\(\int x^2 (a+b \log (c (d+\frac {e}{x^{2/3}})^n)) \, dx\) [509]

Optimal result

Integrand size = 22, antiderivative size = 121 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=-\frac {2 b e^4 n \sqrt [3]{x}}{3 d^4}+\frac {2 b e^3 n x}{9 d^3}-\frac {2 b e^2 n x^{5/3}}{15 d^2}+\frac {2 b e n x^{7/3}}{21 d}+\frac {2 b e^{9/2} n \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{3 d^{9/2}}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \] Output:

-2/3*b*e^4*n*x^(1/3)/d^4+2/9*b*e^3*n*x/d^3-2/15*b*e^2*n*x^(5/3)/d^2+2/21*b 
*e*n*x^(7/3)/d+2/3*b*e^(9/2)*n*arctan(d^(1/2)*x^(1/3)/e^(1/2))/d^(9/2)+1/3 
*x^3*(a+b*ln(c*(d+e/x^(2/3))^n))
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.54 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {a x^3}{3}+\frac {2 b e n x^{7/3} \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},1,-\frac {5}{2},-\frac {e}{d x^{2/3}}\right )}{21 d}+\frac {1}{3} b x^3 \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right ) \] Input:

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

Output:

(a*x^3)/3 + (2*b*e*n*x^(7/3)*Hypergeometric2F1[-7/2, 1, -5/2, -(e/(d*x^(2/ 
3)))])/(21*d) + (b*x^3*Log[c*(d + e/x^(2/3))^n])/3
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.92, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {2905, 795, 864, 254, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx\)

\(\Big \downarrow \) 2905

\(\displaystyle \frac {2}{9} b e n \int \frac {x^{4/3}}{d+\frac {e}{x^{2/3}}}dx+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\)

\(\Big \downarrow \) 795

\(\displaystyle \frac {2}{9} b e n \int \frac {x^2}{x^{2/3} d+e}dx+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\)

\(\Big \downarrow \) 864

\(\displaystyle \frac {2}{3} b e n \int \frac {x^{8/3}}{x^{2/3} d+e}d\sqrt [3]{x}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\)

\(\Big \downarrow \) 254

\(\displaystyle \frac {2}{3} b e n \int \left (\frac {e^4}{d^4 \left (x^{2/3} d+e\right )}-\frac {e^3}{d^4}+\frac {x^{2/3} e^2}{d^3}-\frac {x^{4/3} e}{d^2}+\frac {x^2}{d}\right )d\sqrt [3]{x}+\frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{3} x^3 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right )+\frac {2}{3} b e n \left (\frac {e^{7/2} \arctan \left (\frac {\sqrt {d} \sqrt [3]{x}}{\sqrt {e}}\right )}{d^{9/2}}-\frac {e^3 \sqrt [3]{x}}{d^4}+\frac {e^2 x}{3 d^3}-\frac {e x^{5/3}}{5 d^2}+\frac {x^{7/3}}{7 d}\right )\)

Input:

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

Output:

(2*b*e*n*(-((e^3*x^(1/3))/d^4) + (e^2*x)/(3*d^3) - (e*x^(5/3))/(5*d^2) + x 
^(7/3)/(7*d) + (e^(7/2)*ArcTan[(Sqrt[d]*x^(1/3))/Sqrt[e]])/d^(9/2)))/3 + ( 
x^3*(a + b*Log[c*(d + e/x^(2/3))^n]))/3
 

Defintions of rubi rules used

rule 254

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, 
 a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
 

rule 795

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 864

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denomi 
nator[n]}, Simp[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 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2905

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] - Simp[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]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 399, normalized size of antiderivative = 3.30 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\left [\frac {105 \, b d^{4} x^{3} \log \left (c\right ) + 105 \, a d^{4} x^{3} - 42 \, b d^{2} e^{2} n x^{\frac {5}{3}} + 105 \, b e^{4} n \sqrt {-\frac {e}{d}} \log \left (\frac {d^{3} x^{2} - 2 \, d^{2} e x \sqrt {-\frac {e}{d}} - e^{3} + 2 \, {\left (d^{3} x \sqrt {-\frac {e}{d}} + d e^{2}\right )} x^{\frac {2}{3}} - 2 \, {\left (d^{2} e x - d e^{2} \sqrt {-\frac {e}{d}}\right )} x^{\frac {1}{3}}}{d^{3} x^{2} + e^{3}}\right ) + 70 \, b d e^{3} n x + 105 \, b d^{4} n \log \left (d x^{\frac {2}{3}} + e\right ) - 210 \, b d^{4} n \log \left (x^{\frac {1}{3}}\right ) + 105 \, {\left (b d^{4} n x^{3} - b d^{4} n\right )} \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right ) + 30 \, {\left (b d^{3} e n x^{2} - 7 \, b e^{4} n\right )} x^{\frac {1}{3}}}{315 \, d^{4}}, \frac {105 \, b d^{4} x^{3} \log \left (c\right ) + 105 \, a d^{4} x^{3} - 42 \, b d^{2} e^{2} n x^{\frac {5}{3}} + 210 \, b e^{4} n \sqrt {\frac {e}{d}} \arctan \left (\frac {d x^{\frac {1}{3}} \sqrt {\frac {e}{d}}}{e}\right ) + 70 \, b d e^{3} n x + 105 \, b d^{4} n \log \left (d x^{\frac {2}{3}} + e\right ) - 210 \, b d^{4} n \log \left (x^{\frac {1}{3}}\right ) + 105 \, {\left (b d^{4} n x^{3} - b d^{4} n\right )} \log \left (\frac {d x + e x^{\frac {1}{3}}}{x}\right ) + 30 \, {\left (b d^{3} e n x^{2} - 7 \, b e^{4} n\right )} x^{\frac {1}{3}}}{315 \, d^{4}}\right ] \] Input:

