Integrand size = 21, antiderivative size = 163 \[ \int x (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {8 b d^3 n \sqrt {d+e x}}{35 e^2}+\frac {8 b d^2 n (d+e x)^{3/2}}{105 e^2}+\frac {8 b d n (d+e x)^{5/2}}{175 e^2}-\frac {4 b n (d+e x)^{7/2}}{49 e^2}-\frac {8 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{35 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2} \]
8/105*b*d^2*n*(e*x+d)^(3/2)/e^2+8/175*b*d*n*(e*x+d)^(5/2)/e^2-4/49*b*n*(e* x+d)^(7/2)/e^2-8/35*b*d^(7/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/e^2-2/5*d*( e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^2+2/7*(e*x+d)^(7/2)*(a+b*ln(c*x^n))/e^2+8/3 5*b*d^3*n*(e*x+d)^(1/2)/e^2
Time = 0.12 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.74 \[ \int x (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {2 \left (420 b d^{7/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+\sqrt {d+e x} \left (105 a (2 d-5 e x) (d+e x)^2+2 b n \left (-247 d^3+71 d^2 e x+183 d e^2 x^2+75 e^3 x^3\right )+105 b (2 d-5 e x) (d+e x)^2 \log \left (c x^n\right )\right )\right )}{3675 e^2} \]
(-2*(420*b*d^(7/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(105*a *(2*d - 5*e*x)*(d + e*x)^2 + 2*b*n*(-247*d^3 + 71*d^2*e*x + 183*d*e^2*x^2 + 75*e^3*x^3) + 105*b*(2*d - 5*e*x)*(d + e*x)^2*Log[c*x^n])))/(3675*e^2)
Time = 0.30 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {2792, 27, 90, 60, 60, 60, 73, 221}
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.
| \(\displaystyle \int x (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {2 (2 d-5 e x) (d+e x)^{5/2}}{35 e^2 x}dx+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b n \int \frac {(2 d-5 e x) (d+e x)^{5/2}}{x}dx}{35 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {2 b n \left (2 d \int \frac {(d+e x)^{5/2}}{x}dx-\frac {10}{7} (d+e x)^{7/2}\right )}{35 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {2 b n \left (2 d \left (d \int \frac {(d+e x)^{3/2}}{x}dx+\frac {2}{5} (d+e x)^{5/2}\right )-\frac {10}{7} (d+e x)^{7/2}\right )}{35 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {2 b n \left (2 d \left (d \left (d \int \frac {\sqrt {d+e x}}{x}dx+\frac {2}{3} (d+e x)^{3/2}\right )+\frac {2}{5} (d+e x)^{5/2}\right )-\frac {10}{7} (d+e x)^{7/2}\right )}{35 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {2 b n \left (2 d \left (d \left (d \left (d \int \frac {1}{x \sqrt {d+e x}}dx+2 \sqrt {d+e x}\right )+\frac {2}{3} (d+e x)^{3/2}\right )+\frac {2}{5} (d+e x)^{5/2}\right )-\frac {10}{7} (d+e x)^{7/2}\right )}{35 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 b n \left (2 d \left (d \left (d \left (\frac {2 d \int \frac {1}{\frac {d+e x}{e}-\frac {d}{e}}d\sqrt {d+e x}}{e}+2 \sqrt {d+e x}\right )+\frac {2}{3} (d+e x)^{3/2}\right )+\frac {2}{5} (d+e x)^{5/2}\right )-\frac {10}{7} (d+e x)^{7/2}\right )}{35 e^2}+\frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 (d+e x)^{7/2} \left (a+b \log \left (c x^n\right )\right )}{7 e^2}-\frac {2 d (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}+\frac {2 b n \left (2 d \left (d \left (d \left (2 \sqrt {d+e x}-2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )+\frac {2}{3} (d+e x)^{3/2}\right )+\frac {2}{5} (d+e x)^{5/2}\right )-\frac {10}{7} (d+e x)^{7/2}\right )}{35 e^2}\) |
