Integrand size = 20, antiderivative size = 391 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^3} \, dx=\frac {3 b d^2 p}{2 e \left (b d^3-a e^3\right ) (d+e x)}-\frac {\sqrt {3} \sqrt [3]{a} b^{2/3} \left (2 b d^3-3 \sqrt [3]{a} b^{2/3} d^2 e+a e^3\right ) p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{2 \left (b d^3-a e^3\right )^2}+\frac {\sqrt [3]{a} b^{2/3} \left (2 b d^3+3 \sqrt [3]{a} b^{2/3} d^2 e+a e^3\right ) p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 \left (b d^3-a e^3\right )^2}-\frac {3 b d \left (b d^3+2 a e^3\right ) p \log (d+e x)}{2 e \left (b d^3-a e^3\right )^2}-\frac {\sqrt [3]{a} b^{2/3} \left (2 b d^3+3 \sqrt [3]{a} b^{2/3} d^2 e+a e^3\right ) p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 \left (b d^3-a e^3\right )^2}+\frac {b d \left (b d^3+2 a e^3\right ) p \log \left (a+b x^3\right )}{2 e \left (b d^3-a e^3\right )^2}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 e (d+e x)^2} \] Output:
3/2*b*d^2*p/e/(-a*e^3+b*d^3)/(e*x+d)-1/2*3^(1/2)*a^(1/3)*b^(2/3)*(2*b*d^3- 3*a^(1/3)*b^(2/3)*d^2*e+a*e^3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/ a^(1/3))/(-a*e^3+b*d^3)^2+1/2*a^(1/3)*b^(2/3)*(2*b*d^3+3*a^(1/3)*b^(2/3)*d ^2*e+a*e^3)*p*ln(a^(1/3)+b^(1/3)*x)/(-a*e^3+b*d^3)^2-3/2*b*d*(2*a*e^3+b*d^ 3)*p*ln(e*x+d)/e/(-a*e^3+b*d^3)^2-1/4*a^(1/3)*b^(2/3)*(2*b*d^3+3*a^(1/3)*b ^(2/3)*d^2*e+a*e^3)*p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(-a*e^3+b* d^3)^2+1/2*b*d*(2*a*e^3+b*d^3)*p*ln(b*x^3+a)/e/(-a*e^3+b*d^3)^2-1/2*ln(c*( b*x^3+a)^p)/e/(e*x+d)^2
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.63 (sec) , antiderivative size = 325, normalized size of antiderivative = 0.83 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^3} \, dx=\frac {\frac {b^{2/3} p (d+e x) \left (6 \sqrt [3]{b} d^2 \left (b d^3-a e^3\right )-2 \sqrt {3} \sqrt [3]{a} e \left (2 b d^3+a e^3\right ) (d+e x) \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-9 b^{4/3} d^2 e^2 x^2 (d+e x) \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b x^3}{a}\right )+2 \sqrt [3]{a} e \left (2 b d^3+a e^3\right ) (d+e x) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-6 \sqrt [3]{b} d \left (b d^3+2 a e^3\right ) (d+e x) \log (d+e x)-\sqrt [3]{a} e \left (2 b d^3+a e^3\right ) (d+e x) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+2 \sqrt [3]{b} d \left (b d^3+2 a e^3\right ) (d+e x) \log \left (a+b x^3\right )\right )}{\left (b d^3-a e^3\right )^2}-2 \log \left (c \left (a+b x^3\right )^p\right )}{4 e (d+e x)^2} \] Input:
Integrate[Log[c*(a + b*x^3)^p]/(d + e*x)^3,x]
Output:
((b^(2/3)*p*(d + e*x)*(6*b^(1/3)*d^2*(b*d^3 - a*e^3) - 2*Sqrt[3]*a^(1/3)*e *(2*b*d^3 + a*e^3)*(d + e*x)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 9*b^(4/3)*d^2*e^2*x^2*(d + e*x)*Hypergeometric2F1[2/3, 1, 5/3, -((b*x^3)/ a)] + 2*a^(1/3)*e*(2*b*d^3 + a*e^3)*(d + e*x)*Log[a^(1/3) + b^(1/3)*x] - 6 *b^(1/3)*d*(b*d^3 + 2*a*e^3)*(d + e*x)*Log[d + e*x] - a^(1/3)*e*(2*b*d^3 + a*e^3)*(d + e*x)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2] + 2*b^(1/ 3)*d*(b*d^3 + 2*a*e^3)*(d + e*x)*Log[a + b*x^3]))/(b*d^3 - a*e^3)^2 - 2*Lo g[c*(a + b*x^3)^p])/(4*e*(d + e*x)^2)
Time = 1.52 (sec) , antiderivative size = 379, normalized size of antiderivative = 0.97, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {2913, 7276, 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.
