Integrand size = 25, antiderivative size = 158 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {5 b d n \sqrt {d+e x^2}}{3 e^3}-\frac {b n \left (d+e x^2\right )^{3/2}}{9 e^3}-\frac {8 b d^{3/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{3 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3} \]
-1/9*b*n*(e*x^2+d)^(3/2)/e^3-8/3*b*d^(3/2)*n*arctanh((e*x^2+d)^(1/2)/d^(1/ 2))/e^3+1/3*(e*x^2+d)^(3/2)*(a+b*ln(c*x^n))/e^3-d^2*(a+b*ln(c*x^n))/e^3/(e *x^2+d)^(1/2)+5/3*b*d*n*(e*x^2+d)^(1/2)/e^3-2*d*(a+b*ln(c*x^n))*(e*x^2+d)^ (1/2)/e^3
Time = 0.14 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.01 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\frac {-24 a d^2+14 b d^2 n-12 a d e x^2+13 b d e n x^2+3 a e^2 x^4-b e^2 n x^4+24 b d^{3/2} n \sqrt {d+e x^2} \log (x)-3 b \left (8 d^2+4 d e x^2-e^2 x^4\right ) \log \left (c x^n\right )-24 b d^{3/2} n \sqrt {d+e x^2} \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{9 e^3 \sqrt {d+e x^2}} \]
(-24*a*d^2 + 14*b*d^2*n - 12*a*d*e*x^2 + 13*b*d*e*n*x^2 + 3*a*e^2*x^4 - b* e^2*n*x^4 + 24*b*d^(3/2)*n*Sqrt[d + e*x^2]*Log[x] - 3*b*(8*d^2 + 4*d*e*x^2 - e^2*x^4)*Log[c*x^n] - 24*b*d^(3/2)*n*Sqrt[d + e*x^2]*Log[d + Sqrt[d]*Sq rt[d + e*x^2]])/(9*e^3*Sqrt[d + e*x^2])
Time = 0.48 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.95, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {2792, 27, 1578, 1192, 25, 1467, 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.
| \(\displaystyle \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {-e^2 x^4+4 d e x^2+8 d^2}{3 e^3 x \sqrt {e x^2+d}}dx-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b n \int \frac {-e^2 x^4+4 d e x^2+8 d^2}{x \sqrt {e x^2+d}}dx}{3 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle \frac {b n \int \frac {-e^2 x^4+4 d e x^2+8 d^2}{x^2 \sqrt {e x^2+d}}dx^2}{6 e^3}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle \frac {b n \int -\frac {-e^2 x^8+6 d e^2 x^4+3 d^2 e^2}{d-x^4}d\sqrt {e x^2+d}}{3 e^5}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {b n \int \frac {-e^2 x^8+6 d e^2 x^4+3 d^2 e^2}{d-x^4}d\sqrt {e x^2+d}}{3 e^5}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 1467 |
| \(\displaystyle -\frac {b n \int \left (e^2 x^4-5 d e^2+\frac {8 d^2 e^2}{d-x^4}\right )d\sqrt {e x^2+d}}{3 e^5}-\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {d^2 \left (a+b \log \left (c x^n\right )\right )}{e^3 \sqrt {d+e x^2}}-\frac {2 d \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {b n \left (-8 d^{3/2} e^2 \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )+5 d e^2 \sqrt {d+e x^2}-\frac {1}{3} e^2 x^6\right )}{3 e^5}\) |
(b*n*(-1/3*(e^2*x^6) + 5*d*e^2*Sqrt[d + e*x^2] - 8*d^(3/2)*e^2*ArcTanh[Sqr t[d + e*x^2]/Sqrt[d]]))/(3*e^5) - (d^2*(a + b*Log[c*x^n]))/(e^3*Sqrt[d + e *x^2]) - (2*d*Sqrt[d + e*x^2]*(a + b*Log[c*x^n]))/e^3 + ((d + e*x^2)^(3/2) *(a + b*Log[c*x^n]))/(3*e^3)
3.3.87.3.1 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[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -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[((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])
\[\int \frac {x^{5} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}d x\]
