Integrand size = 23, antiderivative size = 242 \[ \int x^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {64 b d^4 n \sqrt {d+e x}}{315 e^4}+\frac {64 b d^3 n (d+e x)^{3/2}}{945 e^4}-\frac {356 b d^2 n (d+e x)^{5/2}}{1575 e^4}+\frac {80 b d n (d+e x)^{7/2}}{441 e^4}-\frac {4 b n (d+e x)^{9/2}}{81 e^4}-\frac {64 b d^{9/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{315 e^4}-\frac {2 d^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}+\frac {6 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {6 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^4} \]
64/945*b*d^3*n*(e*x+d)^(3/2)/e^4-356/1575*b*d^2*n*(e*x+d)^(5/2)/e^4+80/441 *b*d*n*(e*x+d)^(7/2)/e^4-4/81*b*n*(e*x+d)^(9/2)/e^4-64/315*b*d^(9/2)*n*arc tanh((e*x+d)^(1/2)/d^(1/2))/e^4-2/3*d^3*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^4+ 6/5*d^2*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^4-6/7*d*(e*x+d)^(7/2)*(a+b*ln(c*x^ n))/e^4+2/9*(e*x+d)^(9/2)*(a+b*ln(c*x^n))/e^4+64/315*b*d^4*n*(e*x+d)^(1/2) /e^4
Time = 0.17 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.76 \[ \int x^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {2 \left (10080 b d^{9/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+\sqrt {d+e x} \left (315 a \left (16 d^4-8 d^3 e x+6 d^2 e^2 x^2-5 d e^3 x^3-35 e^4 x^4\right )+2 b n \left (-4388 d^4+934 d^3 e x-543 d^2 e^2 x^2+400 d e^3 x^3+1225 e^4 x^4\right )+315 b \left (16 d^4-8 d^3 e x+6 d^2 e^2 x^2-5 d e^3 x^3-35 e^4 x^4\right ) \log \left (c x^n\right )\right )\right )}{99225 e^4} \]
(-2*(10080*b*d^(9/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(315 *a*(16*d^4 - 8*d^3*e*x + 6*d^2*e^2*x^2 - 5*d*e^3*x^3 - 35*e^4*x^4) + 2*b*n *(-4388*d^4 + 934*d^3*e*x - 543*d^2*e^2*x^2 + 400*d*e^3*x^3 + 1225*e^4*x^4 ) + 315*b*(16*d^4 - 8*d^3*e*x + 6*d^2*e^2*x^2 - 5*d*e^3*x^3 - 35*e^4*x^4)* Log[c*x^n])))/(99225*e^4)
Time = 0.50 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.90, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2792, 27, 2123, 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.
| \(\displaystyle \int x^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {2 (d+e x)^{3/2} \left (16 d^3-24 e x d^2+30 e^2 x^2 d-35 e^3 x^3\right )}{315 e^4 x}dx-\frac {2 d^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}+\frac {6 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^4}-\frac {6 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b n \int \frac {(d+e x)^{3/2} \left (16 d^3-24 e x d^2+30 e^2 x^2 d-35 e^3 x^3\right )}{x}dx}{315 e^4}-\frac {2 d^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}+\frac {6 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^4}-\frac {6 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle \frac {2 b n \int \left (-35 e (d+e x)^{7/2}+100 d e (d+e x)^{5/2}-89 d^2 e (d+e x)^{3/2}+\frac {16 d^3 (d+e x)^{3/2}}{x}\right )dx}{315 e^4}-\frac {2 d^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}+\frac {6 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^4}-\frac {6 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 d^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}+\frac {6 d^2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {2 (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{9 e^4}-\frac {6 d (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {2 b n \left (-32 d^{9/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+32 d^4 \sqrt {d+e x}+\frac {32}{3} d^3 (d+e x)^{3/2}-\frac {178}{5} d^2 (d+e x)^{5/2}+\frac {200}{7} d (d+e x)^{7/2}-\frac {70}{9} (d+e x)^{9/2}\right )}{315 e^4}\) |
