Integrand size = 20, antiderivative size = 250 \[ \int (d+e x)^2 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-3 d^2 p x-\frac {3}{2} d e p x^2-\frac {1}{3} e^2 p x^3-\frac {\sqrt {3} \sqrt [3]{a} d \left (\sqrt [3]{b} d+\sqrt [3]{a} e\right ) p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{b^{2/3}}+\frac {\sqrt [3]{a} d \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}-\frac {\sqrt [3]{a} d \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 b^{2/3}}-\frac {\left (b d^3-a e^3\right ) p \log \left (a+b x^3\right )}{3 b e}+\frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e} \] Output:
-3*d^2*p*x-3/2*d*e*p*x^2-1/3*e^2*p*x^3-3^(1/2)*a^(1/3)*d*(b^(1/3)*d+a^(1/3 )*e)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)*3^(1/2)/a^(1/3))/b^(2/3)+a^(1/3)*d *(b^(1/3)*d-a^(1/3)*e)*p*ln(a^(1/3)+b^(1/3)*x)/b^(2/3)-1/2*a^(1/3)*d*(b^(1 /3)*d-a^(1/3)*e)*p*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(2/3)-1/3*( -a*e^3+b*d^3)*p*ln(b*x^3+a)/b/e+1/3*(e*x+d)^3*ln(c*(b*x^3+a)^p)/e
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.33 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.92 \[ \int (d+e x)^2 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {-\frac {p \left (18 b d^2 e x+9 b d e^2 x^2+2 b e^3 x^3+6 \sqrt {3} \sqrt [3]{a} b^{2/3} d^2 e \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )-9 b d e^2 x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b x^3}{a}\right )-6 \sqrt [3]{a} b^{2/3} d^2 e \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+3 \sqrt [3]{a} b^{2/3} d^2 e \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+2 \left (b d^3-a e^3\right ) \log \left (a+b x^3\right )\right )}{2 b}+(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e} \] Input:
Integrate[(d + e*x)^2*Log[c*(a + b*x^3)^p],x]
Output:
(-1/2*(p*(18*b*d^2*e*x + 9*b*d*e^2*x^2 + 2*b*e^3*x^3 + 6*Sqrt[3]*a^(1/3)*b ^(2/3)*d^2*e*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 9*b*d*e^2*x^2*H ypergeometric2F1[2/3, 1, 5/3, -((b*x^3)/a)] - 6*a^(1/3)*b^(2/3)*d^2*e*Log[ a^(1/3) + b^(1/3)*x] + 3*a^(1/3)*b^(2/3)*d^2*e*Log[a^(2/3) - a^(1/3)*b^(1/ 3)*x + b^(2/3)*x^2] + 2*(b*d^3 - a*e^3)*Log[a + b*x^3]))/b + (d + e*x)^3*L og[c*(a + b*x^3)^p])/(3*e)
Time = 1.20 (sec) , antiderivative size = 262, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2913, 2375, 27, 2426, 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.
