3.3.34 \(\int \frac {x^2 \log (c (a+b x^3)^p)}{d+e x} \, dx\) [234]

3.3.34.1 Optimal result
3.3.34.2 Mathematica [C] (verified)
3.3.34.3 Rubi [A] (verified)
3.3.34.4 Maple [C] (verified)
3.3.34.5 Fricas [F]
3.3.34.6 Sympy [F(-1)]
3.3.34.7 Maxima [F]
3.3.34.8 Giac [F]
3.3.34.9 Mupad [F(-1)]

3.3.34.1 Optimal result

Integrand size = 23, antiderivative size = 643 \[ \int \frac {x^2 \log \left (c \left (a+b x^3\right )^p\right )}{d+e x} \, dx=\frac {3 d p x}{e^2}-\frac {3 p x^2}{4 e}+\frac {\sqrt {3} \sqrt [3]{a} d p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{b} e^2}-\frac {\sqrt {3} a^{2/3} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{2 b^{2/3} e}-\frac {\sqrt [3]{a} d p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} e^2}-\frac {a^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 b^{2/3} e}-\frac {d^2 p \log \left (-\frac {e \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right ) \log (d+e x)}{e^3}-\frac {d^2 p \log \left (-\frac {e \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right ) \log (d+e x)}{e^3}-\frac {d^2 p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right ) \log (d+e x)}{e^3}+\frac {\sqrt [3]{a} d p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b} e^2}+\frac {a^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 b^{2/3} e}-\frac {d x \log \left (c \left (a+b x^3\right )^p\right )}{e^2}+\frac {x^2 \log \left (c \left (a+b x^3\right )^p\right )}{2 e}+\frac {d^2 \log (d+e x) \log \left (c \left (a+b x^3\right )^p\right )}{e^3}-\frac {d^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right )}{e^3}-\frac {d^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right )}{e^3}-\frac {d^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right )}{e^3} \]

output
3*d*p*x/e^2-3/4*p*x^2/e-a^(1/3)*d*p*ln(a^(1/3)+b^(1/3)*x)/b^(1/3)/e^2-1/2* 
a^(2/3)*p*ln(a^(1/3)+b^(1/3)*x)/b^(2/3)/e-d^2*p*ln(-e*(a^(1/3)+b^(1/3)*x)/ 
(b^(1/3)*d-a^(1/3)*e))*ln(e*x+d)/e^3-d^2*p*ln(-e*((-1)^(2/3)*a^(1/3)+b^(1/ 
3)*x)/(b^(1/3)*d-(-1)^(2/3)*a^(1/3)*e))*ln(e*x+d)/e^3-d^2*p*ln((-1)^(1/3)* 
e*(a^(1/3)+(-1)^(2/3)*b^(1/3)*x)/(b^(1/3)*d+(-1)^(1/3)*a^(1/3)*e))*ln(e*x+ 
d)/e^3+1/2*a^(1/3)*d*p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(1/3)/e 
^2+1/4*a^(2/3)*p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(2/3)/e-d*x*l 
n(c*(b*x^3+a)^p)/e^2+1/2*x^2*ln(c*(b*x^3+a)^p)/e+d^2*ln(e*x+d)*ln(c*(b*x^3 
+a)^p)/e^3-d^2*p*polylog(2,b^(1/3)*(e*x+d)/(b^(1/3)*d-a^(1/3)*e))/e^3-d^2* 
p*polylog(2,b^(1/3)*(e*x+d)/(b^(1/3)*d+(-1)^(1/3)*a^(1/3)*e))/e^3-d^2*p*po 
lylog(2,b^(1/3)*(e*x+d)/(b^(1/3)*d-(-1)^(2/3)*a^(1/3)*e))/e^3+a^(1/3)*d*p* 
arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))*3^(1/2)/b^(1/3)/e^2-1/2* 
a^(2/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))*3^(1/2)/b^(2/3 
)/e
 
