Integrand size = 23, antiderivative size = 192 \[ \int x^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {32 b d^3 n \sqrt {d+e x}}{105 e^3}-\frac {32 b d^2 n (d+e x)^{3/2}}{315 e^3}+\frac {36 b d n (d+e x)^{5/2}}{175 e^3}-\frac {4 b n (d+e x)^{7/2}}{49 e^3}+\frac {32 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{105 e^3}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {4 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3} \]
-32/315*b*d^2*n*(e*x+d)^(3/2)/e^3+36/175*b*d*n*(e*x+d)^(5/2)/e^3-4/49*b*n* (e*x+d)^(7/2)/e^3+32/105*b*d^(7/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e^3+2/ 3*d^2*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^3-4/5*d*(e*x+d)^(5/2)*(a+b*ln(c*x^n) )/e^3+2/7*(e*x+d)^(7/2)*(a+b*ln(c*x^n))/e^3-32/105*b*d^3*n*(e*x+d)^(1/2)/e ^3
Time = 0.13 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.79 \[ \int x^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {3360 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+2 \sqrt {d+e x} \left (105 a \left (8 d^3-4 d^2 e x+3 d e^2 x^2+15 e^3 x^3\right )-2 b n \left (778 d^3-179 d^2 e x+108 d e^2 x^2+225 e^3 x^3\right )+105 b \left (8 d^3-4 d^2 e x+3 d e^2 x^2+15 e^3 x^3\right ) \log \left (c x^n\right )\right )}{11025 e^3} \]
(3360*b*d^(7/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 2*Sqrt[d + e*x]*(105*a* (8*d^3 - 4*d^2*e*x + 3*d*e^2*x^2 + 15*e^3*x^3) - 2*b*n*(778*d^3 - 179*d^2* e*x + 108*d*e^2*x^2 + 225*e^3*x^3) + 105*b*(8*d^3 - 4*d^2*e*x + 3*d*e^2*x^ 2 + 15*e^3*x^3)*Log[c*x^n]))/(11025*e^3)
Time = 0.42 (sec) , antiderivative size = 188, normalized size of antiderivative = 0.98, 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.
| \(\displaystyle \int x^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int \frac {2 (d+e x)^{3/2} \left (8 d^2-12 e x d+15 e^2 x^2\right )}{105 e^3 x}dx+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {4 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 b n \int \frac {(d+e x)^{3/2} \left (8 d^2-12 e x d+15 e^2 x^2\right )}{x}dx}{105 e^3}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {4 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 1192 |
| \(\displaystyle -\frac {4 b n \int \frac {(d+e x)^2 \left (35 d^2 e^2+15 (d+e x)^2 e^2-42 d (d+e x) e^2\right )}{e x}d\sqrt {d+e x}}{105 e^5}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {4 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {4 b n \int -\frac {(d+e x)^2 \left (35 d^2 e^2+15 (d+e x)^2 e^2-42 d (d+e x) e^2\right )}{e x}d\sqrt {d+e x}}{105 e^5}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {4 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {4 b n \int \left (-\frac {8 e d^4}{x}-8 e^2 d^3-8 e^2 (d+e x) d^2+27 e^2 (d+e x)^2 d-15 e^2 (d+e x)^3\right )d\sqrt {d+e x}}{105 e^5}+\frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {4 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 d^2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^3}-\frac {4 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^3}-\frac {4 b n \left (-8 d^{7/2} e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+8 d^3 e^2 \sqrt {d+e x}+\frac {8}{3} d^2 e^2 (d+e x)^{3/2}-\frac {27}{5} d e^2 (d+e x)^{5/2}+\frac {15}{7} e^2 (d+e x)^{7/2}\right )}{105 e^5}\) |