(2*b*n*((-10*(d + e*x)^(7/2))/7 + 2*d*((2*(d + e*x)^(5/2))/5 + d*((2*(d + e*x)^(3/2))/3 + d*(2*Sqrt[d + e*x] - 2*Sqrt[d]*ArcTanh[Sqrt[d + e*x]/Sqrt[ d]])))))/(35*e^2) - (2*d*(d + e*x)^(5/2)*(a + b*Log[c*x^n]))/(5*e^2) + (2* (d + e*x)^(7/2)*(a + b*Log[c*x^n]))/(7*e^2)
3.2.39.3.1 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[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p + 2)) Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
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])
\[\int x \left (e x +d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )d x\]
Time = 0.34 (sec) , antiderivative size = 391, normalized size of antiderivative = 2.40 \[ \int x (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {2 \, {\left (210 \, b d^{\frac {7}{2}} n \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (494 \, b d^{3} n - 210 \, a d^{3} - 75 \, {\left (2 \, b e^{3} n - 7 \, a e^{3}\right )} x^{3} - 6 \, {\left (61 \, b d e^{2} n - 140 \, a d e^{2}\right )} x^{2} - {\left (142 \, b d^{2} e n - 105 \, a d^{2} e\right )} x + 105 \, {\left (5 \, b e^{3} x^{3} + 8 \, b d e^{2} x^{2} + b d^{2} e x - 2 \, b d^{3}\right )} \log \left (c\right ) + 105 \, {\left (5 \, b e^{3} n x^{3} + 8 \, b d e^{2} n x^{2} + b d^{2} e n x - 2 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{3675 \, e^{2}}, \frac {2 \, {\left (420 \, b \sqrt {-d} d^{3} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (494 \, b d^{3} n - 210 \, a d^{3} - 75 \, {\left (2 \, b e^{3} n - 7 \, a e^{3}\right )} x^{3} - 6 \, {\left (61 \, b d e^{2} n - 140 \, a d e^{2}\right )} x^{2} - {\left (142 \, b d^{2} e n - 105 \, a d^{2} e\right )} x + 105 \, {\left (5 \, b e^{3} x^{3} + 8 \, b d e^{2} x^{2} + b d^{2} e x - 2 \, b d^{3}\right )} \log \left (c\right ) + 105 \, {\left (5 \, b e^{3} n x^{3} + 8 \, b d e^{2} n x^{2} + b d^{2} e n x - 2 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{3675 \, e^{2}}\right ] \]
[2/3675*(210*b*d^(7/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (4 94*b*d^3*n - 210*a*d^3 - 75*(2*b*e^3*n - 7*a*e^3)*x^3 - 6*(61*b*d*e^2*n - 140*a*d*e^2)*x^2 - (142*b*d^2*e*n - 105*a*d^2*e)*x + 105*(5*b*e^3*x^3 + 8* b*d*e^2*x^2 + b*d^2*e*x - 2*b*d^3)*log(c) + 105*(5*b*e^3*n*x^3 + 8*b*d*e^2 *n*x^2 + b*d^2*e*n*x - 2*b*d^3*n)*log(x))*sqrt(e*x + d))/e^2, 2/3675*(420* b*sqrt(-d)*d^3*n*arctan(sqrt(e*x + d)*sqrt(-d)/d) + (494*b*d^3*n - 210*a*d ^3 - 75*(2*b*e^3*n - 7*a*e^3)*x^3 - 6*(61*b*d*e^2*n - 140*a*d*e^2)*x^2 - ( 142*b*d^2*e*n - 105*a*d^2*e)*x + 105*(5*b*e^3*x^3 + 8*b*d*e^2*x^2 + b*d^2* e*x - 2*b*d^3)*log(c) + 105*(5*b*e^3*n*x^3 + 8*b*d*e^2*n*x^2 + b*d^2*e*n*x - 2*b*d^3*n)*log(x))*sqrt(e*x + d))/e^2]
