3.33 \(\int \frac{\log (c (a+\frac{b}{x})^p)}{x^3} \, dx\)

Optimal. Leaf size=59 \[ \frac{a^2 p \log \left (a+\frac{b}{x}\right )}{2 b^2}-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 x^2}-\frac{a p}{2 b x}+\frac{p}{4 x^2} \]

[Out]

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

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Rubi [A]  time = 0.0354572, antiderivative size = 59, 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} \[ \frac{a^2 p \log \left (a+\frac{b}{x}\right )}{2 b^2}-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 x^2}-\frac{a p}{2 b x}+\frac{p}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b/x)^p]/x^3,x]

[Out]

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

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

Mathematica [A]  time = 0.01413, size = 59, normalized size = 1. \[ \frac{a^2 p \log \left (a+\frac{b}{x}\right )}{2 b^2}-\frac{\log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 x^2}-\frac{a p}{2 b x}+\frac{p}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b/x)^p]/x^3,x]

[Out]

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

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Maple [F]  time = 0.273, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}}\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^3,x)

[Out]

int(ln(c*(a+b/x)^p)/x^3,x)

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Maxima [A]  time = 1.05556, size = 85, normalized size = 1.44 \begin{align*} \frac{1}{4} \, b p{\left (\frac{2 \, a^{2} \log \left (a x + b\right )}{b^{3}} - \frac{2 \, a^{2} \log \left (x\right )}{b^{3}} - \frac{2 \, a x - b}{b^{2} x^{2}}\right )} - \frac{\log \left ({\left (a + \frac{b}{x}\right )}^{p} c\right )}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.22248, size = 124, normalized size = 2.1 \begin{align*} -\frac{2 \, a b p x - b^{2} p + 2 \, b^{2} \log \left (c\right ) - 2 \,{\left (a^{2} p x^{2} - b^{2} p\right )} \log \left (\frac{a x + b}{x}\right )}{4 \, b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 6.97564, size = 66, normalized size = 1.12 \begin{align*} \begin{cases} \frac{a^{2} p \log{\left (a + \frac{b}{x} \right )}}{2 b^{2}} - \frac{a p}{2 b x} - \frac{p \log{\left (a + \frac{b}{x} \right )}}{2 x^{2}} + \frac{p}{4 x^{2}} - \frac{\log{\left (c \right )}}{2 x^{2}} & \text{for}\: b \neq 0 \\- \frac{\log{\left (a^{p} c \right )}}{2 x^{2}} & \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**3,x)

[Out]

Piecewise((a**2*p*log(a + b/x)/(2*b**2) - a*p/(2*b*x) - p*log(a + b/x)/(2*x**2) + p/(4*x**2) - log(c)/(2*x**2)
, Ne(b, 0)), (-log(a**p*c)/(2*x**2), True))

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Giac [A]  time = 1.32137, size = 95, normalized size = 1.61 \begin{align*} \frac{a^{2} p \log \left (a x + b\right )}{2 \, b^{2}} - \frac{a^{2} p \log \left (x\right )}{2 \, b^{2}} - \frac{p \log \left (a x + b\right )}{2 \, x^{2}} + \frac{p \log \left (x\right )}{2 \, x^{2}} - \frac{2 \, a p x - b p + 2 \, b \log \left (c\right )}{4 \, b x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(a+b/x)^p)/x^3,x, algorithm="giac")

[Out]

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