Integrand size = 14, antiderivative size = 147 \[ \int x \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {3 p x^2}{4}-\frac {\sqrt {3} a^{2/3} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{2 b^{2/3}}-\frac {a^{2/3} p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{2 b^{2/3}}+\frac {a^{2/3} p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{4 b^{2/3}}+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right ) \] Output:
-3/4*p*x^2-1/2*3^(1/2)*a^(2/3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/ a^(1/3))/b^(2/3)-1/2*a^(2/3)*p*ln(a^(1/3)+b^(1/3)*x)/b^(2/3)+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)+1/2*x^2*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 = 53, normalized size of antiderivative = 0.36 \[ \int x \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {3 p x^2}{4}+\frac {3}{4} p x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b x^3}{a}\right )+\frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right ) \] Input:
Integrate[x*Log[c*(a + b*x^3)^p],x]
Output:
(-3*p*x^2)/4 + (3*p*x^2*Hypergeometric2F1[2/3, 1, 5/3, -((b*x^3)/a)])/4 + (x^2*Log[c*(a + b*x^3)^p])/2
Time = 0.57 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.08, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {2905, 843, 821, 16, 1142, 25, 27, 1082, 217, 1103}
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.
| \(\displaystyle \int x \log \left (c \left (a+b x^3\right )^p\right ) \, dx\) |
| \(\Big \downarrow \) 2905 |
| \(\displaystyle \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{2} b p \int \frac {x^4}{b x^3+a}dx\) |
| \(\Big \downarrow \) 843 |
| \(\displaystyle \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{2} b p \left (\frac {x^2}{2 b}-\frac {a \int \frac {x}{b x^3+a}dx}{b}\right )\) |
| \(\Big \downarrow \) 821 |
| \(\displaystyle \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{2} b p \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\int \frac {1}{\sqrt [3]{b} x+\sqrt [3]{a}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 16 |
| \(\displaystyle \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{2} b p \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\int \frac {\sqrt [3]{b} x+\sqrt [3]{a}}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 1142 |
| \(\displaystyle \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{2} b p \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx+\frac {\int -\frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{2} b p \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\int \frac {\sqrt [3]{b} \left (\sqrt [3]{a}-2 \sqrt [3]{b} x\right )}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{2 \sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{2} b p \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3}{2} \sqrt [3]{a} \int \frac {1}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 1082 |
| \(\displaystyle \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{2} b p \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {3 \int \frac {1}{-\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )^2-3}d\left (1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{\sqrt [3]{b}}-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{2} b p \left (\frac {x^2}{2 b}-\frac {a \left (\frac {-\frac {1}{2} \int \frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{b^{2/3} x^2-\sqrt [3]{a} \sqrt [3]{b} x+a^{2/3}}dx-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )\) |
| \(\Big \downarrow \) 1103 |
| \(\displaystyle \frac {1}{2} x^2 \log \left (c \left (a+b x^3\right )^p\right )-\frac {3}{2} b p \left (\frac {x^2}{2 b}-\frac {a \left (\frac {\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 \sqrt [3]{b}}-\frac {\sqrt {3} \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{3 \sqrt [3]{a} \sqrt [3]{b}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}\right )}{b}\right )\) |
Input:
Int[x*Log[c*(a + b*x^3)^p],x]
Output:
