3.5.95 \(\int \frac {a+b \log (c (d+\frac {e}{\sqrt [3]{x}})^n)}{x^3} \, dx\) [495]

Optimal. Leaf size=138 \[ \frac {b n}{12 x^2}-\frac {b d n}{10 e x^{5/3}}+\frac {b d^2 n}{8 e^2 x^{4/3}}-\frac {b d^3 n}{6 e^3 x}+\frac {b d^4 n}{4 e^4 x^{2/3}}-\frac {b d^5 n}{2 e^5 \sqrt [3]{x}}+\frac {b d^6 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{2 e^6}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2} \]

[Out]

1/12*b*n/x^2-1/10*b*d*n/e/x^(5/3)+1/8*b*d^2*n/e^2/x^(4/3)-1/6*b*d^3*n/e^3/x+1/4*b*d^4*n/e^4/x^(2/3)-1/2*b*d^5*
n/e^5/x^(1/3)+1/2*b*d^6*n*ln(d+e/x^(1/3))/e^6+1/2*(-a-b*ln(c*(d+e/x^(1/3))^n))/x^2

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Rubi [A]
time = 0.07, antiderivative size = 138, normalized size of antiderivative = 1.00, 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 = {2504, 2442, 45} \begin {gather*} -\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2}+\frac {b d^6 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{2 e^6}-\frac {b d^5 n}{2 e^5 \sqrt [3]{x}}+\frac {b d^4 n}{4 e^4 x^{2/3}}-\frac {b d^3 n}{6 e^3 x}+\frac {b d^2 n}{8 e^2 x^{4/3}}-\frac {b d n}{10 e x^{5/3}}+\frac {b n}{12 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*n)/(12*x^2) - (b*d*n)/(10*e*x^(5/3)) + (b*d^2*n)/(8*e^2*x^(4/3)) - (b*d^3*n)/(6*e^3*x) + (b*d^4*n)/(4*e^4*x
^(2/3)) - (b*d^5*n)/(2*e^5*x^(1/3)) + (b*d^6*n*Log[d + e/x^(1/3)])/(2*e^6) - (a + b*Log[c*(d + e/x^(1/3))^n])/
(2*x^2)

Rule 45

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])

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 \frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{x^3} \, dx &=-\left (3 \text {Subst}\left (\int x^5 \left (a+b \log \left (c (d+e x)^n\right )\right ) \, dx,x,\frac {1}{\sqrt [3]{x}}\right )\right )\\ &=-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2}+\frac {1}{2} (b e n) \text {Subst}\left (\int \frac {x^6}{d+e x} \, dx,x,\frac {1}{\sqrt [3]{x}}\right )\\ &=-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2}+\frac {1}{2} (b e n) \text {Subst}\left (\int \left (-\frac {d^5}{e^6}+\frac {d^4 x}{e^5}-\frac {d^3 x^2}{e^4}+\frac {d^2 x^3}{e^3}-\frac {d x^4}{e^2}+\frac {x^5}{e}+\frac {d^6}{e^6 (d+e x)}\right ) \, dx,x,\frac {1}{\sqrt [3]{x}}\right )\\ &=\frac {b n}{12 x^2}-\frac {b d n}{10 e x^{5/3}}+\frac {b d^2 n}{8 e^2 x^{4/3}}-\frac {b d^3 n}{6 e^3 x}+\frac {b d^4 n}{4 e^4 x^{2/3}}-\frac {b d^5 n}{2 e^5 \sqrt [3]{x}}+\frac {b d^6 n \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{2 e^6}-\frac {a+b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 135, normalized size = 0.98 \begin {gather*} -\frac {a}{2 x^2}+\frac {1}{2} b e n \left (\frac {1}{6 e x^2}-\frac {d}{5 e^2 x^{5/3}}+\frac {d^2}{4 e^3 x^{4/3}}-\frac {d^3}{3 e^4 x}+\frac {d^4}{2 e^5 x^{2/3}}-\frac {d^5}{e^6 \sqrt [3]{x}}+\frac {d^6 \log \left (d+\frac {e}{\sqrt [3]{x}}\right )}{e^7}\right )-\frac {b \log \left (c \left (d+\frac {e}{\sqrt [3]{x}}\right )^n\right )}{2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a/x^2 + (b*e*n*(1/(6*e*x^2) - d/(5*e^2*x^(5/3)) + d^2/(4*e^3*x^(4/3)) - d^3/(3*e^4*x) + d^4/(2*e^5*x^(2/3
)) - d^5/(e^6*x^(1/3)) + (d^6*Log[d + e/x^(1/3)])/e^7))/2 - (b*Log[c*(d + e/x^(1/3))^n])/(2*x^2)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (d +\frac {e}{x^{\frac {1}{3}}}\right )^{n}\right )}{x^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.32, size = 114, normalized size = 0.83 \begin {gather*} \frac {1}{120} \, {\left (60 \, d^{6} e^{\left (-7\right )} \log \left (d x^{\frac {1}{3}} + e\right ) - 20 \, d^{6} e^{\left (-7\right )} \log \left (x\right ) - \frac {{\left (60 \, d^{5} x^{\frac {5}{3}} - 30 \, d^{4} x^{\frac {4}{3}} e + 20 \, d^{3} x e^{2} - 15 \, d^{2} x^{\frac {2}{3}} e^{3} + 12 \, d x^{\frac {1}{3}} e^{4} - 10 \, e^{5}\right )} e^{\left (-6\right )}}{x^{2}}\right )} b n e - \frac {b \log \left (c {\left (d + \frac {e}{x^{\frac {1}{3}}}\right )}^{n}\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(60*d^6*e^(-7)*log(d*x^(1/3) + e) - 20*d^6*e^(-7)*log(x) - (60*d^5*x^(5/3) - 30*d^4*x^(4/3)*e + 20*d^3*x
*e^2 - 15*d^2*x^(2/3)*e^3 + 12*d*x^(1/3)*e^4 - 10*e^5)*e^(-6)/x^2)*b*n*e - 1/2*b*log(c*(d + e/x^(1/3))^n)/x^2
- 1/2*a/x^2

