Integrand size = 22, antiderivative size = 234 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {b d^{11} n \sqrt [3]{x}}{4 e^{11}}-\frac {b d^{10} n x^{2/3}}{8 e^{10}}+\frac {b d^9 n x}{12 e^9}-\frac {b d^8 n x^{4/3}}{16 e^8}+\frac {b d^7 n x^{5/3}}{20 e^7}-\frac {b d^6 n x^2}{24 e^6}+\frac {b d^5 n x^{7/3}}{28 e^5}-\frac {b d^4 n x^{8/3}}{32 e^4}+\frac {b d^3 n x^3}{36 e^3}-\frac {b d^2 n x^{10/3}}{40 e^2}+\frac {b d n x^{11/3}}{44 e}-\frac {1}{48} b n x^4-\frac {b d^{12} n \log \left (d+e \sqrt [3]{x}\right )}{4 e^{12}}+\frac {1}{4} x^4 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \] Output:
1/4*b*d^11*n*x^(1/3)/e^11-1/8*b*d^10*n*x^(2/3)/e^10+1/12*b*d^9*n*x/e^9-1/1 6*b*d^8*n*x^(4/3)/e^8+1/20*b*d^7*n*x^(5/3)/e^7-1/24*b*d^6*n*x^2/e^6+1/28*b *d^5*n*x^(7/3)/e^5-1/32*b*d^4*n*x^(8/3)/e^4+1/36*b*d^3*n*x^3/e^3-1/40*b*d^ 2*n*x^(10/3)/e^2+1/44*b*d*n*x^(11/3)/e-1/48*b*n*x^4-1/4*b*d^12*n*ln(d+e*x^ (1/3))/e^12+1/4*x^4*(a+b*ln(c*(d+e*x^(1/3))^n))
Time = 0.24 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.93 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {a x^4}{4}-\frac {1}{12} b e n \left (-\frac {3 d^{11} \sqrt [3]{x}}{e^{12}}+\frac {3 d^{10} x^{2/3}}{2 e^{11}}-\frac {d^9 x}{e^{10}}+\frac {3 d^8 x^{4/3}}{4 e^9}-\frac {3 d^7 x^{5/3}}{5 e^8}+\frac {d^6 x^2}{2 e^7}-\frac {3 d^5 x^{7/3}}{7 e^6}+\frac {3 d^4 x^{8/3}}{8 e^5}-\frac {d^3 x^3}{3 e^4}+\frac {3 d^2 x^{10/3}}{10 e^3}-\frac {3 d x^{11/3}}{11 e^2}+\frac {x^4}{4 e}+\frac {3 d^{12} \log \left (d+e \sqrt [3]{x}\right )}{e^{13}}\right )+\frac {1}{4} b x^4 \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right ) \] Input:
Integrate[x^3*(a + b*Log[c*(d + e*x^(1/3))^n]),x]
Output:
(a*x^4)/4 - (b*e*n*((-3*d^11*x^(1/3))/e^12 + (3*d^10*x^(2/3))/(2*e^11) - ( d^9*x)/e^10 + (3*d^8*x^(4/3))/(4*e^9) - (3*d^7*x^(5/3))/(5*e^8) + (d^6*x^2 )/(2*e^7) - (3*d^5*x^(7/3))/(7*e^6) + (3*d^4*x^(8/3))/(8*e^5) - (d^3*x^3)/ (3*e^4) + (3*d^2*x^(10/3))/(10*e^3) - (3*d*x^(11/3))/(11*e^2) + x^4/(4*e) + (3*d^12*Log[d + e*x^(1/3)])/e^13))/12 + (b*x^4*Log[c*(d + e*x^(1/3))^n]) /4
Time = 0.67 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2904, 2842, 49, 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.
