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*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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Rubi [A]  time = 0.0385288, antiderivative size = 45, normalized size of antiderivative = 1., 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 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 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 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]

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*}

Mathematica [A]  time = 0.002599, size = 45, normalized size = 1. \[ -\frac{\log \left (c \left (a+b x^3\right )^p\right )}{3 x^3}-\frac{b p \log \left (a+b x^3\right )}{3 a}+\frac{b p \log (x)}{a} \]

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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Maple [C]  time = 0.27, size = 173, normalized size = 3.8 \begin{align*} -{\frac{\ln \left ( \left ( b{x}^{3}+a \right ) ^{p} \right ) }{3\,{x}^{3}}}-{\frac{i\pi \,a{\it csgn} \left ( i \left ( b{x}^{3}+a \right ) ^{p} \right ) \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{2}-i\pi \,a{\it csgn} \left ( i \left ( b{x}^{3}+a \right ) ^{p} \right ){\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ){\it csgn} \left ( ic \right ) -i\pi \,a \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{3}+i\pi \,a \left ({\it csgn} \left ( ic \left ( b{x}^{3}+a \right ) ^{p} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -6\,bp\ln \left ( x \right ){x}^{3}+2\,bp\ln \left ( b{x}^{3}+a \right ){x}^{3}+2\,\ln \left ( c \right ) a}{6\,a{x}^{3}}} \end{align*}

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*ln(c)*a)/a/x^3

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Maxima [A]  time = 1.08538, size = 59, normalized size = 1.31 \begin{align*} -\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}} \end{align*}

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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Fricas [A]  time = 1.90716, size = 103, normalized size = 2.29 \begin{align*} \frac{3 \, b p x^{3} \log \left (x\right ) -{\left (b p x^{3} + a p\right )} \log \left (b x^{3} + a\right ) - a \log \left (c\right )}{3 \, a x^{3}} \end{align*}

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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Sympy [A]  time = 22.911, size = 82, normalized size = 1.82 \begin{align*} \begin{cases} - \frac{p \log{\left (a + b x^{3} \right )}}{3 x^{3}} - \frac{\log{\left (c \right )}}{3 x^{3}} + \frac{b p \log{\left (x \right )}}{a} - \frac{b p \log{\left (a + b x^{3} \right )}}{3 a} & \text{for}\: a \neq 0 \\- \frac{p \log{\left (b \right )}}{3 x^{3}} - \frac{p \log{\left (x \right )}}{x^{3}} - \frac{p}{3 x^{3}} - \frac{\log{\left (c \right )}}{3 x^{3}} & \text{otherwise} \end{cases} \end{align*}

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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Giac [A]  time = 1.22113, size = 78, normalized size = 1.73 \begin{align*} -\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 \left (c\right )}{x^{3}}}{3 \, b} \end{align*}

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