Integrand size = 23, antiderivative size = 169 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {32 b d^2 n \sqrt {d+e x}}{15 e^3}+\frac {28 b d n (d+e x)^{3/2}}{45 e^3}-\frac {4 b n (d+e x)^{5/2}}{25 e^3}+\frac {32 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{15 e^3}+\frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3} \]
28/45*b*d*n*(e*x+d)^(3/2)/e^3-4/25*b*n*(e*x+d)^(5/2)/e^3+32/15*b*d^(5/2)*n *arctanh((e*x+d)^(1/2)/d^(1/2))/e^3-4/3*d*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^ 3+2/5*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^3-32/15*b*d^2*n*(e*x+d)^(1/2)/e^3+2* d^2*(a+b*ln(c*x^n))*(e*x+d)^(1/2)/e^3
Time = 0.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.70 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {480 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+2 \sqrt {d+e x} \left (15 a \left (8 d^2-4 d e x+3 e^2 x^2\right )-2 b n \left (94 d^2-17 d e x+9 e^2 x^2\right )+15 b \left (8 d^2-4 d e x+3 e^2 x^2\right ) \log \left (c x^n\right )\right )}{225 e^3} \]
(480*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 2*Sqrt[d + e*x]*(15*a*(8 *d^2 - 4*d*e*x + 3*e^2*x^2) - 2*b*n*(94*d^2 - 17*d*e*x + 9*e^2*x^2) + 15*b *(8*d^2 - 4*d*e*x + 3*e^2*x^2)*Log[c*x^n]))/(225*e^3)
Time = 0.41 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2792, 27, 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.
| \(\displaystyle \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int \frac {2 \sqrt {d+e x} \left (8 d^2-4 e x d+3 e^2 x^2\right )}{15 e^3 x}dx+\frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 b n \int \frac {\sqrt {d+e x} \left (8 d^2-4 e x d+3 e^2 x^2\right )}{x}dx}{15 e^3}+\frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle -\frac {4 b n \int \frac {(d+e x) \left (15 d^2 e^2+3 (d+e x)^2 e^2-10 d (d+e x) e^2\right )}{e x}d\sqrt {d+e x}}{15 e^5}+\frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4 b n \int -\frac {(d+e x) \left (15 d^2 e^2+3 (d+e x)^2 e^2-10 d (d+e x) e^2\right )}{e x}d\sqrt {d+e x}}{15 e^5}+\frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {4 b n \int \left (-\frac {8 e d^3}{x}-8 e^2 d^2+7 e^2 (d+e x) d-3 e^2 (d+e x)^2\right )d\sqrt {d+e x}}{15 e^5}+\frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^3}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {4 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {4 b n \left (-8 d^{5/2} e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+8 d^2 e^2 \sqrt {d+e x}-\frac {7}{3} d e^2 (d+e x)^{3/2}+\frac {3}{5} e^2 (d+e x)^{5/2}\right )}{15 e^5}\) |
(-4*b*n*(8*d^2*e^2*Sqrt[d + e*x] - (7*d*e^2*(d + e*x)^(3/2))/3 + (3*e^2*(d + e*x)^(5/2))/5 - 8*d^(5/2)*e^2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]))/(15*e^5) + (2*d^2*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/e^3 - (4*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^3) + (2*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3)
3.2.45.3.1 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[((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])
\[\int \frac {x^{2} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {e x +d}}d x\]
