Integrand size = 21, antiderivative size = 142 \[ \int x \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {8 b d^2 n \sqrt {d+e x}}{15 e^2}+\frac {8 b d n (d+e x)^{3/2}}{45 e^2}-\frac {4 b n (d+e x)^{5/2}}{25 e^2}-\frac {8 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{15 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2} \]
8/45*b*d*n*(e*x+d)^(3/2)/e^2-4/25*b*n*(e*x+d)^(5/2)/e^2-8/15*b*d^(5/2)*n*a rctanh((e*x+d)^(1/2)/d^(1/2))/e^2-2/3*d*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^2+ 2/5*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^2+8/15*b*d^2*n*(e*x+d)^(1/2)/e^2
Time = 0.08 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.82 \[ \int x \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {-120 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+2 \sqrt {d+e x} \left (2 b n \left (31 d^2-8 d e x-9 e^2 x^2\right )+15 a \left (-2 d^2+d e x+3 e^2 x^2\right )+15 b \left (-2 d^2+d e x+3 e^2 x^2\right ) \log \left (c x^n\right )\right )}{225 e^2} \]
(-120*b*d^(5/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 2*Sqrt[d + e*x]*(2*b*n* (31*d^2 - 8*d*e*x - 9*e^2*x^2) + 15*a*(-2*d^2 + d*e*x + 3*e^2*x^2) + 15*b* (-2*d^2 + d*e*x + 3*e^2*x^2)*Log[c*x^n]))/(225*e^2)
Time = 0.28 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2792, 27, 90, 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.
| \(\displaystyle \int x \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {2 (2 d-3 e x) (d+e x)^{3/2}}{15 e^2 x}dx+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b n \int \frac {(2 d-3 e x) (d+e x)^{3/2}}{x}dx}{15 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {2 b n \left (2 d \int \frac {(d+e x)^{3/2}}{x}dx-\frac {6}{5} (d+e x)^{5/2}\right )}{15 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {2 b n \left (2 d \left (d \int \frac {\sqrt {d+e x}}{x}dx+\frac {2}{3} (d+e x)^{3/2}\right )-\frac {6}{5} (d+e x)^{5/2}\right )}{15 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {2 b n \left (2 d \left (d \left (d \int \frac {1}{x \sqrt {d+e x}}dx+2 \sqrt {d+e x}\right )+\frac {2}{3} (d+e x)^{3/2}\right )-\frac {6}{5} (d+e x)^{5/2}\right )}{15 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 b n \left (2 d \left (d \left (\frac {2 d \int \frac {1}{\frac {d+e x}{e}-\frac {d}{e}}d\sqrt {d+e x}}{e}+2 \sqrt {d+e x}\right )+\frac {2}{3} (d+e x)^{3/2}\right )-\frac {6}{5} (d+e x)^{5/2}\right )}{15 e^2}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^2}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {2 b n \left (2 d \left (d \left (2 \sqrt {d+e x}-2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )+\frac {2}{3} (d+e x)^{3/2}\right )-\frac {6}{5} (d+e x)^{5/2}\right )}{15 e^2}\) |
(2*b*n*((-6*(d + e*x)^(5/2))/5 + 2*d*((2*(d + e*x)^(3/2))/3 + d*(2*Sqrt[d + e*x] - 2*Sqrt[d]*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]))))/(15*e^2) - (2*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^2) + (2*(d + e*x)^(5/2)*(a + b*Log[c* x^n]))/(5*e^2)
3.2.32.3.1 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])
\[\int x \left (a +b \ln \left (c \,x^{n}\right )\right ) \sqrt {e x +d}d x\]
Time = 0.33 (sec) , antiderivative size = 291, normalized size of antiderivative = 2.05 \[ \int x \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\left [\frac {2 \, {\left (30 \, b d^{\frac {5}{2}} n \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (62 \, b d^{2} n - 30 \, a d^{2} - 9 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - {\left (16 \, b d e n - 15 \, a d e\right )} x + 15 \, {\left (3 \, b e^{2} x^{2} + b d e x - 2 \, b d^{2}\right )} \log \left (c\right ) + 15 \, {\left (3 \, b e^{2} n x^{2} + b d e n x - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{225 \, e^{2}}, \frac {2 \, {\left (60 \, b \sqrt {-d} d^{2} n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (62 \, b d^{2} n - 30 \, a d^{2} - 9 \, {\left (2 \, b e^{2} n - 5 \, a e^{2}\right )} x^{2} - {\left (16 \, b d e n - 15 \, a d e\right )} x + 15 \, {\left (3 \, b e^{2} x^{2} + b d e x - 2 \, b d^{2}\right )} \log \left (c\right ) + 15 \, {\left (3 \, b e^{2} n x^{2} + b d e n x - 2 \, b d^{2} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{225 \, e^{2}}\right ] \]
