Integrand size = 22, antiderivative size = 123 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^4} \, dx=-\frac {2 b e n}{21 d x^{7/3}}+\frac {2 b e^2 n}{15 d^2 x^{5/3}}-\frac {2 b e^3 n}{9 d^3 x}+\frac {2 b e^4 n}{3 d^4 \sqrt [3]{x}}+\frac {2 b e^{9/2} n \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{3 d^{9/2}}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^3} \] Output:
-2/21*b*e*n/d/x^(7/3)+2/15*b*e^2*n/d^2/x^(5/3)-2/9*b*e^3*n/d^3/x+2/3*b*e^4 *n/d^4/x^(1/3)+2/3*b*e^(9/2)*n*arctan(e^(1/2)*x^(1/3)/d^(1/2))/d^(9/2)-1/3 *(a+b*ln(c*(d+e*x^(2/3))^n))/x^3
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 = 0.53 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^4} \, dx=-\frac {a}{3 x^3}-\frac {2 b e n \operatorname {Hypergeometric2F1}\left (-\frac {7}{2},1,-\frac {5}{2},-\frac {e x^{2/3}}{d}\right )}{21 d x^{7/3}}-\frac {b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^3} \] Input:
Integrate[(a + b*Log[c*(d + e*x^(2/3))^n])/x^4,x]
Output:
-1/3*a/x^3 - (2*b*e*n*Hypergeometric2F1[-7/2, 1, -5/2, -((e*x^(2/3))/d)])/ (21*d*x^(7/3)) - (b*Log[c*(d + e*x^(2/3))^n])/(3*x^3)
Time = 0.43 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {2905, 864, 264, 264, 264, 264, 218}
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.
\(\displaystyle \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 2905 |
\(\displaystyle \frac {2}{9} b e n \int \frac {1}{\left (d+e x^{2/3}\right ) x^{10/3}}dx-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^3}\) |
\(\Big \downarrow \) 864 |
\(\displaystyle \frac {2}{3} b e n \int \frac {1}{\left (d+e x^{2/3}\right ) x^{8/3}}d\sqrt [3]{x}-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^3}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2}{3} b e n \left (-\frac {e \int \frac {1}{\left (d+e x^{2/3}\right ) x^2}d\sqrt [3]{x}}{d}-\frac {1}{7 d x^{7/3}}\right )-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^3}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2}{3} b e n \left (-\frac {e \left (-\frac {e \int \frac {1}{\left (d+e x^{2/3}\right ) x^{4/3}}d\sqrt [3]{x}}{d}-\frac {1}{5 d x^{5/3}}\right )}{d}-\frac {1}{7 d x^{7/3}}\right )-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^3}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2}{3} b e n \left (-\frac {e \left (-\frac {e \left (-\frac {e \int \frac {1}{\left (d+e x^{2/3}\right ) x^{2/3}}d\sqrt [3]{x}}{d}-\frac {1}{3 d x}\right )}{d}-\frac {1}{5 d x^{5/3}}\right )}{d}-\frac {1}{7 d x^{7/3}}\right )-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^3}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2}{3} b e n \left (-\frac {e \left (-\frac {e \left (-\frac {e \left (-\frac {e \int \frac {1}{d+e x^{2/3}}d\sqrt [3]{x}}{d}-\frac {1}{d \sqrt [3]{x}}\right )}{d}-\frac {1}{3 d x}\right )}{d}-\frac {1}{5 d x^{5/3}}\right )}{d}-\frac {1}{7 d x^{7/3}}\right )-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^3}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2}{3} b e n \left (-\frac {e \left (-\frac {e \left (-\frac {e \left (-\frac {\sqrt {e} \arctan \left (\frac {\sqrt {e} \sqrt [3]{x}}{\sqrt {d}}\right )}{d^{3/2}}-\frac {1}{d \sqrt [3]{x}}\right )}{d}-\frac {1}{3 d x}\right )}{d}-\frac {1}{5 d x^{5/3}}\right )}{d}-\frac {1}{7 d x^{7/3}}\right )-\frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{3 x^3}\) |
Input:
Int[(a + b*Log[c*(d + e*x^(2/3))^n])/x^4,x]
Output:
(2*b*e*n*(-1/7*1/(d*x^(7/3)) - (e*(-1/5*1/(d*x^(5/3)) - (e*(-1/3*1/(d*x) - (e*(-(1/(d*x^(1/3))) - (Sqrt[e]*ArcTan[(Sqrt[e]*x^(1/3))/Sqrt[d]])/d^(3/2 )))/d))/d))/d))/3 - (a + b*Log[c*(d + e*x^(2/3))^n])/(3*x^3)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
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]
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]
\[\int \frac {a +b \ln \left (c \left (d +e \,x^{\frac {2}{3}}\right )^{n}\right )}{x^{4}}d x\]
Input:
int((a+b*ln(c*(d+e*x^(2/3))^n))/x^4,x)
Output:
int((a+b*ln(c*(d+e*x^(2/3))^n))/x^4,x)
Time = 0.10 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.54 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^4} \, dx=\left [\frac {105 \, b e^{4} n x^{3} \sqrt {-\frac {e}{d}} \log \left (\frac {e^{3} x^{2} - 2 \, d^{2} e x \sqrt {-\frac {e}{d}} - d^{3} + 2 \, {\left (d e^{2} x \sqrt {-\frac {e}{d}} + d^{2} e\right )} x^{\frac {2}{3}} - 2 \, {\left (d e^{2} x - d^{3} \sqrt {-\frac {e}{d}}\right )} x^{\frac {1}{3}}}{e^{3} x^{2} + d^{3}}\right ) - 70 \, b d e^{3} n x^{2} + 42 \, b d^{2} e^{2} n x^{\frac {4}{3}} - 105 \, b d^{4} n \log \left (e x^{\frac {2}{3}} + d\right ) - 105 \, b d^{4} \log \left (c\right ) - 105 \, a d^{4} + 30 \, {\left (7 \, b e^{4} n x^{2} - b d^{3} e n\right )} x^{\frac {2}{3}}}{315 \, d^{4} x^{3}}, \frac {210 \, b e^{4} n x^{3} \sqrt {\frac {e}{d}} \arctan \left (x^{\frac {1}{3}} \sqrt {\frac {e}{d}}\right ) - 70 \, b d e^{3} n x^{2} + 42 \, b d^{2} e^{2} n x^{\frac {4}{3}} - 105 \, b d^{4} n \log \left (e x^{\frac {2}{3}} + d\right ) - 105 \, b d^{4} \log \left (c\right ) - 105 \, a d^{4} + 30 \, {\left (7 \, b e^{4} n x^{2} - b d^{3} e n\right )} x^{\frac {2}{3}}}{315 \, d^{4} x^{3}}\right ] \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))/x^4,x, algorithm="fricas")
Output:
[1/315*(105*b*e^4*n*x^3*sqrt(-e/d)*log((e^3*x^2 - 2*d^2*e*x*sqrt(-e/d) - d ^3 + 2*(d*e^2*x*sqrt(-e/d) + d^2*e)*x^(2/3) - 2*(d*e^2*x - d^3*sqrt(-e/d)) *x^(1/3))/(e^3*x^2 + d^3)) - 70*b*d*e^3*n*x^2 + 42*b*d^2*e^2*n*x^(4/3) - 1 05*b*d^4*n*log(e*x^(2/3) + d) - 105*b*d^4*log(c) - 105*a*d^4 + 30*(7*b*e^4 *n*x^2 - b*d^3*e*n)*x^(2/3))/(d^4*x^3), 1/315*(210*b*e^4*n*x^3*sqrt(e/d)*a rctan(x^(1/3)*sqrt(e/d)) - 70*b*d*e^3*n*x^2 + 42*b*d^2*e^2*n*x^(4/3) - 105 *b*d^4*n*log(e*x^(2/3) + d) - 105*b*d^4*log(c) - 105*a*d^4 + 30*(7*b*e^4*n *x^2 - b*d^3*e*n)*x^(2/3))/(d^4*x^3)]
Timed out. \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^4} \, dx=\text {Timed out} \] Input:
integrate((a+b*ln(c*(d+e*x**(2/3))**n))/x**4,x)
Output:
Timed out
Exception generated. \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^4} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))/x^4,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
Time = 0.15 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^4} \, dx=\frac {1}{315} \, {\left (2 \, e {\left (\frac {105 \, e^{4} \arctan \left (\frac {e x^{\frac {1}{3}}}{\sqrt {d e}}\right )}{\sqrt {d e} d^{4}} + \frac {105 \, e^{3} x^{2} - 35 \, d e^{2} x^{\frac {4}{3}} + 21 \, d^{2} e x^{\frac {2}{3}} - 15 \, d^{3}}{d^{4} x^{\frac {7}{3}}}\right )} - \frac {105 \, \log \left (e x^{\frac {2}{3}} + d\right )}{x^{3}}\right )} b n - \frac {b \log \left (c\right )}{3 \, x^{3}} - \frac {a}{3 \, x^{3}} \] Input:
integrate((a+b*log(c*(d+e*x^(2/3))^n))/x^4,x, algorithm="giac")
Output:
1/315*(2*e*(105*e^4*arctan(e*x^(1/3)/sqrt(d*e))/(sqrt(d*e)*d^4) + (105*e^3 *x^2 - 35*d*e^2*x^(4/3) + 21*d^2*e*x^(2/3) - 15*d^3)/(d^4*x^(7/3))) - 105* log(e*x^(2/3) + d)/x^3)*b*n - 1/3*b*log(c)/x^3 - 1/3*a/x^3
Timed out. \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^4} \, dx=\int \frac {a+b\,\ln \left (c\,{\left (d+e\,x^{2/3}\right )}^n\right )}{x^4} \,d x \] Input:
int((a + b*log(c*(d + e*x^(2/3))^n))/x^4,x)
Output:
int((a + b*log(c*(d + e*x^(2/3))^n))/x^4, x)
Time = 0.16 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.89 \[ \int \frac {a+b \log \left (c \left (d+e x^{2/3}\right )^n\right )}{x^4} \, dx=\frac {210 \sqrt {e}\, \sqrt {d}\, \mathit {atan} \left (\frac {x^{\frac {1}{3}} e}{\sqrt {e}\, \sqrt {d}}\right ) b \,e^{4} n \,x^{3}-30 x^{\frac {2}{3}} b \,d^{4} e n +210 x^{\frac {8}{3}} b d \,e^{4} n +42 x^{\frac {4}{3}} b \,d^{3} e^{2} n -105 \,\mathrm {log}\left (\left (x^{\frac {2}{3}} e +d \right )^{n} c \right ) b \,d^{5}-105 a \,d^{5}-70 b \,d^{2} e^{3} n \,x^{2}}{315 d^{5} x^{3}} \] Input:
int((a+b*log(c*(d+e*x^(2/3))^n))/x^4,x)
Output:
(210*sqrt(e)*sqrt(d)*atan((x**(1/3)*e)/(sqrt(e)*sqrt(d)))*b*e**4*n*x**3 - 30*x**(2/3)*b*d**4*e*n + 210*x**(2/3)*b*d*e**4*n*x**2 + 42*x**(1/3)*b*d**3 *e**2*n*x - 105*log((x**(2/3)*e + d)**n*c)*b*d**5 - 105*a*d**5 - 70*b*d**2 *e**3*n*x**2)/(315*d**5*x**3)