(2*b*n*(32*d^4*Sqrt[d + e*x] + (32*d^3*(d + e*x)^(3/2))/3 - (178*d^2*(d + e*x)^(5/2))/5 + (200*d*(d + e*x)^(7/2))/7 - (70*(d + e*x)^(9/2))/9 - 32*d^ (9/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]))/(315*e^4) - (2*d^3*(d + e*x)^(3/2)* (a + b*Log[c*x^n]))/(3*e^4) + (6*d^2*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/( 5*e^4) - (6*d*(d + e*x)^(7/2)*(a + b*Log[c*x^n]))/(7*e^4) + (2*(d + e*x)^( 9/2)*(a + b*Log[c*x^n]))/(9*e^4)
3.2.30.3.1 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[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -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])
\[\int x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e x +d}d x\]
Time = 0.33 (sec) , antiderivative size = 495, normalized size of antiderivative = 2.05 \[ \int x^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {2 \, {\left (5040 \, b d^{\frac {9}{2}} n \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (8776 \, b d^{4} n - 5040 \, a d^{4} - 1225 \, {\left (2 \, b e^{4} n - 9 \, a e^{4}\right )} x^{4} - 25 \, {\left (32 \, b d e^{3} n - 63 \, a d e^{3}\right )} x^{3} + 6 \, {\left (181 \, b d^{2} e^{2} n - 315 \, a d^{2} e^{2}\right )} x^{2} - 4 \, {\left (467 \, b d^{3} e n - 630 \, a d^{3} e\right )} x + 315 \, {\left (35 \, b e^{4} x^{4} + 5 \, b d e^{3} x^{3} - 6 \, b d^{2} e^{2} x^{2} + 8 \, b d^{3} e x - 16 \, b d^{4}\right )} \log \left (c\right ) + 315 \, {\left (35 \, b e^{4} n x^{4} + 5 \, b d e^{3} n x^{3} - 6 \, b d^{2} e^{2} n x^{2} + 8 \, b d^{3} e n x - 16 \, b d^{4} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{99225 \, e^{4}}, \frac {2 \, {\left (10080 \, b \sqrt {-d} d^{4} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (8776 \, b d^{4} n - 5040 \, a d^{4} - 1225 \, {\left (2 \, b e^{4} n - 9 \, a e^{4}\right )} x^{4} - 25 \, {\left (32 \, b d e^{3} n - 63 \, a d e^{3}\right )} x^{3} + 6 \, {\left (181 \, b d^{2} e^{2} n - 315 \, a d^{2} e^{2}\right )} x^{2} - 4 \, {\left (467 \, b d^{3} e n - 630 \, a d^{3} e\right )} x + 315 \, {\left (35 \, b e^{4} x^{4} + 5 \, b d e^{3} x^{3} - 6 \, b d^{2} e^{2} x^{2} + 8 \, b d^{3} e x - 16 \, b d^{4}\right )} \log \left (c\right ) + 315 \, {\left (35 \, b e^{4} n x^{4} + 5 \, b d e^{3} n x^{3} - 6 \, b d^{2} e^{2} n x^{2} + 8 \, b d^{3} e n x - 16 \, b d^{4} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{99225 \, e^{4}}\right ] \]
[2/99225*(5040*b*d^(9/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (8776*b*d^4*n - 5040*a*d^4 - 1225*(2*b*e^4*n - 9*a*e^4)*x^4 - 25*(32*b*d*e ^3*n - 63*a*d*e^3)*x^3 + 6*(181*b*d^2*e^2*n - 315*a*d^2*e^2)*x^2 - 4*(467* b*d^3*e*n - 630*a*d^3*e)*x + 315*(35*b*e^4*x^4 + 5*b*d*e^3*x^3 - 6*b*d^2*e ^2*x^2 + 8*b*d^3*e*x - 16*b*d^4)*log(c) + 315*(35*b*e^4*n*x^4 + 5*b*d*e^3* n*x^3 - 6*b*d^2*e^2*n*x^2 + 8*b*d^3*e*n*x - 16*b*d^4*n)*log(x))*sqrt(e*x + d))/e^4, 2/99225*(10080*b*sqrt(-d)*d^4*n*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (8776*b*d^4*n - 5040*a*d^4 - 1225*(2*b*e^4*n - 9*a*e^4)*x^4 - 25*(32*b* d*e^3*n - 63*a*d*e^3)*x^3 + 6*(181*b*d^2*e^2*n - 315*a*d^2*e^2)*x^2 - 4*(4 67*b*d^3*e*n - 630*a*d^3*e)*x + 315*(35*b*e^4*x^4 + 5*b*d*e^3*x^3 - 6*b*d^ 2*e^2*x^2 + 8*b*d^3*e*x - 16*b*d^4)*log(c) + 315*(35*b*e^4*n*x^4 + 5*b*d*e ^3*n*x^3 - 6*b*d^2*e^2*n*x^2 + 8*b*d^3*e*n*x - 16*b*d^4*n)*log(x))*sqrt(e* x + d))/e^4]
