Integrand size = 25, antiderivative size = 155 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\frac {b d n}{3 e^3 \sqrt {d+e x^2}}-\frac {b n \sqrt {d+e x^2}}{e^3}+\frac {8 b \sqrt {d} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3} \]
-1/3*d^2*(a+b*ln(c*x^n))/e^3/(e*x^2+d)^(3/2)+8/3*b*n*arctanh((e*x^2+d)^(1/ 2)/d^(1/2))*d^(1/2)/e^3+1/3*b*d*n/e^3/(e*x^2+d)^(1/2)+2*d*(a+b*ln(c*x^n))/ e^3/(e*x^2+d)^(1/2)-b*n*(e*x^2+d)^(1/2)/e^3+(a+b*ln(c*x^n))*(e*x^2+d)^(1/2 )/e^3
Time = 0.16 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.32 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=-\frac {8 b \sqrt {d} n \log (x)}{3 e^3}+\frac {b n \left (8 d^2+12 d e x^2+3 e^2 x^4\right ) \log (x)}{3 e^3 \left (d+e x^2\right )^{3/2}}+\sqrt {d+e x^2} \left (-\frac {d^2 \left (a+b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{3 e^3 \left (d+e x^2\right )^2}+\frac {a-b n+b \left (-n \log (x)+\log \left (c x^n\right )\right )}{e^3}+\frac {d \left (6 a+b n+6 b \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{3 e^3 \left (d+e x^2\right )}\right )+\frac {8 b \sqrt {d} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{3 e^3} \]
(-8*b*Sqrt[d]*n*Log[x])/(3*e^3) + (b*n*(8*d^2 + 12*d*e*x^2 + 3*e^2*x^4)*Lo g[x])/(3*e^3*(d + e*x^2)^(3/2)) + Sqrt[d + e*x^2]*(-1/3*(d^2*(a + b*(-(n*L og[x]) + Log[c*x^n])))/(e^3*(d + e*x^2)^2) + (a - b*n + b*(-(n*Log[x]) + L og[c*x^n]))/e^3 + (d*(6*a + b*n + 6*b*(-(n*Log[x]) + Log[c*x^n])))/(3*e^3* (d + e*x^2))) + (8*b*Sqrt[d]*n*Log[d + Sqrt[d]*Sqrt[d + e*x^2]])/(3*e^3)
Time = 0.48 (sec) , antiderivative size = 147, 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, 1578, 1192, 1584, 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.
| \(\displaystyle \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int \frac {3 e^2 x^4+12 d e x^2+8 d^2}{3 e^3 x \left (e x^2+d\right )^{3/2}}dx-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b n \int \frac {3 e^2 x^4+12 d e x^2+8 d^2}{x \left (e x^2+d\right )^{3/2}}dx}{3 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle -\frac {b n \int \frac {3 e^2 x^4+12 d e x^2+8 d^2}{x^2 \left (e x^2+d\right )^{3/2}}dx^2}{6 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle -\frac {b n \int \frac {-3 e^2 x^8-6 d e^2 x^4+d^2 e^2}{x^4 \left (d-x^4\right )}d\sqrt {e x^2+d}}{3 e^5}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle -\frac {b n \int \left (-\frac {8 d e^2}{d-x^4}+\frac {d e^2}{x^4}+3 e^2\right )d\sqrt {e x^2+d}}{3 e^5}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{3 e^3 \left (d+e x^2\right )^{3/2}}+\frac {2 d \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {b n \left (-8 \sqrt {d} e^2 \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )+3 e^2 \sqrt {d+e x^2}-\frac {d e^2}{x^2}\right )}{3 e^5}\) |
-1/3*(b*n*(-((d*e^2)/x^2) + 3*e^2*Sqrt[d + e*x^2] - 8*Sqrt[d]*e^2*ArcTanh[ Sqrt[d + e*x^2]/Sqrt[d]]))/e^5 - (d^2*(a + b*Log[c*x^n]))/(3*e^3*(d + e*x^ 2)^(3/2)) + (2*d*(a + b*Log[c*x^n]))/(e^3*Sqrt[d + e*x^2]) + (Sqrt[d + e*x ^2]*(a + b*Log[c*x^n]))/e^3
3.3.98.3.1 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[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2/e^(n + 2*p + 1) Subst[Int[x^( 2*m + 1)*(e*f - d*g + g*x^2)^n*(c*d^2 - b*d*e + a*e^2 - (2*c*d - b*e)*x^2 + c*x^4)^p, x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && IGtQ[p, 0] && ILtQ[n, 0] && IntegerQ[m + 1/2]
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -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])
\[\int \frac {x^{5} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}d x\]
