Integrand size = 25, antiderivative size = 182 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=-\frac {8 b d^2 n \sqrt {d+e x^2}}{15 e^3}+\frac {7 b d n \left (d+e x^2\right )^{3/2}}{45 e^3}-\frac {b n \left (d+e x^2\right )^{5/2}}{25 e^3}+\frac {8 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{15 e^3}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3} \]
7/45*b*d*n*(e*x^2+d)^(3/2)/e^3-1/25*b*n*(e*x^2+d)^(5/2)/e^3+8/15*b*d^(5/2) *n*arctanh((e*x^2+d)^(1/2)/d^(1/2))/e^3-2/3*d*(e*x^2+d)^(3/2)*(a+b*ln(c*x^ n))/e^3+1/5*(e*x^2+d)^(5/2)*(a+b*ln(c*x^n))/e^3-8/15*b*d^2*n*(e*x^2+d)^(1/ 2)/e^3+d^2*(a+b*ln(c*x^n))*(e*x^2+d)^(1/2)/e^3
Time = 0.15 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.12 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\frac {120 a d^2 \sqrt {d+e x^2}-94 b d^2 n \sqrt {d+e x^2}-60 a d e x^2 \sqrt {d+e x^2}+17 b d e n x^2 \sqrt {d+e x^2}+45 a e^2 x^4 \sqrt {d+e x^2}-9 b e^2 n x^4 \sqrt {d+e x^2}-120 b d^{5/2} n \log (x)+15 b \sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right ) \log \left (c x^n\right )+120 b d^{5/2} n \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )}{225 e^3} \]
(120*a*d^2*Sqrt[d + e*x^2] - 94*b*d^2*n*Sqrt[d + e*x^2] - 60*a*d*e*x^2*Sqr t[d + e*x^2] + 17*b*d*e*n*x^2*Sqrt[d + e*x^2] + 45*a*e^2*x^4*Sqrt[d + e*x^ 2] - 9*b*e^2*n*x^4*Sqrt[d + e*x^2] - 120*b*d^(5/2)*n*Log[x] + 15*b*Sqrt[d + e*x^2]*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4)*Log[c*x^n] + 120*b*d^(5/2)*n*Log[ d + Sqrt[d]*Sqrt[d + e*x^2]])/(225*e^3)
Time = 0.51 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.90, 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, 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.
| \(\displaystyle \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int \frac {\sqrt {e x^2+d} \left (3 e^2 x^4-4 d e x^2+8 d^2\right )}{15 e^3 x}dx+\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {b n \int \frac {\sqrt {e x^2+d} \left (3 e^2 x^4-4 d e x^2+8 d^2\right )}{x}dx}{15 e^3}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle -\frac {b n \int \frac {\sqrt {e x^2+d} \left (3 e^2 x^4-4 d e x^2+8 d^2\right )}{x^2}dx^2}{30 e^3}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle -\frac {b n \int -\frac {x^4 \left (3 e^2 x^8-10 d e^2 x^4+15 d^2 e^2\right )}{d-x^4}d\sqrt {e x^2+d}}{15 e^5}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {b n \int \frac {x^4 \left (3 e^2 x^8-10 d e^2 x^4+15 d^2 e^2\right )}{d-x^4}d\sqrt {e x^2+d}}{15 e^5}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {b n \int \left (-3 e^2 x^8+7 d e^2 x^4-8 d^2 e^2+\frac {8 d^3 e^2}{d-x^4}\right )d\sqrt {e x^2+d}}{15 e^5}+\frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \sqrt {d+e x^2} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {2 d \left (d+e x^2\right )^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {b n \left (-8 d^{5/2} e^2 \text {arctanh}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )+8 d^2 e^2 \sqrt {d+e x^2}-\frac {7}{3} d e^2 x^6+\frac {3 e^2 x^{10}}{5}\right )}{15 e^5}\) |
-1/15*(b*n*((-7*d*e^2*x^6)/3 + (3*e^2*x^10)/5 + 8*d^2*e^2*Sqrt[d + e*x^2] - 8*d^(5/2)*e^2*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]]))/e^5 + (d^2*Sqrt[d + e*x ^2]*(a + b*Log[c*x^n]))/e^3 - (2*d*(d + e*x^2)^(3/2)*(a + b*Log[c*x^n]))/( 3*e^3) + ((d + e*x^2)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3)
3.3.76.3.1 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])
\[\int \frac {x^{5} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {e \,x^{2}+d}}d x\]
