Integrand size = 25, antiderivative size = 166 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^{5/2}} \, dx=-\frac {b n}{d^2 x \sqrt {d+e x^2}}-\frac {2 b e n x}{3 d^3 \sqrt {d+e x^2}}+\frac {8 b \sqrt {e} n \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 d^3}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}} \]
(-a-b*ln(c*x^n))/d/x/(e*x^2+d)^(3/2)-4/3*e*x*(a+b*ln(c*x^n))/d^2/(e*x^2+d) ^(3/2)+8/3*b*n*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))*e^(1/2)/d^3-b*n/d^2/x/(e *x^2+d)^(1/2)-2/3*b*e*n*x/d^3/(e*x^2+d)^(1/2)-8/3*e*x*(a+b*ln(c*x^n))/d^3/ (e*x^2+d)^(1/2)
Time = 0.13 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.87 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^{5/2}} \, dx=\frac {-3 a d^2-3 b d^2 n-12 a d e x^2-5 b d e n x^2-8 a e^2 x^4-2 b e^2 n x^4-b \left (3 d^2+12 d e x^2+8 e^2 x^4\right ) \log \left (c x^n\right )+8 b \sqrt {e} n x \left (d+e x^2\right )^{3/2} \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{3 d^3 x \left (d+e x^2\right )^{3/2}} \]
(-3*a*d^2 - 3*b*d^2*n - 12*a*d*e*x^2 - 5*b*d*e*n*x^2 - 8*a*e^2*x^4 - 2*b*e ^2*n*x^4 - b*(3*d^2 + 12*d*e*x^2 + 8*e^2*x^4)*Log[c*x^n] + 8*b*Sqrt[e]*n*x *(d + e*x^2)^(3/2)*Log[e*x + Sqrt[e]*Sqrt[d + e*x^2]])/(3*d^3*x*(d + e*x^2 )^(3/2))
Time = 0.38 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.97, 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, 298, 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.
| \(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {8 e^2 x^4+12 d e x^2+3 d^2}{3 d^3 x^2 \left (e x^2+d\right )^{3/2}}dx-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b n \int \frac {8 e^2 x^4+12 d e x^2+3 d^2}{x^2 \left (e x^2+d\right )^{3/2}}dx}{3 d^3}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1588 |
| \(\displaystyle \frac {b n \left (-\frac {\int -\frac {2 d e \left (4 e x^2+3 d\right )}{\left (e x^2+d\right )^{3/2}}dx}{d}-\frac {3 d}{x \sqrt {d+e x^2}}\right )}{3 d^3}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b n \left (2 e \int \frac {4 e x^2+3 d}{\left (e x^2+d\right )^{3/2}}dx-\frac {3 d}{x \sqrt {d+e x^2}}\right )}{3 d^3}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 298 |
| \(\displaystyle \frac {b n \left (2 e \left (4 \int \frac {1}{\sqrt {e x^2+d}}dx-\frac {x}{\sqrt {d+e x^2}}\right )-\frac {3 d}{x \sqrt {d+e x^2}}\right )}{3 d^3}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {b n \left (2 e \left (4 \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}-\frac {x}{\sqrt {d+e x^2}}\right )-\frac {3 d}{x \sqrt {d+e x^2}}\right )}{3 d^3}-\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {8 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \log \left (c x^n\right )\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \log \left (c x^n\right )}{d x \left (d+e x^2\right )^{3/2}}+\frac {b n \left (2 e \left (\frac {4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{\sqrt {e}}-\frac {x}{\sqrt {d+e x^2}}\right )-\frac {3 d}{x \sqrt {d+e x^2}}\right )}{3 d^3}\) |
