3.10 \(\int \frac{\log (c (a+b x^2)^p)}{x^5} \, dx\)

Optimal. Leaf size=64 \[ \frac{b^2 p \log \left (a+b x^2\right )}{4 a^2}-\frac{b^2 p \log (x)}{2 a^2}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4}-\frac{b p}{4 a x^2} \]

[Out]

-(b*p)/(4*a*x^2) - (b^2*p*Log[x])/(2*a^2) + (b^2*p*Log[a + b*x^2])/(4*a^2) - Log[c*(a + b*x^2)^p]/(4*x^4)

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Rubi [A]  time = 0.0520589, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {2454, 2395, 44} \[ \frac{b^2 p \log \left (a+b x^2\right )}{4 a^2}-\frac{b^2 p \log (x)}{2 a^2}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4}-\frac{b p}{4 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^2)^p]/x^5,x]

[Out]

-(b*p)/(4*a*x^2) - (b^2*p*Log[x])/(2*a^2) + (b^2*p*Log[a + b*x^2])/(4*a^2) - Log[c*(a + b*x^2)^p]/(4*x^4)

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 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\log \left (c \left (a+b x^2\right )^p\right )}{x^5} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{\log \left (c (a+b x)^p\right )}{x^3} \, dx,x,x^2\right )\\ &=-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4}+\frac{1}{4} (b p) \operatorname{Subst}\left (\int \frac{1}{x^2 (a+b x)} \, dx,x,x^2\right )\\ &=-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4}+\frac{1}{4} (b p) \operatorname{Subst}\left (\int \left (\frac{1}{a x^2}-\frac{b}{a^2 x}+\frac{b^2}{a^2 (a+b x)}\right ) \, dx,x,x^2\right )\\ &=-\frac{b p}{4 a x^2}-\frac{b^2 p \log (x)}{2 a^2}+\frac{b^2 p \log \left (a+b x^2\right )}{4 a^2}-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4}\\ \end{align*}

Mathematica [A]  time = 0.0368477, size = 56, normalized size = 0.88 \[ \frac{1}{4} b p \left (\frac{b \log \left (a+b x^2\right )}{a^2}-\frac{2 b \log (x)}{a^2}-\frac{1}{a x^2}\right )-\frac{\log \left (c \left (a+b x^2\right )^p\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^2)^p]/x^5,x]

[Out]

(b*p*(-(1/(a*x^2)) - (2*b*Log[x])/a^2 + (b*Log[a + b*x^2])/a^2))/4 - Log[c*(a + b*x^2)^p]/(4*x^4)

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Maple [C]  time = 0.295, size = 198, normalized size = 3.1 \begin{align*} -{\frac{\ln \left ( \left ( b{x}^{2}+a \right ) ^{p} \right ) }{4\,{x}^{4}}}-{\frac{4\,{b}^{2}p\ln \left ( x \right ){x}^{4}-2\,{b}^{2}p\ln \left ( -b{x}^{2}-a \right ){x}^{4}+i\pi \,{a}^{2}{\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 \,{a}^{2}{\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 \,{a}^{2} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{3}+i\pi \,{a}^{2} \left ({\it csgn} \left ( ic \left ( b{x}^{2}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) +2\,abp{x}^{2}+2\,\ln \left ( c \right ){a}^{2}}{8\,{a}^{2}{x}^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^2+a)^p)/x^5,x)

[Out]

-1/4/x^4*ln((b*x^2+a)^p)-1/8*(4*b^2*p*ln(x)*x^4-2*b^2*p*ln(-b*x^2-a)*x^4+I*Pi*a^2*csgn(I*(b*x^2+a)^p)*csgn(I*c
*(b*x^2+a)^p)^2-I*Pi*a^2*csgn(I*(b*x^2+a)^p)*csgn(I*c*(b*x^2+a)^p)*csgn(I*c)-I*Pi*a^2*csgn(I*c*(b*x^2+a)^p)^3+
I*Pi*a^2*csgn(I*c*(b*x^2+a)^p)^2*csgn(I*c)+2*a*b*p*x^2+2*ln(c)*a^2)/a^2/x^4