| \(\displaystyle \int x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx\) |
| \(\Big \downarrow \) 2904 |
| \(\displaystyle 3 \int x^{11/3} \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )d\sqrt [3]{x}\) |
| \(\Big \downarrow \) 2842 |
| \(\displaystyle 3 \left (\frac {1}{12} x^4 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )-\frac {1}{12} b e n \int \frac {x^4}{d+e \sqrt [3]{x}}d\sqrt [3]{x}\right )\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle 3 \left (\frac {1}{12} x^4 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )-\frac {1}{12} b e n \int \left (\frac {d^{12}}{e^{12} \left (d+e \sqrt [3]{x}\right )}-\frac {d^{11}}{e^{12}}+\frac {\sqrt [3]{x} d^{10}}{e^{11}}-\frac {x^{2/3} d^9}{e^{10}}+\frac {x d^8}{e^9}-\frac {x^{4/3} d^7}{e^8}+\frac {x^{5/3} d^6}{e^7}-\frac {x^2 d^5}{e^6}+\frac {x^{7/3} d^4}{e^5}-\frac {x^{8/3} d^3}{e^4}+\frac {x^3 d^2}{e^3}-\frac {x^{10/3} d}{e^2}+\frac {x^{11/3}}{e}\right )d\sqrt [3]{x}\right )\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle 3 \left (\frac {1}{12} x^4 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right )-\frac {1}{12} b e n \left (\frac {d^{12} \log \left (d+e \sqrt [3]{x}\right )}{e^{13}}-\frac {d^{11} \sqrt [3]{x}}{e^{12}}+\frac {d^{10} x^{2/3}}{2 e^{11}}-\frac {d^9 x}{3 e^{10}}+\frac {d^8 x^{4/3}}{4 e^9}-\frac {d^7 x^{5/3}}{5 e^8}+\frac {d^6 x^2}{6 e^7}-\frac {d^5 x^{7/3}}{7 e^6}+\frac {d^4 x^{8/3}}{8 e^5}-\frac {d^3 x^3}{9 e^4}+\frac {d^2 x^{10/3}}{10 e^3}-\frac {d x^{11/3}}{11 e^2}+\frac {x^4}{12 e}\right )\right )\) |
Input:
Int[x^3*(a + b*Log[c*(d + e*x^(1/3))^n]),x]
Output:
3*(-1/12*(b*e*n*(-((d^11*x^(1/3))/e^12) + (d^10*x^(2/3))/(2*e^11) - (d^9*x )/(3*e^10) + (d^8*x^(4/3))/(4*e^9) - (d^7*x^(5/3))/(5*e^8) + (d^6*x^2)/(6* e^7) - (d^5*x^(7/3))/(7*e^6) + (d^4*x^(8/3))/(8*e^5) - (d^3*x^3)/(9*e^4) + (d^2*x^(10/3))/(10*e^3) - (d*x^(11/3))/(11*e^2) + x^4/(12*e) + (d^12*Log[ d + e*x^(1/3)])/e^13)) + (x^4*(a + b*Log[c*(d + e*x^(1/3))^n]))/12)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_ ))^(q_.), x_Symbol] :> Simp[(f + g*x)^(q + 1)*((a + b*Log[c*(d + e*x)^n])/( g*(q + 1))), x] - Simp[b*e*(n/(g*(q + 1))) Int[(f + g*x)^(q + 1)/(d + e*x ), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & & !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
\[\int x^{3} \left (a +b \ln \left (c \left (d +e \,x^{\frac {1}{3}}\right )^{n}\right )\right )d x\]
Input:
int(x^3*(a+b*ln(c*(d+e*x^(1/3))^n)),x)
Output:
int(x^3*(a+b*ln(c*(d+e*x^(1/3))^n)),x)
Time = 0.12 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.86 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {27720 \, b e^{12} x^{4} \log \left (c\right ) + 3080 \, b d^{3} e^{9} n x^{3} - 4620 \, b d^{6} e^{6} n x^{2} + 9240 \, b d^{9} e^{3} n x - 2310 \, {\left (b e^{12} n - 12 \, a e^{12}\right )} x^{4} + 27720 \, {\left (b e^{12} n x^{4} - b d^{12} n\right )} \log \left (e x^{\frac {1}{3}} + d\right ) + 63 \, {\left (40 \, b d e^{11} n x^{3} - 55 \, b d^{4} e^{8} n x^{2} + 88 \, b d^{7} e^{5} n x - 220 \, b d^{10} e^{2} n\right )} x^{\frac {2}{3}} - 198 \, {\left (14 \, b d^{2} e^{10} n x^{3} - 20 \, b d^{5} e^{7} n x^{2} + 35 \, b d^{8} e^{4} n x - 140 \, b d^{11} e n\right )} x^{\frac {1}{3}}}{110880 \, e^{12}} \] Input:
integrate(x^3*(a+b*log(c*(d+e*x^(1/3))^n)),x, algorithm="fricas")
Output:
1/110880*(27720*b*e^12*x^4*log(c) + 3080*b*d^3*e^9*n*x^3 - 4620*b*d^6*e^6* n*x^2 + 9240*b*d^9*e^3*n*x - 2310*(b*e^12*n - 12*a*e^12)*x^4 + 27720*(b*e^ 12*n*x^4 - b*d^12*n)*log(e*x^(1/3) + d) + 63*(40*b*d*e^11*n*x^3 - 55*b*d^4 *e^8*n*x^2 + 88*b*d^7*e^5*n*x - 220*b*d^10*e^2*n)*x^(2/3) - 198*(14*b*d^2* e^10*n*x^3 - 20*b*d^5*e^7*n*x^2 + 35*b*d^8*e^4*n*x - 140*b*d^11*e*n)*x^(1/ 3))/e^12
Time = 34.07 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.92 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {a x^{4}}{4} + b \left (- \frac {e n \left (\frac {3 d^{12} \left (\begin {cases} \frac {\sqrt [3]{x}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e \sqrt [3]{x} \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{12}} - \frac {3 d^{11} \sqrt [3]{x}}{e^{12}} + \frac {3 d^{10} x^{\frac {2}{3}}}{2 e^{11}} - \frac {d^{9} x}{e^{10}} + \frac {3 d^{8} x^{\frac {4}{3}}}{4 e^{9}} - \frac {3 d^{7} x^{\frac {5}{3}}}{5 e^{8}} + \frac {d^{6} x^{2}}{2 e^{7}} - \frac {3 d^{5} x^{\frac {7}{3}}}{7 e^{6}} + \frac {3 d^{4} x^{\frac {8}{3}}}{8 e^{5}} - \frac {d^{3} x^{3}}{3 e^{4}} + \frac {3 d^{2} x^{\frac {10}{3}}}{10 e^{3}} - \frac {3 d x^{\frac {11}{3}}}{11 e^{2}} + \frac {x^{4}}{4 e}\right )}{12} + \frac {x^{4} \log {\left (c \left (d + e \sqrt [3]{x}\right )^{n} \right )}}{4}\right ) \] Input:
integrate(x**3*(a+b*ln(c*(d+e*x**(1/3))**n)),x)
Output:
a*x**4/4 + b*(-e*n*(3*d**12*Piecewise((x**(1/3)/d, Eq(e, 0)), (log(d + e*x **(1/3))/e, True))/e**12 - 3*d**11*x**(1/3)/e**12 + 3*d**10*x**(2/3)/(2*e* *11) - d**9*x/e**10 + 3*d**8*x**(4/3)/(4*e**9) - 3*d**7*x**(5/3)/(5*e**8) + d**6*x**2/(2*e**7) - 3*d**5*x**(7/3)/(7*e**6) + 3*d**4*x**(8/3)/(8*e**5) - d**3*x**3/(3*e**4) + 3*d**2*x**(10/3)/(10*e**3) - 3*d*x**(11/3)/(11*e** 2) + x**4/(4*e))/12 + x**4*log(c*(d + e*x**(1/3))**n)/4)
Time = 0.05 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.74 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {1}{4} \, b x^{4} \log \left ({\left (e x^{\frac {1}{3}} + d\right )}^{n} c\right ) + \frac {1}{4} \, a x^{4} - \frac {1}{110880} \, b e n {\left (\frac {27720 \, d^{12} \log \left (e x^{\frac {1}{3}} + d\right )}{e^{13}} + \frac {2310 \, e^{11} x^{4} - 2520 \, d e^{10} x^{\frac {11}{3}} + 2772 \, d^{2} e^{9} x^{\frac {10}{3}} - 3080 \, d^{3} e^{8} x^{3} + 3465 \, d^{4} e^{7} x^{\frac {8}{3}} - 3960 \, d^{5} e^{6} x^{\frac {7}{3}} + 4620 \, d^{6} e^{5} x^{2} - 5544 \, d^{7} e^{4} x^{\frac {5}{3}} + 6930 \, d^{8} e^{3} x^{\frac {4}{3}} - 9240 \, d^{9} e^{2} x + 13860 \, d^{10} e x^{\frac {2}{3}} - 27720 \, d^{11} x^{\frac {1}{3}}}{e^{12}}\right )} \] Input:
integrate(x^3*(a+b*log(c*(d+e*x^(1/3))^n)),x, algorithm="maxima")
Output:
1/4*b*x^4*log((e*x^(1/3) + d)^n*c) + 1/4*a*x^4 - 1/110880*b*e*n*(27720*d^1 2*log(e*x^(1/3) + d)/e^13 + (2310*e^11*x^4 - 2520*d*e^10*x^(11/3) + 2772*d ^2*e^9*x^(10/3) - 3080*d^3*e^8*x^3 + 3465*d^4*e^7*x^(8/3) - 3960*d^5*e^6*x ^(7/3) + 4620*d^6*e^5*x^2 - 5544*d^7*e^4*x^(5/3) + 6930*d^8*e^3*x^(4/3) - 9240*d^9*e^2*x + 13860*d^10*e*x^(2/3) - 27720*d^11*x^(1/3))/e^12)
Leaf count of result is larger than twice the leaf count of optimal. 516 vs. \(2 (186) = 372\).