Time = 0.36 (sec) , antiderivative size = 401, normalized size of antiderivative = 2.59 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\left [\frac {4 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt {d} \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (3 \, {\left (b e^{2} n - a e^{2}\right )} x^{4} + 2 \, b d^{2} n - 8 \, a d^{2} + {\left (5 \, b d e n - 12 \, a d e\right )} x^{2} - {\left (3 \, b e^{2} x^{4} + 12 \, b d e x^{2} + 8 \, b d^{2}\right )} \log \left (c\right ) - {\left (3 \, b e^{2} n x^{4} + 12 \, b d e n x^{2} + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{3 \, {\left (e^{5} x^{4} + 2 \, d e^{4} x^{2} + d^{2} e^{3}\right )}}, -\frac {8 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (3 \, {\left (b e^{2} n - a e^{2}\right )} x^{4} + 2 \, b d^{2} n - 8 \, a d^{2} + {\left (5 \, b d e n - 12 \, a d e\right )} x^{2} - {\left (3 \, b e^{2} x^{4} + 12 \, b d e x^{2} + 8 \, b d^{2}\right )} \log \left (c\right ) - {\left (3 \, b e^{2} n x^{4} + 12 \, b d e n x^{2} + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{3 \, {\left (e^{5} x^{4} + 2 \, d e^{4} x^{2} + d^{2} e^{3}\right )}}\right ] \]
[1/3*(4*(b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*sqrt(d)*log(-(e*x^2 + 2*sq rt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - (3*(b*e^2*n - a*e^2)*x^4 + 2*b*d^2*n - 8*a*d^2 + (5*b*d*e*n - 12*a*d*e)*x^2 - (3*b*e^2*x^4 + 12*b*d*e*x^2 + 8*b* d^2)*log(c) - (3*b*e^2*n*x^4 + 12*b*d*e*n*x^2 + 8*b*d^2*n)*log(x))*sqrt(e* x^2 + d))/(e^5*x^4 + 2*d*e^4*x^2 + d^2*e^3), -1/3*(8*(b*e^2*n*x^4 + 2*b*d* e*n*x^2 + b*d^2*n)*sqrt(-d)*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (3*(b*e^2*n - a*e^2)*x^4 + 2*b*d^2*n - 8*a*d^2 + (5*b*d*e*n - 12*a*d*e)*x^2 - (3*b*e^ 2*x^4 + 12*b*d*e*x^2 + 8*b*d^2)*log(c) - (3*b*e^2*n*x^4 + 12*b*d*e*n*x^2 + 8*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(e^5*x^4 + 2*d*e^4*x^2 + d^2*e^3)]
Time = 65.07 (sec) , antiderivative size = 415, normalized size of antiderivative = 2.68 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=a \left (\begin {cases} - \frac {d^{2}}{3 e^{3} \left (d + e x^{2}\right )^{\frac {3}{2}}} + \frac {2 d}{e^{3} \sqrt {d + e x^{2}}} + \frac {\sqrt {d + e x^{2}}}{e^{3}} & \text {for}\: e \neq 0 \\\frac {x^{6}}{6 d^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {3 \sqrt {d} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{e^{3}} - \frac {2 d^{5} \sqrt {1 + \frac {e x^{2}}{d}}}{6 d^{\frac {9}{2}} e^{3} + 6 d^{\frac {7}{2}} e^{4} x^{2}} - \frac {d^{5} \log {\left (\frac {e x^{2}}{d} \right )}}{6 d^{\frac {9}{2}} e^{3} + 6 d^{\frac {7}{2}} e^{4} x^{2}} + \frac {2 d^{5} \log {\left (\sqrt {1 + \frac {e x^{2}}{d}} + 1 \right )}}{6 d^{\frac {9}{2}} e^{3} + 6 d^{\frac {7}{2}} e^{4} x^{2}} - \frac {d^{4} x^{2} \log {\left (\frac {e x^{2}}{d} \right )}}{6 d^{\frac {9}{2}} e^{2} + 6 d^{\frac {7}{2}} e^{3} x^{2}} + \frac {2 d^{4} x^{2} \log {\left (\sqrt {1 + \frac {e x^{2}}{d}} + 1 \right )}}{6 d^{\frac {9}{2}} e^{2} + 6 d^{\frac {7}{2}} e^{3} x^{2}} + \frac {d}{e^{\frac {7}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {x}{e^{\frac {5}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{6}}{36 d^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {d^{2}}{3 e^{3} \left (d + e x^{2}\right )^{\frac {3}{2}}} + \frac {2 d}{e^{3} \sqrt {d + e x^{2}}} + \frac {\sqrt {d + e x^{2}}}{e^{3}} & \text {for}\: e \neq 0 \\\frac {x^{6}}{6 d^{\frac {5}{2}}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((-d**2/(3*e**3*(d + e*x**2)**(3/2)) + 2*d/(e**3*sqrt(d + e*x** 2)) + sqrt(d + e*x**2)/e**3, Ne(e, 0)), (x**6/(6*d**(5/2)), True)) - b*n*P iecewise((-3*sqrt(d)*asinh(sqrt(d)/(sqrt(e)*x))/e**3 - 2*d**5*sqrt(1 + e*x **2/d)/(6*d**(9/2)*e**3 + 6*d**(7/2)*e**4*x**2) - d**5*log(e*x**2/d)/(6*d* *(9/2)*e**3 + 6*d**(7/2)*e**4*x**2) + 2*d**5*log(sqrt(1 + e*x**2/d) + 1)/( 6*d**(9/2)*e**3 + 6*d**(7/2)*e**4*x**2) - d**4*x**2*log(e*x**2/d)/(6*d**(9 /2)*e**2 + 6*d**(7/2)*e**3*x**2) + 2*d**4*x**2*log(sqrt(1 + e*x**2/d) + 1) /(6*d**(9/2)*e**2 + 6*d**(7/2)*e**3*x**2) + d/(e**(7/2)*x*sqrt(d/(e*x**2) + 1)) + x/(e**(5/2)*sqrt(d/(e*x**2) + 1)), (e > -oo) & (e < oo) & Ne(e, 0) ), (x**6/(36*d**(5/2)), True)) + b*Piecewise((-d**2/(3*e**3*(d + e*x**2)** (3/2)) + 2*d/(e**3*sqrt(d + e*x**2)) + sqrt(d + e*x**2)/e**3, Ne(e, 0)), ( x**6/(6*d**(5/2)), True))*log(c*x**n)
Exception generated. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{5/2}} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]