(-4*b*n*(8*d^3*e^2*Sqrt[d + e*x] + (8*d^2*e^2*(d + e*x)^(3/2))/3 - (27*d*e ^2*(d + e*x)^(5/2))/5 + (15*e^2*(d + e*x)^(7/2))/7 - 8*d^(7/2)*e^2*ArcTanh [Sqrt[d + e*x]/Sqrt[d]]))/(105*e^5) + (2*d^2*(d + e*x)^(3/2)*(a + b*Log[c* x^n]))/(3*e^3) - (4*d*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^3) + (2*(d + e*x)^(7/2)*(a + b*Log[c*x^n]))/(7*e^3)
3.2.31.3.1 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])
\[\int x^{2} \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e x +d}d x\]
Time = 0.31 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.06 \[ \int x^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {2 \, {\left (840 \, b d^{\frac {7}{2}} n \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (1556 \, b d^{3} n - 840 \, a d^{3} + 225 \, {\left (2 \, b e^{3} n - 7 \, a e^{3}\right )} x^{3} + 9 \, {\left (24 \, b d e^{2} n - 35 \, a d e^{2}\right )} x^{2} - 2 \, {\left (179 \, b d^{2} e n - 210 \, a d^{2} e\right )} x - 105 \, {\left (15 \, b e^{3} x^{3} + 3 \, b d e^{2} x^{2} - 4 \, b d^{2} e x + 8 \, b d^{3}\right )} \log \left (c\right ) - 105 \, {\left (15 \, b e^{3} n x^{3} + 3 \, b d e^{2} n x^{2} - 4 \, b d^{2} e n x + 8 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{11025 \, e^{3}}, -\frac {2 \, {\left (1680 \, b \sqrt {-d} d^{3} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (1556 \, b d^{3} n - 840 \, a d^{3} + 225 \, {\left (2 \, b e^{3} n - 7 \, a e^{3}\right )} x^{3} + 9 \, {\left (24 \, b d e^{2} n - 35 \, a d e^{2}\right )} x^{2} - 2 \, {\left (179 \, b d^{2} e n - 210 \, a d^{2} e\right )} x - 105 \, {\left (15 \, b e^{3} x^{3} + 3 \, b d e^{2} x^{2} - 4 \, b d^{2} e x + 8 \, b d^{3}\right )} \log \left (c\right ) - 105 \, {\left (15 \, b e^{3} n x^{3} + 3 \, b d e^{2} n x^{2} - 4 \, b d^{2} e n x + 8 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{11025 \, e^{3}}\right ] \]
[2/11025*(840*b*d^(7/2)*n*log((e*x + 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) - ( 1556*b*d^3*n - 840*a*d^3 + 225*(2*b*e^3*n - 7*a*e^3)*x^3 + 9*(24*b*d*e^2*n - 35*a*d*e^2)*x^2 - 2*(179*b*d^2*e*n - 210*a*d^2*e)*x - 105*(15*b*e^3*x^3 + 3*b*d*e^2*x^2 - 4*b*d^2*e*x + 8*b*d^3)*log(c) - 105*(15*b*e^3*n*x^3 + 3 *b*d*e^2*n*x^2 - 4*b*d^2*e*n*x + 8*b*d^3*n)*log(x))*sqrt(e*x + d))/e^3, -2 /11025*(1680*b*sqrt(-d)*d^3*n*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (1556*b*d ^3*n - 840*a*d^3 + 225*(2*b*e^3*n - 7*a*e^3)*x^3 + 9*(24*b*d*e^2*n - 35*a* d*e^2)*x^2 - 2*(179*b*d^2*e*n - 210*a*d^2*e)*x - 105*(15*b*e^3*x^3 + 3*b*d *e^2*x^2 - 4*b*d^2*e*x + 8*b*d^3)*log(c) - 105*(15*b*e^3*n*x^3 + 3*b*d*e^2 *n*x^2 - 4*b*d^2*e*n*x + 8*b*d^3*n)*log(x))*sqrt(e*x + d))/e^3]