(-3*b*p*(x^2/(2*b) - (a*(-1/3*Log[a^(1/3) + b^(1/3)*x]/(a^(1/3)*b^(2/3)) + (-((Sqrt[3]*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3)) + Log[a ^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2]/(2*b^(1/3)))/(3*a^(1/3)*b^(1/3)) ))/b))/2 + (x^2*Log[c*(a + b*x^3)^p])/2
Int[(c_.)/((a_.) + (b_.)*(x_)), x_Symbol] :> Simp[c*(Log[RemoveContent[a + b*x, x]]/b), x] /; FreeQ[{a, b, c}, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> Simp[-(3*Rt[a, 3]*Rt[b, 3])^(- 1) Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Simp[1/(3*Rt[a, 3]*Rt[b, 3]) Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2 *x^2), x], x] /; FreeQ[{a, b}, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*S implify[a*(c/b^2)]}, Simp[-2/b Subst[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b )], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /; Fre eQ[{a, b, c}, x]
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[d*(Log[RemoveContent[a + b*x + c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> S imp[(2*c*d - b*e)/(2*c) Int[1/(a + b*x + c*x^2), x], x] + Simp[e/(2*c) Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x]
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 = 0.74 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.87
| method | result | size |
| parts | \(\frac {x^{2} \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )}{2}-\frac {3 p b \left (\frac {x^{2}}{2 b}-\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) a}{b}\right )}{2}\) | \(128\) |
| risch | \(\frac {x^{2} \ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{2}+\frac {i {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) x^{2} \pi }{4}-\frac {i \pi \,x^{2} \operatorname {csgn}\left (i \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right ) \operatorname {csgn}\left (i c \right )}{4}-\frac {i \pi \,x^{2} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{4}+\frac {i \operatorname {csgn}\left (i c \right ) {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} x^{2} \pi }{4}+\frac {\ln \left (c \right ) x^{2}}{2}-\frac {3 p \,x^{2}}{4}+\frac {a p \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (b \,\textit {\_Z}^{3}+a \right )}{\sum }\frac {\ln \left (x -\textit {\_R} \right )}{\textit {\_R}}\right )}{2 b}\) | \(184\) |
Input:
int(x*ln(c*(b*x^3+a)^p),x,method=_RETURNVERBOSE)
Output:
1/2*x^2*ln(c*(b*x^3+a)^p)-3/2*p*b*(1/2*x^2/b-(-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)))/b*a)
Time = 0.08 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.02 \[ \int x \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{2} \, p x^{2} \log \left (b x^{3} + a\right ) - \frac {3}{4} \, p x^{2} + \frac {1}{2} \, x^{2} \log \left (c\right ) + \frac {1}{2} \, \sqrt {3} p \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} + \sqrt {3} a}{3 \, a}\right ) - \frac {1}{4} \, p \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} - a \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + \frac {1}{2} \, p \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (-\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right ) \] Input:
integrate(x*log(c*(b*x^3+a)^p),x, algorithm="fricas")
Output:
1/2*p*x^2*log(b*x^3 + a) - 3/4*p*x^2 + 1/2*x^2*log(c) + 1/2*sqrt(3)*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) - 1/4*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)) + 1/2*p*(-a^2/b^2)^(1/3)*log(a*x + b*(-a^2/b^2)^(2/3))
Time = 56.03 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.21 \[ \int x \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\begin {cases} \frac {x^{2} \log {\left (0^{p} c \right )}}{2} & \text {for}\: a = 0 \wedge b = 0 \\\frac {x^{2} \log {\left (a^{p} c \right )}}{2} & \text {for}\: b = 0 \\- \frac {3 p x^{2}}{4} + \frac {x^{2} \log {\left (c \left (b x^{3}\right )^{p} \right )}}{2} & \text {for}\: a = 0 \\- \frac {3 p x^{2}}{4} + \frac {3 p \left (- \frac {a}{b}\right )^{\frac {2}{3}} \log {\left (4 x^{2} + 4 x \sqrt [3]{- \frac {a}{b}} + 4 \left (- \frac {a}{b}\right )^{\frac {2}{3}} \right )}}{4} - \frac {\sqrt {3} p \left (- \frac {a}{b}\right )^{\frac {2}{3}} \operatorname {atan}{\left (\frac {2 \sqrt {3} x}{3 \sqrt [3]{- \frac {a}{b}}} + \frac {\sqrt {3}}{3} \right )}}{2} + \frac {x^{2} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{2} - \frac {\left (- \frac {a}{b}\right )^{\frac {2}{3}} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{2} & \text {otherwise} \end {cases} \] Input:
