3.2.27 \(\int \frac {x^3}{\log ^3(c (a+b x^2))} \, dx\) [127]

Optimal. Leaf size=127 \[ \frac {\text {Ei}\left (2 \log \left (c \left (a+b x^2\right )\right )\right )}{b^2 c^2}-\frac {x^2 \left (a+b x^2\right )}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a \left (a+b x^2\right )}{4 b^2 \log \left (c \left (a+b x^2\right )\right )}-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}-\frac {a \text {li}\left (c \left (a+b x^2\right )\right )}{4 b^2 c} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 9, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.562, Rules used = {2504, 2447, 2446, 2436, 2335, 2437, 2346, 2209, 2334} \begin {gather*} \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 )}{4 b^2 c}-\frac {a \left (a+b x^2\right )}{4 b^2 \log \left (c \left (a+b x^2\right )\right )}-\frac {x^2 \left (a+b x^2\right )}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3/Log[c*(a + b*x^2)]^3,x]

[Out]

ExpIntegralEi[2*Log[c*(a + b*x^2)]]/(b^2*c^2) - (x^2*(a + b*x^2))/(4*b*Log[c*(a + b*x^2)]^2) - (a*(a + b*x^2))
/(4*b^2*Log[c*(a + b*x^2)]) - (x^2*(a + b*x^2))/(2*b*Log[c*(a + b*x^2)]) - (a*LogIntegral[c*(a + b*x^2)])/(4*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 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 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 2447

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 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 ^3\left (c \left (a+b x^2\right )\right )} \, dx &=\frac {1}{2} \text {Subst}\left (\int \frac {x}{\log ^3(c (a+b x))} \, dx,x,x^2\right )\\ &=-\frac {x^2 \left (a+b x^2\right )}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}+\frac {1}{2} \text {Subst}\left (\int \frac {x}{\log ^2(c (a+b x))} \, dx,x,x^2\right )+\frac {a \text {Subst}\left (\int \frac {1}{\log ^2(c (a+b x))} \, dx,x,x^2\right )}{4 b}\\ &=-\frac {x^2 \left (a+b x^2\right )}{4 b \log ^2\left (c \left (a+b x^2\right )\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 {Subst}\left (\int \frac {1}{\log ^2(c x)} \, dx,x,a+b x^2\right )}{4 b^2}+\frac {a \text {Subst}\left (\int \frac {1}{\log (c (a+b x))} \, dx,x,x^2\right )}{2 b}+\text {Subst}\left (\int \frac {x}{\log (c (a+b x))} \, dx,x,x^2\right )\\ &=-\frac {x^2 \left (a+b x^2\right )}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a \left (a+b x^2\right )}{4 b^2 \log \left (c \left (a+b x^2\right )\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 {Subst}\left (\int \frac {1}{\log (c x)} \, dx,x,a+b x^2\right )}{4 b^2}+\frac {a \text {Subst}\left (\int \frac {1}{\log (c x)} \, dx,x,a+b x^2\right )}{2 b^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 {x^2 \left (a+b x^2\right )}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a \left (a+b x^2\right )}{4 b^2 \log \left (c \left (a+b x^2\right )\right )}-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}+\frac {3 a \text {li}\left (c \left (a+b x^2\right )\right )}{4 b^2 c}+\frac {\text {Subst}\left (\int \frac {a+b x}{\log (c (a+b x))} \, dx,x,x^2\right )}{b}-\frac {a \text {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 )}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a \left (a+b x^2\right )}{4 b^2 \log \left (c \left (a+b x^2\right )\right )}-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}+\frac {3 a \text {li}\left (c \left (a+b x^2\right )\right )}{4 b^2 c}+\frac {\text {Subst}\left (\int \frac {x}{\log (c x)} \, dx,x,a+b x^2\right )}{b^2}-\frac {a \text {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 )}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a \left (a+b x^2\right )}{4 b^2 \log \left (c \left (a+b x^2\right )\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 )}{4 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 )}{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 )}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a \left (a+b x^2\right )}{4 b^2 \log \left (c \left (a+b x^2\right )\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 )}{4 b^2 c}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 87, normalized size = 0.69 \begin {gather*} -\frac {\frac {a \text {Ei}\left (\log \left (c \left (a+b x^2\right )\right )\right )}{c}-\frac {4 \text {Ei}\left (2 \log \left (c \left (a+b x^2\right )\right )\right )}{c^2}+\frac {\left (a+b x^2\right ) \left (b x^2+\left (a+2 b x^2\right ) \log \left (c \left (a+b x^2\right )\right )\right )}{\log ^2\left (c \left (a+b x^2\right )\right )}}{4 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3/Log[c*(a + b*x^2)]^3,x]