| \(\displaystyle \int (d+e x)^2 \log \left (c \left (a+b x^3\right )^p\right ) \, dx\) |
| \(\Big \downarrow \) 2913 |
| \(\displaystyle \frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}-\frac {b p \int \frac {x^2 (d+e x)^3}{b x^3+a}dx}{e}\) |
| \(\Big \downarrow \) 2375 |
| \(\displaystyle \frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}-\frac {b p \left (\frac {\int \frac {3 x^2 \left (b d^3+3 b e x d^2+3 b e^2 x^2 d-a e^3\right )}{b x^3+a}dx}{3 b}+\frac {e^3 x^3}{3 b}\right )}{e}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}-\frac {b p \left (\frac {\int \frac {x^2 \left (b d^3+3 b e x d^2+3 b e^2 x^2 d-a e^3\right )}{b x^3+a}dx}{b}+\frac {e^3 x^3}{3 b}\right )}{e}\) |
| \(\Big \downarrow \) 2426 |
| \(\displaystyle \frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}-\frac {b p \left (\frac {\int \left (3 e d^2+3 e^2 x d-\frac {3 a e d^2+3 a e^2 x d-\left (b d^3-a e^3\right ) x^2}{b x^3+a}\right )dx}{b}+\frac {e^3 x^3}{3 b}\right )}{e}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}-\frac {b p \left (\frac {\frac {\sqrt [3]{a} d e \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 b^{2/3}}+\frac {\sqrt {3} \sqrt [3]{a} d e \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (\sqrt [3]{a} e+\sqrt [3]{b} d\right )}{b^{2/3}}-\frac {\sqrt [3]{a} d e \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}+\frac {\left (b d^3-a e^3\right ) \log \left (a+b x^3\right )}{3 b}+3 d^2 e x+\frac {3}{2} d e^2 x^2}{b}+\frac {e^3 x^3}{3 b}\right )}{e}\) |
Input:
Int[(d + e*x)^2*Log[c*(a + b*x^3)^p],x]
Output:
-((b*p*((e^3*x^3)/(3*b) + (3*d^2*e*x + (3*d*e^2*x^2)/2 + (Sqrt[3]*a^(1/3)* d*e*(b^(1/3)*d + a^(1/3)*e)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3 ))])/b^(2/3) - (a^(1/3)*d*e*(b^(1/3)*d - a^(1/3)*e)*Log[a^(1/3) + b^(1/3)* x])/b^(2/3) + (a^(1/3)*d*e*(b^(1/3)*d - a^(1/3)*e)*Log[a^(2/3) - a^(1/3)*b ^(1/3)*x + b^(2/3)*x^2])/(2*b^(2/3)) + ((b*d^3 - a*e^3)*Log[a + b*x^3])/(3 *b))/b))/e) + ((d + e*x)^3*Log[c*(a + b*x^3)^p])/(3*e)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Wi th[{q = Expon[Pq, x]}, With[{Pqq = Coeff[Pq, x, q]}, Simp[Pqq*(c*x)^(m + q - n + 1)*((a + b*x^n)^(p + 1)/(b*c^(q - n + 1)*(m + q + n*p + 1))), x] + Si mp[1/(b*(m + q + n*p + 1)) Int[(c*x)^m*ExpandToSum[b*(m + q + n*p + 1)*(P q - Pqq*x^q) - a*Pqq*(m + q - n + 1)*x^(q - n), x]*(a + b*x^n)^p, x], x]] / ; NeQ[m + q + n*p + 1, 0] && q - n >= 0 && (IntegerQ[2*p] || IntegerQ[p + ( q + 1)/(2*n)])] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0]
Int[(Pq_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegrand[Pq/(a + b*x^n), x], x] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && IntegerQ[n]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_. )*(x_))^(r_.), x_Symbol] :> Simp[(f + g*x)^(r + 1)*((a + b*Log[c*(d + e*x^n )^p])/(g*(r + 1))), x] - Simp[b*e*n*(p/(g*(r + 1))) Int[x^(n - 1)*((f + g *x)^(r + 1)/(d + e*x^n)), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x ] && (IGtQ[r, 0] || RationalQ[n]) && NeQ[r, -1]
Time = 1.69 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.34
| method | result | size |
| parts | \(\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) e^{2} x^{3}}{3}+\ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) e d \,x^{2}+d^{2} x \ln \left (c \left (b \,x^{3}+a \right )^{p}\right )+\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) d^{3}}{3 e}-\frac {p b \left (\frac {e \left (\frac {1}{3} e^{2} x^{3}+\frac {3}{2} e \,x^{2} d +3 d^{2} x \right )}{b}+\frac {-3 e \,d^{2} a \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )-3 e^{2} d a \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+\frac {\left (-a \,e^{3}+b \,d^{3}\right ) \ln \left (b \,x^{3}+a \right )}{3 b}}{b}\right )}{e}\) | \(336\) |