| \(\displaystyle \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 2913 |
| \(\displaystyle \frac {3 b p \int \frac {x^2}{(d+e x)^2 \left (b x^3+a\right )}dx}{2 e}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 e (d+e x)^2}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \frac {3 b p \int \left (-\frac {e d^2}{\left (b d^3-a e^3\right ) (d+e x)^2}-\frac {e \left (b d^3+2 a e^3\right ) d}{\left (b d^3-a e^3\right )^2 (d+e x)}+\frac {-3 a b d^2 x e^2+a \left (2 b d^3+a e^3\right ) e+b d \left (b d^3+2 a e^3\right ) x^2}{\left (b d^3-a e^3\right )^2 \left (b x^3+a\right )}\right )dx}{2 e}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 e (d+e x)^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 b p \left (-\frac {\sqrt [3]{a} e \left (3 \sqrt [3]{a} b^{2/3} d^2 e+a e^3+2 b d^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{b} \left (b d^3-a e^3\right )^2}-\frac {\sqrt [3]{a} e \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (-3 \sqrt [3]{a} b^{2/3} d^2 e+a e^3+2 b d^3\right )}{\sqrt {3} \sqrt [3]{b} \left (b d^3-a e^3\right )^2}+\frac {\sqrt [3]{a} e \left (3 \sqrt [3]{a} b^{2/3} d^2 e+a e^3+2 b d^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{b} \left (b d^3-a e^3\right )^2}+\frac {d \left (2 a e^3+b d^3\right ) \log \left (a+b x^3\right )}{3 \left (b d^3-a e^3\right )^2}-\frac {d \left (2 a e^3+b d^3\right ) \log (d+e x)}{\left (b d^3-a e^3\right )^2}+\frac {d^2}{(d+e x) \left (b d^3-a e^3\right )}\right )}{2 e}-\frac {\log \left (c \left (a+b x^3\right )^p\right )}{2 e (d+e x)^2}\) |
Input:
Int[Log[c*(a + b*x^3)^p]/(d + e*x)^3,x]
Output:
(3*b*p*(d^2/((b*d^3 - a*e^3)*(d + e*x)) - (a^(1/3)*e*(2*b*d^3 - 3*a^(1/3)* b^(2/3)*d^2*e + a*e^3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/ (Sqrt[3]*b^(1/3)*(b*d^3 - a*e^3)^2) + (a^(1/3)*e*(2*b*d^3 + 3*a^(1/3)*b^(2 /3)*d^2*e + a*e^3)*Log[a^(1/3) + b^(1/3)*x])/(3*b^(1/3)*(b*d^3 - a*e^3)^2) - (d*(b*d^3 + 2*a*e^3)*Log[d + e*x])/(b*d^3 - a*e^3)^2 - (a^(1/3)*e*(2*b* d^3 + 3*a^(1/3)*b^(2/3)*d^2*e + a*e^3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b ^(2/3)*x^2])/(6*b^(1/3)*(b*d^3 - a*e^3)^2) + (d*(b*d^3 + 2*a*e^3)*Log[a + b*x^3])/(3*(b*d^3 - a*e^3)^2)))/(2*e) - Log[c*(a + b*x^3)^p]/(2*e*(d + e*x )^2)
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_. )*(x_))^(r_.), x_Symbol] :> Simp[(f + g*x)^(r + 1)*((a + b*Log[c*(d + e*x^n )^p])/(g*(r + 1))), x] - Simp[b*e*n*(p/(g*(r + 1))) Int[x^(n - 1)*((f + g *x)^(r + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x ] && (IGtQ[r, 0] || RationalQ[n]) && NeQ[r, -1]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Time = 3.30 (sec) , antiderivative size = 347, normalized size of antiderivative = 0.89
| method | result | size |
| parts | \(-\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{2 e \left (e x +d \right )^{2}}+\frac {3 p b \left (-\frac {d^{2}}{\left (a \,e^{3}-b \,d^{3}\right ) \left (e x +d \right )}-\frac {d \left (2 a \,e^{3}+b \,d^{3}\right ) \ln \left (e x +d \right )}{\left (a \,e^{3}-b \,d^{3}\right )^{2}}+\frac {\left (a^{2} e^{4}+2 a b \,d^{3} e \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )-3 a b \,d^{2} e^{2} \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+\frac {\left (2 a b d \,e^{3}+b^{2} d^{4}\right ) \ln \left (b \,x^{3}+a \right )}{3 b}}{\left (a \,e^{3}-b \,d^{3}\right )^{2}}\right )}{2 e}\) | \(347\) |
