Integrand size = 25, antiderivative size = 176 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=-\frac {b n \sqrt {d+e x^2}}{9 d^2 x^3}+\frac {14 b e n \sqrt {d+e x^2}}{9 d^3 x}-\frac {8 b e^{3/2} n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^3}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}} \]
-8/3*b*e^(3/2)*n*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/d^3+1/3*(-a-b*ln(c*x^n ))/d/x^3/(e*x^2+d)^(1/2)+4/3*e*(a+b*ln(c*x^n))/d^2/x/(e*x^2+d)^(1/2)+8/3*e ^2*x*(a+b*ln(c*x^n))/d^3/(e*x^2+d)^(1/2)-1/9*b*n*(e*x^2+d)^(1/2)/d^2/x^3+1 4/9*b*e*n*(e*x^2+d)^(1/2)/d^3/x
Time = 0.11 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.82 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\frac {-3 a d^2-b d^2 n+12 a d e x^2+13 b d e n x^2+24 a e^2 x^4+14 b e^2 n x^4-3 b \left (d^2-4 d e x^2-8 e^2 x^4\right ) \log \left (c x^n\right )-24 b e^{3/2} n x^3 \sqrt {d+e x^2} \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{9 d^3 x^3 \sqrt {d+e x^2}} \]
(-3*a*d^2 - b*d^2*n + 12*a*d*e*x^2 + 13*b*d*e*n*x^2 + 24*a*e^2*x^4 + 14*b* e^2*n*x^4 - 3*b*(d^2 - 4*d*e*x^2 - 8*e^2*x^4)*Log[c*x^n] - 24*b*e^(3/2)*n* x^3*Sqrt[d + e*x^2]*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(9*d^3*x^3*Sqrt[d + e*x^2])
Time = 0.40 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.98, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2792, 27, 1588, 27, 358, 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.
| \(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {-8 e^2 x^4-4 d e x^2+d^2}{3 d^3 x^4 \sqrt {e x^2+d}}dx+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b n \int \frac {-8 e^2 x^4-4 d e x^2+d^2}{x^4 \sqrt {e x^2+d}}dx}{3 d^3}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 1588 |
| \(\displaystyle \frac {b n \left (-\frac {\int \frac {2 d e \left (12 e x^2+7 d\right )}{x^2 \sqrt {e x^2+d}}dx}{3 d}-\frac {d \sqrt {d+e x^2}}{3 x^3}\right )}{3 d^3}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b n \left (-\frac {2}{3} e \int \frac {12 e x^2+7 d}{x^2 \sqrt {e x^2+d}}dx-\frac {d \sqrt {d+e x^2}}{3 x^3}\right )}{3 d^3}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 358 |
| \(\displaystyle \frac {b n \left (-\frac {2}{3} e \left (12 e \int \frac {1}{\sqrt {e x^2+d}}dx-\frac {7 \sqrt {d+e x^2}}{x}\right )-\frac {d \sqrt {d+e x^2}}{3 x^3}\right )}{3 d^3}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {b n \left (-\frac {2}{3} e \left (12 e \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}-\frac {7 \sqrt {d+e x^2}}{x}\right )-\frac {d \sqrt {d+e x^2}}{3 x^3}\right )}{3 d^3}+\frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {8 e^2 x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {4 e \left (a+b \log \left (c x^n\right )\right )}{3 d^2 x \sqrt {d+e x^2}}-\frac {a+b \log \left (c x^n\right )}{3 d x^3 \sqrt {d+e x^2}}+\frac {b n \left (-\frac {2}{3} e \left (12 \sqrt {e} \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {7 \sqrt {d+e x^2}}{x}\right )-\frac {d \sqrt {d+e x^2}}{3 x^3}\right )}{3 d^3}\) |
(b*n*(-1/3*(d*Sqrt[d + e*x^2])/x^3 - (2*e*((-7*Sqrt[d + e*x^2])/x + 12*Sqr t[e]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]))/3))/(3*d^3) - (a + b*Log[c*x^n ])/(3*d*x^3*Sqrt[d + e*x^2]) + (4*e*(a + b*Log[c*x^n]))/(3*d^2*x*Sqrt[d + e*x^2]) + (8*e^2*x*(a + b*Log[c*x^n]))/(3*d^3*Sqrt[d + e*x^2])
3.3.95.3.1 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[((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[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c _.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x, x]}, Simp[R*(f*x)^(m + 1)*((d + e*x^2)^(q + 1)/(d*f*(m + 1))), x] + Simp[1/(d*f ^2*(m + 1)) Int[(f*x)^(m + 2)*(d + e*x^2)^q*ExpandToSum[d*f*(m + 1)*(Qx/x ) - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q}, x] && Ne Q[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[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])
\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{4} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]
Time = 0.38 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.10 \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\left [\frac {12 \, {\left (b e^{2} n x^{5} + b d e n x^{3}\right )} \sqrt {e} \log \left (-2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) + {\left (2 \, {\left (7 \, b e^{2} n + 12 \, a e^{2}\right )} x^{4} - b d^{2} n - 3 \, a d^{2} + {\left (13 \, b d e n + 12 \, a d e\right )} x^{2} + 3 \, {\left (8 \, b e^{2} x^{4} + 4 \, b d e x^{2} - b d^{2}\right )} \log \left (c\right ) + 3 \, {\left (8 \, b e^{2} n x^{4} + 4 \, b d e n x^{2} - b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{9 \, {\left (d^{3} e x^{5} + d^{4} x^{3}\right )}}, \frac {24 \, {\left (b e^{2} n x^{5} + b d e n x^{3}\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (2 \, {\left (7 \, b e^{2} n + 12 \, a e^{2}\right )} x^{4} - b d^{2} n - 3 \, a d^{2} + {\left (13 \, b d e n + 12 \, a d e\right )} x^{2} + 3 \, {\left (8 \, b e^{2} x^{4} + 4 \, b d e x^{2} - b d^{2}\right )} \log \left (c\right ) + 3 \, {\left (8 \, b e^{2} n x^{4} + 4 \, b d e n x^{2} - b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{9 \, {\left (d^{3} e x^{5} + d^{4} x^{3}\right )}}\right ] \]
[1/9*(12*(b*e^2*n*x^5 + b*d*e*n*x^3)*sqrt(e)*log(-2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x - d) + (2*(7*b*e^2*n + 12*a*e^2)*x^4 - b*d^2*n - 3*a*d^2 + ( 13*b*d*e*n + 12*a*d*e)*x^2 + 3*(8*b*e^2*x^4 + 4*b*d*e*x^2 - b*d^2)*log(c) + 3*(8*b*e^2*n*x^4 + 4*b*d*e*n*x^2 - b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d^ 3*e*x^5 + d^4*x^3), 1/9*(24*(b*e^2*n*x^5 + b*d*e*n*x^3)*sqrt(-e)*arctan(sq rt(-e)*x/sqrt(e*x^2 + d)) + (2*(7*b*e^2*n + 12*a*e^2)*x^4 - b*d^2*n - 3*a* d^2 + (13*b*d*e*n + 12*a*d*e)*x^2 + 3*(8*b*e^2*x^4 + 4*b*d*e*x^2 - b*d^2)* log(c) + 3*(8*b*e^2*n*x^4 + 4*b*d*e*n*x^2 - b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d^3*e*x^5 + d^4*x^3)]
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\int \frac {a + b \log {\left (c x^{n} \right )}}{x^{4} \left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \]
Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} x^{4}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^4 \left (d+e x^2\right )^{3/2}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^4\,{\left (e\,x^2+d\right )}^{3/2}} \,d x \]