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

Optimal result

Integrand size = 22, antiderivative size = 77 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^2} \, dx=\frac {2 b n}{3 x}-\frac {2 b d n}{e \sqrt [3]{x}}-\frac {2 b d^{3/2} n \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{e^{3/2}}-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x} \]

[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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2505, 269, 348, 331, 211} \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^2} \, dx=-\frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x}-\frac {2 b d^{3/2} n \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{e^{3/2}}-\frac {2 b d n}{e \sqrt [3]{x}}+\frac {2 b n}{3 x} \]

[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*} \text {integral}& = -\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*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.04 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^2} \, dx=-\frac {a}{x}+\frac {2 b n}{3 x}-\frac {2 b d n}{e \sqrt [3]{x}}+\frac {2 b d^{3/2} n \arctan \left (\frac {\sqrt {e}}{\sqrt {d} \sqrt [3]{x}}\right )}{e^{3/2}}-\frac {b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x} \]

[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

Maple [F]

\[\int \frac {a +b \ln \left (c \left (d +\frac {e}{x^{\frac {2}{3}}}\right )^{n}\right )}{x^{2}}d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 235, normalized size of antiderivative = 3.05 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^2} \, dx=\left [\frac {3 \, b d n x \sqrt {-\frac {d}{e}} \log \left (\frac {d^{3} x^{2} + 2 \, d e^{2} x \sqrt {-\frac {d}{e}} - e^{3} - 2 \, {\left (d^{2} e x \sqrt {-\frac {d}{e}} - d e^{2}\right )} x^{\frac {2}{3}} - 2 \, {\left (d^{2} e x + e^{3} \sqrt {-\frac {d}{e}}\right )} x^{\frac {1}{3}}}{d^{3} x^{2} + e^{3}}\right ) - 3 \, b e n \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right ) - 6 \, b d n x^{\frac {2}{3}} + 2 \, b e n - 3 \, b e \log \left (c\right ) - 3 \, a e}{3 \, e x}, -\frac {6 \, b d n x \sqrt {\frac {d}{e}} \arctan \left (x^{\frac {1}{3}} \sqrt {\frac {d}{e}}\right ) + 3 \, b e n \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right ) + 6 \, b d n x^{\frac {2}{3}} - 2 \, b e n + 3 \, b e \log \left (c\right ) + 3 \, a e}{3 \, e x}\right ] \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.01 \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^2} \, dx=-\frac {1}{3} \, {\left (2 \, e {\left (\frac {3 \, d^{2} \arctan \left (\frac {d x^{\frac {1}{3}}}{\sqrt {d e}}\right )}{\sqrt {d e} e^{2}} + \frac {3 \, d x^{\frac {2}{3}} - e}{e^{2} x}\right )} + \frac {3 \, \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right )}{x}\right )} b n - \frac {b \log \left (c\right )}{x} - \frac {a}{x} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )}{x^2} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+\frac {e}{x^{2/3}}\right )}^n\right )}{x^2} \,d x \]

[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)