Time = 70.60 (sec) , antiderivative size = 347, normalized size of antiderivative = 1.43 \[ \int x^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=a \left (\begin {cases} - \frac {2 d^{3} \left (d + e x\right )^{\frac {3}{2}}}{3 e^{4}} + \frac {6 d^{2} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} - \frac {6 d \left (d + e x\right )^{\frac {7}{2}}}{7 e^{4}} + \frac {2 \left (d + e x\right )^{\frac {9}{2}}}{9 e^{4}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {17552 d^{\frac {9}{2}} \sqrt {1 + \frac {e x}{d}}}{99225 e^{4}} - \frac {32 d^{\frac {9}{2}} \log {\left (\frac {e x}{d} \right )}}{315 e^{4}} + \frac {64 d^{\frac {9}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{315 e^{4}} + \frac {3736 d^{\frac {7}{2}} x \sqrt {1 + \frac {e x}{d}}}{99225 e^{3}} - \frac {724 d^{\frac {5}{2}} x^{2} \sqrt {1 + \frac {e x}{d}}}{33075 e^{2}} + \frac {64 d^{\frac {3}{2}} x^{3} \sqrt {1 + \frac {e x}{d}}}{3969 e} + \frac {4 \sqrt {d} x^{4} \sqrt {1 + \frac {e x}{d}}}{81} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{4}}{16} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {2 d^{3} \left (d + e x\right )^{\frac {3}{2}}}{3 e^{4}} + \frac {6 d^{2} \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} - \frac {6 d \left (d + e x\right )^{\frac {7}{2}}}{7 e^{4}} + \frac {2 \left (d + e x\right )^{\frac {9}{2}}}{9 e^{4}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{4}}{4} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((-2*d**3*(d + e*x)**(3/2)/(3*e**4) + 6*d**2*(d + e*x)**(5/2)/( 5*e**4) - 6*d*(d + e*x)**(7/2)/(7*e**4) + 2*(d + e*x)**(9/2)/(9*e**4), Ne( e, 0)), (sqrt(d)*x**4/4, True)) - b*n*Piecewise((-17552*d**(9/2)*sqrt(1 + e*x/d)/(99225*e**4) - 32*d**(9/2)*log(e*x/d)/(315*e**4) + 64*d**(9/2)*log( sqrt(1 + e*x/d) + 1)/(315*e**4) + 3736*d**(7/2)*x*sqrt(1 + e*x/d)/(99225*e **3) - 724*d**(5/2)*x**2*sqrt(1 + e*x/d)/(33075*e**2) + 64*d**(3/2)*x**3*s qrt(1 + e*x/d)/(3969*e) + 4*sqrt(d)*x**4*sqrt(1 + e*x/d)/81, (e > -oo) & ( e < oo) & Ne(e, 0)), (sqrt(d)*x**4/16, True)) + b*Piecewise((-2*d**3*(d + e*x)**(3/2)/(3*e**4) + 6*d**2*(d + e*x)**(5/2)/(5*e**4) - 6*d*(d + e*x)**( 7/2)/(7*e**4) + 2*(d + e*x)**(9/2)/(9*e**4), Ne(e, 0)), (sqrt(d)*x**4/4, T rue))*log(c*x**n)
Time = 0.28 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.94 \[ \int x^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4}{99225} \, {\left (\frac {2520 \, d^{\frac {9}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{4}} - \frac {1225 \, {\left (e x + d\right )}^{\frac {9}{2}} - 4500 \, {\left (e x + d\right )}^{\frac {7}{2}} d + 5607 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2} - 1680 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3} - 5040 \, \sqrt {e x + d} d^{4}}{e^{4}}\right )} b n + \frac {2}{315} \, b {\left (\frac {35 \, {\left (e x + d\right )}^{\frac {9}{2}}}{e^{4}} - \frac {135 \, {\left (e x + d\right )}^{\frac {7}{2}} d}{e^{4}} + \frac {189 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2}}{e^{4}} - \frac {105 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3}}{e^{4}}\right )} \log \left (c x^{n}\right ) + \frac {2}{315} \, a {\left (\frac {35 \, {\left (e x + d\right )}^{\frac {9}{2}}}{e^{4}} - \frac {135 \, {\left (e x + d\right )}^{\frac {7}{2}} d}{e^{4}} + \frac {189 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{2}}{e^{4}} - \frac {105 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{3}}{e^{4}}\right )} \]
4/99225*(2520*d^(9/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt( d)))/e^4 - (1225*(e*x + d)^(9/2) - 4500*(e*x + d)^(7/2)*d + 5607*(e*x + d) ^(5/2)*d^2 - 1680*(e*x + d)^(3/2)*d^3 - 5040*sqrt(e*x + d)*d^4)/e^4)*b*n + 2/315*b*(35*(e*x + d)^(9/2)/e^4 - 135*(e*x + d)^(7/2)*d/e^4 + 189*(e*x + d)^(5/2)*d^2/e^4 - 105*(e*x + d)^(3/2)*d^3/e^4)*log(c*x^n) + 2/315*a*(35*( e*x + d)^(9/2)/e^4 - 135*(e*x + d)^(7/2)*d/e^4 + 189*(e*x + d)^(5/2)*d^2/e ^4 - 105*(e*x + d)^(3/2)*d^3/e^4)
\[ \int x^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sqrt {e x + d} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \,d x } \]
Timed out. \[ \int x^3 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )\,\sqrt {d+e\,x} \,d x \]