Integrand size = 23, antiderivative size = 83 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\frac {e p \log \left (d+e x^2\right )}{2 g (e f-d g)}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{2 g \left (f+g x^2\right )}-\frac {e p \log \left (f+g x^2\right )}{2 g (e f-d g)} \] Output:
1/2*e*p*ln(e*x^2+d)/g/(-d*g+e*f)-1/2*ln(c*(e*x^2+d)^p)/g/(g*x^2+f)-1/2*e*p *ln(g*x^2+f)/g/(-d*g+e*f)
Time = 0.05 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\frac {\frac {e p \log \left (d+e x^2\right )}{e f-d g}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{f+g x^2}+\frac {e p \log \left (f+g x^2\right )}{-e f+d g}}{2 g} \] Input:
Integrate[(x*Log[c*(d + e*x^2)^p])/(f + g*x^2)^2,x]
Output:
((e*p*Log[d + e*x^2])/(e*f - d*g) - Log[c*(d + e*x^2)^p]/(f + g*x^2) + (e* p*Log[f + g*x^2])/(-(e*f) + d*g))/(2*g)
Time = 0.42 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2925, 2842, 47, 16}
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.
\(\displaystyle \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 2925 |
\(\displaystyle \frac {1}{2} \int \frac {\log \left (c \left (e x^2+d\right )^p\right )}{\left (g x^2+f\right )^2}dx^2\) |
\(\Big \downarrow \) 2842 |
\(\displaystyle \frac {1}{2} \left (\frac {e p \int \frac {1}{\left (e x^2+d\right ) \left (g x^2+f\right )}dx^2}{g}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{g \left (f+g x^2\right )}\right )\) |
\(\Big \downarrow \) 47 |
\(\displaystyle \frac {1}{2} \left (\frac {e p \left (\frac {e \int \frac {1}{e x^2+d}dx^2}{e f-d g}-\frac {g \int \frac {1}{g x^2+f}dx^2}{e f-d g}\right )}{g}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{g \left (f+g x^2\right )}\right )\) |
\(\Big \downarrow \) 16 |
\(\displaystyle \frac {1}{2} \left (\frac {e p \left (\frac {\log \left (d+e x^2\right )}{e f-d g}-\frac {\log \left (f+g x^2\right )}{e f-d g}\right )}{g}-\frac {\log \left (c \left (d+e x^2\right )^p\right )}{g \left (f+g x^2\right )}\right )\) |
Input:
Int[(x*Log[c*(d + e*x^2)^p])/(f + g*x^2)^2,x]
Output:
(-(Log[c*(d + e*x^2)^p]/(g*(f + g*x^2))) + (e*p*(Log[d + e*x^2]/(e*f - d*g ) - Log[f + g*x^2]/(e*f - d*g)))/g)/2
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Simp[b/(b*c - a*d) Int[1/(a + b*x), x], x] - Simp[d/(b*c - a*d) Int[1/(c + d*x), x ], x] /; FreeQ[{a, b, c, d}, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_ ))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/( g*(q + 1))), x] - Simp[b*e*(n/(g*(q + 1))) Int[(f + g*x)^(q + 1)/(d + e*x ), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Si mplify[(m + 1)/n] - 1)*(f + g*x^(s/n))^r*(a + b*Log[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, q, r, s}, x] && Integer Q[r] && IntegerQ[s/n] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0 ] || IGtQ[q, 0])
Time = 2.99 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90
method | result | size |
parts | \(-\frac {\ln \left (c \left (e \,x^{2}+d \right )^{p}\right )}{2 g \left (g \,x^{2}+f \right )}+\frac {e p \left (-\frac {\ln \left (e \,x^{2}+d \right )}{2 \left (d g -e f \right )}+\frac {\ln \left (g \,x^{2}+f \right )}{2 d g -2 e f}\right )}{g}\) | \(75\) |
parallelrisch | \(-\frac {\ln \left (e \,x^{2}+d \right ) x^{2} e^{2} g p -\ln \left (g \,x^{2}+f \right ) x^{2} e^{2} g p +\ln \left (e \,x^{2}+d \right ) e^{2} f p -\ln \left (g \,x^{2}+f \right ) e^{2} f p +\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) d e g -\ln \left (c \left (e \,x^{2}+d \right )^{p}\right ) e^{2} f}{2 \left (d g -e f \right ) \left (g \,x^{2}+f \right ) e g}\) | \(127\) |
risch | \(-\frac {\ln \left (\left (e \,x^{2}+d \right )^{p}\right )}{2 g \left (g \,x^{2}+f \right )}-\frac {i \pi d g \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}-i \pi d g \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3} d g +i \pi {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right ) d g -i \pi e f \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2}+i \pi e f \,\operatorname {csgn}\left (i \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right ) \operatorname {csgn}\left (i c \right )+i \pi e f {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{3}-i \pi e f {\operatorname {csgn}\left (i c \left (e \,x^{2}+d \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )-2 \ln \left (g \,x^{2}+f \right ) e g p \,x^{2}+2 \ln \left (-e \,x^{2}-d \right ) e g p \,x^{2}-2 \ln \left (g \,x^{2}+f \right ) e f p +2 \ln \left (-e \,x^{2}-d \right ) e f p +2 \ln \left (c \right ) d g -2 \ln \left (c \right ) e f}{4 g \left (g \,x^{2}+f \right ) \left (d g -e f \right )}\) | \(371\) |