Time = 0.36 (sec) , antiderivative size = 296, normalized size of antiderivative = 1.75 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\left [\frac {2 \, {\left (120 \, b d^{\frac {5}{2}} n \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (188 \, b d^{2} n - 120 \, a d^{2} + 9 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - 2 \, {\left (17 \, b d e n - 30 \, a d e\right )} x - 15 \, {\left (3 \, b e^{2} x^{2} - 4 \, b d e x + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b e^{2} n x^{2} - 4 \, b d e n x + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{225 \, e^{3}}, -\frac {2 \, {\left (240 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (188 \, b d^{2} n - 120 \, a d^{2} + 9 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - 2 \, {\left (17 \, b d e n - 30 \, a d e\right )} x - 15 \, {\left (3 \, b e^{2} x^{2} - 4 \, b d e x + 8 \, b d^{2}\right )} \log \left (c\right ) - 15 \, {\left (3 \, b e^{2} n x^{2} - 4 \, b d e n x + 8 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{225 \, e^{3}}\right ] \]
[2/225*(120*b*d^(5/2)*n*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - (18 8*b*d^2*n - 120*a*d^2 + 9*(2*b*e^2*n - 5*a*e^2)*x^2 - 2*(17*b*d*e*n - 30*a *d*e)*x - 15*(3*b*e^2*x^2 - 4*b*d*e*x + 8*b*d^2)*log(c) - 15*(3*b*e^2*n*x^ 2 - 4*b*d*e*n*x + 8*b*d^2*n)*log(x))*sqrt(e*x + d))/e^3, -2/225*(240*b*sqr t(-d)*d^2*n*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (188*b*d^2*n - 120*a*d^2 + 9*(2*b*e^2*n - 5*a*e^2)*x^2 - 2*(17*b*d*e*n - 30*a*d*e)*x - 15*(3*b*e^2*x^ 2 - 4*b*d*e*x + 8*b*d^2)*log(c) - 15*(3*b*e^2*n*x^2 - 4*b*d*e*n*x + 8*b*d^ 2*n)*log(x))*sqrt(e*x + d))/e^3]
Time = 42.28 (sec) , antiderivative size = 337, normalized size of antiderivative = 1.99 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=a \left (\begin {cases} \frac {2 d^{2} \sqrt {d + e x}}{e^{3}} - \frac {4 d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{3}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} & \text {for}\: e \neq 0 \\\frac {x^{3}}{3 \sqrt {d}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {524 d^{\frac {5}{2}} \sqrt {1 + \frac {e x}{d}}}{225 e^{3}} - \frac {14 d^{\frac {5}{2}} \log {\left (\frac {e x}{d} \right )}}{15 e^{3}} + \frac {28 d^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{15 e^{3}} - \frac {4 d^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} \sqrt {x}} \right )}}{e^{3}} - \frac {68 d^{\frac {3}{2}} x \sqrt {1 + \frac {e x}{d}}}{225 e^{2}} + \frac {4 \sqrt {d} x^{2} \sqrt {1 + \frac {e x}{d}}}{25 e} + \frac {4 d^{3}}{e^{\frac {7}{2}} \sqrt {x} \sqrt {\frac {d}{e x} + 1}} + \frac {4 d^{2} \sqrt {x}}{e^{\frac {5}{2}} \sqrt {\frac {d}{e x} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{3}}{9 \sqrt {d}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {2 d^{2} \sqrt {d + e x}}{e^{3}} - \frac {4 d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{3}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} & \text {for}\: e \neq 0 \\\frac {x^{3}}{3 \sqrt {d}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((2*d**2*sqrt(d + e*x)/e**3 - 4*d*(d + e*x)**(3/2)/(3*e**3) + 2 *(d + e*x)**(5/2)/(5*e**3), Ne(e, 0)), (x**3/(3*sqrt(d)), True)) - b*n*Pie cewise((-524*d**(5/2)*sqrt(1 + e*x/d)/(225*e**3) - 14*d**(5/2)*log(e*x/d)/ (15*e**3) + 