Time = 0.13 (sec) , antiderivative size = 516, normalized size of antiderivative = 2.21 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx =\text {Too large to display} \] Input:
integrate(x^3*(a+b*log(c*(d+e*x^(1/3))^n)),x, algorithm="giac")
Output:
1/110880*(27720*b*e*x^4*log(c) + 27720*a*e*x^4 + (27720*(e*x^(1/3) + d)^12 *log(e*x^(1/3) + d)/e^11 - 332640*(e*x^(1/3) + d)^11*d*log(e*x^(1/3) + d)/ e^11 + 1829520*(e*x^(1/3) + d)^10*d^2*log(e*x^(1/3) + d)/e^11 - 6098400*(e *x^(1/3) + d)^9*d^3*log(e*x^(1/3) + d)/e^11 + 13721400*(e*x^(1/3) + d)^8*d ^4*log(e*x^(1/3) + d)/e^11 - 21954240*(e*x^(1/3) + d)^7*d^5*log(e*x^(1/3) + d)/e^11 + 25613280*(e*x^(1/3) + d)^6*d^6*log(e*x^(1/3) + d)/e^11 - 21954 240*(e*x^(1/3) + d)^5*d^7*log(e*x^(1/3) + d)/e^11 + 13721400*(e*x^(1/3) + d)^4*d^8*log(e*x^(1/3) + d)/e^11 - 6098400*(e*x^(1/3) + d)^3*d^9*log(e*x^( 1/3) + d)/e^11 + 1829520*(e*x^(1/3) + d)^2*d^10*log(e*x^(1/3) + d)/e^11 - 332640*(e*x^(1/3) + d)*d^11*log(e*x^(1/3) + d)/e^11 - 2310*(e*x^(1/3) + d) ^12/e^11 + 30240*(e*x^(1/3) + d)^11*d/e^11 - 182952*(e*x^(1/3) + d)^10*d^2 /e^11 + 677600*(e*x^(1/3) + d)^9*d^3/e^11 - 1715175*(e*x^(1/3) + d)^8*d^4/ e^11 + 3136320*(e*x^(1/3) + d)^7*d^5/e^11 - 4268880*(e*x^(1/3) + d)^6*d^6/ e^11 + 4390848*(e*x^(1/3) + d)^5*d^7/e^11 - 3430350*(e*x^(1/3) + d)^4*d^8/ e^11 + 2032800*(e*x^(1/3) + d)^3*d^9/e^11 - 914760*(e*x^(1/3) + d)^2*d^10/ e^11 + 332640*(e*x^(1/3) + d)*d^11/e^11)*b*n)/e
Time = 14.81 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.81 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {a\,x^4}{4}-\frac {b\,n\,x^4}{48}+\frac {b\,x^4\,\ln \left (c\,{\left (d+e\,x^{1/3}\right )}^n\right )}{4}+\frac {b\,d\,n\,x^{11/3}}{44\,e}+\frac {b\,d^9\,n\,x}{12\,e^9}-\frac {b\,d^{12}\,n\,\ln \left (d+e\,x^{1/3}\right )}{4\,e^{12}}+\frac {b\,d^3\,n\,x^3}{36\,e^3}-\frac {b\,d^6\,n\,x^2}{24\,e^6}-\frac {b\,d^2\,n\,x^{10/3}}{40\,e^2}-\frac {b\,d^4\,n\,x^{8/3}}{32\,e^4}+\frac {b\,d^5\,n\,x^{7/3}}{28\,e^5}+\frac {b\,d^7\,n\,x^{5/3}}{20\,e^7}-\frac {b\,d^8\,n\,x^{4/3}}{16\,e^8}-\frac {b\,d^{10}\,n\,x^{2/3}}{8\,e^{10}}+\frac {b\,d^{11}\,n\,x^{1/3}}{4\,e^{11}} \] Input:
