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

Optimal. Leaf size=61 \[ -\frac{b^2 p x}{3 a^2}+\frac{b^3 p \log (a x+b)}{3 a^3}+\frac{1}{3} x^3 \log \left (c \left (a+\frac{b}{x}\right )^p\right )+\frac{b p x^2}{6 a} \]

[Out]

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

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Rubi [A]  time = 0.0316943, antiderivative size = 61, 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 = {2455, 263, 43} \[ -\frac{b^2 p x}{3 a^2}+\frac{b^3 p \log (a x+b)}{3 a^3}+\frac{1}{3} x^3 \log \left (c \left (a+\frac{b}{x}\right )^p\right )+\frac{b p x^2}{6 a} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Log[c*(a + b/x)^p],x]

[Out]

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

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 263

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

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

Mathematica [A]  time = 0.0240148, size = 62, normalized size = 1.02 \[ \frac{2 a^3 x^3 \log \left (c \left (a+\frac{b}{x}\right )^p\right )+2 b^3 p \log \left (a+\frac{b}{x}\right )+a b p x (a x-2 b)+2 b^3 p \log (x)}{6 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Log[c*(a + b/x)^p],x]

[Out]

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

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Maple [F]  time = 0.068, size = 0, normalized size = 0. \begin{align*} \int{x}^{2}\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^2*ln(c*(a+b/x)^p),x)

[Out]

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

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Maxima [A]  time = 1.20513, size = 69, normalized size = 1.13 \begin{align*} \frac{1}{3} \, x^{3} \log \left ({\left (a + \frac{b}{x}\right )}^{p} c\right ) + \frac{1}{6} \, b p{\left (\frac{2 \, b^{2} \log \left (a x + b\right )}{a^{3}} + \frac{a x^{2} - 2 \, b x}{a^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.14907, size = 149, normalized size = 2.44 \begin{align*} \frac{2 \, a^{3} p x^{3} \log \left (\frac{a x + b}{x}\right ) + 2 \, a^{3} x^{3} \log \left (c\right ) + a^{2} b p x^{2} - 2 \, a b^{2} p x + 2 \, b^{3} p \log \left (a x + b\right )}{6 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 6.51316, size = 95, normalized size = 1.56 \begin{align*} \begin{cases} \frac{p x^{3} \log{\left (a + \frac{b}{x} \right )}}{3} + \frac{x^{3} \log{\left (c \right )}}{3} + \frac{b p x^{2}}{6 a} - \frac{b^{2} p x}{3 a^{2}} + \frac{b^{3} p \log{\left (a x + b \right )}}{3 a^{3}} & \text{for}\: a \neq 0 \\\frac{p x^{3} \log{\left (b \right )}}{3} - \frac{p x^{3} \log{\left (x \right )}}{3} + \frac{p x^{3}}{9} + \frac{x^{3} \log{\left (c \right )}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28979, size = 85, normalized size = 1.39 \begin{align*} \frac{1}{3} \, p x^{3} \log \left (a x + b\right ) - \frac{1}{3} \, p x^{3} \log \left (x\right ) + \frac{1}{3} \, x^{3} \log \left (c\right ) + \frac{b p x^{2}}{6 \, a} - \frac{b^{2} p x}{3 \, a^{2}} + \frac{b^{3} p \log \left (a x + b\right )}{3 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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