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

Optimal. Leaf size=107 \[ \frac{\left (a+b x^2\right )^2 \left (c \left (a+b x^2\right )^p\right )^{-2/p} \text{Ei}\left (\frac{2 \log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b^2 p}-\frac{a \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b^2 p} \]

[Out]

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

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Rubi [A]  time = 0.154434, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {2454, 2399, 2389, 2300, 2178, 2390, 2310} \[ \frac{\left (a+b x^2\right )^2 \left (c \left (a+b x^2\right )^p\right )^{-2/p} \text{Ei}\left (\frac{2 \log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b^2 p}-\frac{a \left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b^2 p} \]

Antiderivative was successfully verified.

[In]

Int[x^3/Log[c*(a + b*x^2)^p],x]

[Out]

-(a*(a + b*x^2)*ExpIntegralEi[Log[c*(a + b*x^2)^p]/p])/(2*b^2*p*(c*(a + b*x^2)^p)^p^(-1)) + ((a + b*x^2)^2*Exp
IntegralEi[(2*Log[c*(a + b*x^2)^p])/p])/(2*b^2*p*(c*(a + b*x^2)^p)^(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 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 \left (c \left (a+b x^2\right )^p\right )} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x}{\log \left (c (a+b x)^p\right )} \, dx,x,x^2\right )\\ &=\frac{1}{2} \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 )\\ &=\frac{\operatorname{Subst}\left (\int \frac{a+b x}{\log \left (c (a+b x)^p\right )} \, dx,x,x^2\right )}{2 b}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\log \left (c (a+b x)^p\right )} \, dx,x,x^2\right )}{2 b}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x}{\log \left (c x^p\right )} \, dx,x,a+b x^2\right )}{2 b^2}-\frac{a \operatorname{Subst}\left (\int \frac{1}{\log \left (c x^p\right )} \, dx,x,a+b x^2\right )}{2 b^2}\\ &=\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 )}{2 b^2 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}\\ &=-\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}+\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 )}{2 b^2 p}\\ \end{align*}

Mathematica [A]  time = 0.139206, size = 96, normalized size = 0.9 \[ -\frac{\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-2/p} \left (a \left (c \left (a+b x^2\right )^p\right )^{\frac{1}{p}} \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )-\left (a+b x^2\right ) \text{Ei}\left (\frac{2 \log \left (c \left (b x^2+a\right )^p\right )}{p}\right )\right )}{2 b^2 p} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/Log[c*(a + b*x^2)^p],x]

[Out]

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

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Maple [F]  time = 0.484, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{3}}{\ln \left ( c \left ( b{x}^{2}+a \right ) ^{p} \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [A]  time = 2.26807, size = 162, normalized size = 1.51 \begin{align*} -\frac{a c^{\left (\frac{1}{p}\right )} \logintegral \left ({\left (b x^{2} + a\right )} c^{\left (\frac{1}{p}\right )}\right ) - \logintegral \left ({\left (b^{2} x^{4} + 2 \, a b x^{2} + a^{2}\right )} c^{\frac{2}{p}}\right )}{2 \, b^{2} c^{\frac{2}{p}} p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a*c^(1/p)*log_integral((b*x^2 + a)*c^(1/p)) - log_integral((b^2*x^4 + 2*a*b*x^2 + a^2)*c^(2/p)))/(b^2*c^
(2/p)*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 )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27946, size = 99, normalized size = 0.93 \begin{align*} -\frac{\frac{a{\rm Ei}\left (\frac{\log \left (c\right )}{p} + \log \left (b x^{2} + a\right )\right )}{b c^{\left (\frac{1}{p}\right )} p} - \frac{{\rm Ei}\left (\frac{2 \, \log \left (c\right )}{p} + 2 \, \log \left (b x^{2} + a\right )\right )}{b c^{\frac{2}{p}} 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),x, algorithm="giac")

[Out]

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