| risch | \(\text {Expression too large to display}\) | \(4085\) |
Input:
int(ln(c*(b*x^3+a)^p)/(e*x+d)^3,x,method=_RETURNVERBOSE)
Output:
-1/2*ln(c*(b*x^3+a)^p)/e/(e*x+d)^2+3/2*p*b/e*(-d^2/(a*e^3-b*d^3)/(e*x+d)-d *(2*a*e^3+b*d^3)/(a*e^3-b*d^3)^2*ln(e*x+d)+((a^2*e^4+2*a*b*d^3*e)*(1/3/b/( 1/b*a)^(2/3)*ln(x+(1/b*a)^(1/3))-1/6/b/(1/b*a)^(2/3)*ln(x^2-(1/b*a)^(1/3)* x+(1/b*a)^(2/3))+1/3/b/(1/b*a)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/b*a) ^(1/3)*x-1)))-3*a*b*d^2*e^2*(-1/3/b/(1/b*a)^(1/3)*ln(x+(1/b*a)^(1/3))+1/6/ b/(1/b*a)^(1/3)*ln(x^2-(1/b*a)^(1/3)*x+(1/b*a)^(2/3))+1/3*3^(1/2)/b/(1/b*a )^(1/3)*arctan(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*x-1)))+1/3*(2*a*b*d*e^3+b^2*d^ 4)/b*ln(b*x^3+a))/(a*e^3-b*d^3)^2)
Result contains complex when optimal does not.
Time = 8.60 (sec) , antiderivative size = 13236, normalized size of antiderivative = 33.85 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^3} \, dx=\text {Too large to display} \] Input:
integrate(log(c*(b*x^3+a)^p)/(e*x+d)^3,x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^3} \, dx=\text {Timed out} \] Input:
integrate(ln(c*(b*x**3+a)**p)/(e*x+d)**3,x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 517, normalized size of antiderivative = 1.32 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^3} \, dx=-\frac {{\left (\frac {2 \, \sqrt {3} {\left (3 \, a b d^{2} e^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, a b d^{3} e \left (\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} e^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{{\left (b^{3} d^{6} \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, a b^{2} d^{3} e^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}} + a^{2} b e^{6} \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {6 \, d^{2}}{b d^{4} - a d e^{3} + {\left (b d^{3} e - a e^{4}\right )} x} + \frac {6 \, {\left (b d^{4} + 2 \, a d e^{3}\right )} \log \left (e x + d\right )}{b^{2} d^{6} - 2 \, a b d^{3} e^{3} + a^{2} e^{6}} - \frac {{\left (2 \, b^{2} d^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}} + 4 \, a b d e^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}} - 3 \, a b d^{2} e^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}} - 2 \, a b d^{3} e - a^{2} e^{4}\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{3} d^{6} \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, a b^{2} d^{3} e^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}} + a^{2} b e^{6} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {2 \, {\left (b^{2} d^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}} + 2 \, a b d e^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}} + 3 \, a b d^{2} e^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, a b d^{3} e + a^{2} e^{4}\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{3} d^{6} \left (\frac {a}{b}\right )^{\frac {2}{3}} - 2 \, a b^{2} d^{3} e^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}} + a^{2} b e^{6} \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )} b p}{4 \, e} - \frac {\log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{2 \, {\left (e x + d\right )}^{2} e} \] Input:
