3.447 \(\int \frac{a+b \log (c (d+e \sqrt [3]{x})^n)}{x^2} \, dx\)

Optimal. Leaf size=87 \[ -\frac{a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x}+\frac{b e^2 n}{d^2 \sqrt [3]{x}}-\frac{b e^3 n \log \left (d+e \sqrt [3]{x}\right )}{d^3}+\frac{b e^3 n \log (x)}{3 d^3}-\frac{b e n}{2 d x^{2/3}} \]

[Out]

-(b*e*n)/(2*d*x^(2/3)) + (b*e^2*n)/(d^2*x^(1/3)) - (b*e^3*n*Log[d + e*x^(1/3)])/d^3 - (a + b*Log[c*(d + e*x^(1
/3))^n])/x + (b*e^3*n*Log[x])/(3*d^3)

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Rubi [A]  time = 0.0679198, antiderivative size = 87, 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, 44} \[ -\frac{a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x}+\frac{b e^2 n}{d^2 \sqrt [3]{x}}-\frac{b e^3 n \log \left (d+e \sqrt [3]{x}\right )}{d^3}+\frac{b e^3 n \log (x)}{3 d^3}-\frac{b e n}{2 d x^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Log[c*(d + e*x^(1/3))^n])/x^2,x]

[Out]

-(b*e*n)/(2*d*x^(2/3)) + (b*e^2*n)/(d^2*x^(1/3)) - (b*e^3*n*Log[d + e*x^(1/3)])/d^3 - (a + b*Log[c*(d + e*x^(1
/3))^n])/x + (b*e^3*n*Log[x])/(3*d^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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0331193, size = 84, normalized size = 0.97 \[ -\frac{a}{x}-\frac{b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )}{x}+b e n \left (-\frac{e^2 \log \left (d+e \sqrt [3]{x}\right )}{d^3}+\frac{e^2 \log (x)}{3 d^3}+\frac{e}{d^2 \sqrt [3]{x}}-\frac{1}{2 d x^{2/3}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Log[c*(d + e*x^(1/3))^n])/x^2,x]

[Out]

-(a/x) - (b*Log[c*(d + e*x^(1/3))^n])/x + b*e*n*(-1/(2*d*x^(2/3)) + e/(d^2*x^(1/3)) - (e^2*Log[d + e*x^(1/3)])
/d^3 + (e^2*Log[x])/(3*d^3))

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Maple [F]  time = 0.098, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{{x}^{2}} \left ( a+b\ln \left ( c \left ( d+e\sqrt [3]{x} \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^(1/3))^n))/x^2,x)

[Out]

int((a+b*ln(c*(d+e*x^(1/3))^n))/x^2,x)

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Maxima [A]  time = 1.02714, size = 101, normalized size = 1.16 \begin{align*} -\frac{1}{6} \, b e n{\left (\frac{6 \, e^{2} \log \left (e x^{\frac{1}{3}} + d\right )}{d^{3}} - \frac{2 \, e^{2} \log \left (x\right )}{d^{3}} - \frac{3 \,{\left (2 \, e x^{\frac{1}{3}} - d\right )}}{d^{2} x^{\frac{2}{3}}}\right )} - \frac{b \log \left ({\left (e x^{\frac{1}{3}} + d\right )}^{n} c\right )}{x} - \frac{a}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.8584, size = 208, normalized size = 2.39 \begin{align*} \frac{2 \, b e^{3} n x \log \left (x^{\frac{1}{3}}\right ) + 2 \, b d e^{2} n x^{\frac{2}{3}} - b d^{2} e n x^{\frac{1}{3}} - 2 \, b d^{3} \log \left (c\right ) - 2 \, a d^{3} - 2 \,{\left (b e^{3} n x + b d^{3} n\right )} \log \left (e x^{\frac{1}{3}} + d\right )}{2 \, d^{3} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^2,x, algorithm="fricas")

[Out]

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

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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**(1/3))**n))/x**2,x)

[Out]

Timed out

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Giac [B]  time = 1.3379, size = 378, normalized size = 4.34 \begin{align*} -\frac{{\left (2 \,{\left (x^{\frac{1}{3}} e + d\right )}^{3} b n e^{4} \log \left (x^{\frac{1}{3}} e + d\right ) - 6 \,{\left (x^{\frac{1}{3}} e + d\right )}^{2} b d n e^{4} \log \left (x^{\frac{1}{3}} e + d\right ) + 6 \,{\left (x^{\frac{1}{3}} e + d\right )} b d^{2} n e^{4} \log \left (x^{\frac{1}{3}} e + d\right ) - 2 \,{\left (x^{\frac{1}{3}} e + d\right )}^{3} b n e^{4} \log \left (x^{\frac{1}{3}} e\right ) + 6 \,{\left (x^{\frac{1}{3}} e + d\right )}^{2} b d n e^{4} \log \left (x^{\frac{1}{3}} e\right ) - 6 \,{\left (x^{\frac{1}{3}} e + d\right )} b d^{2} n e^{4} \log \left (x^{\frac{1}{3}} e\right ) + 2 \, b d^{3} n e^{4} \log \left (x^{\frac{1}{3}} e\right ) - 2 \,{\left (x^{\frac{1}{3}} e + d\right )}^{2} b d n e^{4} + 5 \,{\left (x^{\frac{1}{3}} e + d\right )} b d^{2} n e^{4} - 3 \, b d^{3} n e^{4} + 2 \, b d^{3} e^{4} \log \left (c\right ) + 2 \, a d^{3} e^{4}\right )} e^{\left (-1\right )}}{2 \,{\left ({\left (x^{\frac{1}{3}} e + d\right )}^{3} d^{3} - 3 \,{\left (x^{\frac{1}{3}} e + d\right )}^{2} d^{4} + 3 \,{\left (x^{\frac{1}{3}} e + d\right )} d^{5} - d^{6}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*(d+e*x^(1/3))^n))/x^2,x, algorithm="giac")

[Out]

-1/2*(2*(x^(1/3)*e + d)^3*b*n*e^4*log(x^(1/3)*e + d) - 6*(x^(1/3)*e + d)^2*b*d*n*e^4*log(x^(1/3)*e + d) + 6*(x
^(1/3)*e + d)*b*d^2*n*e^4*log(x^(1/3)*e + d) - 2*(x^(1/3)*e + d)^3*b*n*e^4*log(x^(1/3)*e) + 6*(x^(1/3)*e + d)^
2*b*d*n*e^4*log(x^(1/3)*e) - 6*(x^(1/3)*e + d)*b*d^2*n*e^4*log(x^(1/3)*e) + 2*b*d^3*n*e^4*log(x^(1/3)*e) - 2*(
x^(1/3)*e + d)^2*b*d*n*e^4 + 5*(x^(1/3)*e + d)*b*d^2*n*e^4 - 3*b*d^3*n*e^4 + 2*b*d^3*e^4*log(c) + 2*a*d^3*e^4)
*e^(-1)/((x^(1/3)*e + d)^3*d^3 - 3*(x^(1/3)*e + d)^2*d^4 + 3*(x^(1/3)*e + d)*d^5 - d^6)