Integrand size = 25, antiderivative size = 170 \[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\frac {2 b e^2 n \sqrt {d+e x^2}}{15 d^2 x}+\frac {2 b e n \left (d+e x^2\right )^{3/2}}{45 d^2 x^3}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 d^2 x^5}-\frac {2 b e^{5/2} n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{15 d^2}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {2 e \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3} \]
2/45*b*e*n*(e*x^2+d)^(3/2)/d^2/x^3-1/25*b*n*(e*x^2+d)^(5/2)/d^2/x^5-2/15*b *e^(5/2)*n*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/d^2-1/5*(e*x^2+d)^(3/2)*(a+b *ln(c*x^n))/d/x^5+2/15*e*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/d^2/x^3+2/15*b*e^ 2*n*(e*x^2+d)^(1/2)/d^2/x
Time = 0.13 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.85 \[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {\sqrt {d+e x^2} \left (b n \left (9 d^2+8 d e x^2-31 e^2 x^4\right )+15 a \left (3 d^2+d e x^2-2 e^2 x^4\right )\right )+15 b \sqrt {d+e x^2} \left (3 d^2+d e x^2-2 e^2 x^4\right ) \log \left (c x^n\right )+30 b e^{5/2} n x^5 \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{225 d^2 x^5} \]
-1/225*(Sqrt[d + e*x^2]*(b*n*(9*d^2 + 8*d*e*x^2 - 31*e^2*x^4) + 15*a*(3*d^ 2 + d*e*x^2 - 2*e^2*x^4)) + 15*b*Sqrt[d + e*x^2]*(3*d^2 + d*e*x^2 - 2*e^2* x^4)*Log[c*x^n] + 30*b*e^(5/2)*n*x^5*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/( d^2*x^5)
Time = 0.35 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.93, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2792, 27, 358, 247, 247, 224, 219}
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.
| \(\displaystyle \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {\left (3 d-2 e x^2\right ) \left (e x^2+d\right )^{3/2}}{15 d^2 x^6}dx+\frac {2 e \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b n \int \frac {\left (3 d-2 e x^2\right ) \left (e x^2+d\right )^{3/2}}{x^6}dx}{15 d^2}+\frac {2 e \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\) |
| \(\Big \downarrow \) 358 |
| \(\displaystyle \frac {b n \left (-2 e \int \frac {\left (e x^2+d\right )^{3/2}}{x^4}dx-\frac {3 \left (d+e x^2\right )^{5/2}}{5 x^5}\right )}{15 d^2}+\frac {2 e \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {b n \left (-2 e \left (e \int \frac {\sqrt {e x^2+d}}{x^2}dx-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )-\frac {3 \left (d+e x^2\right )^{5/2}}{5 x^5}\right )}{15 d^2}+\frac {2 e \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\) |
| \(\Big \downarrow \) 247 |
| \(\displaystyle \frac {b n \left (-2 e \left (e \left (e \int \frac {1}{\sqrt {e x^2+d}}dx-\frac {\sqrt {d+e x^2}}{x}\right )-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )-\frac {3 \left (d+e x^2\right )^{5/2}}{5 x^5}\right )}{15 d^2}+\frac {2 e \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {b n \left (-2 e \left (e \left (e \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}-\frac {\sqrt {d+e x^2}}{x}\right )-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )-\frac {3 \left (d+e x^2\right )^{5/2}}{5 x^5}\right )}{15 d^2}+\frac {2 e \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 e \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{15 d^2 x^3}-\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {b n \left (-2 e \left (e \left (\sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\sqrt {d+e x^2}}{x}\right )-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )-\frac {3 \left (d+e x^2\right )^{5/2}}{5 x^5}\right )}{15 d^2}\) |
(b*n*((-3*(d + e*x^2)^(5/2))/(5*x^5) - 2*e*(-1/3*(d + e*x^2)^(3/2)/x^3 + e *(-(Sqrt[d + e*x^2]/x) + Sqrt[e]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]))))/ (15*d^2) - ((d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/(5*d*x^5) + (2*e*(d + e* x^2)^(3/2)*(a + b*Log[c*x^n]))/(15*d^2*x^3)
3.3.61.3.1 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[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
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_.))*((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])
\[\int \frac {\left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e \,x^{2}+d}}{x^{6}}d x\]
Time = 0.35 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.90 \[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\left [\frac {15 \, b e^{\frac {5}{2}} n x^{5} \log \left (-2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) + {\left ({\left (31 \, b e^{2} n + 30 \, a e^{2}\right )} x^{4} - 9 \, b d^{2} n - 45 \, a d^{2} - {\left (8 \, b d e n + 15 \, a d e\right )} x^{2} + 15 \, {\left (2 \, b e^{2} x^{4} - b d e x^{2} - 3 \, b d^{2}\right )} \log \left (c\right ) + 15 \, {\left (2 \, b e^{2} n x^{4} - b d e n x^{2} - 3 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{225 \, d^{2} x^{5}}, \frac {30 \, b \sqrt {-e} e^{2} n x^{5} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left ({\left (31 \, b e^{2} n + 30 \, a e^{2}\right )} x^{4} - 9 \, b d^{2} n - 45 \, a d^{2} - {\left (8 \, b d e n + 15 \, a d e\right )} x^{2} + 15 \, {\left (2 \, b e^{2} x^{4} - b d e x^{2} - 3 \, b d^{2}\right )} \log \left (c\right ) + 15 \, {\left (2 \, b e^{2} n x^{4} - b d e n x^{2} - 3 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{225 \, d^{2} x^{5}}\right ] \]
[1/225*(15*b*e^(5/2)*n*x^5*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + ((31*b*e^2*n + 30*a*e^2)*x^4 - 9*b*d^2*n - 45*a*d^2 - (8*b*d*e*n + 15*a *d*e)*x^2 + 15*(2*b*e^2*x^4 - b*d*e*x^2 - 3*b*d^2)*log(c) + 15*(2*b*e^2*n* x^4 - b*d*e*n*x^2 - 3*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d^2*x^5), 1/225*( 30*b*sqrt(-e)*e^2*n*x^5*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + ((31*b*e^2*n + 30*a*e^2)*x^4 - 9*b*d^2*n - 45*a*d^2 - (8*b*d*e*n + 15*a*d*e)*x^2 + 15*( 2*b*e^2*x^4 - b*d*e*x^2 - 3*b*d^2)*log(c) + 15*(2*b*e^2*n*x^4 - b*d*e*n*x^ 2 - 3*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d^2*x^5)]
\[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \sqrt {d + e x^{2}}}{x^{6}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\int { \frac {\sqrt {e x^{2} + d} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{6}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\int \frac {\sqrt {e\,x^2+d}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^6} \,d x \]