Time = 51.04 (sec) , antiderivative size = 284, normalized size of antiderivative = 1.48 \[ \int x^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=a \left (\begin {cases} \frac {2 d^{2} \left (d + e x\right )^{\frac {3}{2}}}{3 e^{3}} - \frac {4 d \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} + \frac {2 \left (d + e x\right )^{\frac {7}{2}}}{7 e^{3}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{3}}{3} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {3112 d^{\frac {7}{2}} \sqrt {1 + \frac {e x}{d}}}{11025 e^{3}} + \frac {16 d^{\frac {7}{2}} \log {\left (\frac {e x}{d} \right )}}{105 e^{3}} - \frac {32 d^{\frac {7}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{105 e^{3}} - \frac {716 d^{\frac {5}{2}} x \sqrt {1 + \frac {e x}{d}}}{11025 e^{2}} + \frac {48 d^{\frac {3}{2}} x^{2} \sqrt {1 + \frac {e x}{d}}}{1225 e} + \frac {4 \sqrt {d} x^{3} \sqrt {1 + \frac {e x}{d}}}{49} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{3}}{9} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {2 d^{2} \left (d + e x\right )^{\frac {3}{2}}}{3 e^{3}} - \frac {4 d \left (d + e x\right )^{\frac {5}{2}}}{5 e^{3}} + \frac {2 \left (d + e x\right )^{\frac {7}{2}}}{7 e^{3}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{3}}{3} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((2*d**2*(d + e*x)**(3/2)/(3*e**3) - 4*d*(d + e*x)**(5/2)/(5*e* *3) + 2*(d + e*x)**(7/2)/(7*e**3), Ne(e, 0)), (sqrt(d)*x**3/3, True)) - b* n*Piecewise((3112*d**(7/2)*sqrt(1 + e*x/d)/(11025*e**3) + 16*d**(7/2)*log( e*x/d)/(105*e**3) - 32*d**(7/2)*log(sqrt(1 + e*x/d) + 1)/(105*e**3) - 716* d**(5/2)*x*sqrt(1 + e*x/d)/(11025*e**2) + 48*d**(3/2)*x**2*sqrt(1 + e*x/d) /(1225*e) + 4*sqrt(d)*x**3*sqrt(1 + e*x/d)/49, (e > -oo) & (e < oo) & Ne(e , 0)), (sqrt(d)*x**3/9, True)) + b*Piecewise((2*d**2*(d + e*x)**(3/2)/(3*e **3) - 4*d*(d + e*x)**(5/2)/(5*e**3) + 2*(d + e*x)**(7/2)/(7*e**3), Ne(e, 0)), (sqrt(d)*x**3/3, True))*log(c*x**n)
Time = 0.28 (sec) , antiderivative size = 184, normalized size of antiderivative = 0.96 \[ \int x^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {4}{11025} \, {\left (\frac {420 \, d^{\frac {7}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{3}} + \frac {225 \, {\left (e x + d\right )}^{\frac {7}{2}} - 567 \, {\left (e x + d\right )}^{\frac {5}{2}} d + 280 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} + 840 \, \sqrt {e x + d} d^{3}}{e^{3}}\right )} b n + \frac {2}{105} \, b {\left (\frac {15 \, {\left (e x + d\right )}^{\frac {7}{2}}}{e^{3}} - \frac {42 \, {\left (e x + d\right )}^{\frac {5}{2}} d}{e^{3}} + \frac {35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2}}{e^{3}}\right )} \log \left (c x^{n}\right ) + \frac {2}{105} \, a {\left (\frac {15 \, {\left (e x + d\right )}^{\frac {7}{2}}}{e^{3}} - \frac {42 \, {\left (e x + d\right )}^{\frac {5}{2}} d}{e^{3}} + \frac {35 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2}}{e^{3}}\right )} \]
-4/11025*(420*d^(7/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt( d)))/e^3 + (225*(e*x + d)^(7/2) - 567*(e*x + d)^(5/2)*d + 280*(e*x + d)^(3 /2)*d^2 + 840*sqrt(e*x + d)*d^3)/e^3)*b*n + 2/105*b*(15*(e*x + d)^(7/2)/e^ 3 - 42*(e*x + d)^(5/2)*d/e^3 + 35*(e*x + d)^(3/2)*d^2/e^3)*log(c*x^n) + 2/ 105*a*(15*(e*x + d)^(7/2)/e^3 - 42*(e*x + d)^(5/2)*d/e^3 + 35*(e*x + d)^(3 /2)*d^2/e^3)
\[ \int x^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sqrt {e x + d} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \,d x } \]
Timed out. \[ \int x^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^2\,\left (a+b\,\ln \left (c\,x^n\right )\right )\,\sqrt {d+e\,x} \,d x \]