3.34 \(\int \frac{\log (c (a+\frac{b}{x})^p)}{x^4} \, dx\)

Optimal. Leaf size=73 \[ \frac{a^2 p}{3 b^2 x}-\frac{a^3 p \log \left (a+\frac{b}{x}\right )}{3 b^3}-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{3 x^3}-\frac{a p}{6 b x^2}+\frac{p}{9 x^3} \]

[Out]

p/(9*x^3) - (a*p)/(6*b*x^2) + (a^2*p)/(3*b^2*x) - (a^3*p*Log[a + b/x])/(3*b^3) - Log[c*(a + b/x)^p]/(3*x^3)

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Rubi [A]  time = 0.0501546, antiderivative size = 73, 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, 43} \[ \frac{a^2 p}{3 b^2 x}-\frac{a^3 p \log \left (a+\frac{b}{x}\right )}{3 b^3}-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{3 x^3}-\frac{a p}{6 b x^2}+\frac{p}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b/x)^p]/x^4,x]

[Out]

p/(9*x^3) - (a*p)/(6*b*x^2) + (a^2*p)/(3*b^2*x) - (a^3*p*Log[a + b/x])/(3*b^3) - Log[c*(a + b/x)^p]/(3*x^3)

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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{x^4} \, dx &=-\operatorname{Subst}\left (\int x^2 \log \left (c (a+b x)^p\right ) \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{3 x^3}+\frac{1}{3} (b p) \operatorname{Subst}\left (\int \frac{x^3}{a+b x} \, dx,x,\frac{1}{x}\right )\\ &=-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{3 x^3}+\frac{1}{3} (b p) \operatorname{Subst}\left (\int \left (\frac{a^2}{b^3}-\frac{a x}{b^2}+\frac{x^2}{b}-\frac{a^3}{b^3 (a+b x)}\right ) \, dx,x,\frac{1}{x}\right )\\ &=\frac{p}{9 x^3}-\frac{a p}{6 b x^2}+\frac{a^2 p}{3 b^2 x}-\frac{a^3 p \log \left (a+\frac{b}{x}\right )}{3 b^3}-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0173841, size = 73, normalized size = 1. \[ \frac{a^2 p}{3 b^2 x}-\frac{a^3 p \log \left (a+\frac{b}{x}\right )}{3 b^3}-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{3 x^3}-\frac{a p}{6 b x^2}+\frac{p}{9 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b/x)^p]/x^4,x]

[Out]

p/(9*x^3) - (a*p)/(6*b*x^2) + (a^2*p)/(3*b^2*x) - (a^3*p*Log[a + b/x])/(3*b^3) - Log[c*(a + b/x)^p]/(3*x^3)

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Maple [F]  time = 0.07, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{4}}\ln \left ( c \left ( a+{\frac{b}{x}} \right ) ^{p} \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(a+b/x)^p)/x^4,x)

[Out]

int(ln(c*(a+b/x)^p)/x^4,x)

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Maxima [A]  time = 1.09592, size = 100, normalized size = 1.37 \begin{align*} -\frac{1}{18} \, b p{\left (\frac{6 \, a^{3} \log \left (a x + b\right )}{b^{4}} - \frac{6 \, a^{3} \log \left (x\right )}{b^{4}} - \frac{6 \, a^{2} x^{2} - 3 \, a b x + 2 \, b^{2}}{b^{3} x^{3}}\right )} - \frac{\log \left ({\left (a + \frac{b}{x}\right )}^{p} c\right )}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/x^4,x, algorithm="maxima")

[Out]

-1/18*b*p*(6*a^3*log(a*x + b)/b^4 - 6*a^3*log(x)/b^4 - (6*a^2*x^2 - 3*a*b*x + 2*b^2)/(b^3*x^3)) - 1/3*log((a +
b/x)^p*c)/x^3

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Fricas [A]  time = 2.2719, size = 151, normalized size = 2.07 \begin{align*} \frac{6 \, a^{2} b p x^{2} - 3 \, a b^{2} p x + 2 \, b^{3} p - 6 \, b^{3} \log \left (c\right ) - 6 \,{\left (a^{3} p x^{3} + b^{3} p\right )} \log \left (\frac{a x + b}{x}\right )}{18 \, b^{3} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/x^4,x, algorithm="fricas")

[Out]

1/18*(6*a^2*b*p*x^2 - 3*a*b^2*p*x + 2*b^3*p - 6*b^3*log(c) - 6*(a^3*p*x^3 + b^3*p)*log((a*x + b)/x))/(b^3*x^3)

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Sympy [A]  time = 11.2477, size = 80, normalized size = 1.1 \begin{align*} \begin{cases} - \frac{a^{3} p \log{\left (a + \frac{b}{x} \right )}}{3 b^{3}} + \frac{a^{2} p}{3 b^{2} x} - \frac{a p}{6 b x^{2}} - \frac{p \log{\left (a + \frac{b}{x} \right )}}{3 x^{3}} + \frac{p}{9 x^{3}} - \frac{\log{\left (c \right )}}{3 x^{3}} & \text{for}\: b \neq 0 \\- \frac{\log{\left (a^{p} c \right )}}{3 x^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(a+b/x)**p)/x**4,x)

[Out]

Piecewise((-a**3*p*log(a + b/x)/(3*b**3) + a**2*p/(3*b**2*x) - a*p/(6*b*x**2) - p*log(a + b/x)/(3*x**3) + p/(9
*x**3) - log(c)/(3*x**3), Ne(b, 0)), (-log(a**p*c)/(3*x**3), True))

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Giac [A]  time = 1.28302, size = 113, normalized size = 1.55 \begin{align*} -\frac{a^{3} p \log \left (a x + b\right )}{3 \, b^{3}} + \frac{a^{3} p \log \left (x\right )}{3 \, b^{3}} - \frac{p \log \left (a x + b\right )}{3 \, x^{3}} + \frac{p \log \left (x\right )}{3 \, x^{3}} + \frac{6 \, a^{2} p x^{2} - 3 \, a b p x + 2 \, b^{2} p - 6 \, b^{2} \log \left (c\right )}{18 \, b^{2} x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/x^4,x, algorithm="giac")

[Out]

-1/3*a^3*p*log(a*x + b)/b^3 + 1/3*a^3*p*log(x)/b^3 - 1/3*p*log(a*x + b)/x^3 + 1/3*p*log(x)/x^3 + 1/18*(6*a^2*p
*x^2 - 3*a*b*p*x + 2*b^2*p - 6*b^2*log(c))/(b^2*x^3)