Integrand size = 16, antiderivative size = 159 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}+\frac {\sqrt {3} a^{5/3} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{5 b^{5/3}}+\frac {a^{5/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{5 b^{5/3}}-\frac {a^{5/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{10 b^{5/3}}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right ) \] Output:
3/10*a*p*x^2/b-3/25*p*x^5+1/5*3^(1/2)*a^(5/3)*p*arctan(1/3*(a^(1/3)-2*b^(1 /3)*x)*3^(1/2)/a^(1/3))/b^(5/3)+1/5*a^(5/3)*p*ln(a^(1/3)+b^(1/3)*x)/b^(5/3 )-1/10*a^(5/3)*p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(5/3)+1/5*x^5 *ln(c*(b*x^3+a)^p)
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.01 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.43 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {3 a p x^2}{10 b}-\frac {3 p x^5}{25}-\frac {3 a p x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b x^3}{a}\right )}{10 b}+\frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right ) \] Input:
Integrate[x^4*Log[c*(a + b*x^3)^p],x]
Output:
(3*a*p*x^2)/(10*b) - (3*p*x^5)/25 - (3*a*p*x^2*Hypergeometric2F1[2/3, 1, 5 /3, -((b*x^3)/a)])/(10*b) + (x^5*Log[c*(a + b*x^3)^p])/5
Time = 0.54 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.02, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {2905, 831, 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.
| \(\displaystyle \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx\) |
| \(\Big \downarrow \) 2905 |
| \(\displaystyle \frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{5} b p \int \frac {x^7}{b x^3+a}dx\) |
| \(\Big \downarrow \) 831 |
| \(\displaystyle \frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{5} b p \int \left (\frac {x^4}{b}+\frac {a^2 x}{b^2 \left (b x^3+a\right )}-\frac {a x}{b^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{5} x^5 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{5} b p \left (-\frac {a^{5/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{8/3}}+\frac {a^{5/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{8/3}}-\frac {a^{5/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{8/3}}-\frac {a x^2}{2 b^2}+\frac {x^5}{5 b}\right )\) |
Input:
Int[x^4*Log[c*(a + b*x^3)^p],x]
Output:
(-3*b*p*(-1/2*(a*x^2)/b^2 + x^5/(5*b) - (a^(5/3)*ArcTan[(a^(1/3) - 2*b^(1/ 3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(8/3)) - (a^(5/3)*Log[a^(1/3) + b^(1/ 3)*x])/(3*b^(8/3)) + (a^(5/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^ 2])/(6*b^(8/3))))/5 + (x^5*Log[c*(a + b*x^3)^p])/5
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x ^m, a + b*x^n, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && Gt Q[m, 2*n - 1]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.)*(x_))^ (m_.), x_Symbol] :> Simp[(f*x)^(m + 1)*((a + b*Log[c*(d + e*x^n)^p])/(f*(m + 1))), x] - Simp[b*e*n*(p/(f*(m + 1))) Int[x^(n - 1)*((f*x)^(m + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && NeQ[m, -1]
Time = 1.21 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.87
| method | result | size |
| parts | \(\frac {x^{5} \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{5}-\frac {3 p b \left (-\frac {-\frac {1}{5} b \,x^{5}+\frac {1}{2} a \,x^{2}}{b^{2}}+\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) a^{2}}{b^{2}}\right )}{5}\) | \(139\) |
| risch | \(\frac {x^{5} \ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{5}+\frac {i \pi \,x^{5} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}}{10}-\frac {i \pi \,x^{5} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{10}-\frac {i \pi \,x^{5} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{10}+\frac {i \pi \,x^{5} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{10}+\frac {\ln \left (c \right ) x^{5}}{5}-\frac {3 p \,x^{5}}{25}+\frac {3 a p \,x^{2}}{10 b}-\frac {a^{2} p \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{5 b^{2}}\) | \(196\) |
Input:
int(x^4*ln(c*(b*x^3+a)^p),x,method=_RETURNVERBOSE)
Output:
1/5*x^5*ln(c*(b*x^3+a)^p)-3/5*p*b*(-1/b^2*(-1/5*b*x^5+1/2*a*x^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)))*a^2/b^2)
Time = 0.08 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.01 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {10 \, b p x^{5} \log \left (b x^{3} + a\right ) - 6 \, b p x^{5} + 10 \, b x^{5} \log \left (c\right ) + 15 \, a p x^{2} - 10 \, \sqrt {3} a p \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) - 5 \, a p \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + 10 \, a p \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{50 \, b} \] Input:
