Integrand size = 25, antiderivative size = 209 \[ \int \frac {x^7 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=-\frac {11 b d^2 n \sqrt {d+e x^2}}{5 e^4}+\frac {4 b d n \left (d+e x^2\right )^{3/2}}{15 e^4}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^4}+\frac {16 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{5 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4} \]
4/15*b*d*n*(e*x^2+d)^(3/2)/e^4-1/25*b*n*(e*x^2+d)^(5/2)/e^4+16/5*b*d^(5/2) *n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e^4-d*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/ e^4+1/5*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/e^4+d^3*(a+b*ln(c*x^n))/e^4/(e*x^2 +d)^(1/2)-11/5*b*d^2*n*(e*x^2+d)^(1/2)/e^4+3*d^2*(a+b*ln(c*x^n))*(e*x^2+d) ^(1/2)/e^4
Time = 0.17 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.93 \[ \int \frac {x^7 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {240 a d^3-148 b d^3 n+120 a d^2 e x^2-134 b d^2 e n x^2-30 a d e^2 x^4+11 b d e^2 n x^4+15 a e^3 x^6-3 b e^3 n x^6-240 b d^{5/2} n \sqrt {d+e x^2} \log (x)+15 b \left (16 d^3+8 d^2 e x^2-2 d e^2 x^4+e^3 x^6\right ) \log \left (c x^n\right )+240 b d^{5/2} n \sqrt {d+e x^2} \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{75 e^4 \sqrt {d+e x^2}} \]
(240*a*d^3 - 148*b*d^3*n + 120*a*d^2*e*x^2 - 134*b*d^2*e*n*x^2 - 30*a*d*e^ 2*x^4 + 11*b*d*e^2*n*x^4 + 15*a*e^3*x^6 - 3*b*e^3*n*x^6 - 240*b*d^(5/2)*n* Sqrt[d + e*x^2]*Log[x] + 15*b*(16*d^3 + 8*d^2*e*x^2 - 2*d*e^2*x^4 + e^3*x^ 6)*Log[c*x^n] + 240*b*d^(5/2)*n*Sqrt[d + e*x^2]*Log[d + Sqrt[d]*Sqrt[d + e *x^2]])/(75*e^4*Sqrt[d + e*x^2])
Time = 0.63 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2792, 27, 2331, 2123, 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.
| \(\displaystyle \int \frac {x^7 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int \frac {e^3 x^6-2 d e^2 x^4+8 d^2 e x^2+16 d^3}{5 e^4 x \sqrt {e x^2+d}}dx+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b n \int \frac {e^3 x^6-2 d e^2 x^4+8 d^2 e x^2+16 d^3}{x \sqrt {e x^2+d}}dx}{5 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}\) |
| \(\Big \downarrow \) 2331 |
| \(\displaystyle -\frac {b n \int \frac {e^3 x^6-2 d e^2 x^4+8 d^2 e x^2+16 d^3}{x^2 \sqrt {e x^2+d}}dx^2}{10 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle -\frac {b n \int \left (\frac {16 d^3}{x^2 \sqrt {e x^2+d}}+\frac {11 e d^2}{\sqrt {e x^2+d}}-4 e \sqrt {e x^2+d} d+e \left (e x^2+d\right )^{3/2}\right )dx^2}{10 e^4}+\frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x^2}}+\frac {3 d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {b n \left (-32 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )+22 d^2 \sqrt {d+e x^2}-\frac {8}{3} d \left (d+e x^2\right )^{3/2}+\frac {2}{5} \left (d+e x^2\right )^{5/2}\right )}{10 e^4}\) |
-1/10*(b*n*(22*d^2*Sqrt[d + e*x^2] - (8*d*(d + e*x^2)^(3/2))/3 + (2*(d + e *x^2)^(5/2))/5 - 32*d^(5/2)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]))/e^4 + (d^3* (a + b*Log[c*x^n]))/(e^4*Sqrt[d + e*x^2]) + (3*d^2*Sqrt[d + e*x^2]*(a + b* Log[c*x^n]))/e^4 - (d*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/e^4 + ((d + e* x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*e^4)
3.3.86.3.1 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[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[1/2 S ubst[Int[x^((m - 1)/2)*SubstFor[x^2, Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]
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])
\[\int \frac {x^{7} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]
Time = 0.38 (sec) , antiderivative size = 461, normalized size of antiderivative = 2.21 \[ \int \frac {x^7 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\left [\frac {120 \, {\left (b d^{2} e n x^{2} + b d^{3} n\right )} \sqrt {d} \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (3 \, {\left (b e^{3} n - 5 \, a e^{3}\right )} x^{6} + 148 \, b d^{3} n - {\left (11 \, b d e^{2} n - 30 \, a d e^{2}\right )} x^{4} - 240 \, a d^{3} + 2 \, {\left (67 \, b d^{2} e n - 60 \, a d^{2} e\right )} x^{2} - 15 \, {\left (b e^{3} x^{6} - 2 \, b d e^{2} x^{4} + 8 \, b d^{2} e x^{2} + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b e^{3} n