Integrand size = 23, antiderivative size = 293 \[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {b d n \sqrt {d+e x}}{4 x^2}-\frac {11 b e n \sqrt {d+e x}}{8 x}-\frac {9 b e^2 n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{8 \sqrt {d}}+\frac {3 b e^2 n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 \sqrt {d}}-\frac {3 b e^2 n \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{2 \sqrt {d}}-\frac {3 b e^2 n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{4 \sqrt {d}} \]
-1/2*(e*x+d)^(3/2)*(a+b*ln(c*x^n))/x^2-9/8*b*e^2*n*arctanh((e*x+d)^(1/2)/d ^(1/2))/d^(1/2)+3/4*b*e^2*n*arctanh((e*x+d)^(1/2)/d^(1/2))^2/d^(1/2)-3/4*e ^2*arctanh((e*x+d)^(1/2)/d^(1/2))*(a+b*ln(c*x^n))/d^(1/2)-3/2*b*e^2*n*arct anh((e*x+d)^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)-(e*x+d)^(1/2)))/d^(1/2)-3 /4*b*e^2*n*polylog(2,1-2*d^(1/2)/(d^(1/2)-(e*x+d)^(1/2)))/d^(1/2)-1/4*b*d* n*(e*x+d)^(1/2)/x^2-11/8*b*e*n*(e*x+d)^(1/2)/x-3/4*e*(a+b*ln(c*x^n))*(e*x+ d)^(1/2)/x
Time = 0.35 (sec) , antiderivative size = 501, normalized size of antiderivative = 1.71 \[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=-\frac {8 a d^{3/2} \sqrt {d+e x}+4 b d^{3/2} n \sqrt {d+e x}+20 a \sqrt {d} e x \sqrt {d+e x}+22 b \sqrt {d} e n x \sqrt {d+e x}+18 b e^2 n x^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )+8 b d^{3/2} \sqrt {d+e x} \log \left (c x^n\right )+20 b \sqrt {d} e x \sqrt {d+e x} \log \left (c x^n\right )-6 a e^2 x^2 \log \left (\sqrt {d}-\sqrt {d+e x}\right )-6 b e^2 x^2 \log \left (c x^n\right ) \log \left (\sqrt {d}-\sqrt {d+e x}\right )+3 b e^2 n x^2 \log ^2\left (\sqrt {d}-\sqrt {d+e x}\right )+6 a e^2 x^2 \log \left (\sqrt {d}+\sqrt {d+e x}\right )+6 b e^2 x^2 \log \left (c x^n\right ) \log \left (\sqrt {d}+\sqrt {d+e x}\right )-3 b e^2 n x^2 \log ^2\left (\sqrt {d}+\sqrt {d+e x}\right )-6 b e^2 n x^2 \log \left (\sqrt {d}+\sqrt {d+e x}\right ) \log \left (\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )+6 b e^2 n x^2 \log \left (\sqrt {d}-\sqrt {d+e x}\right ) \log \left (\frac {1}{2} \left (1+\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )+6 b e^2 n x^2 \operatorname {PolyLog}\left (2,\frac {1}{2}-\frac {\sqrt {d+e x}}{2 \sqrt {d}}\right )-6 b e^2 n x^2 \operatorname {PolyLog}\left (2,\frac {1}{2} \left (1+\frac {\sqrt {d+e x}}{\sqrt {d}}\right )\right )}{16 \sqrt {d} x^2} \]
-1/16*(8*a*d^(3/2)*Sqrt[d + e*x] + 4*b*d^(3/2)*n*Sqrt[d + e*x] + 20*a*Sqrt [d]*e*x*Sqrt[d + e*x] + 22*b*Sqrt[d]*e*n*x*Sqrt[d + e*x] + 18*b*e^2*n*x^2* ArcTanh[Sqrt[d + e*x]/Sqrt[d]] + 8*b*d^(3/2)*Sqrt[d + e*x]*Log[c*x^n] + 20 *b*Sqrt[d]*e*x*Sqrt[d + e*x]*Log[c*x^n] - 6*a*e^2*x^2*Log[Sqrt[d] - Sqrt[d + e*x]] - 6*b*e^2*x^2*Log[c*x^n]*Log[Sqrt[d] - Sqrt[d + e*x]] + 3*b*e^2*n *x^2*Log[Sqrt[d] - Sqrt[d + e*x]]^2 + 6*a*e^2*x^2*Log[Sqrt[d] + Sqrt[d + e *x]] + 6*b*e^2*x^2*Log[c*x^n]*Log[Sqrt[d] + Sqrt[d + e*x]] - 3*b*e^2*n*x^2 *Log[Sqrt[d] + Sqrt[d + e*x]]^2 - 6*b*e^2*n*x^2*Log[Sqrt[d] + Sqrt[d + e*x ]]*Log[1/2 - Sqrt[d + e*x]/(2*Sqrt[d])] + 6*b*e^2*n*x^2*Log[Sqrt[d] - Sqrt [d + e*x]]*Log[(1 + Sqrt[d + e*x]/Sqrt[d])/2] + 6*b*e^2*n*x^2*PolyLog[2, 1 /2 - Sqrt[d + e*x]/(2*Sqrt[d])] - 6*b*e^2*n*x^2*PolyLog[2, (1 + Sqrt[d + e *x]/Sqrt[d])/2])/(Sqrt[d]*x^2)
Time = 0.65 (sec) , antiderivative size = 280, normalized size of antiderivative = 0.96, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2792, 27, 2010, 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.
