3.6.10 \(\int x (a+b \log (c (d+\frac {e}{x^{2/3}})^n)) \, dx\) [510]

Optimal. Leaf size=94 \[ -\frac {b e^2 n x^{2/3}}{2 d^2}+\frac {b e n x^{4/3}}{4 d}+\frac {b e^3 n \log \left (d+\frac {e}{x^{2/3}}\right )}{2 d^3}+\frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )+\frac {b e^3 n \log (x)}{3 d^3} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2504, 2442, 46} \begin {gather*} \frac {1}{2} x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )+\frac {b e^3 n \log \left (d+\frac {e}{x^{2/3}}\right )}{2 d^3}+\frac {b e^3 n \log (x)}{3 d^3}-\frac {b e^2 n x^{2/3}}{2 d^2}+\frac {b e n x^{4/3}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(b*e^2*n*x^(2/3))/d^2 + (b*e*n*x^(4/3))/(4*d) + (b*e^3*n*Log[d + e/x^(2/3)])/(2*d^3) + (x^2*(a + b*Log[c*
(d + e/x^(2/3))^n]))/2 + (b*e^3*n*Log[x])/(3*d^3)

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2442

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 2504

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

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Mathematica [A]
time = 0.02, size = 91, normalized size = 0.97 \begin {gather*} \frac {a x^2}{2}+\frac {1}{2} b x^2 \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )-\frac {1}{2} b e n \left (\frac {e x^{2/3}}{d^2}-\frac {x^{4/3}}{2 d}-\frac {e^2 \log \left (d+\frac {e}{x^{2/3}}\right )}{d^3}-\frac {2 e^2 \log (x)}{3 d^3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int x \left (a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.29, size = 66, normalized size = 0.70 \begin {gather*} \frac {1}{4} \, b n {\left (\frac {d x^{\frac {4}{3}} - 2 \, x^{\frac {2}{3}} e}{d^{2}} + \frac {2 \, e^{2} \log \left (d x^{\frac {2}{3}} + e\right )}{d^{3}}\right )} e + \frac {1}{2} \, b x^{2} \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right ) + \frac {1}{2} \, a x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.41, size = 115, normalized size = 1.22 \begin {gather*} \frac {2 \, b d^{3} x^{2} \log \left (c\right ) + b d^{2} n x^{\frac {4}{3}} e + 2 \, a d^{3} x^{2} - 4 \, b d^{3} n \log \left (x^{\frac {1}{3}}\right ) - 2 \, b d n x^{\frac {2}{3}} e^{2} + 2 \, {\left (b d^{3} n + b n e^{3}\right )} \log \left (d x^{\frac {2}{3}} + e\right ) + 2 \, {\left (b d^{3} n x^{2} - b d^{3} n\right )} \log \left (\frac {d x + x^{\frac {1}{3}} e}{x}\right )}{4 \, d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]
time = 4.62, size = 72, normalized size = 0.77 \begin {gather*} \frac {1}{2} \, b x^{2} \log \left (c\right ) + \frac {1}{4} \, {\left (2 \, x^{2} \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right ) + {\left (\frac {d x^{\frac {4}{3}} - 2 \, x^{\frac {2}{3}} e}{d^{2}} + \frac {2 \, e^{2} \log \left ({\left | d x^{\frac {2}{3}} + e \right |}\right )}{d^{3}}\right )} e\right )} b n + \frac {1}{2} \, a x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.59, size = 73, normalized size = 0.78 \begin {gather*} \frac {x^{4/3}\,\left (\frac {b\,e\,n}{2\,d}-\frac {b\,e^2\,n}{d^2\,x^{2/3}}\right )}{2}+\frac {a\,x^2}{2}+\frac {b\,x^2\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}{2}+\frac {b\,e^3\,n\,\mathrm {atanh}\left (\frac {2\,e}{d\,x^{2/3}}+1\right )}{d^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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