x^{6} - 2 \, b d e^{2} n x^{4} + 8 \, b d^{2} e n x^{2} + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{75 \, {\left (e^{5} x^{2} + d e^{4}\right )}}, -\frac {240 \, {\left (b d^{2} e n x^{2} + b d^{3} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (3 \, {\left (b e^{3} n - 5 \, a e^{3}\right )} x^{6} + 148 \, b d^{3} n - {\left (11 \, b d e^{2} n - 30 \, a d e^{2}\right )} x^{4} - 240 \, a d^{3} + 2 \, {\left (67 \, b d^{2} e n - 60 \, a d^{2} e\right )} x^{2} - 15 \, {\left (b e^{3} x^{6} - 2 \, b d e^{2} x^{4} + 8 \, b d^{2} e x^{2} + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b e^{3} n x^{6} - 2 \, b d e^{2} n x^{4} + 8 \, b d^{2} e n x^{2} + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{75 \, {\left (e^{5} x^{2} + d e^{4}\right )}}\right ] \]
[1/75*(120*(b*d^2*e*n*x^2 + b*d^3*n)*sqrt(d)*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - (3*(b*e^3*n - 5*a*e^3)*x^6 + 148*b*d^3*n - (11*b* d*e^2*n - 30*a*d*e^2)*x^4 - 240*a*d^3 + 2*(67*b*d^2*e*n - 60*a*d^2*e)*x^2 - 15*(b*e^3*x^6 - 2*b*d*e^2*x^4 + 8*b*d^2*e*x^2 + 16*b*d^3)*log(c) - 15*(b *e^3*n*x^6 - 2*b*d*e^2*n*x^4 + 8*b*d^2*e*n*x^2 + 16*b*d^3*n)*log(x))*sqrt( e*x^2 + d))/(e^5*x^2 + d*e^4), -1/75*(240*(b*d^2*e*n*x^2 + b*d^3*n)*sqrt(- d)*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (3*(b*e^3*n - 5*a*e^3)*x^6 + 148*b*d ^3*n - (11*b*d*e^2*n - 30*a*d*e^2)*x^4 - 240*a*d^3 + 2*(67*b*d^2*e*n - 60* a*d^2*e)*x^2 - 15*(b*e^3*x^6 - 2*b*d*e^2*x^4 + 8*b*d^2*e*x^2 + 16*b*d^3)*l og(c) - 15*(b*e^3*n*x^6 - 2*b*d*e^2*n*x^4 + 8*b*d^2*e*n*x^2 + 16*b*d^3*n)* log(x))*sqrt(e*x^2 + d))/(e^5*x^2 + d*e^4)]
Time = 44.81 (sec) , antiderivative size = 374, normalized size of antiderivative = 1.79 \[ \int \frac {x^7 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=a \left (\begin {cases} \frac {d^{3}}{e^{4} \sqrt {d + e x^{2}}} + \frac {3 d^{2} \sqrt {d + e x^{2}}}{e^{4}} - \frac {d \left (d + e x^{2}\right )^{\frac {3}{2}}}{e^{4}} + \frac {\left (d + e x^{2}\right )^{\frac {5}{2}}}{5 e^{4}} & \text {for}\: e \neq 0 \\\frac {x^{8}}{8 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {77 d^{\frac {5}{2}} \sqrt {1 + \frac {e x^{2}}{d}}}{75 e^{4}} - \frac {2 d^{\frac {5}{2}} \log {\left (\frac {e x^{2}}{d} \right )}}{5 e^{4}} + \frac {4 d^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {e x^{2}}{d}} + 1 \right )}}{5 e^{4}} - \frac {4 d^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{e^{4}} - \frac {14 d^{\frac {3}{2}} x^{2} \sqrt {1 + \frac {e x^{2}}{d}}}{75 e^{3}} + \frac {\sqrt {d} x^{4} \sqrt {1 + \frac {e x^{2}}{d}}}{25 e^{2}} + \frac {3 d^{3}}{e^{\frac {9}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {3 d^{2} x}{e^{\frac {7}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{8}}{64 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {d^{3}}{e^{4} \sqrt {d + e x^{2}}} + \frac {3 d^{2} \sqrt {d + e x^{2}}}{e^{4}} - \frac {d \left (d + e x^{2}\right )^{\frac {3}{2}}}{e^{4}} + \frac {\left (d + e x^{2}\right )^{\frac {5}{2}}}{5 e^{4}} & \text {for}\: e \neq 0 \\\frac {x^{8}}{8 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((d**3/(e**4*sqrt(d + e*x**2)) + 3*d**2*sqrt(d + e*x**2)/e**4 - d*(d + e*x**2)**(3/2)/e**4 + (d + e*x**2)**(5/2)/(5*e**4), Ne(e, 0)), (x* *8/(8*d**(3/2)), True)) - b*n*Piecewise((-77*d**(5/2)*sqrt(1 + e*x**2/d)/( 75*e**4) - 2*d**(5/2)*log(e*x**2/d)/(5*e**4) + 4*d**(5/2)*log(sqrt(1 + e*x **2/d) + 1)/(5*e**4) - 4*d**(5/2)*asinh(sqrt(d)/(sqrt(e)*x))/e**4 - 14*d** (3/2)*x**2*sqrt(1 + e*x**2/d)/(75*e**3) + sqrt(d)*x**4*sqrt(1 + e*x**2/d)/ (25*e**2) + 3*d**3/(e**(9/2)*x*sqrt(d/(e*x**2) + 1)) + 3*d**2*x/(e**(7/2)* sqrt(d/(e*x**2) + 1)), (e > -oo) & (e < oo) & Ne(e, 0)), (x**8/(64*d**(3/2 )), True)) + b*Piecewise((d**3/(e**4*sqrt(d + e*x**2)) + 3*d**2*sqrt(d + e *x**2)/e**4 - d*(d + e*x**2)**(3/2)/e**4 + (d + e*x**2)**(5/2)/(5*e**4), N e(e, 0)), (x**8/(8*d**(3/2)), True))*log(c*x**n)
Exception generated. \[ \int \frac {x^7 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^7 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{7}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^7 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^7\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]