3.2.26 \(\int \frac {x}{\log ^2(c (a+b x^2))} \, dx\) [126]

Optimal. Leaf size=47 \[ -\frac {a+b x^2}{2 b \log \left (c \left (a+b x^2\right )\right )}+\frac {\text {li}\left (c \left (a+b x^2\right )\right )}{2 b c} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {2504, 2436, 2334, 2335} \begin {gather*} \frac {\text {li}\left (c \left (b x^2+a\right )\right )}{2 b c}-\frac {a+b x^2}{2 b \log \left (c \left (a+b x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/Log[c*(a + b*x^2)]^2,x]

[Out]

-1/2*(a + b*x^2)/(b*Log[c*(a + b*x^2)]) + LogIntegral[c*(a + b*x^2)]/(2*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*} \int \frac {x}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {1}{\log ^2(c (a+b x))} \, dx,x,x^2\right )\\ &=\frac {\text {Subst}\left (\int \frac {1}{\log ^2(c x)} \, dx,x,a+b x^2\right )}{2 b}\\ &=-\frac {a+b x^2}{2 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 )}{2 b}\\ &=-\frac {a+b x^2}{2 b \log \left (c \left (a+b x^2\right )\right )}+\frac {\text {li}\left (c \left (a+b x^2\right )\right )}{2 b c}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 44, normalized size = 0.94 \begin {gather*} \frac {\frac {\text {Ei}\left (\log \left (c \left (a+b x^2\right )\right )\right )}{c}-\frac {a+b x^2}{\log \left (c \left (a+b x^2\right )\right )}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/Log[c*(a + b*x^2)]^2,x]

[Out]

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

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Maple [A]
time = 0.41, size = 48, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {-\frac {c \left (b \,x^{2}+a \right )}{\ln \left (c \left (b \,x^{2}+a \right )\right )}-\expIntegral \left (1, -\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2 b c}\)	\(48\)
default	\(\frac {-\frac {c \left (b \,x^{2}+a \right )}{\ln \left (c \left (b \,x^{2}+a \right )\right )}-\expIntegral \left (1, -\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2 b c}\)	\(48\)
risch	\(-\frac {b \,x^{2}+a}{2 \ln \left (c \left (b \,x^{2}+a \right )\right ) b}-\frac {\expIntegral \left (1, -\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2 b c}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 55, normalized size = 1.17 \begin {gather*} -\frac {b c x^{2} + a c - \log \left (b c x^{2} + a c\right ) \operatorname {log\_integral}\left (b c x^{2} + a c\right )}{2 \, b c \log \left (b c x^{2} + a c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.95, size = 49, normalized size = 1.04 \begin {gather*} \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} + \frac {- a - b x^{2}}{2 b \log {\left (c \left (a + b x^{2}\right ) \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [A]
time = 7.12, size = 47, normalized size = 1.00 \begin {gather*} -\frac {\frac {b c x^{2} + a c}{\log \left (b c x^{2} + a c\right )} - {\rm Ei}\left (\log \left (b c x^{2} + a c\right )\right )}{2 \, b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.36, size = 46, normalized size = 0.98 \begin {gather*} \frac {\mathrm {logint}\left (c\,\left (b\,x^2+a\right )\right )}{2\,b\,c}-\frac {\frac {b\,x^2}{2}+\frac {a}{2}}{b\,\ln \left (c\,\left (b\,x^2+a\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/log(c*(a + b*x^2))^2,x)

[Out]

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

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