Integrand size = 16, antiderivative size = 157 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {3 a p x}{4 b}-\frac {3 p x^4}{16}+\frac {\sqrt {3} a^{4/3} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{4 b^{4/3}}-\frac {a^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}+\frac {a^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 b^{4/3}}+\frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right ) \] Output:
3/4*a*p*x/b-3/16*p*x^4+1/4*3^(1/2)*a^(4/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3) *x)*3^(1/2)/a^(1/3))/b^(4/3)-1/4*a^(4/3)*p*ln(a^(1/3)+b^(1/3)*x)/b^(4/3)+1 /8*a^(4/3)*p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(4/3)+1/4*x^4*ln( c*(b*x^3+a)^p)
Time = 0.07 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.94 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {12 a \sqrt [3]{b} p x-3 b^{4/3} p x^4+4 \sqrt {3} a^{4/3} p \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-4 a^{4/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 a^{4/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+4 b^{4/3} x^4 \log \left (c \left (a+b x^3\right )^p\right )}{16 b^{4/3}} \] Input:
Integrate[x^3*Log[c*(a + b*x^3)^p],x]
Output:
(12*a*b^(1/3)*p*x - 3*b^(4/3)*p*x^4 + 4*Sqrt[3]*a^(4/3)*p*ArcTan[(1 - (2*b ^(1/3)*x)/a^(1/3))/Sqrt[3]] - 4*a^(4/3)*p*Log[a^(1/3) + b^(1/3)*x] + 2*a^( 4/3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2] + 4*b^(4/3)*x^4*Log[ c*(a + b*x^3)^p])/(16*b^(4/3))
Time = 0.49 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.01, 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.
| \(\displaystyle \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx\) |
| \(\Big \downarrow \) 2905 |
| \(\displaystyle \frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{4} b p \int \frac {x^6}{b x^3+a}dx\) |
| \(\Big \downarrow \) 831 |
| \(\displaystyle \frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{4} b p \int \left (\frac {x^3}{b}+\frac {a^2}{b^2 \left (b x^3+a\right )}-\frac {a}{b^2}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{4} x^4 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{4} b p \left (-\frac {a^{4/3} \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{7/3}}-\frac {a^{4/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{7/3}}+\frac {a^{4/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{7/3}}-\frac {a x}{b^2}+\frac {x^4}{4 b}\right )\) |
Input:
Int[x^3*Log[c*(a + b*x^3)^p],x]
Output:
(-3*b*p*(-((a*x)/b^2) + x^4/(4*b) - (a^(4/3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x )/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(7/3)) + (a^(4/3)*Log[a^(1/3) + b^(1/3)*x ])/(3*b^(7/3)) - (a^(4/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/ (6*b^(7/3))))/4 + (x^4*Log[c*(a + b*x^3)^p])/4
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 = 0.98 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.87
| method | result | size |
| parts | \(\frac {x^{4} \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{4}-\frac {3 p b \left (-\frac {-\frac {1}{4} b \,x^{4}+a x}{b^{2}}+\frac {\left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right ) a^{2}}{b^{2}}\right )}{4}\) | \(136\) |
| risch | \(\frac {x^{4} \ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{4}+\frac {i \pi \,x^{4} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2}}{8}-\frac {i \pi \,x^{4} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{8}-\frac {i \pi \,x^{4} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{8}+\frac {i \pi \,x^{4} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{8}+\frac {\ln \left (c \right ) x^{4}}{4}-\frac {3 p \,x^{4}}{16}-\frac {a^{2} p \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{4 b^{2}}+\frac {3 a p x}{4 b}\) | \(194\) |
Input:
int(x^3*ln(c*(b*x^3+a)^p),x,method=_RETURNVERBOSE)
Output:
1/4*x^4*ln(c*(b*x^3+a)^p)-3/4*p*b*(-1/b^2*(-1/4*b*x^4+a*x)+(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)))*a^2/b^2)
Time = 0.09 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.92 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {4 \, b p x^{4} \log \left (b x^{3} + a\right ) - 3 \, b p x^{4} + 4 \, b x^{4} \log \left (c\right ) + 4 \, \sqrt {3} a p \left (-\frac {a}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a}{b}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) - 2 \, a p \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right ) + 4 \, a p \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x - \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right ) + 12 \, a p x}{16 \, b} \] Input:
integrate(x^3*log(c*(b*x^3+a)^p),x, algorithm="fricas")
Output:
1/16*(4*b*p*x^4*log(b*x^3 + a) - 3*b*p*x^4 + 4*b*x^4*log(c) + 4*sqrt(3)*a* p*(-a/b)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*(-a/b)^(2/3) - sqrt(3)*a)/a) - 2* a*p*(-a/b)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3)) + 4*a*p*(-a/b)^( 1/3)*log(x - (-a/b)^(1/3)) + 12*a*p*x)/b
Timed out. \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\text {Timed out} \] Input:
integrate(x**3*ln(c*(b*x**3+a)**p),x)
Output:
Timed out
Time = 0.12 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.92 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{4} \, x^{4} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) - \frac {1}{16} \, b p {\left (\frac {4 \, \sqrt {3} a^{2} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {2 \, a^{2} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {4 \, a^{2} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {3 \, {\left (b x^{4} - 4 \, a x\right )}}{b^{2}}\right )} \] Input:
integrate(x^3*log(c*(b*x^3+a)^p),x, algorithm="maxima")
Output:
1/4*x^4*log((b*x^3 + a)^p*c) - 1/16*b*p*(4*sqrt(3)*a^2*arctan(1/3*sqrt(3)* (2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^3*(a/b)^(2/3)) - 2*a^2*log(x^2 - x*(a/ b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(2/3)) + 4*a^2*log(x + (a/b)^(1/3))/(b^ 3*(a/b)^(2/3)) + 3*(b*x^4 - 4*a*x)/b^2)
Time = 0.12 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.02 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{8} \, a^{2} b^{3} p {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a b^{4}} - \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{5}} - \frac {\left (-a b^{2}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b^{5}}\right )} + \frac {1}{4} \, p x^{4} \log \left (b x^{3} + a\right ) - \frac {1}{16} \, {\left (3 \, p - 4 \, \log \left (c\right )\right )} x^{4} + \frac {3 \, a p x}{4 \, b} \] Input:
integrate(x^3*log(c*(b*x^3+a)^p),x, algorithm="giac")
Output:
1/8*a^2*b^3*p*(2*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^4) - 2*sqrt( 3)*(-a*b^2)^(1/3)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a *b^5) - (-a*b^2)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a*b^5)) + 1/4*p*x^4*log(b*x^3 + a) - 1/16*(3*p - 4*log(c))*x^4 + 3/4*a*p*x/b
Time = 28.73 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.82 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {x^4\,\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{4}-\frac {3\,p\,x^4}{16}+\frac {3\,a\,p\,x}{4\,b}-\frac {a^{4/3}\,p\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{4\,b^{4/3}}+\frac {a^{4/3}\,p\,\ln \left (2\,b^{1/3}\,x-a^{1/3}-\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,b^{4/3}}-\frac {a^{4/3}\,p\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{4\,b^{4/3}} \] Input:
int(x^3*log(c*(a + b*x^3)^p),x)
Output:
(x^4*log(c*(a + b*x^3)^p))/4 - (3*p*x^4)/16 + (3*a*p*x)/(4*b) - (a^(4/3)*p *log(b^(1/3)*x + a^(1/3)))/(4*b^(4/3)) + (a^(4/3)*p*log(2*b^(1/3)*x - 3^(1 /2)*a^(1/3)*1i - a^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(4*b^(4/3)) - (a^(4/3)*p *log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 - 1/2))/( 4*b^(4/3))
Time = 0.18 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.66 \[ \int x^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {4 a^{\frac {4}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) p -6 a^{\frac {4}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) p +2 a^{\frac {4}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right )+4 b^{\frac {4}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) x^{4}+12 b^{\frac {1}{3}} a p x -3 b^{\frac {4}{3}} p \,x^{4}}{16 b^{\frac {4}{3}}} \] Input:
int(x^3*log(c*(b*x^3+a)^p),x)
Output:
(4*a**(1/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*p - 6*a**(1/3)*log(a**(1/3) + b**(1/3)*x)*a*p + 2*a**(1/3)*log((a + b*x**3) **p*c)*a + 4*b**(1/3)*log((a + b*x**3)**p*c)*b*x**4 + 12*b**(1/3)*a*p*x - 3*b**(1/3)*b*p*x**4)/(16*b**(1/3)*b)