\(\int \frac {x^3}{\log ^2(c (a+b x^2))} \, dx\) [125]

Optimal result

Integrand size = 16, antiderivative size = 71 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=\frac {\operatorname {ExpIntegralEi}\left (2 \log \left (c \left (a+b x^2\right )\right )\right )}{b^2 c^2}-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}-\frac {a \operatorname {LogIntegral}\left (c \left (a+b x^2\right )\right )}{2 b^2 c} \] Output:

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

Mathematica [F]

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

Integrate[x^3/Log[c*(a + b*x^2)]^2,x]
 

Output:

Integrate[x^3/Log[c*(a + b*x^2)]^2, x]
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.30, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2904, 2847, 2836, 2735, 2846, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 2904

\(\displaystyle \frac {1}{2} \int \frac {x^2}{\log ^2\left (c \left (b x^2+a\right )\right )}dx^2\)

\(\Big \downarrow \) 2847

\(\displaystyle \frac {1}{2} \left (\frac {a \int \frac {1}{\log \left (c \left (b x^2+a\right )\right )}dx^2}{b}+2 \int \frac {x^2}{\log \left (c \left (b x^2+a\right )\right )}dx^2-\frac {x^2 \left (a+b x^2\right )}{b \log \left (c \left (a+b x^2\right )\right )}\right )\)

\(\Big \downarrow \) 2836

\(\displaystyle \frac {1}{2} \left (\frac {a \int \frac {1}{\log \left (c \left (b x^2+a\right )\right )}d\left (b x^2+a\right )}{b^2}+2 \int \frac {x^2}{\log \left (c \left (b x^2+a\right )\right )}dx^2-\frac {x^2 \left (a+b x^2\right )}{b \log \left (c \left (a+b x^2\right )\right )}\right )\)

\(\Big \downarrow \) 2735

\(\displaystyle \frac {1}{2} \left (2 \int \frac {x^2}{\log \left (c \left (b x^2+a\right )\right )}dx^2+\frac {a \operatorname {LogIntegral}\left (c \left (b x^2+a\right )\right )}{b^2 c}-\frac {x^2 \left (a+b x^2\right )}{b \log \left (c \left (a+b x^2\right )\right )}\right )\)

\(\Big \downarrow \) 2846

\(\displaystyle \frac {1}{2} \left (2 \int \left (\frac {b x^2+a}{b \log \left (c \left (b x^2+a\right )\right )}-\frac {a}{b \log \left (c \left (b x^2+a\right )\right )}\right )dx^2+\frac {a \operatorname {LogIntegral}\left (c \left (b x^2+a\right )\right )}{b^2 c}-\frac {x^2 \left (a+b x^2\right )}{b \log \left (c \left (a+b x^2\right )\right )}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (2 \left (\frac {\operatorname {ExpIntegralEi}\left (2 \log \left (c \left (b x^2+a\right )\right )\right )}{b^2 c^2}-\frac {a \operatorname {LogIntegral}\left (c \left (b x^2+a\right )\right )}{b^2 c}\right )+\frac {a \operatorname {LogIntegral}\left (c \left (b x^2+a\right )\right )}{b^2 c}-\frac {x^2 \left (a+b x^2\right )}{b \log \left (c \left (a+b x^2\right )\right )}\right )\)

Input:

Int[x^3/Log[c*(a + b*x^2)]^2,x]
 

Output:

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

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2735

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

rule 2836

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : 
> Simp[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 2846

Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.) 
]*(b_.)), x_Symbol] :> Int[ExpandIntegrand[(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 2847

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] + (-Simp[(q + 1)/(b*n*(p + 1))   Int[( 
f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x] + Simp[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] && Lt 
Q[p, -1] && GtQ[q, 0]
 

rule 2904

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.), x_Symbol] :> Simp[1/n   Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L 
og[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])
 

Maple [A] (verified)

Time = 2.95 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04

method	result	size
risch	\(-\frac {x^{2} \left (b \,x^{2}+a \right )}{2 b \ln \left (c \left (b \,x^{2}+a \right )\right )}+\frac {a \,\operatorname {expIntegral}_{1}\left (-\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2 b^{2} c}-\frac {\operatorname {expIntegral}_{1}\left (-2 \ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{b^{2} c^{2}}\)	\(74\)
default	\(\frac {-\frac {c^{2} \left (b \,x^{2}+a \right )^{2}}{\ln \left (c \left (b \,x^{2}+a \right )\right )}-2 \,\operatorname {expIntegral}_{1}\left (-2 \ln \left (c \left (b \,x^{2}+a \right )\right )\right )-a c \left (-\frac {c \left (b \,x^{2}+a \right )}{\ln \left (c \left (b \,x^{2}+a \right )\right )}-\operatorname {expIntegral}_{1}\left (-\ln \left (c \left (b \,x^{2}+a \right )\right )\right )\right )}{2 b^{2} c^{2}}\)	\(95\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=-\frac {b^{2} c^{2} x^{4} + a b c^{2} x^{2} + {\left (a c \operatorname {log\_integral}\left (b c x^{2} + a c\right ) - 2 \, \operatorname {log\_integral}\left (b^{2} c^{2} x^{4} + 2 \, a b c^{2} x^{2} + a^{2} c^{2}\right )\right )} \log \left (b c x^{2} + a c\right )}{2 \, b^{2} c^{2} \log \left (b c x^{2} + a c\right )} \] Input:

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

Output:

-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))
 

Sympy [F]

\[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=\frac {- a x^{2} - b x^{4}}{2 b \log {\left (c \left (a + b x^{2}\right ) \right )}} + \frac {\int \frac {a x}{\log {\left (a c + b c x^{2} \right )}}\, dx + \int \frac {2 b x^{3}}{\log {\left (a c + b c x^{2} \right )}}\, dx}{b} \] Input:

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

Output:

(-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
 

Maxima [F]

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

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

Output:

-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)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=\frac {a {\left (\frac {b c x^{2} + a c}{\log \left (b c x^{2} + a c\right )} - {\rm Ei}\left (\log \left (b c x^{2} + a c\right )\right )\right )}}{2 \, b^{2} c} - \frac {\frac {{\left (b c x^{2} + a c\right )}^{2}}{\log \left (b c x^{2} + a c\right )} - 2 \, {\rm Ei}\left (2 \, \log \left (b c x^{2} + a c\right )\right )}{2 \, b^{2} c^{2}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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