3.3.35 \(\int \frac {x \log (c (a+b x^3)^p)}{d+e x} \, dx\) [235]

3.3.35.1 Optimal result
3.3.35.2 Mathematica [A] (verified)
3.3.35.3 Rubi [A] (verified)
3.3.35.4 Maple [C] (verified)
3.3.35.5 Fricas [F]
3.3.35.6 Sympy [F(-1)]
3.3.35.7 Maxima [F]
3.3.35.8 Giac [F]
3.3.35.9 Mupad [F(-1)]

3.3.35.1 Optimal result

Integrand size = 21, antiderivative size = 457 \[ \int \frac {x \log \left (c \left (a+b x^3\right )^p\right )}{d+e x} \, dx=-\frac {3 p x}{e}-\frac {\sqrt {3} \sqrt [3]{a} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt [3]{b} e}+\frac {\sqrt [3]{a} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} e}+\frac {d p \log \left (-\frac {e \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (-\frac {e \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right ) \log (d+e x)}{e^2}+\frac {d p \log \left (\frac {\sqrt [3]{-1} e \left (\sqrt [3]{a}+(-1)^{2/3} \sqrt [3]{b} x\right )}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right ) \log (d+e x)}{e^2}-\frac {\sqrt [3]{a} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b} e}+\frac {x \log \left (c \left (a+b x^3\right )^p\right )}{e}-\frac {d \log (d+e x) \log \left (c \left (a+b x^3\right )^p\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right )}{e^2}+\frac {d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right )}{e^2} \]

output
-3*p*x/e+a^(1/3)*p*ln(a^(1/3)+b^(1/3)*x)/b^(1/3)/e+d*p*ln(-e*(a^(1/3)+b^(1 
/3)*x)/(b^(1/3)*d-a^(1/3)*e))*ln(e*x+d)/e^2+d*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^2+d*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^2-1/2*a^(1/3)*p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(1/3)/ 
e+x*ln(c*(b*x^3+a)^p)/e-d*ln(e*x+d)*ln(c*(b*x^3+a)^p)/e^2+d*p*polylog(2,b^ 
(1/3)*(e*x+d)/(b^(1/3)*d-a^(1/3)*e))/e^2+d*p*polylog(2,b^(1/3)*(e*x+d)/(b^ 
(1/3)*d+(-1)^(1/3)*a^(1/3)*e))/e^2+d*p*polylog(2,b^(1/3)*(e*x+d)/(b^(1/3)* 
d-(-1)^(2/3)*a^(1/3)*e))/e^2-a^(1/3)*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
 
3.3.35.2 Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 430, normalized size of antiderivative = 0.94 \[ \int \frac {x \log \left (c \left (a+b x^3\right )^p\right )}{d+e x} \, dx=\frac {-6 e p x-\frac {2 \sqrt {3} \sqrt [3]{a} e p \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}+\frac {2 \sqrt [3]{a} e p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}+2 d p \log \left (\frac {e \left (\sqrt [3]{-1} \sqrt [3]{a}-\sqrt [3]{b} x\right )}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right ) \log (d+e x)+2 d p \log \left (\frac {e \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} d+\sqrt [3]{a} e}\right ) \log (d+e x)+2 d p \log \left (\frac {e \left ((-1)^{2/3} \sqrt [3]{a}+\sqrt [3]{b} x\right )}{-\sqrt [3]{b} d+(-1)^{2/3} \sqrt [3]{a} e}\right ) \log (d+e x)-\frac {\sqrt [3]{a} e p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{b}}+2 e x \log \left (c \left (a+b x^3\right )^p\right )-2 d \log (d+e x) \log \left (c \left (a+b x^3\right )^p\right )+2 d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-\sqrt [3]{a} e}\right )+2 d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d+\sqrt [3]{-1} \sqrt [3]{a} e}\right )+2 d p \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{b} (d+e x)}{\sqrt [3]{b} d-(-1)^{2/3} \sqrt [3]{a} e}\right )}{2 e^2} \]

input
Integrate[(x*Log[c*(a + b*x^3)^p])/(d + e*x),x]
 
output
(-6*e*p*x - (2*Sqrt[3]*a^(1/3)*e*p*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt 
[3]])/b^(1/3) + (2*a^(1/3)*e*p*Log[a^(1/3) + b^(1/3)*x])/b^(1/3) + 2*d*p*L 
og[(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] + 2*d*p*Log[(e*(a^(1/3) + b^(1/3)*x))/(-(b^(1/3)*d) + a^(1/ 
3)*e)]*Log[d + e*x] + 2*d*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] - (a^(1/3)*e*p*Log[a^(2/3) 
- a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(1/3) + 2*e*x*Log[c*(a + b*x^3)^p] - 
 2*d*Log[d + e*x]*Log[c*(a + b*x^3)^p] + 2*d*p*PolyLog[2, (b^(1/3)*(d + e* 
x))/(b^(1/3)*d - a^(1/3)*e)] + 2*d*p*PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1/ 
3)*d + (-1)^(1/3)*a^(1/3)*e)] + 2*d*p*PolyLog[2, (b^(1/3)*(d + e*x))/(b^(1 
/3)*d - (-1)^(2/3)*a^(1/3)*e)])/(2*e^2)
 
3.3.35.3 Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.

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

\(\Big \downarrow \) 2916

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

\(\Big \downarrow \) 2009

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

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

3.3.35.3.1 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]
 
3.3.35.4 Maple [C] (verified)

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

Time = 1.37 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.45

method result size
parts \(\frac {x \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{e}-\frac {d \ln \left (e x +d \right ) \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{e^{2}}-\frac {3 p b \left (\frac {\left (e x +d \right ) e}{b}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 b d \,\textit {\_Z}^{2}+3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\frac {\ln \left (e x -\textit {\_R} +d \right )}{\textit {\_R}^{2}-2 \textit {\_R} d +d^{2}}\right ) a \,e^{4}}{3 b^{2}}-\frac {d e \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 b d \,\textit {\_Z}^{2}+3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )\right )}{3 b}\right )}{e^{3}}\) \(206\)
risch \(\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right ) x}{e}-\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right ) d \ln \left (e x +d \right )}{e^{2}}-\frac {3 p x}{e}-\frac {3 p d}{e^{2}}+\frac {p e \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 b d \,\textit {\_Z}^{2}+3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\frac {\ln \left (e x -\textit {\_R} +d \right )}{\textit {\_R}^{2}-2 \textit {\_R} d +d^{2}}\right ) a}{b}+\frac {p d \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 b d \,\textit {\_Z}^{2}+3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )\right )}{e^{2}}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {x}{e}-\frac {d \ln \left (e x +d \right )}{e^{2}}\right )\) \(331\)

input
int(x*ln(c*(b*x^3+a)^p)/(e*x+d),x,method=_RETURNVERBOSE)
 
output
x*ln(c*(b*x^3+a)^p)/e-d*ln(e*x+d)*ln(c*(b*x^3+a)^p)/e^2-3*p*b/e^3*(1/b*(e* 
x+d)*e-1/3/b^2*sum(1/(_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^4-1/3*d*e/b*sum(ln(e*x+d)*ln((-e*x+_R1- 
d)/_R1)+dilog((-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)))
 
3.3.35.5 Fricas [F]

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

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

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

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

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

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

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

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

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

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