int(x^3*(a + b*log(c*(d + e*x^(1/3))^n)),x)
Output:
(a*x^4)/4 - (b*n*x^4)/48 + (b*x^4*log(c*(d + e*x^(1/3))^n))/4 + (b*d*n*x^( 11/3))/(44*e) + (b*d^9*n*x)/(12*e^9) - (b*d^12*n*log(d + e*x^(1/3)))/(4*e^ 12) + (b*d^3*n*x^3)/(36*e^3) - (b*d^6*n*x^2)/(24*e^6) - (b*d^2*n*x^(10/3)) /(40*e^2) - (b*d^4*n*x^(8/3))/(32*e^4) + (b*d^5*n*x^(7/3))/(28*e^5) + (b*d ^7*n*x^(5/3))/(20*e^7) - (b*d^8*n*x^(4/3))/(16*e^8) - (b*d^10*n*x^(2/3))/( 8*e^10) + (b*d^11*n*x^(1/3))/(4*e^11)
Time = 0.21 (sec) , antiderivative size = 201, normalized size of antiderivative = 0.86 \[ \int x^3 \left (a+b \log \left (c \left (d+e \sqrt [3]{x}\right )^n\right )\right ) \, dx=\frac {-13860 x^{\frac {2}{3}} b \,d^{10} e^{2} n +5544 x^{\frac {5}{3}} b \,d^{7} e^{5} n -3465 x^{\frac {8}{3}} b \,d^{4} e^{8} n +2520 x^{\frac {11}{3}} b d \,e^{11} n +27720 x^{\frac {1}{3}} b \,d^{11} e n -6930 x^{\frac {4}{3}} b \,d^{8} e^{4} n +3960 x^{\frac {7}{3}} b \,d^{5} e^{7} n -2772 x^{\frac {10}{3}} b \,d^{2} e^{10} n -27720 \,\mathrm {log}\left (\left (x^{\frac {1}{3}} e +d \right )^{n} c \right ) b \,d^{12}+27720 \,\mathrm {log}\left (\left (x^{\frac {1}{3}} e +d \right )^{n} c \right ) b \,e^{12} x^{4}+27720 a \,e^{12} x^{4}+9240 b \,d^{9} e^{3} n x -4620 b \,d^{6} e^{6} n \,x^{2}+3080 b \,d^{3} e^{9} n \,x^{3}-2310 b \,e^{12} n \,x^{4}}{110880 e^{12}} \] Input:
int(x^3*(a+b*log(c*(d+e*x^(1/3))^n)),x)
Output:
( - 13860*x**(2/3)*b*d**10*e**2*n + 5544*x**(2/3)*b*d**7*e**5*n*x - 3465*x **(2/3)*b*d**4*e**8*n*x**2 + 2520*x**(2/3)*b*d*e**11*n*x**3 + 27720*x**(1/ 3)*b*d**11*e*n - 6930*x**(1/3)*b*d**8*e**4*n*x + 3960*x**(1/3)*b*d**5*e**7 *n*x**2 - 2772*x**(1/3)*b*d**2*e**10*n*x**3 - 27720*log((x**(1/3)*e + d)** n*c)*b*d**12 + 27720*log((x**(1/3)*e + d)**n*c)*b*e**12*x**4 + 27720*a*e** 12*x**4 + 9240*b*d**9*e**3*n*x - 4620*b*d**6*e**6*n*x**2 + 3080*b*d**3*e** 9*n*x**3 - 2310*b*e**12*n*x**4)/(110880*e**12)