Integrand size = 16, antiderivative size = 51 \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{2 b p} \] Output:
1/2*(b*x^2+a)*Ei(ln(c*(b*x^2+a)^p)/p)/b/p/((c*(b*x^2+a)^p)^(1/p))
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{2 b p} \] Input:
Integrate[x/Log[c*(a + b*x^2)^p],x]
Output:
((a + b*x^2)*ExpIntegralEi[Log[c*(a + b*x^2)^p]/p])/(2*b*p*(c*(a + b*x^2)^ p)^p^(-1))
Time = 0.46 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2904, 2836, 2737, 2609}
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.
\(\displaystyle \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\log \left (c \left (b x^2+a\right )^p\right )}dx^2\) |
\(\Big \downarrow \) 2836 |
\(\displaystyle \frac {\int \frac {1}{\log \left (c \left (b x^2+a\right )^p\right )}d\left (b x^2+a\right )}{2 b}\) |
\(\Big \downarrow \) 2737 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \int \frac {\left (c \left (b x^2+a\right )^p\right )^{\frac {1}{p}}}{x^2}d\log \left (c \left (b x^2+a\right )^p\right )}{2 b p}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b p}\) |
Input:
Int[x/Log[c*(a + b*x^2)^p],x]
Output:
((a + b*x^2)*ExpIntegralEi[Log[c*(a + b*x^2)^p]/p])/(2*b*p*(c*(a + b*x^2)^ p)^p^(-1))
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F reeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x/(n*(c*x ^n)^(1/n)) Subst[Int[E^(x/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ [{a, b, c, n, p}, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : > Simp[1/e Subst[Int[(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{ a, b, c, d, e, n, p}, x]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & & !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 2.88 (sec) , antiderivative size = 272, normalized size of antiderivative = 5.33
method | result | size |
risch | \(-\frac {\left (b \,x^{2}+a \right ) {\left (\left (b \,x^{2}+a \right )^{p}\right )}^{-\frac {1}{p}} c^{-\frac {1}{p}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \left (-\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )+\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right )\right )}{2 p}} \operatorname {expIntegral}_{1}\left (-\ln \left (b \,x^{2}+a \right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )+2 \ln \left (\left (b \,x^{2}+a \right )^{p}\right )-2 p \ln \left (b \,x^{2}+a \right )}{2 p}\right )}{2 b p}\) | \(272\) |
Input:
int(x/ln(c*(b*x^2+a)^p),x,method=_RETURNVERBOSE)
Output:
-1/2/b/p*(b*x^2+a)*((b*x^2+a)^p)^(-1/p)*c^(-1/p)*exp(1/2*I*Pi*csgn(I*c*(b* x^2+a)^p)*(-csgn(I*c*(b*x^2+a)^p)+csgn(I*c))*(-csgn(I*c*(b*x^2+a)^p)+csgn( I*(b*x^2+a)^p))/p)*Ei(1,-ln(b*x^2+a)-1/2*(I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I* c*(b*x^2+a)^p)^2-I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)- I*Pi*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+2*ln(c )+2*ln((b*x^2+a)^p)-2*p*ln(b*x^2+a))/p)
Time = 0.07 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.57 \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {\operatorname {log\_integral}\left ({\left (b x^{2} + a\right )} c^{\left (\frac {1}{p}\right )}\right )}{2 \, b c^{\left (\frac {1}{p}\right )} p} \] Input:
integrate(x/log(c*(b*x^2+a)^p),x, algorithm="fricas")
Output:
1/2*log_integral((b*x^2 + a)*c^(1/p))/(b*c^(1/p)*p)
\[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {x}{\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}\, dx \] Input:
integrate(x/ln(c*(b*x**2+a)**p),x)
Output:
Integral(x/log(c*(a + b*x**2)**p), x)
\[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\int { \frac {x}{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )} \,d x } \] Input:
integrate(x/log(c*(b*x^2+a)^p),x, algorithm="maxima")
Output:
integrate(x/log((b*x^2 + a)^p*c), x)
Time = 0.12 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.61 \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (b x^{2} + a\right )\right )}{2 \, b c^{\left (\frac {1}{p}\right )} p} \] Input:
integrate(x/log(c*(b*x^2+a)^p),x, algorithm="giac")
Output:
1/2*Ei(log(c)/p + log(b*x^2 + a))/(b*c^(1/p)*p)
Timed out. \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {x}{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )} \,d x \] Input:
int(x/log(c*(a + b*x^2)^p),x)
Output:
int(x/log(c*(a + b*x^2)^p), x)
\[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {2 \left (\int \frac {x^{3}}{\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) a +\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right ) b \,x^{2}}d x \right ) b^{2} p +\mathrm {log}\left (\mathrm {log}\left (\left (b \,x^{2}+a \right )^{p} c \right )\right ) a}{2 b p} \] Input:
int(x/log(c*(b*x^2+a)^p),x)
Output:
(2*int(x**3/(log((a + b*x**2)**p*c)*a + log((a + b*x**2)**p*c)*b*x**2),x)* b**2*p + log(log((a + b*x**2)**p*c))*a)/(2*b*p)