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

Optimal. Leaf size=47 \[ -\frac{b^2 p \log (a x+b)}{2 a^2}+\frac{1}{2} x^2 \log \left (c \left (a+\frac{b}{x}\right )^p\right )+\frac{b p x}{2 a} \]

[Out]

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

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Rubi [A]  time = 0.0210677, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {2455, 193, 43} \[ -\frac{b^2 p \log (a x+b)}{2 a^2}+\frac{1}{2} x^2 \log \left (c \left (a+\frac{b}{x}\right )^p\right )+\frac{b p x}{2 a} \]

Antiderivative was successfully verified.

[In]

Int[x*Log[c*(a + b/x)^p],x]

[Out]

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

Rule 2455

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

Rule 193

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b}, x] && LtQ[n, 0]
&& IntegerQ[p]

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

Mathematica [A]  time = 0.0168691, size = 40, normalized size = 0.85 \[ \frac{1}{2} \left (\frac{b p (a x-b \log (a x+b))}{a^2}+x^2 \log \left (c \left (a+\frac{b}{x}\right )^p\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x*Log[c*(a + b/x)^p],x]

[Out]

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

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Maple [F]  time = 0.081, size = 0, normalized size = 0. \begin{align*} \int x\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(x*ln(c*(a+b/x)^p),x)

[Out]

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

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Maxima [A]  time = 1.1689, size = 54, normalized size = 1.15 \begin{align*} \frac{1}{2} \, b p{\left (\frac{x}{a} - \frac{b \log \left (a x + b\right )}{a^{2}}\right )} + \frac{1}{2} \, x^{2} \log \left ({\left (a + \frac{b}{x}\right )}^{p} c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.12569, size = 116, normalized size = 2.47 \begin{align*} \frac{a^{2} p x^{2} \log \left (\frac{a x + b}{x}\right ) + a^{2} x^{2} \log \left (c\right ) + a b p x - b^{2} p \log \left (a x + b\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.21775, size = 82, normalized size = 1.74 \begin{align*} \begin{cases} \frac{p x^{2} \log{\left (a + \frac{b}{x} \right )}}{2} + \frac{x^{2} \log{\left (c \right )}}{2} + \frac{b p x}{2 a} - \frac{b^{2} p \log{\left (a x + b \right )}}{2 a^{2}} & \text{for}\: a \neq 0 \\\frac{p x^{2} \log{\left (b \right )}}{2} - \frac{p x^{2} \log{\left (x \right )}}{2} + \frac{p x^{2}}{4} + \frac{x^{2} \log{\left (c \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.31848, size = 69, normalized size = 1.47 \begin{align*} \frac{1}{2} \, p x^{2} \log \left (a x + b\right ) - \frac{1}{2} \, p x^{2} \log \left (x\right ) + \frac{1}{2} \, x^{2} \log \left (c\right ) + \frac{b p x}{2 \, a} - \frac{b^{2} p \log \left (a x + b\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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