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

Optimal. Leaf size=138 \[ \frac{\left (a+b x^2\right )^2 \left (c \left (a+b x^2\right )^p\right )^{-2/p} \text{Ei}\left (\frac{2 \log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{b^2 p^2}-\frac{a \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b^2 p^2}-\frac{x^2 \left (a+b x^2\right )}{2 b p \log \left (c \left (a+b x^2\right )^p\right )} \]

[Out]

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

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Rubi [A]  time = 0.206615, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 8, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.444, Rules used = {2454, 2400, 2399, 2389, 2300, 2178, 2390, 2310} \[ \frac{\left (a+b x^2\right )^2 \left (c \left (a+b x^2\right )^p\right )^{-2/p} \text{Ei}\left (\frac{2 \log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{b^2 p^2}-\frac{a \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b^2 p^2}-\frac{x^2 \left (a+b x^2\right )}{2 b p \log \left (c \left (a+b x^2\right )^p\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Log[c*(a + b*x^2)^p]^2,x]

[Out]

-(a*(a + b*x^2)*ExpIntegralEi[Log[c*(a + b*x^2)^p]/p])/(2*b^2*p^2*(c*(a + b*x^2)^p)^p^(-1)) + ((a + b*x^2)^2*E
xpIntegralEi[(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 2454

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])

Rule 2400

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 2399

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 2389

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 2300

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 2178

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[(F^(g*(e - (c*f)/d))*ExpIntegral
Ei[(f*g*(c + d*x)*Log[F])/d])/d, x] /; FreeQ[{F, c, d, e, f, g}, x] &&  !$UseGamma === True

Rule 2390

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 2310

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)*x)/n)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}
, x]

Rubi steps

\begin{align*} \int \frac{x^3}{\log ^2\left (c \left (a+b x^2\right )^p\right )} \, dx &=\frac{1}{2} \operatorname{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{\operatorname{Subst}\left (\int \frac{x}{\log \left (c (a+b x)^p\right )} \, dx,x,x^2\right )}{p}+\frac{a \operatorname{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{\operatorname{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 \operatorname{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{\operatorname{Subst}\left (\int \frac{a+b x}{\log \left (c (a+b x)^p\right )} \, dx,x,x^2\right )}{b p}-\frac{a \operatorname{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 ) \operatorname{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{\operatorname{Subst}\left (\int \frac{x}{\log \left (c x^p\right )} \, dx,x,a+b x^2\right )}{b^2 p}-\frac{a \operatorname{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 ) \operatorname{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 ) \operatorname{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]  time = 0.152928, size = 157, normalized size = 1.14 \[ -\frac{\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-2/p} \left (a \left (c \left (a+b x^2\right )^p\right )^{\frac{1}{p}} \log \left (c \left (a+b x^2\right )^p\right ) \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )-2 \left (a+b x^2\right ) \log \left (c \left (a+b x^2\right )^p\right ) \text{Ei}\left (\frac{2 \log \left (c \left (b x^2+a\right )^p\right )}{p}\right )+b p x^2 \left (c \left (a+b x^2\right )^p\right )^{2/p}\right )}{2 b^2 p^2 \log \left (c \left (a+b x^2\right )^p\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 5.273, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}}{ \left ( \ln \left ( c \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{b x^{4} + a x^{2}}{2 \,{\left (b p \log \left ({\left (b x^{2} + a\right )}^{p}\right ) + b p \log \left (c\right )\right )}} + \int \frac{2 \, b x^{3} + a x}{b p \log \left ({\left (b x^{2} + a\right )}^{p}\right ) + b p \log \left (c\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

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

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Fricas [A]  time = 2.28195, size = 338, normalized size = 2.45 \begin{align*} -\frac{{\left (a p \log \left (b x^{2} + a\right ) + a \log \left (c\right )\right )} c^{\left (\frac{1}{p}\right )} \logintegral \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 )} \logintegral \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}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Giac [B]  time = 1.35384, size = 402, normalized size = 2.91 \begin{align*} -\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{{\left (b x^{2} + a\right )} a p}{b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \left (c\right )} + \frac{a 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 p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \left (c\right )\right )} c^{\left (\frac{1}{p}\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{a{\rm Ei}\left (\frac{\log \left (c\right )}{p} + \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^{\left (\frac{1}{p}\right )}} - \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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