Input:
int(x*ln(c*(e*x^2+d)^p)/(g*x^2+f)^2,x,method=_RETURNVERBOSE)
Output:
-1/2*ln(c*(e*x^2+d)^p)/g/(g*x^2+f)+e*p/g*(-1/2/(d*g-e*f)*ln(e*x^2+d)+1/2/( d*g-e*f)*ln(g*x^2+f))
Time = 0.12 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.10 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\frac {{\left (e g p x^{2} + d g p\right )} \log \left (e x^{2} + d\right ) - {\left (e g p x^{2} + e f p\right )} \log \left (g x^{2} + f\right ) - {\left (e f - d g\right )} \log \left (c\right )}{2 \, {\left (e f^{2} g - d f g^{2} + {\left (e f g^{2} - d g^{3}\right )} x^{2}\right )}} \] Input:
integrate(x*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="fricas")
Output:
1/2*((e*g*p*x^2 + d*g*p)*log(e*x^2 + d) - (e*g*p*x^2 + e*f*p)*log(g*x^2 + f) - (e*f - d*g)*log(c))/(e*f^2*g - d*f*g^2 + (e*f*g^2 - d*g^3)*x^2)
Timed out. \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\text {Timed out} \] Input:
integrate(x*ln(c*(e*x**2+d)**p)/(g*x**2+f)**2,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.89 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\frac {e p {\left (\frac {\log \left (e x^{2} + d\right )}{e f - d g} - \frac {\log \left (g x^{2} + f\right )}{e f - d g}\right )}}{2 \, g} - \frac {\log \left ({\left (e x^{2} + d\right )}^{p} c\right )}{2 \, {\left (g x^{2} + f\right )} g} \] Input:
integrate(x*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="maxima")
Output:
1/2*e*p*(log(e*x^2 + d)/(e*f - d*g) - log(g*x^2 + f)/(e*f - d*g))/g - 1/2* log((e*x^2 + d)^p*c)/((g*x^2 + f)*g)
Time = 0.13 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.52 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=-\frac {e p \log \left (e x^{2} + d\right )}{2 \, {\left (e f g + {\left (e x^{2} + d\right )} g^{2} - d g^{2}\right )}} + \frac {e p \log \left (e x^{2} + d\right )}{2 \, {\left (e f g - d g^{2}\right )}} - \frac {e p \log \left (e f + {\left (e x^{2} + d\right )} g - d g\right )}{2 \, {\left (e f g - d g^{2}\right )}} - \frac {e \log \left (c\right )}{2 \, {\left (e f g + {\left (e x^{2} + d\right )} g^{2} - d g^{2}\right )}} \] Input:
integrate(x*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x, algorithm="giac")
Output:
-1/2*e*p*log(e*x^2 + d)/(e*f*g + (e*x^2 + d)*g^2 - d*g^2) + 1/2*e*p*log(e* x^2 + d)/(e*f*g - d*g^2) - 1/2*e*p*log(e*f + (e*x^2 + d)*g - d*g)/(e*f*g - d*g^2) - 1/2*e*log(c)/(e*f*g + (e*x^2 + d)*g^2 - d*g^2)
Time = 17.98 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=-\frac {\ln \left (c\,{\left (e\,x^2+d\right )}^p\right )}{2\,g\,\left (g\,x^2+f\right )}-\frac {e\,p\,\mathrm {atan}\left (\frac {x^2\,\left (d\,g\,1{}\mathrm {i}-e\,f\,1{}\mathrm {i}\right )}{2\,d\,f+d\,g\,x^2+e\,f\,x^2}\right )\,1{}\mathrm {i}}{d\,g^2-e\,f\,g} \] Input:
int((x*log(c*(d + e*x^2)^p))/(f + g*x^2)^2,x)
Output:
- log(c*(d + e*x^2)^p)/(2*g*(f + g*x^2)) - (e*p*atan((x^2*(d*g*1i - e*f*1i ))/(2*d*f + d*g*x^2 + e*f*x^2))*1i)/(d*g^2 - e*f*g)
Time = 0.16 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.69 \[ \int \frac {x \log \left (c \left (d+e x^2\right )^p\right )}{\left (f+g x^2\right )^2} \, dx=\frac {-\mathrm {log}\left (e \,x^{2}+d \right ) d f g p -\mathrm {log}\left (e \,x^{2}+d \right ) d \,g^{2} p \,x^{2}+\mathrm {log}\left (g \,x^{2}+f \right ) e \,f^{2} p +\mathrm {log}\left (g \,x^{2}+f \right ) e f g p \,x^{2}+\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) d \,g^{2} x^{2}-\mathrm {log}\left (\left (e \,x^{2}+d \right )^{p} c \right ) e f g \,x^{2}}{2 f g \left (d \,g^{2} x^{2}-e f g \,x^{2}+d f g -e \,f^{2}\right )} \] Input:
int(x*log(c*(e*x^2+d)^p)/(g*x^2+f)^2,x)
Output:
( - log(d + e*x**2)*d*f*g*p - log(d + e*x**2)*d*g**2*p*x**2 + log(f + g*x* *2)*e*f**2*p + log(f + g*x**2)*e*f*g*p*x**2 + log((d + e*x**2)**p*c)*d*g** 2*x**2 - log((d + e*x**2)**p*c)*e*f*g*x**2)/(2*f*g*(d*f*g + d*g**2*x**2 - e*f**2 - e*f*g*x**2))