integrate(log(c*(b*x^3+a)^p)/(e*x+d)^3,x, algorithm="maxima")
Output:
-1/4*(2*sqrt(3)*(3*a*b*d^2*e^2*(a/b)^(2/3) - 2*a*b*d^3*e*(a/b)^(1/3) - a^2 *e^4*(a/b)^(1/3))*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/((b^ 3*d^6*(a/b)^(2/3) - 2*a*b^2*d^3*e^3*(a/b)^(2/3) + a^2*b*e^6*(a/b)^(2/3))*( a/b)^(1/3)) - 6*d^2/(b*d^4 - a*d*e^3 + (b*d^3*e - a*e^4)*x) + 6*(b*d^4 + 2 *a*d*e^3)*log(e*x + d)/(b^2*d^6 - 2*a*b*d^3*e^3 + a^2*e^6) - (2*b^2*d^4*(a /b)^(2/3) + 4*a*b*d*e^3*(a/b)^(2/3) - 3*a*b*d^2*e^2*(a/b)^(1/3) - 2*a*b*d^ 3*e - a^2*e^4)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*d^6*(a/b)^(2/3) - 2*a*b^2*d^3*e^3*(a/b)^(2/3) + a^2*b*e^6*(a/b)^(2/3)) - 2*(b^2*d^4*(a/b) ^(2/3) + 2*a*b*d*e^3*(a/b)^(2/3) + 3*a*b*d^2*e^2*(a/b)^(1/3) + 2*a*b*d^3*e + a^2*e^4)*log(x + (a/b)^(1/3))/(b^3*d^6*(a/b)^(2/3) - 2*a*b^2*d^3*e^3*(a /b)^(2/3) + a^2*b*e^6*(a/b)^(2/3)))*b*p/e - 1/2*log((b*x^3 + a)^p*c)/((e*x + d)^2*e)
Time = 0.25 (sec) , antiderivative size = 660, normalized size of antiderivative = 1.69 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^3} \, dx =\text {Too large to display} \] Input:
integrate(log(c*(b*x^3+a)^p)/(e*x+d)^3,x, algorithm="giac")
Output:
1/2*(3*a*b^5*d^8*e^3*p*(-a/b)^(1/3) - 6*a^2*b^4*d^5*e^6*p*(-a/b)^(1/3) + 3 *a^3*b^3*d^2*e^9*p*(-a/b)^(1/3) - 2*a*b^5*d^9*e^2*p + 3*a^2*b^4*d^6*e^5*p - a^4*b^2*e^11*p)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^5*d^12*e^2 - 4*a^2*b^4*d^9*e^5 + 6*a^3*b^3*d^6*e^8 - 4*a^4*b^2*d^3*e^11 + a^5*b*e^14) + 3/2*(2*(-a*b^2)^(1/3)*b*d*p - (-a*b^2)^(2/3)*e*p)*arctan(1/3*sqrt(3)*(2 *x + (-a/b)^(1/3))/(-a/b)^(1/3))/(sqrt(3)*b^2*d^4 + 2*sqrt(3)*a*b*d*e^3 - 2*sqrt(3)*(-a*b^2)^(1/3)*b*d^3*e - sqrt(3)*(-a*b^2)^(1/3)*a*e^4 + 3*sqrt(3 )*(-a*b^2)^(2/3)*d^2*e^2) - 1/2*p*log(b*x^3 + a)/(e^3*x^2 + 2*d*e^2*x + d^ 2*e) - 3/2*(b^2*d^4*p + 2*a*b*d*e^3*p)*log(e*x + d)/(b^2*d^6*e - 2*a*b*d^3 *e^4 + a^2*e^7) + 1/4*(2*(-a*b^2)^(1/3)*b*d^3*p + (-a*b^2)^(1/3)*a*e^3*p - 3*(-a*b^2)^(2/3)*d^2*e*p)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(b^2*d ^6 - 2*a*b*d^3*e^3 + a^2*e^6) + 1/2*(b^2*d^4*p + 2*a*b*d*e^3*p)*log(abs(b* x^3 + a))/(b^2*d^6*e - 2*a*b*d^3*e^4 + a^2*e^7) + 1/2*(3*b*d^2*e*p*x + 3*b *d^3*p - b*d^3*log(c) + a*e^3*log(c))/(b*d^3*e^3*x^2 - a*e^6*x^2 + 2*b*d^4 *e^2*x - 2*a*d*e^5*x + b*d^5*e - a*d^2*e^4)
Time = 26.27 (sec) , antiderivative size = 2227, normalized size of antiderivative = 5.70 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^3} \, dx=\text {Too large to display} \] Input:
int(log(c*(a + b*x^3)^p)/(d + e*x)^3,x)
Output:
