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

Optimal. Leaf size=83 \[ \frac{\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b p^2}-\frac{a+b x^2}{2 b p \log \left (c \left (a+b x^2\right )^p\right )} \]

[Out]

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

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Rubi [A]  time = 0.0720991, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {2454, 2389, 2297, 2300, 2178} \[ \frac{\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b p^2}-\frac{a+b x^2}{2 b p \log \left (c \left (a+b x^2\right )^p\right )} \]

Antiderivative was successfully verified.

[In]

Int[x/Log[c*(a + b*x^2)^p]^2,x]

[Out]

((a + b*x^2)*ExpIntegralEi[Log[c*(a + b*x^2)^p]/p])/(2*b*p^2*(c*(a + b*x^2)^p)^p^(-1)) - (a + b*x^2)/(2*b*p*Lo
g[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 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 2297

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[(x*(a + b*Log[c*x^n])^(p + 1))/(b*n*(p + 1))
, x] - Dist[1/(b*n*(p + 1)), Int[(a + b*Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] &&
IntegerQ[2*p]

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

Rubi steps

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

Mathematica [A]  time = 0.0436608, size = 97, normalized size = 1.17 \[ -\frac{\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \left (p \left (c \left (a+b x^2\right )^p\right )^{\frac{1}{p}}-\log \left (c \left (a+b x^2\right )^p\right ) \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )\right )}{2 b p^2 \log \left (c \left (a+b x^2\right )^p\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Log[c*(a + b*x^2)^p]^2,x]

[Out]

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

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Maple [C]  time = 1.129, size = 466, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.2156, size = 194, normalized size = 2.34 \begin{align*} -\frac{{\left (b p x^{2} + a p\right )} c^{\left (\frac{1}{p}\right )} -{\left (p \log \left (b x^{2} + a\right ) + \log \left (c\right )\right )} \logintegral \left ({\left (b x^{2} + a\right )} c^{\left (\frac{1}{p}\right )}\right )}{2 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x}{\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/ln(c*(b*x**2+a)**p)**2,x)

[Out]

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

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Giac [A]  time = 1.26515, size = 190, normalized size = 2.29 \begin{align*} -\frac{{\left (b x^{2} + a\right )} p}{2 \,{\left (b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \left (c\right )\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 )}{2 \,{\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{{\rm Ei}\left (\frac{\log \left (c\right )}{p} + \log \left (b x^{2} + a\right )\right ) \log \left (c\right )}{2 \,{\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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