Time = 113.00 (sec) , antiderivative size = 517, normalized size of antiderivative = 3.17 \[ \int x (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=a d \left (\begin {cases} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{2}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{2}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right ) + a e \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 d n \left (\begin {cases} - \frac {124 d^{\frac {5}{2}} \sqrt {1 + \frac {e x}{d}}}{225 e^{2}} - \frac {4 d^{\frac {5}{2}} \log {\left (\frac {e x}{d} \right )}}{15 e^{2}} + \frac {8 d^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{15 e^{2}} + \frac {32 d^{\frac {3}{2}} x \sqrt {1 + \frac {e x}{d}}}{225 e} + \frac {4 \sqrt {d} x^{2} \sqrt {1 + \frac {e x}{d}}}{25} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{2}}{4} & \text {otherwise} \end {cases}\right ) + b d \left (\begin {cases} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{2}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{2}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} - b e 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 e \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*d*Piecewise((-2*d*(d + e*x)**(3/2)/(3*e**2) + 2*(d + e*x)**(5/2)/(5*e**2 ), Ne(e, 0)), (sqrt(d)*x**2/2, True)) + a*e*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*d*n*Piecewise((-124*d**(5/2)*sqr t(1 + e*x/d)/(225*e**2) - 4*d**(5/2)*log(e*x/d)/(15*e**2) + 8*d**(5/2)*log (sqrt(1 + e*x/d) + 1)/(15*e**2) + 32*d**(3/2)*x*sqrt(1 + e*x/d)/(225*e) + 4*sqrt(d)*x**2*sqrt(1 + e*x/d)/25, (e > -oo) & (e < oo) & Ne(e, 0)), (sqrt (d)*x**2/4, True)) + b*d*Piecewise((-2*d*(d + e*x)**(3/2)/(3*e**2) + 2*(d + e*x)**(5/2)/(5*e**2), Ne(e, 0)), (sqrt(d)*x**2/2, True))*log(c*x**n) - b *e*n*Piecewise((3112*d**(7/2)*sqrt(1 + e*x/d)/(11025*e**3) + 16*d**(7/2)*l og(e*x/d)/(105*e**3) - 32*d**(7/2)*log(sqrt(1 + e*x/d) + 1)/(105*e**3) - 7 16*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) & N e(e, 0)), (sqrt(d)*x**3/9, True)) + b*e*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), N e(e, 0)), (sqrt(d)*x**3/3, True))*log(c*x**n)
Time = 0.31 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.95 \[ \int x (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4}{3675} \, {\left (\frac {105 \, d^{\frac {7}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{2}} - \frac {75 \, {\left (e x + d\right )}^{\frac {7}{2}} - 42 \, {\left (e x + d\right )}^{\frac {5}{2}} d - 70 \, {\left (e x + d\right )}^{\frac {3}{2}} d^{2} - 210 \, \sqrt {e x + d} d^{3}}{e^{2}}\right )} b n + \frac {2}{35} \, {\left (\frac {5 \, {\left (e x + d\right )}^{\frac {7}{2}}}{e^{2}} - \frac {7 \, {\left (e x + d\right )}^{\frac {5}{2}} d}{e^{2}}\right )} b \log \left (c x^{n}\right ) + \frac {2}{35} \, {\left (\frac {5 \, {\left (e x + d\right )}^{\frac {7}{2}}}{e^{2}} - \frac {7 \, {\left (e x + d\right )}^{\frac {5}{2}} d}{e^{2}}\right )} a \]
4/3675*(105*d^(7/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d) ))/e^2 - (75*(e*x + d)^(7/2) - 42*(e*x + d)^(5/2)*d - 70*(e*x + d)^(3/2)*d ^2 - 210*sqrt(e*x + d)*d^3)/e^2)*b*n + 2/35*(5*(e*x + d)^(7/2)/e^2 - 7*(e* x + d)^(5/2)*d/e^2)*b*log(c*x^n) + 2/35*(5*(e*x + d)^(7/2)/e^2 - 7*(e*x + d)^(5/2)*d/e^2)*a
\[ \int x (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { {\left (e x + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )} x \,d x } \]
Timed out. \[ \int x (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x\,\left (a+b\,\ln \left (c\,x^n\right )\right )\,{\left (d+e\,x\right )}^{3/2} \,d x \]