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

Optimal. Leaf size=71 \[ \frac {\text {Ei}\left (2 \log \left (c \left (b x^2+a\right )\right )\right )}{b^2 c^2}-\frac {a \text {li}\left (c \left (b x^2+a\right )\right )}{2 b^2 c}-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )} \]

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {2454, 2400, 2399, 2389, 2298, 2390, 2309, 2178} \[ \frac {\text {Ei}\left (2 \log \left (c \left (b x^2+a\right )\right )\right )}{b^2 c^2}-\frac {a \text {li}\left (c \left (b x^2+a\right )\right )}{2 b^2 c}-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Log[c*(a + b*x^2)]^2,x]

[Out]

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

Rule 2178

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

Rule 2298

Int[Log[(c_.)*(x_)]^(-1), x_Symbol] :> Simp[LogIntegral[c*x]/c, x] /; FreeQ[c, x]

Rule 2309

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))^(p_)*(x_)^(m_.), x_Symbol] :> Dist[1/c^(m + 1), Subst[Int[E^((m + 1)*x)*(a
+ b*x)^p, x], x, Log[c*x]], x] /; FreeQ[{a, b, c, p}, x] && IntegerQ[m]

Rule 2389

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 2390

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_))^(q_.), x_Symbol] :> Dist[1/
e, Subst[Int[((f*x)/d)^q*(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, q}, x]
&& EqQ[e*f - d*g, 0]

Rule 2399

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.)), x_Symbol] :> Int[ExpandIn
tegrand[(f + g*x)^q/(a + b*Log[c*(d + e*x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && IGtQ[q, 0]

Rule 2400

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((
d + e*x)*(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1))/(b*e*n*(p + 1)), x] + (-Dist[(q + 1)/(b*n*(p + 1)), I
nt[(f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x] + Dist[(q*(e*f - d*g))/(b*e*n*(p + 1)), Int[(f + g*x
)^(q - 1)*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x]) /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g,
0] && LtQ[p, -1] && GtQ[q, 0]

Rule 2454

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^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {x}{\log ^2(c (a+b x))} \, dx,x,x^2\right )\\ &=-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}+\frac {a \operatorname {Subst}\left (\int \frac {1}{\log (c (a+b x))} \, dx,x,x^2\right )}{2 b}+\operatorname {Subst}\left (\int \frac {x}{\log (c (a+b x))} \, dx,x,x^2\right )\\ &=-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}+\frac {a \operatorname {Subst}\left (\int \frac {1}{\log (c x)} \, dx,x,a+b x^2\right )}{2 b^2}+\operatorname {Subst}\left (\int \left (-\frac {a}{b \log (c (a+b x))}+\frac {a+b x}{b \log (c (a+b x))}\right ) \, dx,x,x^2\right )\\ &=-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}+\frac {a \text {li}\left (c \left (a+b x^2\right )\right )}{2 b^2 c}+\frac {\operatorname {Subst}\left (\int \frac {a+b x}{\log (c (a+b x))} \, dx,x,x^2\right )}{b}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\log (c (a+b x))} \, dx,x,x^2\right )}{b}\\ &=-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}+\frac {a \text {li}\left (c \left (a+b x^2\right )\right )}{2 b^2 c}+\frac {\operatorname {Subst}\left (\int \frac {x}{\log (c x)} \, dx,x,a+b x^2\right )}{b^2}-\frac {a \operatorname {Subst}\left (\int \frac {1}{\log (c x)} \, dx,x,a+b x^2\right )}{b^2}\\ &=-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}-\frac {a \text {li}\left (c \left (a+b x^2\right )\right )}{2 b^2 c}+\frac {\operatorname {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log \left (c \left (a+b x^2\right )\right )\right )}{b^2 c^2}\\ &=\frac {\text {Ei}\left (2 \log \left (c \left (a+b x^2\right )\right )\right )}{b^2 c^2}-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}-\frac {a \text {li}\left (c \left (a+b x^2\right )\right )}{2 b^2 c}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 66, normalized size = 0.93 \[ -\frac {-\frac {2 \text {Ei}\left (2 \log \left (c \left (b x^2+a\right )\right )\right )}{c^2}+\frac {a \text {Ei}\left (\log \left (c \left (b x^2+a\right )\right )\right )}{c}+\frac {b x^2 \left (a+b x^2\right )}{\log \left (c \left (a+b x^2\right )\right )}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Log[c*(a + b*x^2)]^2,x]

[Out]

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

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fricas [A]  time = 0.43, size = 99, normalized size = 1.39 \[ -\frac {b^{2} c^{2} x^{4} + a b c^{2} x^{2} + {\left (a c \operatorname {log\_integral}\left (b c x^{2} + a c\right ) - 2 \, \operatorname {log\_integral}\left (b^{2} c^{2} x^{4} + 2 \, a b c^{2} x^{2} + a^{2} c^{2}\right )\right )} \log \left (b c x^{2} + a c\right )}{2 \, b^{2} c^{2} \log \left (b c x^{2} + a c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 89, normalized size = 1.25 \[ -\frac {a c {\rm Ei}\left (\log \left ({\left (b x^{2} + a\right )} c\right )\right ) - \frac {{\left (b c x^{2} + a c\right )} a c}{\log \left ({\left (b x^{2} + a\right )} c\right )} + \frac {{\left (b c x^{2} + a c\right )}^{2}}{\log \left ({\left (b x^{2} + a\right )} c\right )} - 2 \, {\rm Ei}\left (2 \, \log \left ({\left (b x^{2} + a\right )} c\right )\right )}{2 \, b^{2} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.07, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{\ln \left (\left (b \,x^{2}+a \right ) c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b x^{4} + a x^{2}}{2 \, {\left (b \log \left (b x^{2} + a\right ) + b \log \relax (c)\right )}} + \int \frac {2 \, b x^{3} + a x}{b \log \left (b x^{2} + a\right ) + b \log \relax (c)}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b*x^4 + a*x^2)/(b*log(b*x^2 + a) + b*log(c)) + integrate((2*b*x^3 + a*x)/(b*log(b*x^2 + a) + b*log(c)),
x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3}{{\ln \left (c\,\left (b\,x^2+a\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/log(c*(a + b*x^2))^2,x)

[Out]

int(x^3/log(c*(a + b*x^2))^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {- a x^{2} - b x^{4}}{2 b \log {\left (c \left (a + b x^{2}\right ) \right )}} + \frac {\int \frac {a x}{\log {\left (a c + b c x^{2} \right )}}\, dx + \int \frac {2 b x^{3}}{\log {\left (a c + b c x^{2} \right )}}\, dx}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-a*x**2 - b*x**4)/(2*b*log(c*(a + b*x**2))) + (Integral(a*x/log(a*c + b*c*x**2), x) + Integral(2*b*x**3/log(a
*c + b*c*x**2), x))/b

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