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]

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

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Rubi [A]  time = 0.07, antiderivative size = 83, normalized size of antiderivative = 1.00, 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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.05, 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]

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

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fricas [A]  time = 0.46, size = 78, normalized size = 0.94 \[ -\frac {{\left (b p x^{2} + a p\right )} c^{\left (\frac {1}{p}\right )} - {\left (p \log \left (b x^{2} + a\right ) + \log \relax (c)\right )} \operatorname {log\_integral}\left ({\left (b x^{2} + a\right )} c^{\left (\frac {1}{p}\right )}\right )}{2 \, {\left (b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \relax (c)\right )} c^{\left (\frac {1}{p}\right )}} \]

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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giac [A]  time = 0.18, size = 141, normalized size = 1.70 \[ -\frac {{\left (b x^{2} + a\right )} p}{2 \, {\left (b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \relax (c)\right )}} + \frac {p {\rm Ei}\left (\frac {\log \relax (c)}{p} + \log \left (b x^{2} + a\right )\right ) \log \left (b x^{2} + a\right )}{2 \, {\left (b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \relax (c)\right )} c^{\left (\frac {1}{p}\right )}} + \frac {{\rm Ei}\left (\frac {\log \relax (c)}{p} + \log \left (b x^{2} + a\right )\right ) \log \relax (c)}{2 \, {\left (b p^{3} \log \left (b x^{2} + a\right ) + b p^{2} \log \relax (c)\right )} c^{\left (\frac {1}{p}\right )}} \]

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))

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maple [C]  time = 1.22, size = 421, normalized size = 5.07 \[ -\frac {\left (b \,x^{2}+a \right ) c^{-\frac {1}{p}} \left (\left (b \,x^{2}+a \right )^{p}\right )^{-\frac {1}{p}} \Ei \left (1, -\ln \left (b \,x^{2}+a \right )-\frac {-i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )+i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}-i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3}-2 p \ln \left (b \,x^{2}+a \right )+2 \ln \relax (c )+2 \ln \left (\left (b \,x^{2}+a \right )^{p}\right )}{2 p}\right ) {\mathrm e}^{\frac {i \pi \left (\mathrm {csgn}\left (i c \right )-\mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )\right ) \left (\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right )-\mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )}{2 p}}}{2 b \,p^{2}}-\frac {b \,x^{2}+a}{\left (-i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )+i \pi \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (b \,x^{2}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{2}-i \pi \mathrm {csgn}\left (i c \left (b \,x^{2}+a \right )^{p}\right )^{3}+2 \ln \relax (c )+2 \ln \left (\left (b \,x^{2}+a \right )^{p}\right )\right ) b p} \]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {b x^{2} + a}{2 \, {\left (b p^{2} \log \left (b x^{2} + a\right ) + b p \log \relax (c)\right )}} + \int \frac {x}{p^{2} \log \left (b x^{2} + a\right ) + p \log \relax (c)}\,{d x} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x}{{\ln \left (c\,{\left (b\,x^2+a\right )}^p\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/log(c*(a + b*x^2)^p)^2,x)

[Out]

int(x/log(c*(a + b*x^2)^p)^2, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x}{\log {\left (c \left (a + b x^{2}\right )^{p} \right )}^{2}}\, dx \]

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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