Integrand size = 23, antiderivative size = 263 \[ \int x^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {64 b d^5 n \sqrt {d+e x}}{1155 e^4}+\frac {64 b d^4 n (d+e x)^{3/2}}{3465 e^4}+\frac {64 b d^3 n (d+e x)^{5/2}}{5775 e^4}-\frac {172 b d^2 n (d+e x)^{7/2}}{1617 e^4}+\frac {32 b d n (d+e x)^{9/2}}{297 e^4}-\frac {4 b n (d+e x)^{11/2}}{121 e^4}-\frac {64 b d^{11/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{1155 e^4}-\frac {2 d^3 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {6 d^2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}-\frac {2 d (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}+\frac {2 (d+e x)^{11/2} \left (a+b \log \left (c x^n\right )\right )}{11 e^4} \]
64/3465*b*d^4*n*(e*x+d)^(3/2)/e^4+64/5775*b*d^3*n*(e*x+d)^(5/2)/e^4-172/16 17*b*d^2*n*(e*x+d)^(7/2)/e^4+32/297*b*d*n*(e*x+d)^(9/2)/e^4-4/121*b*n*(e*x +d)^(11/2)/e^4-64/1155*b*d^(11/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e^4-2/5 *d^3*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^4+6/7*d^2*(e*x+d)^(7/2)*(a+b*ln(c*x^n ))/e^4-2/3*d*(e*x+d)^(9/2)*(a+b*ln(c*x^n))/e^4+2/11*(e*x+d)^(11/2)*(a+b*ln (c*x^n))/e^4+64/1155*b*d^5*n*(e*x+d)^(1/2)/e^4
Time = 0.19 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.71 \[ \int x^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {-221760 b d^{11/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+2 \sqrt {d+e x} \left (-3465 a (d+e x)^2 \left (16 d^3-40 d^2 e x+70 d e^2 x^2-105 e^3 x^3\right )+2 b n \left (53308 d^5-12794 d^4 e x+7863 d^3 e^2 x^2-5975 d^2 e^3 x^3-57575 d e^4 x^4-33075 e^5 x^5\right )-3465 b (d+e x)^2 \left (16 d^3-40 d^2 e x+70 d e^2 x^2-105 e^3 x^3\right ) \log \left (c x^n\right )\right )}{4002075 e^4} \]
(-221760*b*d^(11/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 2*Sqrt[d + e*x]*(-3 465*a*(d + e*x)^2*(16*d^3 - 40*d^2*e*x + 70*d*e^2*x^2 - 105*e^3*x^3) + 2*b *n*(53308*d^5 - 12794*d^4*e*x + 7863*d^3*e^2*x^2 - 5975*d^2*e^3*x^3 - 5757 5*d*e^4*x^4 - 33075*e^5*x^5) - 3465*b*(d + e*x)^2*(16*d^3 - 40*d^2*e*x + 7 0*d*e^2*x^2 - 105*e^3*x^3)*Log[c*x^n]))/(4002075*e^4)
Time = 0.52 (sec) , antiderivative size = 234, normalized size of antiderivative = 0.89, 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.
| \(\displaystyle \int x^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {2 (d+e x)^{5/2} \left (16 d^3-40 e x d^2+70 e^2 x^2 d-105 e^3 x^3\right )}{1155 e^4 x}dx-\frac {2 d^3 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {6 d^2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {2 (d+e x)^{11/2} \left (a+b \log \left (c x^n\right )\right )}{11 e^4}-\frac {2 d (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b n \int \frac {(d+e x)^{5/2} \left (16 d^3-40 e x d^2+70 e^2 x^2 d-105 e^3 x^3\right )}{x}dx}{1155 e^4}-\frac {2 d^3 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {6 d^2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {2 (d+e x)^{11/2} \left (a+b \log \left (c x^n\right )\right )}{11 e^4}-\frac {2 d (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle \frac {2 b n \int \left (-105 e (d+e x)^{9/2}+280 d e (d+e x)^{7/2}-215 d^2 e (d+e x)^{5/2}+\frac {16 d^3 (d+e x)^{5/2}}{x}\right )dx}{1155 e^4}-\frac {2 d^3 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {6 d^2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {2 (d+e x)^{11/2} \left (a+b \log \left (c x^n\right )\right )}{11 e^4}-\frac {2 d (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {2 d^3 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}+\frac {6 d^2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^4}+\frac {2 (d+e x)^{11/2} \left (a+b \log \left (c x^n\right )\right )}{11 e^4}-\frac {2 d (d+e x)^{9/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^4}+\frac {2 b n \left (-32 d^{11/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+32 d^5 \sqrt {d+e x}+\frac {32}{3} d^4 (d+e x)^{3/2}+\frac {32}{5} d^3 (d+e x)^{5/2}-\frac {430}{7} d^2 (d+e x)^{7/2}+\frac {560}{9} d (d+e x)^{9/2}-\frac {210}{11} (d+e x)^{11/2}\right )}{1155 e^4}\) |
(2*b*n*(32*d^5*Sqrt[d + e*x] + (32*d^4*(d + e*x)^(3/2))/3 + (32*d^3*(d + e *x)^(5/2))/5 - (430*d^2*(d + e*x)^(7/2))/7 + (560*d*(d + e*x)^(9/2))/9 - ( 210*(d + e*x)^(11/2))/11 - 32*d^(11/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]))/(1 155*e^4) - (2*d^3*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^4) + (6*d^2*(d + e*x)^(7/2)*(a + b*Log[c*x^n]))/(7*e^4) - (2*d*(d + e*x)^(9/2)*(a + b*Log [c*x^n]))/(3*e^4) + (2*(d + e*x)^(11/2)*(a + b*Log[c*x^n]))/(11*e^4)
3.2.37.3.1 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])
\[\int x^{3} \left (e x +d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]
Time = 0.34 (sec) , antiderivative size = 595, normalized size of antiderivative = 2.26 \[ \int x^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {2 \, {\left (55440 \, b d^{\frac {11}{2}} n \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (106616 \, b d^{5} n - 55440 \, a d^{5} - 33075 \, {\left (2 \, b e^{5} n - 11 \, a e^{5}\right )} x^{5} - 2450 \, {\left (47 \, b d e^{4} n - 198 \, a d e^{4}\right )} x^{4} - 25 \, {\left (478 \, b d^{2} e^{3} n - 693 \, a d^{2} e^{3}\right )} x^{3} + 6 \, {\left (2621 \, b d^{3} e^{2} n - 3465 \, a d^{3} e^{2}\right )} x^{2} - 4 \, {\left (6397 \, b d^{4} e n - 6930 \, a d^{4} e\right )} x + 3465 \, {\left (105 \, b e^{5} x^{5} + 140 \, b d e^{4} x^{4} + 5 \, b d^{2} e^{3} x^{3} - 6 \, b d^{3} e^{2} x^{2} + 8 \, b d^{4} e x - 16 \, b d^{5}\right )} \log \left (c\right ) + 3465 \, {\left (105 \, b e^{5} n x^{5} + 140 \, b d e^{4} n x^{4} + 5 \, b d^{2} e^{3} n x^{3} - 6 \, b d^{3} e^{2} n x^{2} + 8 \, b d^{4} e n x - 16 \, b d^{5} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{4002075 \, e^{4}}, \frac {2 \, {\left (110880 \, b \sqrt {-d} d^{5} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (106616 \, b d^{5} n - 55440 \, a d^{5} - 33075 \, {\left (2 \, b e^{5} n - 11 \, a e^{5}\right )} x^{5} - 2450 \, {\left (47 \, b d e^{4} n - 198 \, a d e^{4}\right )} x^{4} - 25 \, {\left (478 \, b d^{2} e^{3} n - 693 \, a d^{2} e^{3}\right )} x^{3} + 6 \, {\left (2621 \, b d^{3} e^{2} n - 3465 \, a d^{3} e^{2}\right )} x^{2} - 4 \, {\left (6397 \, b d^{4} e n - 6930 \, a d^{4} e\right )} x + 3465 \, {\left (105 \, b e^{5} x^{5} + 140 \, b d e^{4} x^{4} + 5 \, b d^{2} e^{3} x^{3} - 6 \, b d^{3} e^{2} x^{2} + 8 \, b d^{4} e x - 16 \, b d^{5}\right )} \log \left (c\right ) + 3465 \, {\left (105 \, b e^{5} n x^{5} + 140 \, b d e^{4} n x^{4} + 5 \, b d^{2} e^{3} n x^{3} - 6 \, b d^{3} e^{2} n x^{2} + 8 \, b d^{4} e n x - 16 \, b d^{5} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{4002075 \, e^{4}}\right ] \]