symsum(log(-(27*a*b^6*d^4*p^3 + 216*root(16*a*b*d^3*e^6*z^3 - 8*b^2*d^6*e^ 3*z^3 - 8*a^2*e^9*z^3 + 24*a*b*d*e^5*p*z^2 + 12*b^2*d^4*e^2*p*z^2 - 6*b^2* d^2*e*p^2*z + b^2*p^3, z, k)^3*a^2*b^5*d^7*e^6 - 648*root(16*a*b*d^3*e^6*z ^3 - 8*b^2*d^6*e^3*z^3 - 8*a^2*e^9*z^3 + 24*a*b*d*e^5*p*z^2 + 12*b^2*d^4*e ^2*p*z^2 - 6*b^2*d^2*e*p^2*z + b^2*p^3, z, k)^3*a^3*b^4*d^4*e^9 + 72*root( 16*a*b*d^3*e^6*z^3 - 8*b^2*d^6*e^3*z^3 - 8*a^2*e^9*z^3 + 24*a*b*d*e^5*p*z^ 2 + 12*b^2*d^4*e^2*p*z^2 - 6*b^2*d^2*e*p^2*z + b^2*p^3, z, k)^3*a*b^6*d^10 *e^3 + 360*root(16*a*b*d^3*e^6*z^3 - 8*b^2*d^6*e^3*z^3 - 8*a^2*e^9*z^3 + 2 4*a*b*d*e^5*p*z^2 + 12*b^2*d^4*e^2*p*z^2 - 6*b^2*d^2*e*p^2*z + b^2*p^3, z, k)^3*a^4*b^3*d*e^12 + 18*root(16*a*b*d^3*e^6*z^3 - 8*b^2*d^6*e^3*z^3 - 8* a^2*e^9*z^3 + 24*a*b*d*e^5*p*z^2 + 12*b^2*d^4*e^2*p*z^2 - 6*b^2*d^2*e*p^2* z + b^2*p^3, z, k)*a^3*b^4*e^7*p^2 + 288*root(16*a*b*d^3*e^6*z^3 - 8*b^2*d ^6*e^3*z^3 - 8*a^2*e^9*z^3 + 24*a*b*d*e^5*p*z^2 + 12*b^2*d^4*e^2*p*z^2 - 6 *b^2*d^2*e*p^2*z + b^2*p^3, z, k)^3*a^4*b^3*e^13*x + 27*a^2*b^5*d*e^3*p^3 - 27*a^2*b^5*e^4*p^3*x + 36*root(16*a*b*d^3*e^6*z^3 - 8*b^2*d^6*e^3*z^3 - 8*a^2*e^9*z^3 + 24*a*b*d*e^5*p*z^2 + 12*b^2*d^4*e^2*p*z^2 - 6*b^2*d^2*e*p^ 2*z + b^2*p^3, z, k)^2*a*b^6*d^8*e^2*p + 144*root(16*a*b*d^3*e^6*z^3 - 8*b ^2*d^6*e^3*z^3 - 8*a^2*e^9*z^3 + 24*a*b*d*e^5*p*z^2 + 12*b^2*d^4*e^2*p*z^2 - 6*b^2*d^2*e*p^2*z + b^2*p^3, z, k)^3*a*b^6*d^9*e^4*x - 90*root(16*a*b*d ^3*e^6*z^3 - 8*b^2*d^6*e^3*z^3 - 8*a^2*e^9*z^3 + 24*a*b*d*e^5*p*z^2 + 1...
Time = 0.31 (sec) , antiderivative size = 16189, normalized size of antiderivative = 41.40 \[ \int \frac {\log \left (c \left (a+b x^3\right )^p\right )}{(d+e x)^3} \, dx =\text {Too large to display} \] Input:
int(log(c*(b*x^3+a)^p)/(e*x+d)^3,x)
Output:
(30*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a* *10*b*d**5*e**30*p + 60*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a **(1/3)*sqrt(3)))*a**10*b*d**4*e**31*p*x + 30*a**(2/3)*sqrt(3)*atan((a**(1 /3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**10*b*d**3*e**32*p*x**2 + 756*a* *(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**9*b** 2*d**8*e**27*p + 1512*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a** (1/3)*sqrt(3)))*a**9*b**2*d**7*e**28*p*x + 756*a**(2/3)*sqrt(3)*atan((a**( 1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**9*b**2*d**6*e**29*p*x**2 - 242 46*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a** 8*b**3*d**11*e**24*p - 48492*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)* x)/(a**(1/3)*sqrt(3)))*a**8*b**3*d**10*e**25*p*x - 24246*a**(2/3)*sqrt(3)* atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**8*b**3*d**9*e**26*p* x**2 - 760518*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sq rt(3)))*a**7*b**4*d**14*e**21*p - 1521036*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**7*b**4*d**13*e**22*p*x - 760518*a** (2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**7*b**4 *d**12*e**23*p*x**2 - 3657690*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3) *x)/(a**(1/3)*sqrt(3)))*a**6*b**5*d**17*e**18*p - 7315380*a**(2/3)*sqrt(3) *atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**6*b**5*d**16*e**19* p*x - 3657690*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)...