\(\int \frac {x}{\log ^3(c (a+b x^2))} \, dx\) [128]

Optimal result

Integrand size = 14, antiderivative size = 73 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=-\frac {a+b x^2}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a+b x^2}{4 b \log \left (c \left (a+b x^2\right )\right )}+\frac {\operatorname {LogIntegral}\left (c \left (a+b x^2\right )\right )}{4 b c} \]

[Out]

1/4*Li(c*(b*x^2+a))/b/c+1/4*(-b*x^2-a)/b/ln(c*(b*x^2+a))^2+1/4*(-b*x^2-a)/b/ln(c*(b*x^2+a))

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2504, 2436, 2334, 2335} \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=\frac {\operatorname {LogIntegral}\left (c \left (b x^2+a\right )\right )}{4 b c}-\frac {a+b x^2}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a+b x^2}{4 b \log \left (c \left (a+b x^2\right )\right )} \]

[In]

Int[x/Log[c*(a + b*x^2)]^3,x]

[Out]

-1/4*(a + b*x^2)/(b*Log[c*(a + b*x^2)]^2) - (a + b*x^2)/(4*b*Log[c*(a + b*x^2)]) + LogIntegral[c*(a + b*x^2)]/
(4*b*c)

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b*Log[c*x^n])^(p + 1)/(b*n*(p + 1)))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

Rule 2335

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, 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 ^3(c (a+b x))} \, dx,x,x^2\right ) \\ & = \frac {\text {Subst}\left (\int \frac {1}{\log ^3(c x)} \, dx,x,a+b x^2\right )}{2 b} \\ & = -\frac {a+b x^2}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}+\frac {\text {Subst}\left (\int \frac {1}{\log ^2(c x)} \, dx,x,a+b x^2\right )}{4 b} \\ & = -\frac {a+b x^2}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a+b x^2}{4 b \log \left (c \left (a+b x^2\right )\right )}+\frac {\text {Subst}\left (\int \frac {1}{\log (c x)} \, dx,x,a+b x^2\right )}{4 b} \\ & = -\frac {a+b x^2}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a+b x^2}{4 b \log \left (c \left (a+b x^2\right )\right )}+\frac {\text {li}\left (c \left (a+b x^2\right )\right )}{4 b c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=\frac {-\frac {\left (a+b x^2\right ) \left (1+\log \left (c \left (a+b x^2\right )\right )\right )}{\log ^2\left (c \left (a+b x^2\right )\right )}+\frac {\operatorname {LogIntegral}\left (c \left (a+b x^2\right )\right )}{c}}{4 b} \]

[In]

Integrate[x/Log[c*(a + b*x^2)]^3,x]

[Out]

(-(((a + b*x^2)*(1 + Log[c*(a + b*x^2)]))/Log[c*(a + b*x^2)]^2) + LogIntegral[c*(a + b*x^2)]/c)/(4*b)

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {-\frac {c \left (b \,x^{2}+a \right )}{2 {\ln \left (c \left (b \,x^{2}+a \right )\right )}^{2}}-\frac {c \left (b \,x^{2}+a \right )}{2 \ln \left (c \left (b \,x^{2}+a \right )\right )}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2}}{2 b c}\)	\(70\)
default	\(\frac {-\frac {c \left (b \,x^{2}+a \right )}{2 {\ln \left (c \left (b \,x^{2}+a \right )\right )}^{2}}-\frac {c \left (b \,x^{2}+a \right )}{2 \ln \left (c \left (b \,x^{2}+a \right )\right )}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2}}{2 b c}\)	\(70\)
risch	\(-\frac {\ln \left (c \left (b \,x^{2}+a \right )\right ) b \,x^{2}+b \,x^{2}+\ln \left (c \left (b \,x^{2}+a \right )\right ) a +a}{4 b {\ln \left (c \left (b \,x^{2}+a \right )\right )}^{2}}-\frac {\operatorname {Ei}_{1}\left (-\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{4 b c}\)	\(75\)

[In]

int(x/ln(c*(b*x^2+a))^3,x,method=_RETURNVERBOSE)

[Out]

