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

Optimal. Leaf size=89 \[ \frac {b^5 p \log (a x+b)}{5 a^5}-\frac {b^4 p x}{5 a^4}+\frac {b^3 p x^2}{10 a^3}-\frac {b^2 p x^3}{15 a^2}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b p x^4}{20 a} \]

[Out]

-1/5*b^4*p*x/a^4+1/10*b^3*p*x^2/a^3-1/15*b^2*p*x^3/a^2+1/20*b*p*x^4/a+1/5*x^5*ln(c*(a+b/x)^p)+1/5*b^5*p*ln(a*x
+b)/a^5

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Rubi [A]  time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, 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} \[ \frac {b^3 p x^2}{10 a^3}-\frac {b^2 p x^3}{15 a^2}-\frac {b^4 p x}{5 a^4}+\frac {b^5 p \log (a x+b)}{5 a^5}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b p x^4}{20 a} \]

Antiderivative was successfully verified.

[In]

Int[x^4*Log[c*(a + b/x)^p],x]

[Out]

-(b^4*p*x)/(5*a^4) + (b^3*p*x^2)/(10*a^3) - (b^2*p*x^3)/(15*a^2) + (b*p*x^4)/(20*a) + (x^5*Log[c*(a + b/x)^p])
/5 + (b^5*p*Log[b + a*x])/(5*a^5)

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])

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 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]

Rubi steps

\begin {align*} \int x^4 \log \left (c \left (a+\frac {b}{x}\right )^p\right ) \, dx &=\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{5} (b p) \int \frac {x^3}{a+\frac {b}{x}} \, dx\\ &=\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{5} (b p) \int \frac {x^4}{b+a x} \, dx\\ &=\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {1}{5} (b p) \int \left (-\frac {b^3}{a^4}+\frac {b^2 x}{a^3}-\frac {b x^2}{a^2}+\frac {x^3}{a}+\frac {b^4}{a^4 (b+a x)}\right ) \, dx\\ &=-\frac {b^4 p x}{5 a^4}+\frac {b^3 p x^2}{10 a^3}-\frac {b^2 p x^3}{15 a^2}+\frac {b p x^4}{20 a}+\frac {1}{5} x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+\frac {b^5 p \log (b+a x)}{5 a^5}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 85, normalized size = 0.96 \[ \frac {12 a^5 x^5 \log \left (c \left (a+\frac {b}{x}\right )^p\right )+a b p x \left (3 a^3 x^3-4 a^2 b x^2+6 a b^2 x-12 b^3\right )+12 b^5 p \log \left (a+\frac {b}{x}\right )+12 b^5 p \log (x)}{60 a^5} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Log[c*(a + b/x)^p],x]

[Out]

(a*b*p*x*(-12*b^3 + 6*a*b^2*x - 4*a^2*b*x^2 + 3*a^3*x^3) + 12*b^5*p*Log[a + b/x] + 12*a^5*x^5*Log[c*(a + b/x)^
p] + 12*b^5*p*Log[x])/(60*a^5)

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fricas [A]  time = 0.45, size = 89, normalized size = 1.00 \[ \frac {12 \, a^{5} p x^{5} \log \left (\frac {a x + b}{x}\right ) + 12 \, a^{5} x^{5} \log \relax (c) + 3 \, a^{4} b p x^{4} - 4 \, a^{3} b^{2} p x^{3} + 6 \, a^{2} b^{3} p x^{2} - 12 \, a b^{4} p x + 12 \, b^{5} p \log \left (a x + b\right )}{60 \, a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*a^5*p*x^5*log((a*x + b)/x) + 12*a^5*x^5*log(c) + 3*a^4*b*p*x^4 - 4*a^3*b^2*p*x^3 + 6*a^2*b^3*p*x^2 -
12*a*b^4*p*x + 12*b^5*p*log(a*x + b))/a^5