integrate(x*ln(c*(b*x**3+a)**p),x)
Output:
Piecewise((x**2*log(0**p*c)/2, Eq(a, 0) & Eq(b, 0)), (x**2*log(a**p*c)/2, Eq(b, 0)), (-3*p*x**2/4 + x**2*log(c*(b*x**3)**p)/2, Eq(a, 0)), (-3*p*x**2 /4 + 3*p*(-a/b)**(2/3)*log(4*x**2 + 4*x*(-a/b)**(1/3) + 4*(-a/b)**(2/3))/4 - sqrt(3)*p*(-a/b)**(2/3)*atan(2*sqrt(3)*x/(3*(-a/b)**(1/3)) + sqrt(3)/3) /2 + x**2*log(c*(a + b*x**3)**p)/2 - (-a/b)**(2/3)*log(c*(a + b*x**3)**p)/ 2, True))
Time = 0.12 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.89 \[ \int x \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {1}{4} \, b p {\left (\frac {3 \, x^{2}}{b} - \frac {2 \, \sqrt {3} a \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {a \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {2 \, a \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )} + \frac {1}{2} \, x^{2} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) \] Input:
integrate(x*log(c*(b*x^3+a)^p),x, algorithm="maxima")
Output:
-1/4*b*p*(3*x^2/b - 2*sqrt(3)*a*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/ b)^(1/3))/(b^2*(a/b)^(1/3)) - a*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^ 2*(a/b)^(1/3)) + 2*a*log(x + (a/b)^(1/3))/(b^2*(a/b)^(1/3))) + 1/2*x^2*log ((b*x^3 + a)^p*c)
Time = 0.13 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.02 \[ \int x \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {1}{4} \, a b^{2} p {\left (\frac {2 \, \left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a b^{2}} + \frac {2 \, \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{4}} - \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{a b^{4}}\right )} + \frac {1}{2} \, p x^{2} \log \left (b x^{3} + a\right ) - \frac {1}{4} \, {\left (3 \, p - 2 \, \log \left (c\right )\right )} x^{2} \] Input:
integrate(x*log(c*(b*x^3+a)^p),x, algorithm="giac")
Output:
-1/4*a*b^2*p*(2*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^2) + 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^4) - (-a*b^2)^(2/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a*b^4)) + 1/2*p*x^2*log(b*x^3 + a) - 1/4*(3*p - 2*log(c))*x^2
Time = 28.55 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.82 \[ \int x \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {x^2\,\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )}{2}-\frac {3\,p\,x^2}{4}-\frac {a^{2/3}\,p\,\ln \left (b^{1/3}\,x+a^{1/3}\right )}{2\,b^{2/3}}-\frac {a^{2/3}\,p\,\ln \left (4\,b^{1/3}\,x-2\,a^{1/3}-\sqrt {3}\,a^{1/3}\,2{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,b^{2/3}}+\frac {a^{2/3}\,p\,\ln \left (4\,b^{1/3}\,x-2\,a^{1/3}+\sqrt {3}\,a^{1/3}\,2{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{2\,b^{2/3}} \] Input:
int(x*log(c*(a + b*x^3)^p),x)
Output:
(x^2*log(c*(a + b*x^3)^p))/2 - (3*p*x^2)/4 - (a^(2/3)*p*log(b^(1/3)*x + a^ (1/3)))/(2*b^(2/3)) - (a^(2/3)*p*log(4*b^(1/3)*x - 3^(1/2)*a^(1/3)*2i - 2* a^(1/3))*((3^(1/2)*1i)/2 - 1/2))/(2*b^(2/3)) + (a^(2/3)*p*log(3^(1/2)*a^(1 /3)*2i + 4*b^(1/3)*x - 2*a^(1/3))*((3^(1/2)*1i)/2 + 1/2))/(2*b^(2/3))
Time = 0.17 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.66 \[ \int x \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) a p +2 b^{\frac {2}{3}} a^{\frac {1}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) x^{2}-3 b^{\frac {2}{3}} a^{\frac {1}{3}} p \,x^{2}-3 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a p +\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) a}{4 b^{\frac {2}{3}} a^{\frac {1}{3}}} \] Input:
int(x*log(c*(b*x^3+a)^p),x)
Output:
( - 2*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqrt(3)))*a*p + 2*b **(2/3)*a**(1/3)*log((a + b*x**3)**p*c)*x**2 - 3*b**(2/3)*a**(1/3)*p*x**2 - 3*log(a**(1/3) + b**(1/3)*x)*a*p + log((a + b*x**3)**p*c)*a)/(4*b**(2/3) *a**(1/3))