Integrand size = 21, antiderivative size = 119 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {8 b d n \sqrt {d+e x}}{3 e^2}-\frac {4 b n (d+e x)^{3/2}}{9 e^2}-\frac {8 b d^{3/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{3 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2} \]
-4/9*b*n*(e*x+d)^(3/2)/e^2-8/3*b*d^(3/2)*n*arctanh((e*x+d)^(1/2)/d^(1/2))/ e^2+2/3*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^2+8/3*b*d*n*(e*x+d)^(1/2)/e^2-2*d* (a+b*ln(c*x^n))*(e*x+d)^(1/2)/e^2
Time = 0.07 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.67 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=-\frac {2 \left (12 b d^{3/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+\sqrt {d+e x} \left (6 a d-10 b d n-3 a e x+2 b e n x+b (6 d-3 e x) \log \left (c x^n\right )\right )\right )}{9 e^2} \]
(-2*(12*b*d^(3/2)*n*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + Sqrt[d + e*x]*(6*a*d - 10*b*d*n - 3*a*e*x + 2*b*e*n*x + b*(6*d - 3*e*x)*Log[c*x^n])))/(9*e^2)
Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.95, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2792, 27, 90, 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.
| \(\displaystyle \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {2 (2 d-e x) \sqrt {d+e x}}{3 e^2 x}dx+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 b n \int \frac {(2 d-e x) \sqrt {d+e x}}{x}dx}{3 e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 90 |
| \(\displaystyle \frac {2 b n \left (2 d \int \frac {\sqrt {d+e x}}{x}dx-\frac {2}{3} (d+e x)^{3/2}\right )}{3 e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 60 |
| \(\displaystyle \frac {2 b n \left (2 d \left (d \int \frac {1}{x \sqrt {d+e x}}dx+2 \sqrt {d+e x}\right )-\frac {2}{3} (d+e x)^{3/2}\right )}{3 e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {2 b n \left (2 d \left (\frac {2 d \int \frac {1}{\frac {d+e x}{e}-\frac {d}{e}}d\sqrt {d+e x}}{e}+2 \sqrt {d+e x}\right )-\frac {2}{3} (d+e x)^{3/2}\right )}{3 e^2}+\frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {2 (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{3 e^2}-\frac {2 d \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^2}+\frac {2 b n \left (2 d \left (2 \sqrt {d+e x}-2 \sqrt {d} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )-\frac {2}{3} (d+e x)^{3/2}\right )}{3 e^2}\) |
(2*b*n*((-2*(d + e*x)^(3/2))/3 + 2*d*(2*Sqrt[d + e*x] - 2*Sqrt[d]*ArcTanh[ Sqrt[d + e*x]/Sqrt[d]])))/(3*e^2) - (2*d*Sqrt[d + e*x]*(a + b*Log[c*x^n])) /e^2 + (2*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/(3*e^2)
3.2.46.3.1 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])
\[\int \frac {x \left (a +b \ln \left (c \,x^{n}\right )\right )}{\sqrt {e x +d}}d x\]
Time = 0.31 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.59 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\left [\frac {2 \, {\left (6 \, b d^{\frac {3}{2}} n \log \left (\frac {e x - 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) + {\left (10 \, b d n - 6 \, a d - {\left (2 \, b e n - 3 \, a e\right )} x + 3 \, {\left (b e x - 2 \, b d\right )} \log \left (c\right ) + 3 \, {\left (b e n x - 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{9 \, e^{2}}, \frac {2 \, {\left (12 \, b \sqrt {-d} d n \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (10 \, b d n - 6 \, a d - {\left (2 \, b e n - 3 \, a e\right )} x + 3 \, {\left (b e x - 2 \, b d\right )} \log \left (c\right ) + 3 \, {\left (b e n x - 2 \, b d n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{9 \, e^{2}}\right ] \]
[2/9*(6*b*d^(3/2)*n*log((e*x - 2*sqrt(e*x + d)*sqrt(d) + 2*d)/x) + (10*b*d *n - 6*a*d - (2*b*e*n - 3*a*e)*x + 3*(b*e*x - 2*b*d)*log(c) + 3*(b*e*n*x - 2*b*d*n)*log(x))*sqrt(e*x + d))/e^2, 2/9*(12*b*sqrt(-d)*d*n*arctan(sqrt(e *x + d)*sqrt(-d)/d) + (10*b*d*n - 6*a*d - (2*b*e*n - 3*a*e)*x + 3*(b*e*x - 2*b*d)*log(c) + 3*(b*e*n*x - 2*b*d*n)*log(x))*sqrt(e*x + d))/e^2]