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

Output:

[1/315*(105*b*d^4*x^3*log(c) + 105*a*d^4*x^3 - 42*b*d^2*e^2*n*x^(5/3) + 10 
5*b*e^4*n*sqrt(-e/d)*log((d^3*x^2 - 2*d^2*e*x*sqrt(-e/d) - e^3 + 2*(d^3*x* 
sqrt(-e/d) + d*e^2)*x^(2/3) - 2*(d^2*e*x - d*e^2*sqrt(-e/d))*x^(1/3))/(d^3 
*x^2 + e^3)) + 70*b*d*e^3*n*x + 105*b*d^4*n*log(d*x^(2/3) + e) - 210*b*d^4 
*n*log(x^(1/3)) + 105*(b*d^4*n*x^3 - b*d^4*n)*log((d*x + e*x^(1/3))/x) + 3 
0*(b*d^3*e*n*x^2 - 7*b*e^4*n)*x^(1/3))/d^4, 1/315*(105*b*d^4*x^3*log(c) + 
105*a*d^4*x^3 - 42*b*d^2*e^2*n*x^(5/3) + 210*b*e^4*n*sqrt(e/d)*arctan(d*x^ 
(1/3)*sqrt(e/d)/e) + 70*b*d*e^3*n*x + 105*b*d^4*n*log(d*x^(2/3) + e) - 210 
*b*d^4*n*log(x^(1/3)) + 105*(b*d^4*n*x^3 - b*d^4*n)*log((d*x + e*x^(1/3))/ 
x) + 30*(b*d^3*e*n*x^2 - 7*b*e^4*n)*x^(1/3))/d^4]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.85 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {1}{3} \, b x^{3} \log \left (c\right ) + \frac {1}{3} \, a x^{3} + \frac {1}{315} \, {\left (105 \, x^{3} \log \left (d + \frac {e}{x^{\frac {2}{3}}}\right ) + 2 \, e {\left (\frac {105 \, e^{4} \arctan \left (\frac {d x^{\frac {1}{3}}}{\sqrt {d e}}\right )}{\sqrt {d e} d^{4}} + \frac {15 \, d^{6} x^{\frac {7}{3}} - 21 \, d^{5} e x^{\frac {5}{3}} + 35 \, d^{4} e^{2} x - 105 \, d^{3} e^{3} x^{\frac {1}{3}}}{d^{7}}\right )}\right )} b n \] Input:

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

Output:

1/3*b*x^3*log(c) + 1/3*a*x^3 + 1/315*(105*x^3*log(d + e/x^(2/3)) + 2*e*(10 
5*e^4*arctan(d*x^(1/3)/sqrt(d*e))/(sqrt(d*e)*d^4) + (15*d^6*x^(7/3) - 21*d 
^5*e*x^(5/3) + 35*d^4*e^2*x - 105*d^3*e^3*x^(1/3))/d^7))*b*n
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.94 \[ \int x^2 \left (a+b \log \left (c \left (d+\frac {e}{x^{2/3}}\right )^n\right )\right ) \, dx=\frac {210 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} d}{\sqrt {e}\, \sqrt {d}}\right ) b \,e^{4} n -42 x^{\frac {5}{3}} b \,d^{3} e^{2} n +30 x^{\frac {7}{3}} b \,d^{4} e n -210 x^{\frac {1}{3}} b d \,e^{4} n +105 \,\mathrm {log}\left (\frac {\left (x^{\frac {2}{3}} d +e \right )^{n} c}{x^{\frac {2 n}{3}}}\right ) b \,d^{5} x^{3}+105 a \,d^{5} x^{3}+70 b \,d^{2} e^{3} n x}{315 d^{5}} \] Input:

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

Output:

(210*sqrt(e)*sqrt(d)*atan((x**(1/3)*d)/(sqrt(e)*sqrt(d)))*b*e**4*n - 42*x* 
*(2/3)*b*d**3*e**2*n*x + 30*x**(1/3)*b*d**4*e*n*x**2 - 210*x**(1/3)*b*d*e* 
*4*n + 105*log(((x**(2/3)*d + e)**n*c)/x**((2*n)/3))*b*d**5*x**3 + 105*a*d 
**5*x**3 + 70*b*d**2*e**3*n*x)/(315*d**5)