| \(\displaystyle \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx\) |
| \(\Big \downarrow \) 2792 |
| \(\displaystyle -b n \int -\frac {\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) x^2}{\sqrt {d}}+\sqrt {d+e x} (2 d+5 e x)}{4 x^3}dx-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{4} b n \int \frac {\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) x^2}{\sqrt {d}}+\sqrt {d+e x} (2 d+5 e x)}{x^3}dx-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\) |
| \(\Big \downarrow \) 2010 |
| \(\displaystyle \frac {1}{4} b n \int \left (\frac {3 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) e^2}{\sqrt {d} x}+\frac {5 \sqrt {d+e x} e}{x^2}+\frac {2 d \sqrt {d+e x}}{x^3}\right )dx-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{4 \sqrt {d}}-\frac {3 e \sqrt {d+e x} \left (a+b \log \left (c x^n\right )\right )}{4 x}-\frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {1}{4} b n \left (\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )^2}{\sqrt {d}}-\frac {9 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right )}{2 \sqrt {d}}-\frac {6 e^2 \text {arctanh}\left (\frac {\sqrt {d+e x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}-\frac {3 e^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x}}\right )}{\sqrt {d}}-\frac {d \sqrt {d+e x}}{x^2}-\frac {11 e \sqrt {d+e x}}{2 x}\right )\) |
(-3*e*Sqrt[d + e*x]*(a + b*Log[c*x^n]))/(4*x) - ((d + e*x)^(3/2)*(a + b*Lo g[c*x^n]))/(2*x^2) - (3*e^2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]]*(a + b*Log[c*x^ n]))/(4*Sqrt[d]) + (b*n*(-((d*Sqrt[d + e*x])/x^2) - (11*e*Sqrt[d + e*x])/( 2*x) - (9*e^2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]])/(2*Sqrt[d]) + (3*e^2*ArcTanh [Sqrt[d + e*x]/Sqrt[d]]^2)/Sqrt[d] - (6*e^2*ArcTanh[Sqrt[d + e*x]/Sqrt[d]] *Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])/Sqrt[d] - (3*e^2*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x])])/Sqrt[d]))/4
3.2.43.3.1 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[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
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])
\[\int \frac {\left (e x +d \right )^{\frac {3}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )}{x^{3}}d x\]
\[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{3}} \,d x } \]
\[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {\left (a + b \log {\left (c x^{n} \right )}\right ) \left (d + e x\right )^{\frac {3}{2}}}{x^{3}}\, dx \]
\[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{3}} \,d x } \]
1/8*(3*e^2*log((sqrt(e*x + d) - sqrt(d))/(sqrt(e*x + d) + sqrt(d)))/sqrt(d ) - 2*(5*(e*x + d)^(3/2)*e^2 - 3*sqrt(e*x + d)*d*e^2)/((e*x + d)^2 - 2*(e* x + d)*d + d^2))*a + b*integrate((e*x*log(c) + d*log(c) + (e*x + d)*log(x^ n))*sqrt(e*x + d)/x^3, x)
\[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int { \frac {{\left (e x + d\right )}^{\frac {3}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x^{3}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^{3/2} \left (a+b \log \left (c x^n\right )\right )}{x^3} \, dx=\int \frac {\left (a+b\,\ln \left (c\,x^n\right )\right )\,{\left (d+e\,x\right )}^{3/2}}{x^3} \,d x \]