[Out]

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

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Maple [A]
time = 2.12, size = 143, normalized size = 1.13

method	result	size
risch	\(-\frac {\left (b \,x^{2}+a \right ) \left (2 \ln \left (c \left (b \,x^{2}+a \right )\right ) b \,x^{2}+b \,x^{2}+\ln \left (c \left (b \,x^{2}+a \right )\right ) a \right )}{4 b^{2} \ln \left (c \left (b \,x^{2}+a \right )\right )^{2}}+\frac {a \expIntegral \left (1, -\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{4 c \,b^{2}}-\frac {\expIntegral \left (1, -2 \ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{c^{2} b^{2}}\)	\(105\)
default	\(\frac {-\frac {c^{2} \left (b \,x^{2}+a \right )^{2}}{2 \ln \left (c \left (b \,x^{2}+a \right )\right )^{2}}-\frac {c^{2} \left (b \,x^{2}+a \right )^{2}}{\ln \left (c \left (b \,x^{2}+a \right )\right )}-2 \expIntegral \left (1, -2 \ln \left (c \left (b \,x^{2}+a \right )\right )\right )-c a \left (-\frac {c \left (b \,x^{2}+a \right )}{2 \ln \left (c \left (b \,x^{2}+a \right )\right )^{2}}-\frac {c \left (b \,x^{2}+a \right )}{2 \ln \left (c \left (b \,x^{2}+a \right )\right )}-\frac {\expIntegral \left (1, -\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2}\right )}{2 c^{2} b^{2}}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.38, size = 142, normalized size = 1.12 \begin {gather*} -\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 ) - 4 \, \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} + {\left (2 \, b^{2} c^{2} x^{4} + 3 \, a b c^{2} x^{2} + a^{2} c^{2}\right )} \log \left (b c x^{2} + a c\right )}{4 \, b^{2} c^{2} \log \left (b c x^{2} + a c\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b^2*c^2*x^4 + a*b*c^2*x^2 + (a*c*log_integral(b*c*x^2 + a*c) - 4*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)^2 + (2*b^2*c^2*x^4 + 3*a*b*c^2*x^2 + a^2*c^2)*log(b*c*x^2 + a*c))/(b^2*c^2*lo
g(b*c*x^2 + a*c)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 5.31, size = 151, normalized size = 1.19 \begin {gather*} \frac {a {\left (\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 )\right )}}{4 \, b^{2} c} - \frac {\frac {2 \, {\left (b c x^{2} + a c\right )}^{2}}{\log \left (b c x^{2} + a c\right )} + \frac {{\left (b c x^{2} + a c\right )}^{2}}{\log \left (b c x^{2} + a c\right )^{2}} - 4 \, {\rm Ei}\left (2 \, \log \left (b c x^{2} + a c\right )\right )}{4 \, 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))^3,x, algorithm="giac")

[Out]

1/4*a*((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^
2*c) - 1/4*(2*(b*c*x^2 + a*c)^2/log(b*c*x^2 + a*c) + (b*c*x^2 + a*c)^2/log(b*c*x^2 + a*c)^2 - 4*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 = -0.01 \begin {gather*} \int \frac {x^3}{{\ln \left (c\,\left (b\,x^2+a\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/log(c*(a + b*x^2))^3,x)

[Out]

int(x^3/log(c*(a + b*x^2))^3, x)

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