Integrand size = 25, antiderivative size = 144 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d+e x^2}} \, dx=\frac {2 b e n \sqrt {d+e x^2}}{3 d^2 x}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 d^2 x^3}-\frac {2 b e^{3/2} n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^2}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x} \]
-1/9*b*n*(e*x^2+d)^(3/2)/d^2/x^3-2/3*b*e^(3/2)*n*arctanh(x*e^(1/2)/(e*x^2+ d)^(1/2))/d^2+2/3*b*e*n*(e*x^2+d)^(1/2)/d^2/x-1/3*(a+b*ln(c*x^n))*(e*x^2+d )^(1/2)/d/x^3+2/3*e*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/d^2/x
Time = 0.10 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.76 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d+e x^2}} \, dx=\frac {\sqrt {d+e x^2} \left (-3 a d-b d n+6 a e x^2+5 b e n x^2\right )-3 b \left (d-2 e x^2\right ) \sqrt {d+e x^2} \log \left (c x^n\right )-6 b e^{3/2} n x^3 \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{9 d^2 x^3} \]
(Sqrt[d + e*x^2]*(-3*a*d - b*d*n + 6*a*e*x^2 + 5*b*e*n*x^2) - 3*b*(d - 2*e *x^2)*Sqrt[d + e*x^2]*Log[c*x^n] - 6*b*e^(3/2)*n*x^3*Log[e*x + Sqrt[e]*Sqr t[d + e*x^2]])/(9*d^2*x^3)
Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {2792, 27, 358, 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.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d+e x^2}} \, dx\) |
\(\Big \downarrow \) 2792 |
\(\displaystyle -b n \int -\frac {\left (d-2 e x^2\right ) \sqrt {e x^2+d}}{3 d^2 x^4}dx+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b n \int \frac {\left (d-2 e x^2\right ) \sqrt {e x^2+d}}{x^4}dx}{3 d^2}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}\) |
\(\Big \downarrow \) 358 |
\(\displaystyle \frac {b n \left (-2 e \int \frac {\sqrt {e x^2+d}}{x^2}dx-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )}{3 d^2}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {b n \left (-2 e \left (e \int \frac {1}{\sqrt {e x^2+d}}dx-\frac {\sqrt {d+e x^2}}{x}\right )-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )}{3 d^2}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {b n \left (-2 e \left (e \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}-\frac {\sqrt {d+e x^2}}{x}\right )-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )}{3 d^2}+\frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2 e \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}+\frac {b n \left (-2 e \left (\sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\sqrt {d+e x^2}}{x}\right )-\frac {\left (d+e x^2\right )^{3/2}}{3 x^3}\right )}{3 d^2}\) |
(b*n*(-1/3*(d + e*x^2)^(3/2)/x^3 - 2*e*(-(Sqrt[d + e*x^2]/x) + Sqrt[e]*Arc Tanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])))/(3*d^2) - (Sqrt[d + e*x^2]*(a + b*Log [c*x^n]))/(3*d*x^3) + (2*e*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/(3*d^2*x)
3.3.84.3.1 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])
\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{4} \sqrt {e \,x^{2}+d}}d x\]
Time = 0.36 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.55 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d+e x^2}} \, dx=\left [\frac {3 \, b e^{\frac {3}{2}} n x^{3} \log \left (-2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - {\left (b d n - {\left (5 \, b e n + 6 \, a e\right )} x^{2} + 3 \, a d - 3 \, {\left (2 \, b e x^{2} - b d\right )} \log \left (c\right ) - 3 \, {\left (2 \, b e n x^{2} - b d n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{9 \, d^{2} x^{3}}, \frac {6 \, b \sqrt {-e} e n x^{3} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) - {\left (b d n - {\left (5 \, b e n + 6 \, a e\right )} x^{2} + 3 \, a d - 3 \, {\left (2 \, b e x^{2} - b d\right )} \log \left (c\right ) - 3 \, {\left (2 \, b e n x^{2} - b d n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{9 \, d^{2} x^{3}}\right ] \]
[1/9*(3*b*e^(3/2)*n*x^3*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) - (b*d*n - (5*b*e*n + 6*a*e)*x^2 + 3*a*d - 3*(2*b*e*x^2 - b*d)*log(c) - 3*(2 *b*e*n*x^2 - b*d*n)*log(x))*sqrt(e*x^2 + d))/(d^2*x^3), 1/9*(6*b*sqrt(-e)* e*n*x^3*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) - (b*d*n - (5*b*e*n + 6*a*e)*x^ 2 + 3*a*d - 3*(2*b*e*x^2 - b*d)*log(c) - 3*(2*b*e*n*x^2 - b*d*n)*log(x))*s qrt(e*x^2 + d))/(d^2*x^3)]
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d+e x^2}} \, dx=\int \frac {a + b \log {\left (c x^{n} \right )}}{x^{4} \sqrt {d + e x^{2}}}\, dx \]
Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d+e x^2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d+e x^2}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{\sqrt {e x^{2} + d} x^{4}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \sqrt {d+e x^2}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^4\,\sqrt {e\,x^2+d}} \,d x \]