3.5.77 \(\int (a+b \log (c (d+e x^{2/3})^n))^2 \, dx\) [477]

3.5.77.1 Optimal result

Integrand size = 20, antiderivative size = 364 \[ \int \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {4 a b d n \sqrt [3]{x}}{e}-\frac {32 b^2 d n^2 \sqrt [3]{x}}{3 e}+\frac {8}{9} b^2 n^2 x+\frac {32 b^2 d^{3/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{3 e^{3/2}}-\frac {4 i b^2 d^{3/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{e^{3/2}}-\frac {8 b^2 d^{3/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{e^{3/2}}+\frac {4 b^2 d n \sqrt [3]{x} \log \left (c \left (d+e x^{2/3}\right )^n\right )}{e}-\frac {4}{3} b n x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )-\frac {4 b d^{3/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{e^{3/2}}+x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2-\frac {4 i b^2 d^{3/2} n^2 \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{e^{3/2}} \]

output

4*a*b*d*n*x^(1/3)/e-32/3*b^2*d*n^2*x^(1/3)/e+8/9*b^2*n^2*x+32/3*b^2*d^(3/2 
)*n^2*arctan(x^(1/3)*e^(1/2)/d^(1/2))/e^(3/2)-4*I*b^2*d^(3/2)*n^2*arctan(x 
^(1/3)*e^(1/2)/d^(1/2))^2/e^(3/2)+4*b^2*d*n*x^(1/3)*ln(c*(d+e*x^(2/3))^n)/ 
e-4/3*b*n*x*(a+b*ln(c*(d+e*x^(2/3))^n))-4*b*d^(3/2)*n*arctan(x^(1/3)*e^(1/ 
2)/d^(1/2))*(a+b*ln(c*(d+e*x^(2/3))^n))/e^(3/2)+x*(a+b*ln(c*(d+e*x^(2/3))^ 
n))^2-8*b^2*d^(3/2)*n^2*arctan(x^(1/3)*e^(1/2)/d^(1/2))*ln(2*d^(1/2)/(d^(1 
/2)+I*x^(1/3)*e^(1/2)))/e^(3/2)-4*I*b^2*d^(3/2)*n^2*polylog(2,1-2*d^(1/2)/ 
(d^(1/2)+I*x^(1/3)*e^(1/2)))/e^(3/2)
 

3.5.77.2 Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 319, normalized size of antiderivative = 0.88 \[ \int \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2 \, dx=\frac {-36 i b^2 d^{3/2} n^2 \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2-12 b d^{3/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (3 a-8 b n+6 b n \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )+3 b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )+\sqrt {e} \sqrt [3]{x} \left (12 a b n \left (3 d-e x^{2/3}\right )+8 b^2 n^2 \left (-12 d+e x^{2/3}\right )+9 a^2 e x^{2/3}+6 b \left (6 b d n+3 a e x^{2/3}-2 b e n x^{2/3}\right ) \log \left (c \left (d+e x^{2/3}\right )^n\right )+9 b^2 e x^{2/3} \log ^2\left (c \left (d+e x^{2/3}\right )^n\right )\right )-36 i b^2 d^{3/2} n^2 \operatorname {PolyLog}\left (2,\frac {i \sqrt {d}+\sqrt {e} \sqrt [3]{x}}{-i \sqrt {d}+\sqrt {e} \sqrt [3]{x}}\right )}{9 e^{3/2}} \]

input

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

output

((-36*I)*b^2*d^(3/2)*n^2*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]]^2 - 12*b*d^(3/2 
)*n*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]]*(3*a - 8*b*n + 6*b*n*Log[(2*Sqrt[d]) 
/(Sqrt[d] + I*Sqrt[e]*x^(1/3))] + 3*b*Log[c*(d + e*x^(2/3))^n]) + Sqrt[e]* 
x^(1/3)*(12*a*b*n*(3*d - e*x^(2/3)) + 8*b^2*n^2*(-12*d + e*x^(2/3)) + 9*a^ 
2*e*x^(2/3) + 6*b*(6*b*d*n + 3*a*e*x^(2/3) - 2*b*e*n*x^(2/3))*Log[c*(d + e 
*x^(2/3))^n] + 9*b^2*e*x^(2/3)*Log[c*(d + e*x^(2/3))^n]^2) - (36*I)*b^2*d^ 
(3/2)*n^2*PolyLog[2, (I*Sqrt[d] + Sqrt[e]*x^(1/3))/((-I)*Sqrt[d] + Sqrt[e] 
*x^(1/3))])/(9*e^(3/2))
 

3.5.77.3 Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 349, normalized size of antiderivative = 0.96, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2901, 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.

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

\(\Big \downarrow \) 2901

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

\(\Big \downarrow \) 2907

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

\(\Big \downarrow \) 2926

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

\(\Big \downarrow \) 2009

\(\displaystyle 3 \left (\frac {1}{3} x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )^2-\frac {4}{3} b e n \left (\frac {d^{3/2} \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{e^{5/2}}+\frac {x \left (a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )\right )}{3 e}-\frac {a d \sqrt [3]{x}}{e^2}+\frac {i b d^{3/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )^2}{e^{5/2}}-\frac {8 b d^{3/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{3 e^{5/2}}+\frac {2 b d^{3/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right ) \log \left (\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{e^{5/2}}-\frac {b d \sqrt [3]{x} \log \left (c \left (d+e x^{2/3}\right )^n\right )}{e^2}+\frac {i b d^{3/2} n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {d}}{\sqrt {d}+i \sqrt {e} \sqrt [3]{x}}\right )}{e^{5/2}}+\frac {8 b d n \sqrt [3]{x}}{3 e^2}-\frac {2 b n x}{9 e}\right )\right )\)

input

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

output

3*((x*(a + b*Log[c*(d + e*x^(2/3))^n])^2)/3 - (4*b*e*n*(-((a*d*x^(1/3))/e^ 
2) + (8*b*d*n*x^(1/3))/(3*e^2) - (2*b*n*x)/(9*e) - (8*b*d^(3/2)*n*ArcTan[( 
Sqrt[e]*x^(1/3))/Sqrt[d]])/(3*e^(5/2)) + (I*b*d^(3/2)*n*ArcTan[(Sqrt[e]*x^ 
(1/3))/Sqrt[d]]^2)/e^(5/2) + (2*b*d^(3/2)*n*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[ 
d]]*Log[(2*Sqrt[d])/(Sqrt[d] + I*Sqrt[e]*x^(1/3))])/e^(5/2) - (b*d*x^(1/3) 
*Log[c*(d + e*x^(2/3))^n])/e^2 + (x*(a + b*Log[c*(d + e*x^(2/3))^n]))/(3*e 
) + (d^(3/2)*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]]*(a + b*Log[c*(d + e*x^(2/3) 
)^n]))/e^(5/2) + (I*b*d^(3/2)*n*PolyLog[2, 1 - (2*Sqrt[d])/(Sqrt[d] + I*Sq 
rt[e]*x^(1/3))])/e^(5/2)))/3)
 

3.5.77.3.1 Defintions of rubi rules used

rule 2009

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

rule 2901

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

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 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]
 

3.5.77.4 Maple [F]

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

input

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

output

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

3.5.77.5 Fricas [F]

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

input

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

3.5.77.6 Sympy [F]

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

input

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

output

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

3.5.77.7 Maxima [F(-2)]

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

input

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

3.5.77.8 Giac [F]

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

input

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

output

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

3.5.77.9 Mupad [F(-1)]

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

input

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

output

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