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Fricas [A]
time = 0.39, size = 149, normalized size = 1.08 \begin {gather*} \frac {{\left (60 \, {\left (b x^{2} - b\right )} e^{6} \log \left (c\right ) - 10 \, {\left ({\left (b n - 6 \, a\right )} x^{2} - b n + 6 \, a\right )} e^{6} + 20 \, {\left (b d^{3} n x^{2} - b d^{3} n x\right )} e^{3} + 60 \, {\left (b d^{6} n x^{2} - b n e^{6}\right )} \log \left (\frac {d x + x^{\frac {2}{3}} e}{x}\right ) - 15 \, {\left (4 \, b d^{5} n x e - b d^{2} n e^{4}\right )} x^{\frac {2}{3}} + 6 \, {\left (5 \, b d^{4} n x e^{2} - 2 \, b d n e^{5}\right )} x^{\frac {1}{3}}\right )} e^{\left (-6\right )}}{120 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(60*(b*x^2 - b)*e^6*log(c) - 10*((b*n - 6*a)*x^2 - b*n + 6*a)*e^6 + 20*(b*d^3*n*x^2 - b*d^3*n*x)*e^3 + 6
0*(b*d^6*n*x^2 - b*n*e^6)*log((d*x + x^(2/3)*e)/x) - 15*(4*b*d^5*n*x*e - b*d^2*n*e^4)*x^(2/3) + 6*(5*b*d^4*n*x
*e^2 - 2*b*d*n*e^5)*x^(1/3))*e^(-6)/x^2

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3062 deep

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Giac [A]
time = 4.58, size = 123, normalized size = 0.89 \begin {gather*} \frac {1}{120} \, {\left ({\left (60 \, d^{6} e^{\left (-7\right )} \log \left ({\left | d x^{\frac {1}{3}} + e \right |}\right ) - 20 \, d^{6} e^{\left (-7\right )} \log \left ({\left | x \right |}\right ) - \frac {{\left (60 \, d^{5} x^{\frac {5}{3}} e - 30 \, d^{4} x^{\frac {4}{3}} e^{2} + 20 \, d^{3} x e^{3} - 15 \, d^{2} x^{\frac {2}{3}} e^{4} + 12 \, d x^{\frac {1}{3}} e^{5} - 10 \, e^{6}\right )} e^{\left (-7\right )}}{x^{2}}\right )} e - \frac {60 \, \log \left (d + \frac {e}{x^{\frac {1}{3}}}\right )}{x^{2}}\right )} b n - \frac {b \log \left (c\right )}{2 \, x^{2}} - \frac {a}{2 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*((60*d^6*e^(-7)*log(abs(d*x^(1/3) + e)) - 20*d^6*e^(-7)*log(abs(x)) - (60*d^5*x^(5/3)*e - 30*d^4*x^(4/3)
*e^2 + 20*d^3*x*e^3 - 15*d^2*x^(2/3)*e^4 + 12*d*x^(1/3)*e^5 - 10*e^6)*e^(-7)/x^2)*e - 60*log(d + e/x^(1/3))/x^
2)*b*n - 1/2*b*log(c)/x^2 - 1/2*a/x^2

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Mupad [B]
time = 0.46, size = 113, normalized size = 0.82 \begin {gather*} \frac {b\,n}{12\,x^2}-\frac {a}{2\,x^2}-\frac {b\,\ln \left (c\,{\left (d+\frac {e}{x^{1/3}}\right )}^n\right )}{2\,x^2}-\frac {b\,d\,n}{10\,e\,x^{5/3}}+\frac {b\,d^6\,n\,\ln \left (d+\frac {e}{x^{1/3}}\right )}{2\,e^6}-\frac {b\,d^3\,n}{6\,e^3\,x}+\frac {b\,d^2\,n}{8\,e^2\,x^{4/3}}+\frac {b\,d^4\,n}{4\,e^4\,x^{2/3}}-\frac {b\,d^5\,n}{2\,e^5\,x^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*n)/(12*x^2) - a/(2*x^2) - (b*log(c*(d + e/x^(1/3))^n))/(2*x^2) - (b*d*n)/(10*e*x^(5/3)) + (b*d^6*n*log(d +
e/x^(1/3)))/(2*e^6) - (b*d^3*n)/(6*e^3*x) + (b*d^2*n)/(8*e^2*x^(4/3)) + (b*d^4*n)/(4*e^4*x^(2/3)) - (b*d^5*n)/
(2*e^5*x^(1/3))

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