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

Optimal. Leaf size=51 \[ \frac{\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b p} \]

[Out]

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

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Rubi [A]  time = 0.0576011, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2454, 2389, 2300, 2178} \[ \frac{\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b p} \]

Antiderivative was successfully verified.

[In]

Int[x/Log[c*(a + b*x^2)^p],x]

[Out]

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

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

Mathematica [A]  time = 0.0414916, size = 51, normalized size = 1. \[ \frac{\left (a+b x^2\right ) \left (c \left (a+b x^2\right )^p\right )^{-1/p} \text{Ei}\left (\frac{\log \left (c \left (b x^2+a\right )^p\right )}{p}\right )}{2 b p} \]

Antiderivative was successfully verified.

[In]

Integrate[x/Log[c*(a + b*x^2)^p],x]

[Out]

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

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Maple [C]  time = 1.23, size = 317, normalized size = 6.2 \begin{align*} -{\frac{1}{2\,bp}{{\rm e}^{{\frac{i\pi \,{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) -i\pi \, \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -i\pi \,{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}+i\pi \, \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}+2\,p\ln \left ( b{x}^{2}+a \right ) -2\,\ln \left ( c \right ) -2\,\ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) }{2\,p}}}}{\it Ei} \left ( 1,-\ln \left ( b{x}^{2}+a \right ) -{\frac{i\pi \,{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}-i\pi \,{\it csgn} \left ( i \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) -i\pi \, \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}+i\pi \, \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +2\,\ln \left ( c \right ) +2\,\ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) -2\,p\ln \left ( b{x}^{2}+a \right ) }{2\,p}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b/p*exp(1/2*(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)^2*csgn(I
*c)-I*Pi*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)^2+I*Pi*csgn(I*c*(b*x^2+a)^p)^3+2*p*ln(b*x^2+a)-2*ln(c)-2*ln
((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)

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

[Out]

integrate(x/log((b*x^2 + a)^p*c), x)

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Fricas [A]  time = 2.23548, size = 72, normalized size = 1.41 \begin{align*} \frac{\logintegral \left ({\left (b x^{2} + a\right )} c^{\left (\frac{1}{p}\right )}\right )}{2 \, b c^{\left (\frac{1}{p}\right )} p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/log(c*(a + b*x**2)**p), x)

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Giac [A]  time = 1.28448, size = 42, normalized size = 0.82 \begin{align*} \frac{{\rm Ei}\left (\frac{\log \left (c\right )}{p} + \log \left (b x^{2} + a\right )\right )}{2 \, b c^{\left (\frac{1}{p}\right )} p} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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