Time = 32.67 (sec) , antiderivative size = 272, normalized size of antiderivative = 2.29 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=a \left (\begin {cases} - \frac {2 d \sqrt {d + e x}}{e^{2}} + \frac {2 \left (d + e x\right )^{\frac {3}{2}}}{3 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{2}}{2 \sqrt {d}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} \frac {16 d^{\frac {3}{2}} \sqrt {1 + \frac {e x}{d}}}{9 e^{2}} + \frac {2 d^{\frac {3}{2}} \log {\left (\frac {e x}{d} \right )}}{3 e^{2}} - \frac {4 d^{\frac {3}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{3 e^{2}} + \frac {4 d^{\frac {3}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} \sqrt {x}} \right )}}{e^{2}} + \frac {4 \sqrt {d} x \sqrt {1 + \frac {e x}{d}}}{9 e} - \frac {4 d^{2}}{e^{\frac {5}{2}} \sqrt {x} \sqrt {\frac {d}{e x} + 1}} - \frac {4 d \sqrt {x}}{e^{\frac {3}{2}} \sqrt {\frac {d}{e x} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{2}}{4 \sqrt {d}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} - \frac {2 d \sqrt {d + e x}}{e^{2}} + \frac {2 \left (d + e x\right )^{\frac {3}{2}}}{3 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{2}}{2 \sqrt {d}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((-2*d*sqrt(d + e*x)/e**2 + 2*(d + e*x)**(3/2)/(3*e**2), Ne(e, 0)), (x**2/(2*sqrt(d)), True)) - b*n*Piecewise((16*d**(3/2)*sqrt(1 + e*x/d )/(9*e**2) + 2*d**(3/2)*log(e*x/d)/(3*e**2) - 4*d**(3/2)*log(sqrt(1 + e*x/ d) + 1)/(3*e**2) + 4*d**(3/2)*asinh(sqrt(d)/(sqrt(e)*sqrt(x)))/e**2 + 4*sq rt(d)*x*sqrt(1 + e*x/d)/(9*e) - 4*d**2/(e**(5/2)*sqrt(x)*sqrt(d/(e*x) + 1) ) - 4*d*sqrt(x)/(e**(3/2)*sqrt(d/(e*x) + 1)), (e > -oo) & (e < oo) & Ne(e, 0)), (x**2/(4*sqrt(d)), True)) + b*Piecewise((-2*d*sqrt(d + e*x)/e**2 + 2 *(d + e*x)**(3/2)/(3*e**2), Ne(e, 0)), (x**2/(2*sqrt(d)), True))*log(c*x** n)
Time = 0.27 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.07 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {4}{9} \, b n {\left (\frac {3 \, d^{\frac {3}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{2}} - \frac {{\left (e x + d\right )}^{\frac {3}{2}} - 6 \, \sqrt {e x + d} d}{e^{2}}\right )} + \frac {2}{3} \, b {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}}}{e^{2}} - \frac {3 \, \sqrt {e x + d} d}{e^{2}}\right )} \log \left (c x^{n}\right ) + \frac {2}{3} \, a {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}}}{e^{2}} - \frac {3 \, \sqrt {e x + d} d}{e^{2}}\right )} \]
4/9*b*n*(3*d^(3/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)) )/e^2 - ((e*x + d)^(3/2) - 6*sqrt(e*x + d)*d)/e^2) + 2/3*b*((e*x + d)^(3/2 )/e^2 - 3*sqrt(e*x + d)*d/e^2)*log(c*x^n) + 2/3*a*((e*x + d)^(3/2)/e^2 - 3 *sqrt(e*x + d)*d/e^2)
Time = 0.36 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.10 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\frac {8 \, b d^{2} n \arctan \left (\frac {\sqrt {e x + d}}{\sqrt {-d}}\right )}{3 \, \sqrt {-d} e^{2}} + \frac {2}{3} \, {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}} b n}{e^{2}} - \frac {3 \, \sqrt {e x + d} b d n}{e^{2}}\right )} \log \left (e x\right ) - \frac {2 \, {\left (3 \, b n \log \left (e\right ) + 2 \, b n - 3 \, b \log \left (c\right ) - 3 \, a\right )} {\left (e x + d\right )}^{\frac {3}{2}}}{9 \, e^{2}} + \frac {2 \, {\left (3 \, b d n \log \left (e\right ) + 4 \, b d n - 3 \, b d \log \left (c\right ) - 3 \, a d\right )} \sqrt {e x + d}}{3 \, e^{2}} \]
8/3*b*d^2*n*arctan(sqrt(e*x + d)/sqrt(-d))/(sqrt(-d)*e^2) + 2/3*((e*x + d) ^(3/2)*b*n/e^2 - 3*sqrt(e*x + d)*b*d*n/e^2)*log(e*x) - 2/9*(3*b*n*log(e) + 2*b*n - 3*b*log(c) - 3*a)*(e*x + d)^(3/2)/e^2 + 2/3*(3*b*d*n*log(e) + 4*b *d*n - 3*b*d*log(c) - 3*a*d)*sqrt(e*x + d)/e^2
Timed out. \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\sqrt {d+e x}} \, dx=\int \frac {x\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{\sqrt {d+e\,x}} \,d x \]