3.514 \(\int \frac{a+b \log (c (d+\frac{e}{x^{2/3}})^n)}{x^3} \, dx\)

Optimal. Leaf size=89 \[ -\frac{a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )}{2 x^2}+\frac{b d^2 n}{2 e^2 x^{2/3}}-\frac{b d^3 n \log \left (d+\frac{e}{x^{2/3}}\right )}{2 e^3}-\frac{b d n}{4 e x^{4/3}}+\frac{b n}{6 x^2} \]

[Out]

(b*n)/(6*x^2) - (b*d*n)/(4*e*x^(4/3)) + (b*d^2*n)/(2*e^2*x^(2/3)) - (b*d^3*n*Log[d + e/x^(2/3)])/(2*e^3) - (a
+ b*Log[c*(d + e/x^(2/3))^n])/(2*x^2)

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Rubi [A]  time = 0.0680862, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {2454, 2395, 43} \[ -\frac{a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )}{2 x^2}+\frac{b d^2 n}{2 e^2 x^{2/3}}-\frac{b d^3 n \log \left (d+\frac{e}{x^{2/3}}\right )}{2 e^3}-\frac{b d n}{4 e x^{4/3}}+\frac{b n}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e/x^(2/3))^n])/x^3,x]

[Out]

(b*n)/(6*x^2) - (b*d*n)/(4*e*x^(4/3)) + (b*d^2*n)/(2*e^2*x^(2/3)) - (b*d^3*n*Log[d + e/x^(2/3)])/(2*e^3) - (a
+ b*Log[c*(d + e/x^(2/3))^n])/(2*x^2)

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 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 \frac{a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )}{x^3} \, dx &=-\left (\frac{3}{2} \operatorname{Subst}\left (\int x^2 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx,x,\frac{1}{x^{2/3}}\right )\right )\\ &=-\frac{a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )}{2 x^2}+\frac{1}{2} (b e n) \operatorname{Subst}\left (\int \frac{x^3}{d+e x} \, dx,x,\frac{1}{x^{2/3}}\right )\\ &=-\frac{a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )}{2 x^2}+\frac{1}{2} (b e n) \operatorname{Subst}\left (\int \left (\frac{d^2}{e^3}-\frac{d x}{e^2}+\frac{x^2}{e}-\frac{d^3}{e^3 (d+e x)}\right ) \, dx,x,\frac{1}{x^{2/3}}\right )\\ &=\frac{b n}{6 x^2}-\frac{b d n}{4 e x^{4/3}}+\frac{b d^2 n}{2 e^2 x^{2/3}}-\frac{b d^3 n \log \left (d+\frac{e}{x^{2/3}}\right )}{2 e^3}-\frac{a+b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0325246, size = 94, normalized size = 1.06 \[ -\frac{a}{2 x^2}-\frac{b \log \left (c \left (d+\frac{e}{x^{2/3}}\right )^n\right )}{2 x^2}+\frac{b d^2 n}{2 e^2 x^{2/3}}-\frac{b d^3 n \log \left (d+\frac{e}{x^{2/3}}\right )}{2 e^3}-\frac{b d n}{4 e x^{4/3}}+\frac{b n}{6 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e/x^(2/3))^n])/x^3,x]

[Out]

-a/(2*x^2) + (b*n)/(6*x^2) - (b*d*n)/(4*e*x^(4/3)) + (b*d^2*n)/(2*e^2*x^(2/3)) - (b*d^3*n*Log[d + e/x^(2/3)])/
(2*e^3) - (b*Log[c*(d + e/x^(2/3))^n])/(2*x^2)

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Maple [F]  time = 0.353, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{3}} \left ( a+b\ln \left ( c \left ( d+{e{x}^{-{\frac{2}{3}}}} \right ) ^{n} \right ) \right ) }\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*(d+e/x^(2/3))^n))/x^3,x)

[Out]

int((a+b*ln(c*(d+e/x^(2/3))^n))/x^3,x)

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Maxima [A]  time = 1.0359, size = 119, normalized size = 1.34 \begin{align*} -\frac{1}{12} \, b e n{\left (\frac{6 \, d^{3} \log \left (d x^{\frac{2}{3}} + e\right )}{e^{4}} - \frac{6 \, d^{3} \log \left (x^{\frac{2}{3}}\right )}{e^{4}} - \frac{6 \, d^{2} x^{\frac{4}{3}} - 3 \, d e x^{\frac{2}{3}} + 2 \, e^{2}}{e^{3} x^{2}}\right )} - \frac{b \log \left (c{\left (d + \frac{e}{x^{\frac{2}{3}}}\right )}^{n}\right )}{2 \, x^{2}} - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(2/3))^n))/x^3,x, algorithm="maxima")

[Out]

-1/12*b*e*n*(6*d^3*log(d*x^(2/3) + e)/e^4 - 6*d^3*log(x^(2/3))/e^4 - (6*d^2*x^(4/3) - 3*d*e*x^(2/3) + 2*e^2)/(
e^3*x^2)) - 1/2*b*log(c*(d + e/x^(2/3))^n)/x^2 - 1/2*a/x^2

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Fricas [A]  time = 1.78914, size = 205, normalized size = 2.3 \begin{align*} \frac{6 \, b d^{2} e n x^{\frac{4}{3}} - 3 \, b d e^{2} n x^{\frac{2}{3}} + 2 \, b e^{3} n - 6 \, b e^{3} \log \left (c\right ) - 6 \, a e^{3} - 6 \,{\left (b d^{3} n x^{2} + b e^{3} n\right )} \log \left (\frac{d x + e x^{\frac{1}{3}}}{x}\right )}{12 \, e^{3} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(2/3))^n))/x^3,x, algorithm="fricas")

[Out]

1/12*(6*b*d^2*e*n*x^(4/3) - 3*b*d*e^2*n*x^(2/3) + 2*b*e^3*n - 6*b*e^3*log(c) - 6*a*e^3 - 6*(b*d^3*n*x^2 + b*e^
3*n)*log((d*x + e*x^(1/3))/x))/(e^3*x^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*(d+e/x**(2/3))**n))/x**3,x)

[Out]

Timed out

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Giac [A]  time = 1.45338, size = 140, normalized size = 1.57 \begin{align*} \frac{1}{12} \,{\left ({\left (12 \, d^{3} e^{\left (-4\right )} \log \left (x^{\frac{1}{3}}\right ) - 6 \, d^{3} e^{\left (-4\right )} \log \left ({\left | d x^{\frac{2}{3}} + e \right |}\right ) - \frac{{\left (11 \, d^{3} x^{2} - 6 \, d^{2} x^{\frac{4}{3}} e + 3 \, d x^{\frac{2}{3}} e^{2} - 2 \, e^{3}\right )} e^{\left (-4\right )}}{x^{2}}\right )} e - \frac{6 \, \log \left (d + \frac{e}{x^{\frac{2}{3}}}\right )}{x^{2}}\right )} b n - \frac{b \log \left (c\right )}{2 \, x^{2}} - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e/x^(2/3))^n))/x^3,x, algorithm="giac")

[Out]

1/12*((12*d^3*e^(-4)*log(x^(1/3)) - 6*d^3*e^(-4)*log(abs(d*x^(2/3) + e)) - (11*d^3*x^2 - 6*d^2*x^(4/3)*e + 3*d
*x^(2/3)*e^2 - 2*e^3)*e^(-4)/x^2)*e - 6*log(d + e/x^(2/3))/x^2)*b*n - 1/2*b*log(c)/x^2 - 1/2*a/x^2