\(\int \frac {x}{\log (c (a+b x^2)^p)} \, dx\) [103]

Optimal result

Integrand size = 16, antiderivative size = 51 \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{2 b p} \]

[Out]

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))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2504, 2436, 2337, 2209} \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b p} \]

[In]

Int[x/Log[c*(a + b*x^2)^p],x]

[Out]

((a + b*x^2)*ExpIntegralEi[Log[c*(a + b*x^2)^p]/p])/(2*b*p*(c*(a + b*x^2)^p)^p^(-1))

Rule 2209

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - c*(f/d)))/d)*ExpInteg
ralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]

Rule 2337

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[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]

Rule 2436

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[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]

Rule 2504

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[I
nt[x^(Simplify[(m + 1)/n] - 1)*(a + b*Log[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])

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int \frac {1}{\log \left (c (a+b x)^p\right )} \, dx,x,x^2\right ) \\ & = \frac {\text {Subst}\left (\int \frac {1}{\log \left (c x^p\right )} \, dx,x,a+b x^2\right )}{2 b} \\ & = \frac {\left (\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p}\right ) \text {Subst}\left (\int \frac {e^{\frac {x}{p}}}{x} \, dx,x,\log \left (c \left (a+b x^2\right )^p\right )\right )}{2 b p} \\ & = \frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text {Ei}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{2 b p} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{2 b p} \]

[In]

Integrate[x/Log[c*(a + b*x^2)^p],x]

[Out]

((a + b*x^2)*ExpIntegralEi[Log[c*(a + b*x^2)^p]/p])/(2*b*p*(c*(a + b*x^2)^p)^p^(-1))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.95 (sec) , antiderivative size = 272, normalized size of antiderivative = 5.33

method	result	size
risch	\(-\frac {\left (b \,x^{2}+a \right ) c^{-\frac {1}{p}} {\left (\left (b \,x^{2}+a \right )^{p}\right )}^{-\frac {1}{p}} {\mathrm e}^{\frac {i \pi \,\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \left (-\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )+\operatorname {csgn}\left (i c \right )\right ) \left (-\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )+\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right )\right )}{2 p}} \operatorname {Ei}_{1}\left (-\ln \left (b \,x^{2}+a \right )-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2}-i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )-i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{3}+i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )+2 \ln \left (c \right )+2 \ln \left (\left (b \,x^{2}+a \right )^{p}\right )-2 p \ln \left (b \,x^{2}+a \right )}{2 p}\right )}{2 b p}\)	\(272\)

[In]

int(x/ln(c*(b*x^2+a)^p),x,method=_RETURNVERBOSE)

[Out]

-1/2/b/p*(b*x^2+a)*c^(-1/p)*((b*x^2+a)^p)^(-1/p)*exp(1/2*I*Pi*csgn(I*c*(b*x^2+a)^p)*(-csgn(I*c*(b*x^2+a)^p)+cs
gn(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)

Fricas [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 29, normalized size of antiderivative = 0.57 \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {\operatorname {log\_integral}\left ({\left (b x^{2} + a\right )} c^{\left (\frac {1}{p}\right )}\right )}{2 \, b c^{\left (\frac {1}{p}\right )} p} \]

[In]

integrate(x/log(c*(b*x^2+a)^p),x, algorithm="fricas")

[Out]

1/2*log_integral((b*x^2 + a)*c^(1/p))/(b*c^(1/p)*p)

Sympy [F]

\[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {x}{\log {\left (c \left (a + b x^{2}\right )^{p} \right )}}\, dx \]

[In]

integrate(x/ln(c*(b*x**2+a)**p),x)

[Out]

Integral(x/log(c*(a + b*x**2)**p), x)

Maxima [F]

\[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\int { \frac {x}{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )} \,d x } \]

[In]

integrate(x/log(c*(b*x^2+a)^p),x, algorithm="maxima")

[Out]

integrate(x/log((b*x^2 + a)^p*c), x)

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.61 \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (b x^{2} + a\right )\right )}{2 \, b c^{\left (\frac {1}{p}\right )} p} \]

[In]

integrate(x/log(c*(b*x^2+a)^p),x, algorithm="giac")

[Out]

1/2*Ei(log(c)/p + log(b*x^2 + a))/(b*c^(1/p)*p)

Mupad [F(-1)]

Timed out. \[ \int \frac {x}{\log \left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {x}{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )} \,d x \]

[In]

int(x/log(c*(a + b*x^2)^p),x)

[Out]

int(x/log(c*(a + b*x^2)^p), x)