3.6.13 \(\int \frac {a+b \log (c (d+\frac {e}{x^{2/3}})^n)}{x^2} \, dx\) [513]

Optimal. Leaf size=77 \[ \frac {2 b n}{3 x}-\frac {2 b d n}{e \sqrt [3]{x}}-\frac {2 b d^{3/2} n \tan ^{-1}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{e^{3/2}}-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2505, 269, 348, 331, 211} \begin {gather*} -\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x}-\frac {2 b d^{3/2} n \text {ArcTan}\left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{e^{3/2}}-\frac {2 b d n}{e \sqrt [3]{x}}+\frac {2 b n}{3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 269

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)*(b + a/x^n)^p, x] /; FreeQ[{a, b, m
, n}, x] && IntegerQ[p] && NegQ[n]

Rule 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 348

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[n]}, Dist[k, Subst[Int[x^(k*(
m + 1) - 1)*(a + b*x^(k*n))^p, x], x, x^(1/k)], x]] /; FreeQ[{a, b, m, p}, x] && FractionQ[n]

Rule 2505

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^(m_.), x_Symbol] :> Simp[(f*x)^(m +
1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Dist[b*e*n*(p/(f*(m + 1))), Int[x^(n - 1)*((f*x)^(m + 1)/
(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 80, normalized size = 1.04 \begin {gather*} -\frac {a}{x}+\frac {2 b n}{3 x}-\frac {2 b d n}{e \sqrt [3]{x}}+\frac {2 b d^{3/2} n \tan ^{-1}\left (\frac {\sqrt {e}}{\sqrt {d} \sqrt [3]{x}}\right )}{e^{3/2}}-\frac {b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )}{x^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 67, normalized size = 0.87 \begin {gather*} -\frac {2}{3} \, {\left (3 \, d^{\frac {3}{2}} \arctan \left (\sqrt {d} x^{\frac {1}{3}} e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {5}{2}\right )} + \frac {{\left (3 \, d x^{\frac {2}{3}} - e\right )} e^{\left (-2\right )}}{x}\right )} b n e - \frac {b \log \left (c {\left (d + \frac {e}{x^{\frac {2}{3}}}\right )}^{n}\right )}{x} - \frac {a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.48, size = 232, normalized size = 3.01 \begin {gather*} \left [\frac {{\left (3 \, \sqrt {-d e^{\left (-1\right )}} b d n x \log \left (\frac {d^{3} x^{2} + 2 \, \sqrt {-d e^{\left (-1\right )}} d x e^{2} - 2 \, {\left (\sqrt {-d e^{\left (-1\right )}} d^{2} x e - d e^{2}\right )} x^{\frac {2}{3}} - 2 \, {\left (d^{2} x e + \sqrt {-d e^{\left (-1\right )}} e^{3}\right )} x^{\frac {1}{3}} - e^{3}}{d^{3} x^{2} + e^{3}}\right ) - 3 \, b n e \log \left (\frac {d x + x^{\frac {1}{3}} e}{x}\right ) - 6 \, b d n x^{\frac {2}{3}} - 3 \, b e \log \left (c\right ) + {\left (2 \, b n - 3 \, a\right )} e\right )} e^{\left (-1\right )}}{3 \, x}, -\frac {{\left (6 \, b d^{\frac {3}{2}} n x \arctan \left (\sqrt {d} x^{\frac {1}{3}} e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {1}{2}\right )} + 3 \, b n e \log \left (\frac {d x + x^{\frac {1}{3}} e}{x}\right ) + 6 \, b d n x^{\frac {2}{3}} + 3 \, b e \log \left (c\right ) - {\left (2 \, b n - 3 \, a\right )} e\right )} e^{\left (-1\right )}}{3 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/3*(3*sqrt(-d*e^(-1))*b*d*n*x*log((d^3*x^2 + 2*sqrt(-d*e^(-1))*d*x*e^2 - 2*(sqrt(-d*e^(-1))*d^2*x*e - d*e^2)
*x^(2/3) - 2*(d^2*x*e + sqrt(-d*e^(-1))*e^3)*x^(1/3) - e^3)/(d^3*x^2 + e^3)) - 3*b*n*e*log((d*x + x^(1/3)*e)/x
) - 6*b*d*n*x^(2/3) - 3*b*e*log(c) + (2*b*n - 3*a)*e)*e^(-1)/x, -1/3*(6*b*d^(3/2)*n*x*arctan(sqrt(d)*x^(1/3)*e
^(-1/2))*e^(-1/2) + 3*b*n*e*log((d*x + x^(1/3)*e)/x) + 6*b*d*n*x^(2/3) + 3*b*e*log(c) - (2*b*n - 3*a)*e)*e^(-1
)/x]

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

[Out]

Timed out

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Giac [A]
time = 4.58, size = 73, normalized size = 0.95 \begin {gather*} -\frac {1}{3} \, {\left (2 \, {\left (3 \, d^{\frac {3}{2}} \arctan \left (\sqrt {d} x^{\frac {1}{3}} e^{\left (-\frac {1}{2}\right )}\right ) e^{\left (-\frac {5}{2}\right )} + \frac {{\left (3 \, d x^{\frac {2}{3}} - e\right )} e^{\left (-2\right )}}{x}\right )} e + \frac {3 \, \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right )}{x}\right )} b n - \frac {b \log \left (c\right )}{x} - \frac {a}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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