\(\int \frac {x^3}{\log ^2(c (a+b x^2)^p)} \, dx\) [109]

Optimal result

Integrand size = 18, antiderivative size = 138 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )^p\right )} \, dx=-\frac {a \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{2 b^2 p^2}+\frac {\left (a+b x^2\right )^2 \left (c \left (a+b x^2\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (a+b x^2\right )^p\right )}{p}\right )}{b^2 p^2}-\frac {x^2 \left (a+b x^2\right )}{2 b p \log \left (c \left (a+b x^2\right )^p\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.14 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )^p\right )} \, dx=-\frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-2/p} \left (b p x^2 \left (c \left (a+b x^2\right )^p\right )^{2/p}+a \left (c \left (a+b x^2\right )^p\right )^{\frac {1}{p}} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right ) \log \left (c \left (a+b x^2\right )^p\right )-2 \left (a+b x^2\right ) \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (a+b x^2\right )^p\right )}{p}\right ) \log \left (c \left (a+b x^2\right )^p\right )\right )}{2 b^2 p^2 \log \left (c \left (a+b x^2\right )^p\right )} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.97 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.40, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {2904, 2847, 2836, 2737, 2609, 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.

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

\(\Big \downarrow \) 2904

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

\(\Big \downarrow \) 2847

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

\(\Big \downarrow \) 2836

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

\(\Big \downarrow \) 2737

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

\(\Big \downarrow \) 2609

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

\(\Big \downarrow \) 2846

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 2009

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

rule 2609

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si 
mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F 
reeQ[{F, c, d, e, f, g}, x] &&  !TrueQ[$UseGamma]
 

rule 2737

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x/(n*(c*x 
^n)^(1/n))   Subst[Int[E^(x/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ 
[{a, b, c, n, p}, 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 [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.06 (sec) , antiderivative size = 1487, normalized size of antiderivative = 10.78

method	result	size
risch	\(\text {Expression too large to display}\)	\(1487\)

Input:

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

Output:

-1/p/b*x^2*(b*x^2+a)/(I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-I*P 
i*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*csgn(I*c*(b*x^2 
+a)^p)^3+I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+2*ln(c)+2*ln((b*x^2+a)^p)) 
-1/p^2*c^(-2/p)*((b*x^2+a)^p)^(-2/p)*exp(I*Pi*csgn(I*c*(b*x^2+a)^p)*(-csgn 
(I*c*(b*x^2+a)^p)+csgn(I*c))*(-csgn(I*c*(b*x^2+a)^p)+csgn(I*(b*x^2+a)^p))/ 
p)*Ei(1,-2*ln(b*x^2+a)-(I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-I 
*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*csgn(I*c*(b*x 
^2+a)^p)^3+I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+2*ln(c)+2*ln((b*x^2+a)^p 
)-2*p*ln(b*x^2+a))/p)*x^4-2/p^2/b*c^(-2/p)*((b*x^2+a)^p)^(-2/p)*exp(I*Pi*c 
sgn(I*c*(b*x^2+a)^p)*(-csgn(I*c*(b*x^2+a)^p)+csgn(I*c))*(-csgn(I*c*(b*x^2+ 
a)^p)+csgn(I*(b*x^2+a)^p))/p)*Ei(1,-2*ln(b*x^2+a)-(I*Pi*csgn(I*(b*x^2+a)^p 
)*csgn(I*c*(b*x^2+a)^p)^2-I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*c 
sgn(I*c)-I*Pi*csgn(I*c*(b*x^2+a)^p)^3+I*Pi*csgn(I*c*(b*x^2+a)^p)^2*csgn(I* 
c)+2*ln(c)+2*ln((b*x^2+a)^p)-2*p*ln(b*x^2+a))/p)*a*x^2-1/p^2/b^2*c^(-2/p)* 
((b*x^2+a)^p)^(-2/p)*exp(I*Pi*csgn(I*c*(b*x^2+a)^p)*(-csgn(I*c*(b*x^2+a)^p 
)+csgn(I*c))*(-csgn(I*c*(b*x^2+a)^p)+csgn(I*(b*x^2+a)^p))/p)*Ei(1,-2*ln(b* 
x^2+a)-(I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2-I*Pi*csgn(I*(b*x^ 
2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*csgn(I*c*(b*x^2+a)^p)^3+I*Pi* 
csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+2*ln(c)+2*ln((b*x^2+a)^p)-2*p*ln(b*x^2+a 
))/p)*a^2+1/2/p^2/b*a*c^(-1/p)*((b*x^2+a)^p)^(-1/p)*exp(1/2*I*Pi*csgn(I...
 

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )^p\right )} \, dx=-\frac {{\left (a p \log \left (b x^{2} + a\right ) + a \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )} \operatorname {log\_integral}\left ({\left (b x^{2} + a\right )} c^{\left (\frac {1}{p}\right )}\right ) + {\left (b^{2} p x^{4} + a b p x^{2}\right )} c^{\frac {2}{p}} - 2 \, {\left (p \log \left (b x^{2} + a\right ) + \log \left (c\right )\right )} \operatorname {log\_integral}\left ({\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} c^{\frac {2}{p}}\right )}{2 \, {\left (b^{2} p^{3} \log \left (b x^{2} + a\right ) + b^{2} p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (136) = 272\).

Time = 0.12 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.27 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {1}{2} \, a {\left (\frac {{\left (b x^{2} + a\right )} p}{b^{2} p^{3} \log \left (b x^{2} + a\right ) + b^{2} p^{2} \log \left (c\right )} - \frac {p {\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (b x^{2} + a\right )\right ) \log \left (b x^{2} + a\right )}{{\left (b^{2} p^{3} \log \left (b x^{2} + a\right ) + b^{2} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}} - \frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (b x^{2} + a\right )\right ) \log \left (c\right )}{{\left (b^{2} p^{3} \log \left (b x^{2} + a\right ) + b^{2} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}}\right )} - \frac {\frac {{\left (b x^{2} + a\right )}^{2} p}{b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \left (c\right )} - \frac {2 \, p {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (b x^{2} + a\right )\right ) \log \left (b x^{2} + a\right )}{{\left (b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} - \frac {2 \, {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (b x^{2} + a\right )\right ) \log \left (c\right )}{{\left (b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}}}{2 \, b} \] Input:

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

Output:

1/2*a*((b*x^2 + a)*p/(b^2*p^3*log(b*x^2 + a) + b^2*p^2*log(c)) - p*Ei(log( 
c)/p + log(b*x^2 + a))*log(b*x^2 + a)/((b^2*p^3*log(b*x^2 + a) + b^2*p^2*l 
og(c))*c^(1/p)) - Ei(log(c)/p + log(b*x^2 + a))*log(c)/((b^2*p^3*log(b*x^2 
 + a) + b^2*p^2*log(c))*c^(1/p))) - 1/2*((b*x^2 + a)^2*p/(b*p^3*log(b*x^2 
+ a) + b*p^2*log(c)) - 2*p*Ei(2*log(c)/p + 2*log(b*x^2 + a))*log(b*x^2 + a 
)/((b*p^3*log(b*x^2 + a) + b*p^2*log(c))*c^(2/p)) - 2*Ei(2*log(c)/p + 2*lo 
g(b*x^2 + a))*log(c)/((b*p^3*log(b*x^2 + a) + b*p^2*log(c))*c^(2/p)))/b
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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