 e \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^2+d}}\right )}{4 d^{7/2}}+\frac {\sqrt {d+e x^2}}{4 d^3 x^2}-\frac {e}{3 d^3 \sqrt {d+e x^2}}\right )\) |
(-5*e*(a + b*Log[c*x^n]))/(6*d^2*(d + e*x^2)^(3/2)) - (a + b*Log[c*x^n])/( 2*d*x^2*(d + e*x^2)^(3/2)) - (5*e*(a + b*Log[c*x^n]))/(2*d^3*Sqrt[d + e*x^ 2]) + (5*e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*(a + b*Log[c*x^n]))/(2*d^(7/2) ) - b*n*(-1/3*e/(d^3*Sqrt[d + e*x^2]) + Sqrt[d + e*x^2]/(4*d^3*x^2) + (31* e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(12*d^(7/2)) + (5*e*ArcTanh[Sqrt[d + e *x^2]/Sqrt[d]]^2)/(4*d^(7/2)) - (5*e*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]*Log[ (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(2*d^(7/2)) - (5*e*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^2])])/(4*d^(7/2)))
3.4.2.3.1 Defintions of rubi rules used
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])
\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{3} \left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{3}} \,d x } \]
integral((sqrt(e*x^2 + d)*b*log(c*x^n) + sqrt(e*x^2 + d)*a)/(e^3*x^9 + 3*d *e^2*x^7 + 3*d^2*e*x^5 + d^3*x^3), x)
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{3}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^3 \left (d+e x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^3\,{\left (e\,x^2+d\right )}^{5/2}} \,d x \]