Integrand size = 23, antiderivative size = 194 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=-\frac {44 b d^2 n \sqrt {d+e x}}{5 e^4}+\frac {16 b d n (d+e x)^{3/2}}{15 e^4}-\frac {4 b n (d+e x)^{5/2}}{25 e^4}+\frac {64 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{5 e^4}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4} \]
16/15*b*d*n*(e*x+d)^(3/2)/e^4-4/25*b*n*(e*x+d)^(5/2)/e^4+64/5*b*d^(5/2)*n* arctanh((e*x+d)^(1/2)/d^(1/2))/e^4-2*d*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/e^4+2 /5*(e*x+d)^(5/2)*(a+b*ln(c*x^n))/e^4+2*d^3*(a+b*ln(c*x^n))/e^4/(e*x+d)^(1/ 2)-44/5*b*d^2*n*(e*x+d)^(1/2)/e^4+6*d^2*(a+b*ln(c*x^n))*(e*x+d)^(1/2)/e^4
Time = 0.09 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.82 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=\frac {480 a d^3-592 b d^3 n+240 a d^2 e x-536 b d^2 e n x-60 a d e^2 x^2+44 b d e^2 n x^2+30 a e^3 x^3-12 b e^3 n x^3+960 b d^{5/2} n \sqrt {d+e x} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+30 b \left (16 d^3+8 d^2 e x-2 d e^2 x^2+e^3 x^3\right ) \log \left (c x^n\right )}{75 e^4 \sqrt {d+e x}} \]
(480*a*d^3 - 592*b*d^3*n + 240*a*d^2*e*x - 536*b*d^2*e*n*x - 60*a*d*e^2*x^ 2 + 44*b*d*e^2*n*x^2 + 30*a*e^3*x^3 - 12*b*e^3*n*x^3 + 960*b*d^(5/2)*n*Sqr t[d + e*x]*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 30*b*(16*d^3 + 8*d^2*e*x - 2*d *e^2*x^2 + e^3*x^3)*Log[c*x^n])/(75*e^4*Sqrt[d + e*x])
Time = 0.47 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.93, 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.
| \(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int \frac {2 \left (16 d^3+8 e x d^2-2 e^2 x^2 d+e^3 x^3\right )}{5 e^4 x \sqrt {d+e x}}dx+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 b n \int \frac {16 d^3+8 e x d^2-2 e^2 x^2 d+e^3 x^3}{x \sqrt {d+e x}}dx}{5 e^4}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}\) |
| \(\Big \downarrow \) 2123 |
| \(\displaystyle -\frac {2 b n \int \left (\frac {16 d^3}{x \sqrt {d+e x}}+\frac {11 e d^2}{\sqrt {d+e x}}-4 e \sqrt {d+e x} d+e (d+e x)^{3/2}\right )dx}{5 e^4}+\frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {2 d^3 \left (a+b \log \left (c x^n\right )\right )}{e^4 \sqrt {d+e x}}+\frac {6 d^2 \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{e^4}-\frac {2 d (d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{e^4}+\frac {2 (d+e x)^{5/2} \left (a+b \log \left (c x^n\right )\right )}{5 e^4}-\frac {2 b n \left (-32 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+22 d^2 \sqrt {d+e x}-\frac {8}{3} d (d+e x)^{3/2}+\frac {2}{5} (d+e x)^{5/2}\right )}{5 e^4}\) |
(-2*b*n*(22*d^2*Sqrt[d + e*x] - (8*d*(d + e*x)^(3/2))/3 + (2*(d + e*x)^(5/ 2))/5 - 32*d^(5/2)*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]))/(5*e^4) + (2*d^3*(a + b*Log[c*x^n]))/(e^4*Sqrt[d + e*x]) + (6*d^2*Sqrt[d + e*x]*(a + b*Log[c*x^n ]))/e^4 - (2*d*(d + e*x)^(3/2)*(a + b*Log[c*x^n]))/e^4 + (2*(d + e*x)^(5/2 )*(a + b*Log[c*x^n]))/(5*e^4)
3.2.51.3.1 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])
\[\int \frac {x^{3} \left (a +b \ln \left (c \,x^{n}\right )\right )}{\left (e x +d \right )^{\frac {3}{2}}}d x\]
