Integrand size = 25, antiderivative size = 314 \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^r\right )^{7/2}} \, dx=-\frac {4 b n}{15 d^2 r^2 \left (d+e x^r\right )^{3/2}}-\frac {32 b n}{15 d^3 r^2 \sqrt {d+e x^r}}+\frac {92 b n \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 d^{7/2} r^2}+\frac {2 b n \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{d^{7/2} r^2}+\frac {2}{15} \left (\frac {3}{d r \left (d+e x^r\right )^{5/2}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {15}{d^3 r \sqrt {d+e x^r}}-\frac {15 \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {4 b n \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{d^{7/2} r^2}-\frac {2 b n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{d^{7/2} r^2} \]
-4/15*b*n/d^2/r^2/(d+e*x^r)^(3/2)+92/15*b*n*arctanh((d+e*x^r)^(1/2)/d^(1/2 ))/d^(7/2)/r^2+2*b*n*arctanh((d+e*x^r)^(1/2)/d^(1/2))^2/d^(7/2)/r^2-4*b*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)))/d ^(7/2)/r^2-2*b*n*polylog(2,1-2*d^(1/2)/(d^(1/2)-(d+e*x^r)^(1/2)))/d^(7/2)/ r^2+2/15*(a+b*ln(c*x^n))*(3/d/r/(d+e*x^r)^(5/2)+5/d^2/r/(d+e*x^r)^(3/2)-15 *arctanh((d+e*x^r)^(1/2)/d^(1/2))/d^(7/2)/r+15/d^3/r/(d+e*x^r)^(1/2))-32/1 5*b*n/d^3/r^2/(d+e*x^r)^(1/2)
\[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^r\right )^{7/2}} \, dx=\int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^r\right )^{7/2}} \, dx \]
Time = 0.75 (sec) , antiderivative size = 307, normalized size of antiderivative = 0.98, 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.
\(\displaystyle \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^r\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 2790 |
\(\displaystyle \frac {2}{15} \left (-\frac {15 \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}+\frac {15}{d^3 r \sqrt {d+e x^r}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {3}{d r \left (d+e x^r\right )^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-b n \int \left (-\frac {2 \text {arctanh}\left (\frac {\sqrt {e x^r+d}}{\sqrt {d}}\right )}{d^{7/2} r x}+\frac {2}{d^3 r x \sqrt {e x^r+d}}+\frac {2}{3 d^2 r x \left (e x^r+d\right )^{3/2}}+\frac {2}{5 d r x \left (e x^r+d\right )^{5/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {2}{15} \left (-\frac {15 \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{d^{7/2} r}+\frac {15}{d^3 r \sqrt {d+e x^r}}+\frac {5}{d^2 r \left (d+e x^r\right )^{3/2}}+\frac {3}{d r \left (d+e x^r\right )^{5/2}}\right ) \left (a+b \log \left (c x^n\right )\right )-b n \left (-\frac {2 \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )^2}{d^{7/2} r^2}-\frac {92 \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right )}{15 d^{7/2} r^2}+\frac {4 \text {arctanh}\left (\frac {\sqrt {d+e x^r}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {d+e x^r}}\right )}{d^{7/2} r^2}+\frac {2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}-\sqrt {e x^r+d}}\right )}{d^{7/2} r^2}+\frac {32}{15 d^3 r^2 \sqrt {d+e x^r}}+\frac {4}{15 d^2 r^2 \left (d+e x^r\right )^{3/2}}\right )\) |
(2*(3/(d*r*(d + e*x^r)^(5/2)) + 5/(d^2*r*(d + e*x^r)^(3/2)) + 15/(d^3*r*Sq rt[d + e*x^r]) - (15*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]])/(d^(7/2)*r))*(a + b *Log[c*x^n]))/15 - b*n*(4/(15*d^2*r^2*(d + e*x^r)^(3/2)) + 32/(15*d^3*r^2* Sqrt[d + e*x^r]) - (92*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]])/(15*d^(7/2)*r^2) - (2*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]]^2)/(d^(7/2)*r^2) + (4*ArcTanh[Sqrt[d + e*x^r]/Sqrt[d]]*Log[(2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^r])])/(d^(7/2)* r^2) + (2*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] - Sqrt[d + e*x^r])])/(d^(7/2 )*r^2))
3.5.39.3.1 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]
\[\int \frac {a +b \ln \left (c \,x^{n}\right )}{x \left (d +e \,x^{r}\right )^{\frac {7}{2}}}d x\]
Exception generated. \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^r\right )^{7/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> Error detected within library code: inte grate: implementation incomplete (constant residues)
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^r\right )^{7/2}} \, dx=\text {Timed out} \]
\[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^r\right )^{7/2}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{r} + d\right )}^{\frac {7}{2}} x} \,d x } \]
1/15*a*(15*log((sqrt(e*x^r + d) - sqrt(d))/(sqrt(e*x^r + d) + sqrt(d)))/(d ^(7/2)*r) + 2*(15*(e*x^r + d)^2 + 5*(e*x^r + d)*d + 3*d^2)/((e*x^r + d)^(5 /2)*d^3*r)) + b*integrate((log(c) + log(x^n))/((e^3*x*x^(3*r) + 3*d*e^2*x* x^(2*r) + 3*d^2*e*x*x^r + d^3*x)*sqrt(e*x^r + d)), x)
\[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^r\right )^{7/2}} \, dx=\int { \frac {b \log \left (c x^{n}\right ) + a}{{\left (e x^{r} + d\right )}^{\frac {7}{2}} x} \,d x } \]
Timed out. \[ \int \frac {a+b \log \left (c x^n\right )}{x \left (d+e x^r\right )^{7/2}} \, dx=\int \frac {a+b\,\ln \left (c\,x^n\right )}{x\,{\left (d+e\,x^r\right )}^{7/2}} \,d x \]