3.3.34.2 Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.26 (sec) , antiderivative size = 504, normalized size of antiderivative = 0.78 \[ \int \frac {x^2 \log \left (c \left (a+b x^3\right )^p\right )}{d+e x} \, dx=-\frac {-12 d e p x+3 e^2 p x^2-\frac {4 \sqrt {3} \sqrt [3]{a} d e p \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}-3 e^2 p x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b x^3}{a}\right )+\frac {4 \sqrt [3]{a} d e p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}+4 d^2 p \log \left (\frac {e \left (\sqrt [3]{-1} \sqrt [3]{a}-\sqrt [3]{b} x\right )}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right ) \log (d+e x)+4 d^2 p \log \left (\frac {e \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} d+\sqrt [3]{a} e}\right ) \log (d+e x)+4 d^2 p \log \left (\frac {e \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} d+(-1)^{2/3} \sqrt [3]{a} e}\right ) \log (d+e x)-\frac {2 \sqrt [3]{a} d e p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{b}}+4 d e x \log \left (c \left (a+b x^3\right )^p\right )-2 e^2 x^2 \log \left (c \left (a+b x^3\right )^p\right )-4 d^2 \log (d+e x) \log \left (c \left (a+b x^3\right )^p\right )+4 d^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right )+4 d^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right )+4 d^2 p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right )}{4 e^3} \]

input
Integrate[(x^2*Log[c*(a + b*x^3)^p])/(d + e*x),x]
 
output
-1/4*(-12*d*e*p*x + 3*e^2*p*x^2 - (4*Sqrt[3]*a^(1/3)*d*e*p*ArcTan[(1 - (2* 
b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3) - 3*e^2*p*x^2*Hypergeometric2F1[2/3, 
 1, 5/3, -((b*x^3)/a)] + (4*a^(1/3)*d*e*p*Log[a^(1/3) + b^(1/3)*x])/b^(1/3 
) + 4*d^2*p*Log[(e*((-1)^(1/3)*a^(1/3) - b^(1/3)*x))/(b^(1/3)*d + (-1)^(1/ 
3)*a^(1/3)*e)]*Log[d + e*x] + 4*d^2*p*Log[(e*(a^(1/3) + b^(1/3)*x))/(-(b^( 
1/3)*d) + a^(1/3)*e)]*Log[d + e*x] + 4*d^2*p*Log[(e*((-1)^(2/3)*a^(1/3) + 
b^(1/3)*x))/(-(b^(1/3)*d) + (-1)^(2/3)*a^(1/3)*e)]*Log[d + e*x] - (2*a^(1/ 
3)*d*e*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(1/3) + 4*d*e*x 
*Log[c*(a + b*x^3)^p] - 2*e^2*x^2*Log[c*(a + b*x^3)^p] - 4*d^2*Log[d + e*x 
]*Log[c*(a + b*x^3)^p] + 4*d^2*p*PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1/3)*d 
 - a^(1/3)*e)] + 4*d^2*p*PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1/3)*d + (-1)^ 
(1/3)*a^(1/3)*e)] + 4*d^2*p*PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1/3)*d - (- 
1)^(2/3)*a^(1/3)*e)])/e^3
 
3.3.34.3 Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 643, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2916, 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.

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

\(\Big \downarrow \) 2916

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

\(\Big \downarrow \) 2009

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

input
Int[(x^2*Log[c*(a + b*x^3)^p])/(d + e*x),x]
 
output
(3*d*p*x)/e^2 - (3*p*x^2)/(4*e) + (Sqrt[3]*a^(1/3)*d*p*ArcTan[(a^(1/3) - 2 
*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(b^(1/3)*e^2) - (Sqrt[3]*a^(2/3)*p*ArcTan[ 
(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(2*b^(2/3)*e) - (a^(1/3)*d*p*L 
og[a^(1/3) + b^(1/3)*x])/(b^(1/3)*e^2) - (a^(2/3)*p*Log[a^(1/3) + b^(1/3)* 
x])/(2*b^(2/3)*e) - (d^2*p*Log[-((e*(a^(1/3) + b^(1/3)*x))/(b^(1/3)*d - a^ 
(1/3)*e))]*Log[d + e*x])/e^3 - (d^2*p*Log[-((e*((-1)^(2/3)*a^(1/3) + b^(1/ 
3)*x))/(b^(1/3)*d - (-1)^(2/3)*a^(1/3)*e))]*Log[d + e*x])/e^3 - (d^2*p*Log 
[((-1)^(1/3)*e*(a^(1/3) + (-1)^(2/3)*b^(1/3)*x))/(b^(1/3)*d + (-1)^(1/3)*a 
^(1/3)*e)]*Log[d + e*x])/e^3 + (a^(1/3)*d*p*Log[a^(2/3) - a^(1/3)*b^(1/3)* 
x + b^(2/3)*x^2])/(2*b^(1/3)*e^2) + (a^(2/3)*p*Log[a^(2/3) - a^(1/3)*b^(1/ 
3)*x + b^(2/3)*x^2])/(4*b^(2/3)*e) - (d*x*Log[c*(a + b*x^3)^p])/e^2 + (x^2 
*Log[c*(a + b*x^3)^p])/(2*e) + (d^2*Log[d + e*x]*Log[c*(a + b*x^3)^p])/e^3 
 - (d^2*p*PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1/3)*d - a^(1/3)*e)])/e^3 - ( 
d^2*p*PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1/3)*d + (-1)^(1/3)*a^(1/3)*e)])/ 
e^3 - (d^2*p*PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1/3)*d - (-1)^(2/3)*a^(1/3 
)*e)])/e^3
 

