3.2.23 \(\int \frac {x^3}{\log (c (a+b x^2))} \, dx\) [123]

Optimal. Leaf size=45 \[ \frac {\text {Ei}\left (2 \log \left (c \left (a+b x^2\right )\right )\right )}{2 b^2 c^2}-\frac {a \text {li}\left (c \left (a+b x^2\right )\right )}{2 b^2 c} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.438, Rules used = {2504, 2446, 2436, 2335, 2437, 2346, 2209} \begin {gather*} \frac {\text {Ei}\left (2 \log \left (c \left (b x^2+a\right )\right )\right )}{2 b^2 c^2}-\frac {a \text {li}\left (c \left (b x^2+a\right )\right )}{2 b^2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Log[c*(a + b*x^2)],x]

[Out]

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

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 2335

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

Rule 2346

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 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 2437

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 2446

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 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^3}{\log \left (c \left (a+b x^2\right )\right )} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x}{\log (c (a+b x))} \, dx,x,x^2\right )\\ &=\frac {1}{2} \text {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 {\text {Subst}\left (\int \frac {a+b x}{\log (c (a+b x))} \, dx,x,x^2\right )}{2 b}-\frac {a \text {Subst}\left (\int \frac {1}{\log (c (a+b x))} \, dx,x,x^2\right )}{2 b}\\ &=\frac {\text {Subst}\left (\int \frac {x}{\log (c x)} \, dx,x,a+b x^2\right )}{2 b^2}-\frac {a \text {Subst}\left (\int \frac {1}{\log (c x)} \, dx,x,a+b x^2\right )}{2 b^2}\\ &=-\frac {a \text {li}\left (c \left (a+b x^2\right )\right )}{2 b^2 c}+\frac {\text {Subst}\left (\int \frac {e^{2 x}}{x} \, dx,x,\log \left (c \left (a+b x^2\right )\right )\right )}{2 b^2 c^2}\\ &=\frac {\text {Ei}\left (2 \log \left (c \left (a+b x^2\right )\right )\right )}{2 b^2 c^2}-\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 = 0.07, size = 41, normalized size = 0.91 \begin {gather*} \frac {-a c \text {Ei}\left (\log \left (a c+b c x^2\right )\right )+\text {Ei}\left (2 \log \left (a c+b c x^2\right )\right )}{2 b^2 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Log[c*(a + b*x^2)],x]

[Out]

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

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Maple [A]
time = 1.71, size = 43, normalized size = 0.96

method	result	size
default	\(\frac {-\expIntegral \left (1, -2 \ln \left (c \left (b \,x^{2}+a \right )\right )\right )+c a \expIntegral \left (1, -\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2 c^{2} b^{2}}\)	\(43\)
risch	\(\frac {a \expIntegral \left (1, -\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2 c \,b^{2}}-\frac {\expIntegral \left (1, -2 \ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2 c^{2} b^{2}}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/c^2/b^2*(-Ei(1,-2*ln(c*(b*x^2+a)))+c*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^3/log(c*(b*x^2+a)),x, algorithm="maxima")

[Out]

integrate(x^3/log((b*x^2 + a)*c), x)

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Fricas [A]
time = 0.36, size = 54, normalized size = 1.20 \begin {gather*} -\frac {a c \operatorname {log\_integral}\left (b c x^{2} + a c\right ) - \operatorname {log\_integral}\left (b^{2} c^{2} x^{4} + 2 \, a b c^{2} x^{2} + a^{2} c^{2}\right )}{2 \, b^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3}}{\log {\left (a c + b c x^{2} \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/log(a*c + b*c*x**2), x)

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Giac [A]
time = 3.82, size = 44, normalized size = 0.98 \begin {gather*} -\frac {a {\rm Ei}\left (\log \left (b c x^{2} + a c\right )\right )}{2 \, b^{2} c} + \frac {{\rm Ei}\left (2 \, \log \left (b c x^{2} + a c\right )\right )}{2 \, b^{2} c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^3}{\ln \left (c\,\left (b\,x^2+a\right )\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/log(c*(a + b*x^2)),x)

[Out]

int(x^3/log(c*(a + b*x^2)), x)

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