28*d**(5/2)*log(sqrt(1 + e*x/d) + 1)/(15*e**3) - 4*d**(5/2)*as inh(sqrt(d)/(sqrt(e)*sqrt(x)))/e**3 - 68*d**(3/2)*x*sqrt(1 + e*x/d)/(225*e **2) + 4*sqrt(d)*x**2*sqrt(1 + e*x/d)/(25*e) + 4*d**3/(e**(7/2)*sqrt(x)*sq rt(d/(e*x) + 1)) + 4*d**2*sqrt(x)/(e**(5/2)*sqrt(d/(e*x) + 1)), (e > -oo) & (e < oo) & Ne(e, 0)), (x**3/(9*sqrt(d)), True)) + b*Piecewise((2*d**2*sq rt(d + e*x)/e**3 - 4*d*(d + e*x)**(3/2)/(3*e**3) + 2*(d + e*x)**(5/2)/(5*e **3), Ne(e, 0)), (x**3/(3*sqrt(d)), True))*log(c*x**n)
Time = 0.27 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.02 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {4}{225} \, b n {\left (\frac {60 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{3}} + \frac {9 \, {\left (e x + d\right )}^{\frac {5}{2}} - 35 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 120 \, \sqrt {e x + d} d^{2}}{e^{3}}\right )} + \frac {2}{15} \, b {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}}}{e^{3}} - \frac {10 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{3}} + \frac {15 \, \sqrt {e x + d} d^{2}}{e^{3}}\right )} \log \left (c x^{n}\right ) + \frac {2}{15} \, a {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}}}{e^{3}} - \frac {10 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{3}} + \frac {15 \, \sqrt {e x + d} d^{2}}{e^{3}}\right )} \]
-4/225*b*n*(60*d^(5/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt (d)))/e^3 + (9*(e*x + d)^(5/2) - 35*(e*x + d)^(3/2)*d + 120*sqrt(e*x + d)* d^2)/e^3) + 2/15*b*(3*(e*x + d)^(5/2)/e^3 - 10*(e*x + d)^(3/2)*d/e^3 + 15* sqrt(e*x + d)*d^2/e^3)*log(c*x^n) + 2/15*a*(3*(e*x + d)^(5/2)/e^3 - 10*(e* x + d)^(3/2)*d/e^3 + 15*sqrt(e*x + d)*d^2/e^3)
Time = 0.39 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.14 \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {32 \, b d^{3} n \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{15 \, \sqrt {-d} e^{3}} + \frac {2}{15} \, {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} b n}{e^{3}} - \frac {10 \, {\left (e x + d\right )}^{\frac {3}{2}} b d n}{e^{3}} + \frac {15 \, \sqrt {e x + d} b d^{2} n}{e^{3}}\right )} \log \left (e x\right ) - \frac {2 \, {\left (5 \, b n \log \left (e\right ) + 2 \, b n - 5 \, b \log \left (c\right ) - 5 \, a\right )} {\left (e x + d\right )}^{\frac {5}{2}}}{25 \, e^{3}} + \frac {4 \, {\left (15 \, b d n \log \left (e\right ) + 7 \, b d n - 15 \, b d \log \left (c\right ) - 15 \, a d\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{45 \, e^{3}} - \frac {2 \, {\left (15 \, b d^{2} n \log \left (e\right ) + 16 \, b d^{2} n - 15 \, b d^{2} \log \left (c\right ) - 15 \, a d^{2}\right )} \sqrt {e x + d}}{15 \, e^{3}} \]
-32/15*b*d^3*n*arctan(sqrt(e*x + d)/sqrt(-d))/(sqrt(-d)*e^3) + 2/15*(3*(e* x + d)^(5/2)*b*n/e^3 - 10*(e*x + d)^(3/2)*b*d*n/e^3 + 15*sqrt(e*x + d)*b*d ^2*n/e^3)*log(e*x) - 2/25*(5*b*n*log(e) + 2*b*n - 5*b*log(c) - 5*a)*(e*x + d)^(5/2)/e^3 + 4/45*(15*b*d*n*log(e) + 7*b*d*n - 15*b*d*log(c) - 15*a*d)* (e*x + d)^(3/2)/e^3 - 2/15*(15*b*d^2*n*log(e) + 16*b*d^2*n - 15*b*d^2*log( c) - 15*a*d^2)*sqrt(e*x + d)/e^3
Timed out. \[ \int \frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\int \frac {x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {d+e\,x}} \,d x \]