Time = 0.34 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.25 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\left [\frac {12 \, {\left (b d e n x^{2} + b d^{2} n\right )} \sqrt {d} \log \left (-\frac {e x^{2} - 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left ({\left (b e^{2} n - 3 \, a e^{2}\right )} x^{4} - 14 \, b d^{2} n + 24 \, a d^{2} - {\left (13 \, b d e n - 12 \, a d e\right )} x^{2} - 3 \, {\left (b e^{2} x^{4} - 4 \, b d e x^{2} - 8 \, b d^{2}\right )} \log \left (c\right ) - 3 \, {\left (b e^{2} n x^{4} - 4 \, b d e n x^{2} - 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{9 \, {\left (e^{4} x^{2} + d e^{3}\right )}}, \frac {24 \, {\left (b d e n x^{2} + b d^{2} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) - {\left ({\left (b e^{2} n - 3 \, a e^{2}\right )} x^{4} - 14 \, b d^{2} n + 24 \, a d^{2} - {\left (13 \, b d e n - 12 \, a d e\right )} x^{2} - 3 \, {\left (b e^{2} x^{4} - 4 \, b d e x^{2} - 8 \, b d^{2}\right )} \log \left (c\right ) - 3 \, {\left (b e^{2} n x^{4} - 4 \, b d e n x^{2} - 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{9 \, {\left (e^{4} x^{2} + d e^{3}\right )}}\right ] \]
[1/9*(12*(b*d*e*n*x^2 + b*d^2*n)*sqrt(d)*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*s qrt(d) + 2*d)/x^2) - ((b*e^2*n - 3*a*e^2)*x^4 - 14*b*d^2*n + 24*a*d^2 - (1 3*b*d*e*n - 12*a*d*e)*x^2 - 3*(b*e^2*x^4 - 4*b*d*e*x^2 - 8*b*d^2)*log(c) - 3*(b*e^2*n*x^4 - 4*b*d*e*n*x^2 - 8*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(e^4 *x^2 + d*e^3), 1/9*(24*(b*d*e*n*x^2 + b*d^2*n)*sqrt(-d)*arctan(sqrt(-d)/sq rt(e*x^2 + d)) - ((b*e^2*n - 3*a*e^2)*x^4 - 14*b*d^2*n + 24*a*d^2 - (13*b* d*e*n - 12*a*d*e)*x^2 - 3*(b*e^2*x^4 - 4*b*d*e*x^2 - 8*b*d^2)*log(c) - 3*( b*e^2*n*x^4 - 4*b*d*e*n*x^2 - 8*b*d^2*n)*log(x))*sqrt(e*x^2 + d))/(e^4*x^2 + d*e^3)]
Time = 37.36 (sec) , antiderivative size = 308, normalized size of antiderivative = 1.95 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=a \left (\begin {cases} - \frac {d^{2}}{e^{3} \sqrt {d + e x^{2}}} - \frac {2 d \sqrt {d + e x^{2}}}{e^{3}} + \frac {\left (d + e x^{2}\right )^{\frac {3}{2}}}{3 e^{3}} & \text {for}\: e \neq 0 \\\frac {x^{6}}{6 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {4 d^{\frac {3}{2}} \sqrt {1 + \frac {e x^{2}}{d}}}{9 e^{3}} + \frac {d^{\frac {3}{2}} \log {\left (\frac {e x^{2}}{d} \right )}}{6 e^{3}} - \frac {d^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {e x^{2}}{d}} + 1 \right )}}{3 e^{3}} + \frac {3 d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{e^{3}} + \frac {\sqrt {d} x^{2} \sqrt {1 + \frac {e x^{2}}{d}}}{9 e^{2}} - \frac {2 d^{2}}{e^{\frac {7}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} - \frac {2 d x}{e^{\frac {5}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{6}}{36 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {d^{2}}{e^{3} \sqrt {d + e x^{2}}} - \frac {2 d \sqrt {d + e x^{2}}}{e^{3}} + \frac {\left (d + e x^{2}\right )^{\frac {3}{2}}}{3 e^{3}} & \text {for}\: e \neq 0 \\\frac {x^{6}}{6 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((-d**2/(e**3*sqrt(d + e*x**2)) - 2*d*sqrt(d + e*x**2)/e**3 + ( d + e*x**2)**(3/2)/(3*e**3), Ne(e, 0)), (x**6/(6*d**(3/2)), True)) - b*n*P iecewise((4*d**(3/2)*sqrt(1 + e*x**2/d)/(9*e**3) + d**(3/2)*log(e*x**2/d)/ (6*e**3) - d**(3/2)*log(sqrt(1 + e*x**2/d) + 1)/(3*e**3) + 3*d**(3/2)*asin h(sqrt(d)/(sqrt(e)*x))/e**3 + sqrt(d)*x**2*sqrt(1 + e*x**2/d)/(9*e**2) - 2 *d**2/(e**(7/2)*x*sqrt(d/(e*x**2) + 1)) - 2*d*x/(e**(5/2)*sqrt(d/(e*x**2) + 1)), (e > -oo) & (e < oo) & Ne(e, 0)), (x**6/(36*d**(3/2)), True)) + b*P iecewise((-d**2/(e**3*sqrt(d + e*x**2)) - 2*d*sqrt(d + e*x**2)/e**3 + (d + e*x**2)**(3/2)/(3*e**3), Ne(e, 0)), (x**6/(6*d**(3/2)), True))*log(c*x**n )
Exception generated. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^{3/2}} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]