Integrand size = 25, antiderivative size = 327 \[ \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-\frac {92 b d^2 n \sqrt {d+e x^r}}{15 r^2}-\frac {32 b d n \left (d+e x^r\right )^{3/2}}{45 r^2}-\frac {4 b n \left (d+e x^r\right )^{5/2}}{25 r^2}+\frac {92 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 r^2}+\frac {2 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{r^2}+\frac {2}{15} \left (\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}-\frac {15 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b d^{5/2} n \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{r^2}-\frac {2 b d^{5/2} n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{r^2} \]
-32/45*b*d*n*(d+e*x^r)^(3/2)/r^2-4/25*b*n*(d+e*x^r)^(5/2)/r^2+92/15*b*d^(5 /2)*n*arctanh((d+e*x^r)^(1/2)/d^(1/2))/r^2+2*b*d^(5/2)*n*arctanh((d+e*x^r) ^(1/2)/d^(1/2))^2/r^2-4*b*d^(5/2)*n*arctanh((d+e*x^r)^(1/2)/d^(1/2))*ln(2* d^(1/2)/(d^(1/2)-(d+e*x^r)^(1/2)))/r^2-2*b*d^(5/2)*n*polylog(2,1-2*d^(1/2) /(d^(1/2)-(d+e*x^r)^(1/2)))/r^2-92/15*b*d^2*n*(d+e*x^r)^(1/2)/r^2+2/15*(a+ b*ln(c*x^n))*(5*d*(d+e*x^r)^(3/2)/r+3*(d+e*x^r)^(5/2)/r-15*d^(5/2)*arctanh ((d+e*x^r)^(1/2)/d^(1/2))/r+15*d^2*(d+e*x^r)^(1/2)/r)
\[ \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx \]
Time = 0.77 (sec) , antiderivative size = 318, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2790, 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.
\(\displaystyle \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx\) |
\(\Big \downarrow \) 2790 |
\(\displaystyle \frac {2}{15} \left (-\frac {15 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}+\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-b n \int \left (-\frac {2 \text {arctanh}\left (\frac {\sqrt {e x^r+d}}{\sqrt {d}}\right ) d^{5/2}}{r x}+\frac {2 \sqrt {e x^r+d} d^2}{r x}+\frac {2 \left (e x^r+d\right )^{3/2} d}{3 r x}+\frac {2 \left (e x^r+d\right )^{5/2}}{5 r x}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{15} \left (-\frac {15 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{r}+\frac {15 d^2 \sqrt {d+e x^r}}{r}+\frac {5 d \left (d+e x^r\right )^{3/2}}{r}+\frac {3 \left (d+e x^r\right )^{5/2}}{r}\right ) \left (a+b \log \left (c x^n\right )\right )-b n \left (-\frac {2 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{r^2}-\frac {92 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 r^2}+\frac {4 d^{5/2} \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{r^2}+\frac {2 d^{5/2} \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^r+d}}\right )}{r^2}+\frac {92 d^2 \sqrt {d+e x^r}}{15 r^2}+\frac {32 d \left (d+e x^r\right )^{3/2}}{45 r^2}+\frac {4 \left (d+e x^r\right )^{5/2}}{25 r^2}\right )\) |
(2*((15*d^2*Sqrt[d + e*x^r])/r + (5*d*(d + e*x^r)^(3/2))/r + (3*(d + e*x^r )^(5/2))/r - (15*d^(5/2)*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]])/r)*(a + b*Log[c *x^n]))/15 - b*n*((92*d^2*Sqrt[d + e*x^r])/(15*r^2) + (32*d*(d + e*x^r)^(3 /2))/(45*r^2) + (4*(d + e*x^r)^(5/2))/(25*r^2) - (92*d^(5/2)*ArcTanh[Sqrt[ d + e*x^r]/Sqrt[d]])/(15*r^2) - (2*d^(5/2)*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d] ]^2)/r^2 + (4*d^(5/2)*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sq rt[d] - Sqrt[d + e*x^r])])/r^2 + (2*d^(5/2)*PolyLog[2, 1 - (2*Sqrt[d])/(Sq rt[d] - Sqrt[d + e*x^r])])/r^2)
3.5.33.3.1 Defintions of rubi rules used
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.)) /(x_), x_Symbol] :> With[{u = IntHide[(d + e*x^r)^q/x, x]}, Simp[u*(a + b*L og[c*x^n]), x] - Simp[b*n Int[1/x u, x], x]] /; FreeQ[{a, b, c, d, e, n , r}, x] && IntegerQ[q - 1/2]
\[\int \frac {\left (d +e \,x^{r}\right )^{\frac {5}{2}} \left (a +b \ln \left (c \,x^{n}\right )\right )}{x}d x\]
Exception generated. \[ \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (has polynomial part)
Timed out. \[ \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\text {Timed out} \]
\[ \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int { \frac {{\left (e x^{r} + d\right )}^{\frac {5}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x} \,d x } \]
1/15*(15*d^(5/2)*log((sqrt(e*x^r + d) - sqrt(d))/(sqrt(e*x^r + d) + sqrt(d )))/r + 2*(3*(e*x^r + d)^(5/2) + 5*(e*x^r + d)^(3/2)*d + 15*sqrt(e*x^r + d )*d^2)/r)*a + b*integrate((e^2*x^(2*r)*log(c) + 2*d*e*x^r*log(c) + d^2*log (c) + (e^2*x^(2*r) + 2*d*e*x^r + d^2)*log(x^n))*sqrt(e*x^r + d)/x, x)
\[ \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int { \frac {{\left (e x^{r} + d\right )}^{\frac {5}{2}} {\left (b \log \left (c x^{n}\right ) + a\right )}}{x} \,d x } \]
Timed out. \[ \int \frac {\left (d+e x^r\right )^{5/2} \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\int \frac {{\left (d+e\,x^r\right )}^{5/2}\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x} \,d x \]