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-3^(1/2)*a^(1/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3 ))/b^(1/3)/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
Time = 0.33 (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)
Time = 1.30 (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
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]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 1.80 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.45
| method | result | size |
| parts | \(\frac {x \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{e}-\frac {d \ln \left (e x +d \right ) \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{e^{2}}-\frac {3 p b \left (\frac {\left (e x +d \right ) e}{b}-\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 \textit {\_Z}^{2} b d +3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\frac {\ln \left (e x -\textit {\_R} +d \right )}{\textit {\_R}^{2}-2 \textit {\_R} d +d^{2}}\right ) a \,e^{4}}{3 b^{2}}-\frac {d e \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 \textit {\_Z}^{2} b d +3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )\right )}{3 b}\right )}{e^{3}}\) | \(206\) |
| risch | \(\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right ) x}{e}-\frac {\ln \left (\left (b \,x^{3}+a \right )^{p}\right ) d \ln \left (e x +d \right )}{e^{2}}-\frac {3 p x}{e}-\frac {3 p d}{e^{2}}+\frac {p e \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 \textit {\_Z}^{2} b d +3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\frac {\ln \left (e x -\textit {\_R} +d \right )}{\textit {\_R}^{2}-2 \textit {\_R} d +d^{2}}\right ) a}{b}+\frac {p d \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b -3 \textit {\_Z}^{2} b d +3 b \,d^{2} \textit {\_Z} +a \,e^{3}-b \,d^{3}\right )}{\sum }\left (\ln \left (e x +d \right ) \ln \left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-e x +\textit {\_R1} -d}{\textit {\_R1}}\right )\right )\right )}{e^{2}}+\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \left (\frac {x}{e}-\frac {d \ln \left (e x +d \right )}{e^{2}}\right )\) | \(331\) |
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)))
\[ \int \frac {x \log \left (c \left (a+b x^3\right )^p\right )}{d+e x} \, dx=\int { \frac {x \log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{e x + d} \,d x } \] 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)
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
\[ \int \frac {x \log \left (c \left (a+b x^3\right )^p\right )}{d+e x} \, dx=\int { \frac {x \log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{e x + d} \,d x } \] 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)
\[ \int \frac {x \log \left (c \left (a+b x^3\right )^p\right )}{d+e x} \, dx=\int { \frac {x \log \left ({\left (b x^{3} + a\right )}^{p} c\right )}{e x + d} \,d x } \] 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)
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)
\[ \int \frac {x \log \left (c \left (a+b x^3\right )^p\right )}{d+e x} \, dx=\frac {-6 a^{\frac {1}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) e \,p^{2}+9 a^{\frac {1}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) e \,p^{2}-3 a^{\frac {1}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) e p -6 b^{\frac {1}{3}} \left (\int \frac {\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right )}{b e \,x^{4}+b d \,x^{3}+a e x +a d}d x \right ) a d e p +6 b^{\frac {4}{3}} \left (\int \frac {\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) x^{2}}{b e \,x^{4}+b d \,x^{3}+a e x +a d}d x \right ) d^{2} p -b^{\frac {1}{3}} {\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right )}^{2} d +6 b^{\frac {1}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) e p x -18 b^{\frac {1}{3}} e \,p^{2} x}{6 b^{\frac {1}{3}} e^{2} p} \] Input:
int(x*log(c*(b*x^3+a)^p)/(e*x+d),x)
Output:
( - 6*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))* e*p**2 + 9*a**(1/3)*log(a**(1/3) + b**(1/3)*x)*e*p**2 - 3*a**(1/3)*log((a + b*x**3)**p*c)*e*p - 6*b**(1/3)*int(log((a + b*x**3)**p*c)/(a*d + a*e*x + b*d*x**3 + b*e*x**4),x)*a*d*e*p + 6*b**(1/3)*int((log((a + b*x**3)**p*c)* x**2)/(a*d + a*e*x + b*d*x**3 + b*e*x**4),x)*b*d**2*p - b**(1/3)*log((a + b*x**3)**p*c)**2*d + 6*b**(1/3)*log((a + b*x**3)**p*c)*e*p*x - 18*b**(1/3) *e*p**2*x)/(6*b**(1/3)*e**2*p)