integrate(x^4*log(c*(b*x^3+a)^p),x, algorithm="fricas")
Output:
1/50*(10*b*p*x^5*log(b*x^3 + a) - 6*b*p*x^5 + 10*b*x^5*log(c) + 15*a*p*x^2 - 10*sqrt(3)*a*p*(a^2/b^2)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*(a^2/b^2)^(1/3 ) - sqrt(3)*a)/a) - 5*a*p*(a^2/b^2)^(1/3)*log(a*x^2 - b*x*(a^2/b^2)^(2/3) + a*(a^2/b^2)^(1/3)) + 10*a*p*(a^2/b^2)^(1/3)*log(a*x + b*(a^2/b^2)^(2/3)) )/b
Timed out. \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\text {Timed out} \] Input:
integrate(x**4*ln(c*(b*x**3+a)**p),x)
Output:
Timed out
Time = 0.11 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.92 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{5} \, x^{5} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) - \frac {1}{50} \, b p {\left (\frac {10 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {5 \, a^{2} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {10 \, a^{2} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {3 \, {\left (2 \, b x^{5} - 5 \, a x^{2}\right )}}{b^{2}}\right )} \] Input:
integrate(x^4*log(c*(b*x^3+a)^p),x, algorithm="maxima")
Output:
1/5*x^5*log((b*x^3 + a)^p*c) - 1/50*b*p*(10*sqrt(3)*a^2*arctan(1/3*sqrt(3) *(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^3*(a/b)^(1/3)) + 5*a^2*log(x^2 - x*(a /b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(1/3)) - 10*a^2*log(x + (a/b)^(1/3))/( b^3*(a/b)^(1/3)) + 3*(2*b*x^5 - 5*a*x^2)/b^2)
Time = 0.14 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.02 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{10} \, a^{2} b^{4} p {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a b^{5}} + \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{7}} - \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b^{7}}\right )} + \frac {1}{5} \, p x^{5} \log \left (b x^{3} + a\right ) - \frac {1}{25} \, {\left (3 \, p - 5 \, \log \left (c\right )\right )} x^{5} + \frac {3 \, a p x^{2}}{10 \, b} \] Input:
integrate(x^4*log(c*(b*x^3+a)^p),x, algorithm="giac")
Output:
1/10*a^2*b^4*p*(2*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^5) + 2*sqrt (3)*(-a*b^2)^(2/3)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/( a*b^7) - (-a*b^2)^(2/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a*b^7)) + 1/5*p*x^5*log(b*x^3 + a) - 1/25*(3*p - 5*log(c))*x^5 + 3/10*a*p*x^2/b
Time = 28.68 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.99 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {x^5\,\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{5}-\frac {3\,p\,x^5}{25}+\frac {a^{5/3}\,p\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{5\,b^{5/3}}+\frac {3\,a\,p\,x^2}{10\,b}+\frac {a^{5/3}\,p\,\ln \left (\frac {9\,a^4\,p^2\,x}{25\,b}+\frac {9\,a^{13/3}\,p^2\,{\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{25\,b^{4/3}}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{5\,b^{5/3}}-\frac {a^{5/3}\,p\,\ln \left (\frac {9\,a^4\,p^2\,x}{25\,b}+\frac {9\,a^{13/3}\,p^2\,{\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}^2}{25\,b^{4/3}}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{5\,b^{5/3}} \] Input:
int(x^4*log(c*(a + b*x^3)^p),x)
Output:
(x^5*log(c*(a + b*x^3)^p))/5 - (3*p*x^5)/25 + (a^(5/3)*p*log(b^(1/3)*x + a ^(1/3)))/(5*b^(5/3)) + (3*a*p*x^2)/(10*b) + (a^(5/3)*p*log((9*a^4*p^2*x)/( 25*b) + (9*a^(13/3)*p^2*((3^(1/2)*1i)/2 - 1/2)^2)/(25*b^(4/3)))*((3^(1/2)* 1i)/2 - 1/2))/(5*b^(5/3)) - (a^(5/3)*p*log((9*a^4*p^2*x)/(25*b) + (9*a^(13 /3)*p^2*((3^(1/2)*1i)/2 + 1/2)^2)/(25*b^(4/3)))*((3^(1/2)*1i)/2 + 1/2))/(5 *b^(5/3))
Time = 0.18 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.73 \[ \int x^4 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {10 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) a^{2} p +10 b^{\frac {5}{3}} a^{\frac {1}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) x^{5}+15 b^{\frac {2}{3}} a^{\frac {4}{3}} p \,x^{2}-6 b^{\frac {5}{3}} a^{\frac {1}{3}} p \,x^{5}+15 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a^{2} p -5 \,\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) a^{2}}{50 b^{\frac {5}{3}} a^{\frac {1}{3}}} \] Input:
int(x^4*log(c*(b*x^3+a)^p),x)
Output:
(10*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a**2*p + 10 *b**(2/3)*a**(1/3)*log((a + b*x**3)**p*c)*b*x**5 + 15*b**(2/3)*a**(1/3)*a* p*x**2 - 6*b**(2/3)*a**(1/3)*b*p*x**5 + 15*log(a**(1/3) + b**(1/3)*x)*a**2 *p - 5*log((a + b*x**3)**p*c)*a**2)/(50*b**(2/3)*a**(1/3)*b)