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

Optimal result

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

[Out]

-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)

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {2504, 2447, 2446, 2436, 2337, 2209, 2437, 2347} \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {\left (a+b x^2\right )^2 \left (c \left (a+b x^2\right )^p\right )^{-2/p} \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{b^2 p^2}-\frac {a \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b^2 p^2}-\frac {x^2 \left (a+b x^2\right )}{2 b p \log \left (c \left (a+b x^2\right )^p\right )} \]

[In]

Int[x^3/Log[c*(a + b*x^2)^p]^2,x]

[Out]

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

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 2337

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Dist[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 2347

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.14 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )^p\right )} \, dx=-\frac {\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-2/p} \left (b p x^2 \left (c \left (a+b x^2\right )^p\right )^{2/p}+a \left (c \left (a+b x^2\right )^p\right )^{\frac {1}{p}} \operatorname {ExpIntegralEi}\left (\frac {\log \left (c \left (a+b x^2\right )^p\right )}{p}\right ) \log \left (c \left (a+b x^2\right )^p\right )-2 \left (a+b x^2\right ) \operatorname {ExpIntegralEi}\left (\frac {2 \log \left (c \left (a+b x^2\right )^p\right )}{p}\right ) \log \left (c \left (a+b x^2\right )^p\right )\right )}{2 b^2 p^2 \log \left (c \left (a+b x^2\right )^p\right )} \]

[In]

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

[Out]

-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])

Maple [C] (warning: unable to verify)

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

Time = 0.85 (sec) , antiderivative size = 1487, normalized size of antiderivative = 10.78

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

[In]

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

[Out]

-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*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
))-1/p^2*((b*x^2+a)^p)^(-2/p)*c^(-2/p)*exp(I*Pi*csgn(I*c*(b*x^2+a)^p)*(-csgn(I*c*(b*x^2+a)^p)+csgn(I*c))*(-csg
n(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*((b*x^2+a)^p)^(-2/p)*c^(-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*x^2-1/p^2/b^2*((b*x^2+a)^p)^(-2/p)*c^(-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*((
b*x^2+a)^p)^(-1/p)*c^(-1/p)*exp(1/2*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,-ln(b*x^2+a)-1/2*(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^2+1/2/p^2/b^2*a^2*((b*x^2+a)^p)^(-1/p)*c^(-1/p)*e
xp(1/2*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,-ln(b*x^2+a)-1/2*(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)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.02 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )^p\right )} \, dx=-\frac {{\left (a p \log \left (b x^{2} + a\right ) + a \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )} \operatorname {log\_integral}\left ({\left (b x^{2} + a\right )} c^{\left (\frac {1}{p}\right )}\right ) + {\left (b^{2} p x^{4} + a b p x^{2}\right )} c^{\frac {2}{p}} - 2 \, {\left (p \log \left (b x^{2} + a\right ) + \log \left (c\right )\right )} \operatorname {log\_integral}\left ({\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} c^{\frac {2}{p}}\right )}{2 \, {\left (b^{2} p^{3} \log \left (b x^{2} + a\right ) + b^{2} p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} \]

[In]

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

[Out]

-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))*log_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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

-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 = 0.30 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.27 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )^p\right )} \, dx=\frac {1}{2} \, a {\left (\frac {{\left (b x^{2} + a\right )} p}{b^{2} p^{3} \log \left (b x^{2} + a\right ) + b^{2} p^{2} \log \left (c\right )} - \frac {p {\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (b x^{2} + a\right )\right ) \log \left (b x^{2} + a\right )}{{\left (b^{2} p^{3} \log \left (b x^{2} + a\right ) + b^{2} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}} - \frac {{\rm Ei}\left (\frac {\log \left (c\right )}{p} + \log \left (b x^{2} + a\right )\right ) \log \left (c\right )}{{\left (b^{2} p^{3} \log \left (b x^{2} + a\right ) + b^{2} p^{2} \log \left (c\right )\right )} c^{\left (\frac {1}{p}\right )}}\right )} - \frac {\frac {{\left (b x^{2} + a\right )}^{2} p}{b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \left (c\right )} - \frac {2 \, p {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (b x^{2} + a\right )\right ) \log \left (b x^{2} + a\right )}{{\left (b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}} - \frac {2 \, {\rm Ei}\left (\frac {2 \, \log \left (c\right )}{p} + 2 \, \log \left (b x^{2} + a\right )\right ) \log \left (c\right )}{{\left (b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \left (c\right )\right )} c^{\frac {2}{p}}}}{2 \, b} \]

[In]

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

[Out]

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*log(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*lo
g(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*log(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. \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )^p\right )} \, dx=\int \frac {x^3}{{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2} \,d x \]

[In]

int(x^3/log(c*(a + b*x^2)^p)^2,x)

[Out]

int(x^3/log(c*(a + b*x^2)^p)^2, x)