1/2/b/c*(-1/2/ln(c*(b*x^2+a))^2*c*(b*x^2+a)-1/2/ln(c*(b*x^2+a))*c*(b*x^2+a)-1/2*Ei(1,-ln(c*(b*x^2+a))))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=-\frac {b c x^{2} - \log \left (b c x^{2} + a c\right )^{2} \operatorname {log\_integral}\left (b c x^{2} + a c\right ) + a c + {\left (b c x^{2} + a c\right )} \log \left (b c x^{2} + a c\right )}{4 \, b c \log \left (b c x^{2} + a c\right )^{2}} \]

[In]

integrate(x/log(c*(b*x^2+a))^3,x, algorithm="fricas")

[Out]

-1/4*(b*c*x^2 - log(b*c*x^2 + a*c)^2*log_integral(b*c*x^2 + a*c) + a*c + (b*c*x^2 + a*c)*log(b*c*x^2 + a*c))/(
b*c*log(b*c*x^2 + a*c)^2)

Sympy [A] (verification not implemented)

Time = 0.73 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=\frac {\begin {cases} \frac {x^{2}}{2 \log {\left (a c \right )}} & \text {for}\: b = 0 \\0 & \text {for}\: c = 0 \\\frac {\operatorname {Ei}{\left (\log {\left (a c + b c x^{2} \right )} \right )}}{2 b c} & \text {otherwise} \end {cases}}{2} + \frac {- a - b x^{2} + \left (- a - b x^{2}\right ) \log {\left (c \left (a + b x^{2}\right ) \right )}}{4 b \log {\left (c \left (a + b x^{2}\right ) \right )}^{2}} \]

[In]

integrate(x/ln(c*(b*x**2+a))**3,x)

[Out]

Piecewise((x**2/(2*log(a*c)), Eq(b, 0)), (0, Eq(c, 0)), (Ei(log(a*c + b*c*x**2))/(2*b*c), True))/2 + (-a - b*x
**2 + (-a - b*x**2)*log(c*(a + b*x**2)))/(4*b*log(c*(a + b*x**2))**2)

Maxima [F]

\[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=\int { \frac {x}{\log \left ({\left (b x^{2} + a\right )} c\right )^{3}} \,d x } \]

[In]

integrate(x/log(c*(b*x^2+a))^3,x, algorithm="maxima")

[Out]

-1/4*(b*x^2*(log(c) + 1) + a*(log(c) + 1) + (b*x^2 + a)*log(b*x^2 + a))/(b*log(b*x^2 + a)^2 + 2*b*log(b*x^2 +
a)*log(c) + b*log(c)^2) + integrate(1/2*x/(log(b*x^2 + a) + log(c)), x)

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=-\frac {\frac {b c x^{2} + a c}{\log \left (b c x^{2} + a c\right )} + \frac {b c x^{2} + a c}{\log \left (b c x^{2} + a c\right )^{2}} - {\rm Ei}\left (\log \left (b c x^{2} + a c\right )\right )}{4 \, b c} \]

[In]

integrate(x/log(c*(b*x^2+a))^3,x, algorithm="giac")

[Out]

-1/4*((b*c*x^2 + a*c)/log(b*c*x^2 + a*c) + (b*c*x^2 + a*c)/log(b*c*x^2 + a*c)^2 - Ei(log(b*c*x^2 + a*c)))/(b*c
)

Mupad [B] (verification not implemented)

Time = 1.48 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=\frac {\mathrm {logint}\left (c\,\left (b\,x^2+a\right )\right )}{4\,b\,c}-\frac {\frac {a\,c}{4}+\ln \left (c\,\left (b\,x^2+a\right )\right )\,\left (\frac {b\,c\,x^2}{4}+\frac {a\,c}{4}\right )+\frac {b\,c\,x^2}{4}}{b\,c\,{\ln \left (c\,\left (b\,x^2+a\right )\right )}^2} \]

[In]

int(x/log(c*(a + b*x^2))^3,x)

[Out]

logint(c*(a + b*x^2))/(4*b*c) - ((a*c)/4 + log(c*(a + b*x^2))*((a*c)/4 + (b*c*x^2)/4) + (b*c*x^2)/4)/(b*c*log(
c*(a + b*x^2))^2)