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Maxima [A]  time = 1.0698, size = 73, normalized size = 1.14 \begin{align*} \frac{1}{4} \, b p{\left (\frac{b \log \left (b x^{2} + a\right )}{a^{2}} - \frac{b \log \left (x^{2}\right )}{a^{2}} - \frac{1}{a x^{2}}\right )} - \frac{\log \left ({\left (b x^{2} + a\right )}^{p} c\right )}{4 \, x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^5,x, algorithm="maxima")

[Out]

1/4*b*p*(b*log(b*x^2 + a)/a^2 - b*log(x^2)/a^2 - 1/(a*x^2)) - 1/4*log((b*x^2 + a)^p*c)/x^4

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Fricas [A]  time = 2.08974, size = 134, normalized size = 2.09 \begin{align*} -\frac{2 \, b^{2} p x^{4} \log \left (x\right ) + a b p x^{2} + a^{2} \log \left (c\right ) -{\left (b^{2} p x^{4} - a^{2} p\right )} \log \left (b x^{2} + a\right )}{4 \, a^{2} x^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^5,x, algorithm="fricas")

[Out]

-1/4*(2*b^2*p*x^4*log(x) + a*b*p*x^2 + a^2*log(c) - (b^2*p*x^4 - a^2*p)*log(b*x^2 + a))/(a^2*x^4)

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Sympy [A]  time = 21.4743, size = 102, normalized size = 1.59 \begin{align*} \begin{cases} - \frac{p \log{\left (a + b x^{2} \right )}}{4 x^{4}} - \frac{\log{\left (c \right )}}{4 x^{4}} - \frac{b p}{4 a x^{2}} - \frac{b^{2} p \log{\left (x \right )}}{2 a^{2}} + \frac{b^{2} p \log{\left (a + b x^{2} \right )}}{4 a^{2}} & \text{for}\: a \neq 0 \\- \frac{p \log{\left (b \right )}}{4 x^{4}} - \frac{p \log{\left (x \right )}}{2 x^{4}} - \frac{p}{8 x^{4}} - \frac{\log{\left (c \right )}}{4 x^{4}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(b*x**2+a)**p)/x**5,x)

[Out]

Piecewise((-p*log(a + b*x**2)/(4*x**4) - log(c)/(4*x**4) - b*p/(4*a*x**2) - b**2*p*log(x)/(2*a**2) + b**2*p*lo
g(a + b*x**2)/(4*a**2), Ne(a, 0)), (-p*log(b)/(4*x**4) - p*log(x)/(2*x**4) - p/(8*x**4) - log(c)/(4*x**4), Tru
e))

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Giac [B]  time = 1.28153, size = 178, normalized size = 2.78 \begin{align*} -\frac{\frac{b^{3} p \log \left (b x^{2} + a\right )}{{\left (b x^{2} + a\right )}^{2} - 2 \,{\left (b x^{2} + a\right )} a + a^{2}} - \frac{b^{3} p \log \left (b x^{2} + a\right )}{a^{2}} + \frac{b^{3} p \log \left (b x^{2}\right )}{a^{2}} + \frac{{\left (b x^{2} + a\right )} b^{3} p - a b^{3} p + a b^{3} \log \left (c\right )}{{\left (b x^{2} + a\right )}^{2} a - 2 \,{\left (b x^{2} + a\right )} a^{2} + a^{3}}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^2+a)^p)/x^5,x, algorithm="giac")

[Out]

-1/4*(b^3*p*log(b*x^2 + a)/((b*x^2 + a)^2 - 2*(b*x^2 + a)*a + a^2) - b^3*p*log(b*x^2 + a)/a^2 + b^3*p*log(b*x^
2)/a^2 + ((b*x^2 + a)*b^3*p - a*b^3*p + a*b^3*log(c))/((b*x^2 + a)^2*a - 2*(b*x^2 + a)*a^2 + a^3))/b