Time = 0.34 (sec) , antiderivative size = 435, normalized size of antiderivative = 2.24 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=\left [\frac {2 \, {\left (240 \, {\left (b d^{2} e n x + b d^{3} n\right )} \sqrt {d} \log \left (\frac {e x + 2 \, \sqrt {e x + d} \sqrt {d} + 2 \, d}{x}\right ) - {\left (296 \, b d^{3} n - 240 \, a d^{3} + 3 \, {\left (2 \, b e^{3} n - 5 \, a e^{3}\right )} x^{3} - 2 \, {\left (11 \, b d e^{2} n - 15 \, a d e^{2}\right )} x^{2} + 4 \, {\left (67 \, b d^{2} e n - 30 \, a d^{2} e\right )} x - 15 \, {\left (b e^{3} x^{3} - 2 \, b d e^{2} x^{2} + 8 \, b d^{2} e x + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b e^{3} n x^{3} - 2 \, b d e^{2} n x^{2} + 8 \, b d^{2} e n x + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{75 \, {\left (e^{5} x + d e^{4}\right )}}, -\frac {2 \, {\left (480 \, {\left (b d^{2} e n x + b d^{3} n\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {e x + d} \sqrt {-d}}{d}\right ) + {\left (296 \, b d^{3} n - 240 \, a d^{3} + 3 \, {\left (2 \, b e^{3} n - 5 \, a e^{3}\right )} x^{3} - 2 \, {\left (11 \, b d e^{2} n - 15 \, a d e^{2}\right )} x^{2} + 4 \, {\left (67 \, b d^{2} e n - 30 \, a d^{2} e\right )} x - 15 \, {\left (b e^{3} x^{3} - 2 \, b d e^{2} x^{2} + 8 \, b d^{2} e x + 16 \, b d^{3}\right )} \log \left (c\right ) - 15 \, {\left (b e^{3} n x^{3} - 2 \, b d e^{2} n x^{2} + 8 \, b d^{2} e n x + 16 \, b d^{3} n\right )} \log \left (x\right )\right )} \sqrt {e x + d}\right )}}{75 \, {\left (e^{5} x + d e^{4}\right )}}\right ] \]
[2/75*(240*(b*d^2*e*n*x + b*d^3*n)*sqrt(d)*log((e*x + 2*sqrt(e*x + d)*sqrt (d) + 2*d)/x) - (296*b*d^3*n - 240*a*d^3 + 3*(2*b*e^3*n - 5*a*e^3)*x^3 - 2 *(11*b*d*e^2*n - 15*a*d*e^2)*x^2 + 4*(67*b*d^2*e*n - 30*a*d^2*e)*x - 15*(b *e^3*x^3 - 2*b*d*e^2*x^2 + 8*b*d^2*e*x + 16*b*d^3)*log(c) - 15*(b*e^3*n*x^ 3 - 2*b*d*e^2*n*x^2 + 8*b*d^2*e*n*x + 16*b*d^3*n)*log(x))*sqrt(e*x + d))/( e^5*x + d*e^4), -2/75*(480*(b*d^2*e*n*x + b*d^3*n)*sqrt(-d)*arctan(sqrt(e* x + d)*sqrt(-d)/d) + (296*b*d^3*n - 240*a*d^3 + 3*(2*b*e^3*n - 5*a*e^3)*x^ 3 - 2*(11*b*d*e^2*n - 15*a*d*e^2)*x^2 + 4*(67*b*d^2*e*n - 30*a*d^2*e)*x - 15*(b*e^3*x^3 - 2*b*d*e^2*x^2 + 8*b*d^2*e*x + 16*b*d^3)*log(c) - 15*(b*e^3 *n*x^3 - 2*b*d*e^2*n*x^2 + 8*b*d^2*e*n*x + 16*b*d^3*n)*log(x))*sqrt(e*x + d))/(e^5*x + d*e^4)]
Time = 116.99 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.90 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=a \left (\begin {cases} \frac {2 d^{3}}{e^{4} \sqrt {d + e x}} + \frac {6 d^{2} \sqrt {d + e x}}{e^{4}} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{e^{4}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) - b n \left (\begin {cases} - \frac {308 d^{\frac {5}{2}} \sqrt {1 + \frac {e x}{d}}}{75 e^{4}} - \frac {8 d^{\frac {5}{2}} \log {\left (\frac {e x}{d} \right )}}{5 e^{4}} + \frac {16 d^{\frac {5}{2}} \log {\left (\sqrt {1 + \frac {e x}{d}} + 1 \right )}}{5 e^{4}} - \frac {16 d^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {d}}{\sqrt {e} \sqrt {x}} \right )}}{e^{4}} - \frac {56 d^{\frac {3}{2}} x \sqrt {1 + \frac {e x}{d}}}{75 e^{3}} + \frac {4 \sqrt {d} x^{2} \sqrt {1 + \frac {e x}{d}}}{25 e^{2}} + \frac {12 d^{3}}{e^{\frac {9}{2}} \sqrt {x} \sqrt {\frac {d}{e x} + 1}} + \frac {12 d^{2} \sqrt {x}}{e^{\frac {7}{2}} \sqrt {\frac {d}{e x} + 1}} & \text {for}\: e > -\infty \wedge e < \infty \wedge e \neq 0 \\\frac {x^{4}}{16 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + b \left (\begin {cases} \frac {2 d^{3}}{e^{4} \sqrt {d + e x}} + \frac {6 d^{2} \sqrt {d + e x}}{e^{4}} - \frac {2 d \left (d + e x\right )^{\frac {3}{2}}}{e^{4}} + \frac {2 \left (d + e x\right )^{\frac {5}{2}}}{5 e^{4}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 d^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} \]
a*Piecewise((2*d**3/(e**4*sqrt(d + e*x)) + 6*d**2*sqrt(d + e*x)/e**4 - 2*d *(d + e*x)**(3/2)/e**4 + 2*(d + e*x)**(5/2)/(5*e**4), Ne(e, 0)), (x**4/(4* d**(3/2)), True)) - b*n*Piecewise((-308*d**(5/2)*sqrt(1 + e*x/d)/(75*e**4) - 8*d**(5/2)*log(e*x/d)/(5*e**4) + 16*d**(5/2)*log(sqrt(1 + e*x/d) + 1)/( 5*e**4) - 16*d**(5/2)*asinh(sqrt(d)/(sqrt(e)*sqrt(x)))/e**4 - 56*d**(3/2)* x*sqrt(1 + e*x/d)/(75*e**3) + 4*sqrt(d)*x**2*sqrt(1 + e*x/d)/(25*e**2) + 1 2*d**3/(e**(9/2)*sqrt(x)*sqrt(d/(e*x) + 1)) + 12*d**2*sqrt(x)/(e**(7/2)*sq rt(d/(e*x) + 1)), (e > -oo) & (e < oo) & Ne(e, 0)), (x**4/(16*d**(3/2)), T rue)) + b*Piecewise((2*d**3/(e**4*sqrt(d + e*x)) + 6*d**2*sqrt(d + e*x)/e* *4 - 2*d*(d + e*x)**(3/2)/e**4 + 2*(d + e*x)**(5/2)/(5*e**4), Ne(e, 0)), ( x**4/(4*d**(3/2)), True))*log(c*x**n)
Time = 0.28 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.03 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=-\frac {4}{75} \, b n {\left (\frac {120 \, d^{\frac {5}{2}} \log \left (\frac {\sqrt {e x + d} - \sqrt {d}}{\sqrt {e x + d} + \sqrt {d}}\right )}{e^{4}} + \frac {3 \, {\left (e x + d\right )}^{\frac {5}{2}} - 20 \, {\left (e x + d\right )}^{\frac {3}{2}} d + 165 \, \sqrt {e x + d} d^{2}}{e^{4}}\right )} + \frac {2}{5} \, b {\left (\frac {{\left (e x + d\right )}^{\frac {5}{2}}}{e^{4}} - \frac {5 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{4}} + \frac {15 \, \sqrt {e x + d} d^{2}}{e^{4}} + \frac {5 \, d^{3}}{\sqrt {e x + d} e^{4}}\right )} \log \left (c x^{n}\right ) + \frac {2}{5} \, a {\left (\frac {{\left (e x + d\right )}^{\frac {5}{2}}}{e^{4}} - \frac {5 \, {\left (e x + d\right )}^{\frac {3}{2}} d}{e^{4}} + \frac {15 \, \sqrt {e x + d} d^{2}}{e^{4}} + \frac {5 \, d^{3}}{\sqrt {e x + d} e^{4}}\right )} \]
-4/75*b*n*(120*d^(5/2)*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt (d)))/e^4 + (3*(e*x + d)^(5/2) - 20*(e*x + d)^(3/2)*d + 165*sqrt(e*x + d)* d^2)/e^4) + 2/5*b*((e*x + d)^(5/2)/e^4 - 5*(e*x + d)^(3/2)*d/e^4 + 15*sqrt (e*x + d)*d^2/e^4 + 5*d^3/(sqrt(e*x + d)*e^4))*log(c*x^n) + 2/5*a*((e*x + d)^(5/2)/e^4 - 5*(e*x + d)^(3/2)*d/e^4 + 15*sqrt(e*x + d)*d^2/e^4 + 5*d^3/ (sqrt(e*x + d)*e^4))
\[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{(d+e x)^{3/2}} \, dx=\int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (d+e\,x\right )}^{3/2}} \,d x \]