3.22 \(\int \frac {\log (c (a+b x^3)^p)}{x^4} \, dx\)

Optimal. Leaf size=45 \[ -\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3}-\frac {b p \log \left (a+b x^3\right )}{3 a}+\frac {b p \log (x)}{a} \]

[Out]

b*p*ln(x)/a-1/3*b*p*ln(b*x^3+a)/a-1/3*ln(c*(b*x^3+a)^p)/x^3

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Rubi [A]  time = 0.04, antiderivative size = 45, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {2454, 2395, 36, 29, 31} \[ -\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3}-\frac {b p \log \left (a+b x^3\right )}{3 a}+\frac {b p \log (x)}{a} \]

Antiderivative was successfully verified.

[In]

Int[Log[c*(a + b*x^3)^p]/x^4,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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 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])

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 45, normalized size = 1.00 \[ -\frac {\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3}-\frac {b p \log \left (a+b x^3\right )}{3 a}+\frac {b p \log (x)}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Log[c*(a + b*x^3)^p]/x^4,x]

[Out]

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

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fricas [A]  time = 0.47, size = 43, normalized size = 0.96 \[ \frac {3 \, b p x^{3} \log \relax (x) - {\left (b p x^{3} + a p\right )} \log \left (b x^{3} + a\right ) - a \log \relax (c)}{3 \, a x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^3+a)^p)/x^4,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 58, normalized size = 1.29 \[ -\frac {\frac {b^{2} p \log \left (b x^{3} + a\right )}{a} - \frac {b^{2} p \log \left (b x^{3}\right )}{a} + \frac {b p \log \left (b x^{3} + a\right )}{x^{3}} + \frac {b \log \relax (c)}{x^{3}}}{3 \, b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^3+a)^p)/x^4,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.25, size = 173, normalized size = 3.84 \[ -\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{3 x^{3}}-\frac {-6 b p \,x^{3} \ln \relax (x )+2 b p \,x^{3} \ln \left (b \,x^{3}+a \right )-i \pi a \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )+i \pi a \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )^{2}+i \pi a \,\mathrm {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )^{2}-i \pi a \mathrm {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )^{3}+2 a \ln \relax (c )}{6 a \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(ln(c*(b*x^3+a)^p)/x^4,x)

[Out]

-1/3/x^3*ln((b*x^3+a)^p)-1/6*(I*Pi*a*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)^2-I*Pi*a*csgn(I*(b*x^3+a)^p)*cs
gn(I*c*(b*x^3+a)^p)*csgn(I*c)-I*Pi*a*csgn(I*c*(b*x^3+a)^p)^3+I*Pi*a*csgn(I*c*(b*x^3+a)^p)^2*csgn(I*c)-6*b*p*ln
(x)*x^3+2*b*p*ln(b*x^3+a)*x^3+2*a*ln(c))/a/x^3

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maxima [A]  time = 0.64, size = 44, normalized size = 0.98 \[ -\frac {1}{3} \, b p {\left (\frac {\log \left (b x^{3} + a\right )}{a} - \frac {\log \left (x^{3}\right )}{a}\right )} - \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(log(c*(b*x^3+a)^p)/x^4,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.26, size = 41, normalized size = 0.91 \[ \frac {b\,p\,\ln \relax (x)}{a}-\frac {b\,p\,\ln \left (b\,x^3+a\right )}{3\,a}-\frac {\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x^3)^p)/x^4,x)

[Out]

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

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sympy [A]  time = 10.81, size = 82, normalized size = 1.82 \[ \begin {cases} - \frac {p \log {\left (a + b x^{3} \right )}}{3 x^{3}} - \frac {\log {\relax (c )}}{3 x^{3}} + \frac {b p \log {\relax (x )}}{a} - \frac {b p \log {\left (a + b x^{3} \right )}}{3 a} & \text {for}\: a \neq 0 \\- \frac {p \log {\relax (b )}}{3 x^{3}} - \frac {p \log {\relax (x )}}{x^{3}} - \frac {p}{3 x^{3}} - \frac {\log {\relax (c )}}{3 x^{3}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(ln(c*(b*x**3+a)**p)/x**4,x)

[Out]

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

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