(b*n*((-3*d)/(x*Sqrt[d + e*x^2]) + 2*e*(-(x/Sqrt[d + e*x^2]) + (4*ArcTanh[ (Sqrt[e]*x)/Sqrt[d + e*x^2]])/Sqrt[e])))/(3*d^3) - (a + b*Log[c*x^n])/(d*x *(d + e*x^2)^(3/2)) - (4*e*x*(a + b*Log[c*x^n]))/(3*d^2*(d + e*x^2)^(3/2)) - (8*e*x*(a + b*Log[c*x^n]))/(3*d^3*Sqrt[d + e*x^2])
3.4.7.3.1 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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
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])
\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x^{2} \left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]
Time = 0.37 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.40 \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^{5/2}} \, dx=\left [\frac {4 \, {\left (b e^{2} n x^{5} + 2 \, b d e n x^{3} + b d^{2} n x\right )} \sqrt {e} \log \left (-2 \, e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {e} x - d\right ) - {\left (2 \, {\left (b e^{2} n + 4 \, a e^{2}\right )} x^{4} + 3 \, b d^{2} n + 3 \, a d^{2} + {\left (5 \, b d e n + 12 \, a d e\right )} x^{2} + {\left (8 \, b e^{2} x^{4} + 12 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \left (c\right ) + {\left (8 \, b e^{2} n x^{4} + 12 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{3 \, {\left (d^{3} e^{2} x^{5} + 2 \, d^{4} e x^{3} + d^{5} x\right )}}, -\frac {8 \, {\left (b e^{2} n x^{5} + 2 \, b d e n x^{3} + b d^{2} n x\right )} \sqrt {-e} \arctan \left (\frac {\sqrt {-e} x}{\sqrt {e x^{2} + d}}\right ) + {\left (2 \, {\left (b e^{2} n + 4 \, a e^{2}\right )} x^{4} + 3 \, b d^{2} n + 3 \, a d^{2} + {\left (5 \, b d e n + 12 \, a d e\right )} x^{2} + {\left (8 \, b e^{2} x^{4} + 12 \, b d e x^{2} + 3 \, b d^{2}\right )} \log \left (c\right ) + {\left (8 \, b e^{2} n x^{4} + 12 \, b d e n x^{2} + 3 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{3 \, {\left (d^{3} e^{2} x^{5} + 2 \, d^{4} e x^{3} + d^{5} x\right )}}\right ] \]
[1/3*(4*(b*e^2*n*x^5 + 2*b*d*e*n*x^3 + b*d^2*n*x)*sqrt(e)*log(-2*e*x^2 - 2 *sqrt(e*x^2 + d)*sqrt(e)*x - d) - (2*(b*e^2*n + 4*a*e^2)*x^4 + 3*b*d^2*n + 3*a*d^2 + (5*b*d*e*n + 12*a*d*e)*x^2 + (8*b*e^2*x^4 + 12*b*d*e*x^2 + 3*b* d^2)*log(c) + (8*b*e^2*n*x^4 + 12*b*d*e*n*x^2 + 3*b*d^2*n)*log(x))*sqrt(e* x^2 + d))/(d^3*e^2*x^5 + 2*d^4*e*x^3 + d^5*x), -1/3*(8*(b*e^2*n*x^5 + 2*b* d*e*n*x^3 + b*d^2*n*x)*sqrt(-e)*arctan(sqrt(-e)*x/sqrt(e*x^2 + d)) + (2*(b *e^2*n + 4*a*e^2)*x^4 + 3*b*d^2*n + 3*a*d^2 + (5*b*d*e*n + 12*a*d*e)*x^2 + (8*b*e^2*x^4 + 12*b*d*e*x^2 + 3*b*d^2)*log(c) + (8*b*e^2*n*x^4 + 12*b*d*e *n*x^2 + 3*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(d^3*e^2*x^5 + 2*d^4*e*x^3 + d^5*x)]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{2} + d\right )}^{\frac {5}{2}} x^{2}} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x^2 \left (d+e x^2\right )^{5/2}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x^2\,{\left (e\,x^2+d\right )}^{5/2}} \,d x \]