\(\int \frac {(a+b \log (c (d+e x^{2/3})^n))^2}{x^2} \, dx\) [478]

Optimal result

Integrand size = 24, antiderivative size = 298 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^2} \, dx=\frac {8 b^2 e^{3/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {4 i b^2 e^{3/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{d^{3/2}}-\frac {8 b^2 e^{3/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{d^{3/2}}-\frac {4 b e n \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d \sqrt [3]{x}}-\frac {4 b e^{3/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{d^{3/2}}-\frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x}-\frac {4 i b^2 e^{3/2} n^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{d^{3/2}} \] Output:

8*b^2*e^(3/2)*n^2*arctan(e^(1/2)*x^(1/3)/d^(1/2))/d^(3/2)-4*I*b^2*e^(3/2)* 
n^2*arctan(e^(1/2)*x^(1/3)/d^(1/2))^2/d^(3/2)-8*b^2*e^(3/2)*n^2*arctan(e^( 
1/2)*x^(1/3)/d^(1/2))*ln(2*d^(1/2)/(d^(1/2)+I*e^(1/2)*x^(1/3)))/d^(3/2)-4* 
b*e*n*(a+b*ln(c*(d+e*x^(2/3))^n))/d/x^(1/3)-4*b*e^(3/2)*n*arctan(e^(1/2)*x 
^(1/3)/d^(1/2))*(a+b*ln(c*(d+e*x^(2/3))^n))/d^(3/2)-(a+b*ln(c*(d+e*x^(2/3) 
)^n))^2/x-4*I*b^2*e^(3/2)*n^2*polylog(2,1-2*d^(1/2)/(d^(1/2)+I*e^(1/2)*x^( 
1/3)))/d^(3/2)
 

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.83 \[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^2} \, dx=\frac {-4 i b^2 e^{3/2} n^2 x \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2-4 b e^{3/2} n x \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a-2 b n+2 b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\sqrt {d} \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right ) \left (a d+4 b e n x^{2/3}+b d \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-4 i b^2 e^{3/2} n^2 x \operatorname {PolyLog}\left (2,\frac {i \sqrt {d}+\sqrt {e} \sqrt [3]{x}}{-i \sqrt {d}+\sqrt {e} \sqrt [3]{x}}\right )}{d^{3/2} x} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2908, 2907, 2926, 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.

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

\(\Big \downarrow \) 2908

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

\(\Big \downarrow \) 2907

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

\(\Big \downarrow \) 2926

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

3*(-1/3*(a + b*Log[c*(d + e*x^(2/3))^n])^2/x + (4*b*e*n*((2*b*Sqrt[e]*n*Ar 
cTan[(Sqrt[e]*x^(1/3))/Sqrt[d]])/d^(3/2) - (I*b*Sqrt[e]*n*ArcTan[(Sqrt[e]* 
x^(1/3))/Sqrt[d]]^2)/d^(3/2) - (2*b*Sqrt[e]*n*ArcTan[(Sqrt[e]*x^(1/3))/Sqr 
t[d]]*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x^(1/3))])/d^(3/2) - (a + b*Log 
[c*(d + e*x^(2/3))^n])/(d*x^(1/3)) - (Sqrt[e]*ArcTan[(Sqrt[e]*x^(1/3))/Sqr 
t[d]]*(a + b*Log[c*(d + e*x^(2/3))^n]))/d^(3/2) - (I*b*Sqrt[e]*n*PolyLog[2 
, 1 - (2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x^(1/3))])/d^(3/2)))/3)
 

Defintions of rubi rules used

rule 2009

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

rule 2907

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

rule 2908

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_)*(x_)^(m_ 
.), x_Symbol] :> With[{k = Denominator[n]}, Simp[k   Subst[Int[x^(k*(m + 1) 
 - 1)*(a + b*Log[c*(d + e*x^(k*n))^p])^q, x], x, x^(1/k)], x]] /; FreeQ[{a, 
 b, c, d, e, m, p, q}, x] && FractionQ[n]
 

rule 2926

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_) + (g_.)*(x_)^(s_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b 
*Log[c*(d + e*x^n)^p])^q, x^m*(f + g*x^s)^r, x], x] /; FreeQ[{a, b, c, d, e 
, f, g, m, n, p, q, r, s}, x] && IGtQ[q, 0] && IntegerQ[m] && IntegerQ[r] & 
& IntegerQ[s]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F(-2)]

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

integrate((a+b*log(c*(d+e*x^(2/3))^n))^2/x^2,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 [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2}{x^2} \, dx=\frac {-12 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} e}{\sqrt {e}\, \sqrt {d}}\right ) a b e n x -8 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} e}{\sqrt {e}\, \sqrt {d}}\right ) b^{2} e \,n^{2} x -12 x^{\frac {2}{3}} a b d e n -8 x^{\frac {2}{3}} b^{2} d e \,n^{2}-4 \left (\int \frac {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}{x^{\frac {8}{3}} e +d \,x^{2}}d x \right ) b^{2} d^{3} n x -3 {\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right )}^{2} b^{2} d^{2}-6 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) a b \,d^{2}-4 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b^{2} d^{2} n -3 a^{2} d^{2}}{3 d^{2} x} \] Input:

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

Output:

( - 12*sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d)))*a*b*e*n*x - 8* 
sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d)))*b**2*e*n**2*x - 12*x* 
*(2/3)*a*b*d*e*n - 8*x**(2/3)*b**2*d*e*n**2 - 4*int(log((x**(2/3)*e + d)** 
n*c)/(x**(2/3)*e*x**2 + d*x**2),x)*b**2*d**3*n*x - 3*log((x**(2/3)*e + d)* 
*n*c)**2*b**2*d**2 - 6*log((x**(2/3)*e + d)**n*c)*a*b*d**2 - 4*log((x**(2/ 
3)*e + d)**n*c)*b**2*d**2*n - 3*a**2*d**2)/(3*d**2*x)