3.3.34.3.1 Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2916
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m 
_.)*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log 
[c*(d + e*x^n)^p])^q, x^m*(f + g*x)^r, x], x] /; FreeQ[{a, b, c, d, e, f, g 
, n, p, q}, x] && IntegerQ[m] && IntegerQ[r]
 
3.3.34.4 Maple [C] (verified)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 1.53 (sec) , antiderivative size = 250, normalized size of antiderivative = 0.39

method result size
parts \(\frac {x^{2} \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{2 e}-\frac {d x \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{e^{2}}+\frac {d^{2} \ln \left (e x +d \right ) \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{e^{3}}-\frac {3 p b \left (\frac {d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 b d \,\textit {\_Z}^{2}+3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )\right )}{3 b}+\frac {\left (e x +d \right )^{2}}{4 b}-\frac {3 \left (e x +d \right ) d}{2 b}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 b d \,\textit {\_Z}^{2}+3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\frac {\left (-\textit {\_R} +3 d \right ) \ln \left (e x -\textit {\_R} +d \right )}{\textit {\_R}^{2}-2 \textit {\_R} d +d^{2}}\right ) a \,e^{3}}{6 b^{2}}\right )}{e^{3}}\) \(250\)
risch \(\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right ) x^{2}}{2 e}-\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right ) d x}{e^{2}}+\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right ) d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {p \,d^{2} \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 b d \,\textit {\_Z}^{2}+3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )\right )}{e^{3}}-\frac {3 p \,x^{2}}{4 e}+\frac {3 d p x}{e^{2}}+\frac {15 p \,d^{2}}{4 e^{3}}+\frac {p \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 b d \,\textit {\_Z}^{2}+3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\frac {\left (\textit {\_R} -3 d \right ) \ln \left (e x -\textit {\_R} +d \right )}{\textit {\_R}^{2}-2 \textit {\_R} d +d^{2}}\right ) a}{2 b}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {\frac {1}{2} e \,x^{2}-d x}{e^{2}}+\frac {d^{2} \ln \left (e x +d \right )}{e^{3}}\right )\) \(383\)

input
int(x^2*ln(c*(b*x^3+a)^p)/(e*x+d),x,method=_RETURNVERBOSE)
 
output
1/2*x^2*ln(c*(b*x^3+a)^p)/e-d*x*ln(c*(b*x^3+a)^p)/e^2+d^2*ln(e*x+d)*ln(c*( 
b*x^3+a)^p)/e^3-3*p*b/e^3*(1/3*d^2/b*sum(ln(e*x+d)*ln((-e*x+_R1-d)/_R1)+di 
log((-e*x+_R1-d)/_R1),_R1=RootOf(_Z^3*b-3*_Z^2*b*d+3*_Z*b*d^2+a*e^3-b*d^3) 
)+1/4/b*(e*x+d)^2-3/2/b*(e*x+d)*d+1/6/b^2*sum((-_R+3*d)/(_R^2-2*_R*d+d^2)* 
ln(e*x-_R+d),_R=RootOf(_Z^3*b-3*_Z^2*b*d+3*_Z*b*d^2+a*e^3-b*d^3))*a*e^3)
 
3.3.34.5 Fricas [F]

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

input
integrate(x^2*log(c*(b*x^3+a)^p)/(e*x+d),x, algorithm="fricas")
 
output
integral(x^2*log((b*x^3 + a)^p*c)/(e*x + d), x)
 
3.3.34.6 Sympy [F(-1)]

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

input
integrate(x**2*ln(c*(b*x**3+a)**p)/(e*x+d),x)
 
output
Timed out
 
3.3.34.7 Maxima [F]

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

input
integrate(x^2*log(c*(b*x^3+a)^p)/(e*x+d),x, algorithm="maxima")
 
output
integrate(x^2*log((b*x^3 + a)^p*c)/(e*x + d), x)
 
3.3.34.8 Giac [F]

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

input
integrate(x^2*log(c*(b*x^3+a)^p)/(e*x+d),x, algorithm="giac")
 
output
integrate(x^2*log((b*x^3 + a)^p*c)/(e*x + d), x)
 
3.3.34.9 Mupad [F(-1)]

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

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