[2/4002075*(55440*b*d^(11/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x ) + (106616*b*d^5*n - 55440*a*d^5 - 33075*(2*b*e^5*n - 11*a*e^5)*x^5 - 245 0*(47*b*d*e^4*n - 198*a*d*e^4)*x^4 - 25*(478*b*d^2*e^3*n - 693*a*d^2*e^3)* x^3 + 6*(2621*b*d^3*e^2*n - 3465*a*d^3*e^2)*x^2 - 4*(6397*b*d^4*e*n - 6930 *a*d^4*e)*x + 3465*(105*b*e^5*x^5 + 140*b*d*e^4*x^4 + 5*b*d^2*e^3*x^3 - 6* b*d^3*e^2*x^2 + 8*b*d^4*e*x - 16*b*d^5)*log(c) + 3465*(105*b*e^5*n*x^5 + 1 40*b*d*e^4*n*x^4 + 5*b*d^2*e^3*n*x^3 - 6*b*d^3*e^2*n*x^2 + 8*b*d^4*e*n*x - 16*b*d^5*n)*log(x))*sqrt(e*x + d))/e^4, 2/4002075*(110880*b*sqrt(-d)*d^5* n*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (106616*b*d^5*n - 55440*a*d^5 - 33075 *(2*b*e^5*n - 11*a*e^5)*x^5 - 2450*(47*b*d*e^4*n - 198*a*d*e^4)*x^4 - 25*( 478*b*d^2*e^3*n - 693*a*d^2*e^3)*x^3 + 6*(2621*b*d^3*e^2*n - 3465*a*d^3*e^ 2)*x^2 - 4*(6397*b*d^4*e*n - 6930*a*d^4*e)*x + 3465*(105*b*e^5*x^5 + 140*b *d*e^4*x^4 + 5*b*d^2*e^3*x^3 - 6*b*d^3*e^2*x^2 + 8*b*d^4*e*x - 16*b*d^5)*l og(c) + 3465*(105*b*e^5*n*x^5 + 140*b*d*e^4*n*x^4 + 5*b*d^2*e^3*n*x^3 - 6* b*d^3*e^2*n*x^2 + 8*b*d^4*e*n*x - 16*b*d^5*n)*log(x))*sqrt(e*x + d))/e^4]
Timed out. \[ \int x^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 239, normalized size of antiderivative = 0.91 \[ \int x^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4}{4002075} \, {\left (\frac {27720 \, d^{\frac {11}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{4}} - \frac {33075 \, {\left (e x + d\right )}^{\frac {11}{2}} - 107800 \, {\left (e x + d\right )}^{\frac {9}{2}} d + 106425 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{2} - 11088 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{3} - 18480 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{4} - 55440 \, \sqrt {e x + d} d^{5}}{e^{4}}\right )} b n + \frac {2}{1155} \, {\left (\frac {105 \, {\left (e x + d\right )}^{\frac {11}{2}}}{e^{4}} - \frac {385 \, {\left (e x + d\right )}^{\frac {9}{2}} d}{e^{4}} + \frac {495 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{2}}{e^{4}} - \frac {231 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{3}}{e^{4}}\right )} b \log \left (c x^{n}\right ) + \frac {2}{1155} \, {\left (\frac {105 \, {\left (e x + d\right )}^{\frac {11}{2}}}{e^{4}} - \frac {385 \, {\left (e x + d\right )}^{\frac {9}{2}} d}{e^{4}} + \frac {495 \, {\left (e x + d\right )}^{\frac {7}{2}} d^{2}}{e^{4}} - \frac {231 \, {\left (e x + d\right )}^{\frac {5}{2}} d^{3}}{e^{4}}\right )} a \]
4/4002075*(27720*d^(11/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + s qrt(d)))/e^4 - (33075*(e*x + d)^(11/2) - 107800*(e*x + d)^(9/2)*d + 106425 *(e*x + d)^(7/2)*d^2 - 11088*(e*x + d)^(5/2)*d^3 - 18480*(e*x + d)^(3/2)*d ^4 - 55440*sqrt(e*x + d)*d^5)/e^4)*b*n + 2/1155*(105*(e*x + d)^(11/2)/e^4 - 385*(e*x + d)^(9/2)*d/e^4 + 495*(e*x + d)^(7/2)*d^2/e^4 - 231*(e*x + d)^ (5/2)*d^3/e^4)*b*log(c*x^n) + 2/1155*(105*(e*x + d)^(11/2)/e^4 - 385*(e*x + d)^(9/2)*d/e^4 + 495*(e*x + d)^(7/2)*d^2/e^4 - 231*(e*x + d)^(5/2)*d^3/e ^4)*a
\[ \int x^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (e x + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )} x^{3} \,d x } \]
Timed out. \[ \int x^3 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )\,{\left (d+e\,x\right )}^{3/2} \,d x \]