Time = 0.34 (sec) , antiderivative size = 314, normalized size of antiderivative = 1.73 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\left [\frac {60 \, b d^{\frac {5}{2}} n \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - {\left (9 \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 94 \, b d^{2} n - 120 \, a d^{2} - {\left (17 \, b d e n - 60 \, a d e\right )} x^{2} - 15 \, {\left (3 \, b e^{2} x^{4} - 4 \, b d e x^{2} + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b e^{2} n x^{4} - 4 \, b d e n x^{2} + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{225 \, e^{3}}, -\frac {120 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (9 \, {\left (b e^{2} n - 5 \, a e^{2}\right )} x^{4} + 94 \, b d^{2} n - 120 \, a d^{2} - {\left (17 \, b d e n - 60 \, a d e\right )} x^{2} - 15 \, {\left (3 \, b e^{2} x^{4} - 4 \, b d e x^{2} + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b e^{2} n x^{4} - 4 \, b d e n x^{2} + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x^{2} + d}}{225 \, e^{3}}\right ] \]
[1/225*(60*b*d^(5/2)*n*log(-(e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - (9*(b*e^2*n - 5*a*e^2)*x^4 + 94*b*d^2*n - 120*a*d^2 - (17*b*d*e*n - 60* a*d*e)*x^2 - 15*(3*b*e^2*x^4 - 4*b*d*e*x^2 + 8*b*d^2)*log(c) - 15*(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^3, -1/225*( 120*b*sqrt(-d)*d^2*n*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + (9*(b*e^2*n - 5*a* e^2)*x^4 + 94*b*d^2*n - 120*a*d^2 - (17*b*d*e*n - 60*a*d*e)*x^2 - 15*(3*b* e^2*x^4 - 4*b*d*e*x^2 + 8*b*d^2)*log(c) - 15*(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^3]
Time = 17.14 (sec) , antiderivative size = 359, normalized size of antiderivative = 1.97 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=a \left (\begin {cases} \frac {8 d^{2} \sqrt {d + e x^{2}}}{15 e^{3}} - \frac {4 d x^{2} \sqrt {d + e x^{2}}}{15 e^{2}} + \frac {x^{4} \sqrt {d + e x^{2}}}{5 e} & \text {for}\: e \neq 0 \\\frac {x^{6}}{6 \sqrt {d}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {8 d^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} x} \right )}}{15 e^{3}} + \frac {8 d^{3}}{15 e^{\frac {7}{2}} x \sqrt {\frac {d}{e x^{2}} + 1}} + \frac {8 d^{2} x}{15 e^{\frac {5}{2}} \sqrt {\frac {d}{e x^{2}} + 1}} - \frac {4 d \left (\begin {cases} \frac {d \sqrt {d + e x^{2}}}{3 e} + \frac {x^{2} \sqrt {d + e x^{2}}}{3} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right )}{15 e^{2}} + \frac {\begin {cases} - \frac {2 d^{2} \sqrt {d + e x^{2}}}{15 e^{2}} + \frac {d x^{2} \sqrt {d + e x^{2}}}{15 e} + \frac {x^{4} \sqrt {d + e x^{2}}}{5} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}}{5 e} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{6}}{36 \sqrt {d}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {8 d^{2} \sqrt {d + e x^{2}}}{15 e^{3}} - \frac {4 d x^{2} \sqrt {d + e x^{2}}}{15 e^{2}} + \frac {x^{4} \sqrt {d + e x^{2}}}{5 e} & \text {for}\: e \neq 0 \\\frac {x^{6}}{6 \sqrt {d}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((8*d**2*sqrt(d + e*x**2)/(15*e**3) - 4*d*x**2*sqrt(d + e*x**2) /(15*e**2) + x**4*sqrt(d + e*x**2)/(5*e), Ne(e, 0)), (x**6/(6*sqrt(d)), Tr ue)) - b*n*Piecewise((-8*d**(5/2)*asinh(sqrt(d)/(sqrt(e)*x))/(15*e**3) + 8 *d**3/(15*e**(7/2)*x*sqrt(d/(e*x**2) + 1)) + 8*d**2*x/(15*e**(5/2)*sqrt(d/ (e*x**2) + 1)) - 4*d*Piecewise((d*sqrt(d + e*x**2)/(3*e) + x**2*sqrt(d + e *x**2)/3, Ne(e, 0)), (sqrt(d)*x**2/2, True))/(15*e**2) + Piecewise((-2*d** 2*sqrt(d + e*x**2)/(15*e**2) + d*x**2*sqrt(d + e*x**2)/(15*e) + x**4*sqrt( d + e*x**2)/5, Ne(e, 0)), (sqrt(d)*x**4/4, True))/(5*e), (e > -oo) & (e < oo) & Ne(e, 0)), (x**6/(36*sqrt(d)), True)) + b*Piecewise((8*d**2*sqrt(d + e*x**2)/(15*e**3) - 4*d*x**2*sqrt(d + e*x**2)/(15*e**2) + x**4*sqrt(d + e *x**2)/(5*e), Ne(e, 0)), (x**6/(6*sqrt(d)), True))*log(c*x**n)
Exception generated. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\text {Exception raised: ValueError} \]
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
\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{\sqrt {e x^{2} + d}} \,d x } \]
Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x^2}} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {e\,x^2+d}} \,d x \]