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giac [B]  time = 0.19, size = 308, normalized size = 3.46 \[ -\frac {\frac {12 \, b^{6} p \log \left (\frac {a x + b}{x}\right )}{a^{5} - \frac {5 \, {\left (a x + b\right )} a^{4}}{x} + \frac {10 \, {\left (a x + b\right )}^{2} a^{3}}{x^{2}} - \frac {10 \, {\left (a x + b\right )}^{3} a^{2}}{x^{3}} + \frac {5 \, {\left (a x + b\right )}^{4} a}{x^{4}} - \frac {{\left (a x + b\right )}^{5}}{x^{5}}} + \frac {12 \, b^{6} p \log \left (-a + \frac {a x + b}{x}\right )}{a^{5}} - \frac {12 \, b^{6} p \log \left (\frac {a x + b}{x}\right )}{a^{5}} - \frac {25 \, a^{4} b^{6} p - 12 \, a^{4} b^{6} \log \relax (c) - \frac {77 \, {\left (a x + b\right )} a^{3} b^{6} p}{x} + \frac {94 \, {\left (a x + b\right )}^{2} a^{2} b^{6} p}{x^{2}} - \frac {54 \, {\left (a x + b\right )}^{3} a b^{6} p}{x^{3}} + \frac {12 \, {\left (a x + b\right )}^{4} b^{6} p}{x^{4}}}{a^{9} - \frac {5 \, {\left (a x + b\right )} a^{8}}{x} + \frac {10 \, {\left (a x + b\right )}^{2} a^{7}}{x^{2}} - \frac {10 \, {\left (a x + b\right )}^{3} a^{6}}{x^{3}} + \frac {5 \, {\left (a x + b\right )}^{4} a^{5}}{x^{4}} - \frac {{\left (a x + b\right )}^{5} a^{4}}{x^{5}}}}{60 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(12*b^6*p*log((a*x + b)/x)/(a^5 - 5*(a*x + b)*a^4/x + 10*(a*x + b)^2*a^3/x^2 - 10*(a*x + b)^3*a^2/x^3 +
5*(a*x + b)^4*a/x^4 - (a*x + b)^5/x^5) + 12*b^6*p*log(-a + (a*x + b)/x)/a^5 - 12*b^6*p*log((a*x + b)/x)/a^5 -
(25*a^4*b^6*p - 12*a^4*b^6*log(c) - 77*(a*x + b)*a^3*b^6*p/x + 94*(a*x + b)^2*a^2*b^6*p/x^2 - 54*(a*x + b)^3*a
*b^6*p/x^3 + 12*(a*x + b)^4*b^6*p/x^4)/(a^9 - 5*(a*x + b)*a^8/x + 10*(a*x + b)^2*a^7/x^2 - 10*(a*x + b)^3*a^6/
x^3 + 5*(a*x + b)^4*a^5/x^4 - (a*x + b)^5*a^4/x^5))/b

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int x^{4} \ln \left (c \left (a +\frac {b}{x}\right )^{p}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.67, size = 74, normalized size = 0.83 \[ \frac {1}{5} \, x^{5} \log \left ({\left (a + \frac {b}{x}\right )}^{p} c\right ) + \frac {1}{60} \, b p {\left (\frac {12 \, b^{4} \log \left (a x + b\right )}{a^{5}} + \frac {3 \, a^{3} x^{4} - 4 \, a^{2} b x^{3} + 6 \, a b^{2} x^{2} - 12 \, b^{3} x}{a^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*x^5*log((a + b/x)^p*c) + 1/60*b*p*(12*b^4*log(a*x + b)/a^5 + (3*a^3*x^4 - 4*a^2*b*x^3 + 6*a*b^2*x^2 - 12*b
^3*x)/a^4)

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mupad [B]  time = 0.24, size = 77, normalized size = 0.87 \[ \frac {x^5\,\ln \left (c\,{\left (a+\frac {b}{x}\right )}^p\right )}{5}-\frac {b^2\,p\,x^3}{15\,a^2}+\frac {b^3\,p\,x^2}{10\,a^3}+\frac {b^5\,p\,\ln \left (b+a\,x\right )}{5\,a^5}+\frac {b\,p\,x^4}{20\,a}-\frac {b^4\,p\,x}{5\,a^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*log(c*(a + b/x)^p),x)

[Out]

(x^5*log(c*(a + b/x)^p))/5 - (b^2*p*x^3)/(15*a^2) + (b^3*p*x^2)/(10*a^3) + (b^5*p*log(b + a*x))/(5*a^5) + (b*p
*x^4)/(20*a) - (b^4*p*x)/(5*a^4)

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sympy [A]  time = 12.38, size = 122, normalized size = 1.37 \[ \begin {cases} \frac {p x^{5} \log {\left (a + \frac {b}{x} \right )}}{5} + \frac {x^{5} \log {\relax (c )}}{5} + \frac {b p x^{4}}{20 a} - \frac {b^{2} p x^{3}}{15 a^{2}} + \frac {b^{3} p x^{2}}{10 a^{3}} - \frac {b^{4} p x}{5 a^{4}} + \frac {b^{5} p \log {\left (a x + b \right )}}{5 a^{5}} & \text {for}\: a \neq 0 \\\frac {p x^{5} \log {\relax (b )}}{5} - \frac {p x^{5} \log {\relax (x )}}{5} + \frac {p x^{5}}{25} + \frac {x^{5} \log {\relax (c )}}{5} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((p*x**5*log(a + b/x)/5 + x**5*log(c)/5 + b*p*x**4/(20*a) - b**2*p*x**3/(15*a**2) + b**3*p*x**2/(10*a
**3) - b**4*p*x/(5*a**4) + b**5*p*log(a*x + b)/(5*a**5), Ne(a, 0)), (p*x**5*log(b)/5 - p*x**5*log(x)/5 + p*x**
5/25 + x**5*log(c)/5, True))

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