[2/225*(30*b*d^(5/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (62* b*d^2*n - 30*a*d^2 - 9*(2*b*e^2*n - 5*a*e^2)*x^2 - (16*b*d*e*n - 15*a*d*e) *x + 15*(3*b*e^2*x^2 + b*d*e*x - 2*b*d^2)*log(c) + 15*(3*b*e^2*n*x^2 + b*d *e*n*x - 2*b*d^2*n)*log(x))*sqrt(e*x + d))/e^2, 2/225*(60*b*sqrt(-d)*d^2*n *arctan(sqrt(e*x + d)*sqrt(-d)/d) + (62*b*d^2*n - 30*a*d^2 - 9*(2*b*e^2*n - 5*a*e^2)*x^2 - (16*b*d*e*n - 15*a*d*e)*x + 15*(3*b*e^2*x^2 + b*d*e*x - 2 *b*d^2)*log(c) + 15*(3*b*e^2*n*x^2 + b*d*e*n*x - 2*b*d^2*n)*log(x))*sqrt(e *x + d))/e^2]
Time = 56.22 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.56 \[ \int x \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=a \left (\begin {cases} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{2}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{2}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {124 d^{\frac {5}{2}} \sqrt {1 + \frac {e x}{d}}}{225 e^{2}} - \frac {4 d^{\frac {5}{2}} \log {\left (\frac {e x}{d} \right )}}{15 e^{2}} + \frac {8 d^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{15 e^{2}} + \frac {32 d^{\frac {3}{2}} x \sqrt {1 + \frac {e x}{d}}}{225 e} + \frac {4 \sqrt {d} x^{2} \sqrt {1 + \frac {e x}{d}}}{25} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {\sqrt {d} x^{2}}{4} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{3 e^{2}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{2}} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} x^{2}}{2} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*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)) - b*n*Piecewise((-124*d**(5/2)*sqrt(1 + e*x/d)/(225*e**2) - 4*d**(5/2)*log(e*x/d)/(15*e**2) + 8*d**(5/2)*log(sqr t(1 + e*x/d) + 1)/(15*e**2) + 32*d**(3/2)*x*sqrt(1 + e*x/d)/(225*e) + 4*sq rt(d)*x**2*sqrt(1 + e*x/d)/25, (e > -oo) & (e < oo) & Ne(e, 0)), (sqrt(d)* x**2/4, True)) + b*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)
Time = 0.29 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.01 \[ \int x \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {4}{225} \, {\left (\frac {15 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{2}} - \frac {9 \, {\left (e x + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x + d\right )}^{\frac {3}{2}} d - 30 \, \sqrt {e x + d} d^{2}}{e^{2}}\right )} b n + \frac {2}{15} \, b {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}}}{e^{2}} - \frac {5 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{2}}\right )} \log \left (c x^{n}\right ) + \frac {2}{15} \, a {\left (\frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}}}{e^{2}} - \frac {5 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{2}}\right )} \]
4/225*(15*d^(5/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d))) /e^2 - (9*(e*x + d)^(5/2) - 10*(e*x + d)^(3/2)*d - 30*sqrt(e*x + d)*d^2)/e ^2)*b*n + 2/15*b*(3*(e*x + d)^(5/2)/e^2 - 5*(e*x + d)^(3/2)*d/e^2)*log(c*x ^n) + 2/15*a*(3*(e*x + d)^(5/2)/e^2 - 5*(e*x + d)^(3/2)*d/e^2)
\[ \int x \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int { \sqrt {e x + d} {\left (b \log \left (c x^{n}\right ) + a\right )} x \,d x } \]
Timed out. \[ \int x \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right ) \, dx=\int x\,\left (a+b\,\ln \left (c\,x^n\right )\right )\,\sqrt {d+e\,x} \,d x \]