| risch | \(\frac {\left (e x +d \right )^{3} \ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{3 e}-\frac {i e^{2} \pi \,x^{3} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{6}-\frac {i x \pi \,d^{2} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{2}+\frac {i e \pi d \,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 )}^{2}}{2}-\frac {i e \pi d \,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 )}{2}+\frac {i x \pi \,d^{2} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}-\frac {i e^{2} \pi \,x^{3} \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 )}{6}+\frac {i e \pi d \,x^{2} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {i e^{2} \pi \,x^{3} \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}}{6}+\frac {i e^{2} \pi \,x^{3} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{6}-\frac {i e \pi d \,x^{2} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{2}+\frac {i x \pi \,d^{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 )}^{2}}{2}-\frac {i x \pi \,d^{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 )}{2}+\frac {\ln \left (c \right ) e^{2} x^{3}}{3}-\frac {e^{2} p \,x^{3}}{3}+\ln \left (c \right ) d e \,x^{2}-\frac {3 d e p \,x^{2}}{2}+x \ln \left (c \right ) d^{2}-3 d^{2} p x +\frac {p \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (\left (a \,e^{3}-b \,d^{3}\right ) \textit {\_R}^{2}+3 e^{2} d a \textit {\_R} +3 e \,d^{2} a \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{3 b e}\) | \(537\) |
Input:
int((e*x+d)^2*ln(c*(b*x^3+a)^p),x,method=_RETURNVERBOSE)
Output:
1/3*ln(c*(b*x^3+a)^p)*e^2*x^3+ln(c*(b*x^3+a)^p)*e*d*x^2+d^2*x*ln(c*(b*x^3+ a)^p)+1/3*ln(c*(b*x^3+a)^p)/e*d^3-p*b/e*(e/b*(1/3*e^2*x^3+3/2*e*x^2*d+3*d^ 2*x)+(-3*e*d^2*a*(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)*arct an(1/3*3^(1/2)*(2/(1/b*a)^(1/3)*x-1)))-3*e^2*d*a*(-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)))+1/ 3*(-a*e^3+b*d^3)/b*ln(b*x^3+a))/b)
Result contains complex when optimal does not.
Time = 2.29 (sec) , antiderivative size = 5799, normalized size of antiderivative = 23.20 \[ \int (d+e x)^2 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\text {Too large to display} \] Input:
integrate((e*x+d)^2*log(c*(b*x^3+a)^p),x, algorithm="fricas")
Output:
Too large to include
Time = 15.09 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.69 \[ \int (d+e x)^2 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=3 a d^{2} p \operatorname {RootSum} {\left (27 t^{3} a^{2} b - 1, \left ( t \mapsto t \log {\left (3 t a + x \right )} \right )\right )} + 3 a d e p \operatorname {RootSum} {\left (27 t^{3} a b^{2} + 1, \left ( t \mapsto t \log {\left (9 t^{2} a b + x \right )} \right )\right )} + \frac {a e^{2} p \left (\begin {cases} \frac {x^{3}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x^{3} \right )}}{b} & \text {otherwise} \end {cases}\right )}{3} - 3 d^{2} p x + d^{2} x \log {\left (c \left (a + b x^{3}\right )^{p} \right )} - \frac {3 d e p x^{2}}{2} + d e x^{2} \log {\left (c \left (a + b x^{3}\right )^{p} \right )} - \frac {e^{2} p x^{3}}{3} + \frac {e^{2} x^{3} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{3} \] Input:
integrate((e*x+d)**2*ln(c*(b*x**3+a)**p),x)
Output:
3*a*d**2*p*RootSum(27*_t**3*a**2*b - 1, Lambda(_t, _t*log(3*_t*a + x))) + 3*a*d*e*p*RootSum(27*_t**3*a*b**2 + 1, Lambda(_t, _t*log(9*_t**2*a*b + x)) ) + a*e**2*p*Piecewise((x**3/a, Eq(b, 0)), (log(a + b*x**3)/b, True))/3 - 3*d**2*p*x + d**2*x*log(c*(a + b*x**3)**p) - 3*d*e*p*x**2/2 + d*e*x**2*log (c*(a + b*x**3)**p) - e**2*p*x**3/3 + e**2*x**3*log(c*(a + b*x**3)**p)/3
Time = 0.12 (sec) , antiderivative size = 249, normalized size of antiderivative = 1.00 \[ \int (d+e x)^2 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {1}{6} \, b p {\left (\frac {2 \, e^{2} x^{3} + 9 \, d e x^{2} + 18 \, d^{2} x}{b} - \frac {6 \, \sqrt {3} {\left (a b d e \left (\frac {a}{b}\right )^{\frac {2}{3}} + a b d^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{2}} - \frac {{\left (2 \, a e^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} + 3 \, a d e \left (\frac {a}{b}\right )^{\frac {1}{3}} - 3 \, a d^{2}\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {2 \, {\left (a e^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} - 3 \, a d e \left (\frac {a}{b}\right )^{\frac {1}{3}} + 3 \, a d^{2}\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )} + \frac {1}{3} \, {\left (e^{2} x^{3} + 3 \, d e x^{2} + 3 \, d^{2} x\right )} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) \] Input:
integrate((e*x+d)^2*log(c*(b*x^3+a)^p),x, algorithm="maxima")
Output:
-1/6*b*p*((2*e^2*x^3 + 9*d*e*x^2 + 18*d^2*x)/b - 6*sqrt(3)*(a*b*d*e*(a/b)^ (2/3) + a*b*d^2*(a/b)^(1/3))*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^ (1/3))/(a*b^2) - (2*a*e^2*(a/b)^(2/3) + 3*a*d*e*(a/b)^(1/3) - 3*a*d^2)*log (x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^2*(a/b)^(2/3)) - 2*(a*e^2*(a/b)^(2/ 3) - 3*a*d*e*(a/b)^(1/3) + 3*a*d^2)*log(x + (a/b)^(1/3))/(b^2*(a/b)^(2/3)) ) + 1/3*(e^2*x^3 + 3*d*e*x^2 + 3*d^2*x)*log((b*x^3 + a)^p*c)
Time = 0.15 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.10 \[ \int (d+e x)^2 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {1}{3} \, {\left (e^{2} p - e^{2} \log \left (c\right )\right )} x^{3} + \frac {a e^{2} p \log \left ({\left | b x^{3} + a \right |}\right )}{3 \, b} - \frac {1}{2} \, {\left (3 \, d e p - 2 \, d e \log \left (c\right )\right )} x^{2} - {\left (3 \, d^{2} p - d^{2} \log \left (c\right )\right )} x + \frac {1}{3} \, {\left (e^{2} p x^{3} + 3 \, d e p x^{2} + 3 \, d^{2} p x\right )} \log \left (b x^{3} + a\right ) + \frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {1}{3}} b d^{2} p - \left (-a b^{2}\right )^{\frac {2}{3}} d e p\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{b^{2}} - \frac {{\left (a b d e p \left (-\frac {a}{b}\right )^{\frac {1}{3}} + a b d^{2} p\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{a b} + \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} b d^{2} p + \left (-a b^{2}\right )^{\frac {2}{3}} d e p\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{2 \, b^{2}} \] Input:
integrate((e*x+d)^2*log(c*(b*x^3+a)^p),x, algorithm="giac")
Output:
-1/3*(e^2*p - e^2*log(c))*x^3 + 1/3*a*e^2*p*log(abs(b*x^3 + a))/b - 1/2*(3 *d*e*p - 2*d*e*log(c))*x^2 - (3*d^2*p - d^2*log(c))*x + 1/3*(e^2*p*x^3 + 3 *d*e*p*x^2 + 3*d^2*p*x)*log(b*x^3 + a) + sqrt(3)*((-a*b^2)^(1/3)*b*d^2*p - (-a*b^2)^(2/3)*d*e*p)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3 ))/b^2 - (a*b*d*e*p*(-a/b)^(1/3) + a*b*d^2*p)*(-a/b)^(1/3)*log(abs(x - (-a /b)^(1/3)))/(a*b) + 1/2*((-a*b^2)^(1/3)*b*d^2*p + (-a*b^2)^(2/3)*d*e*p)*lo g(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/b^2
Time = 25.62 (sec) , antiderivative size = 358, normalized size of antiderivative = 1.43 \[ \int (d+e x)^2 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\left (\sum _{k=1}^3\ln \left (\mathrm {root}\left (27\,b^3\,c^3-27\,a\,b^2\,c^2\,e^2\,p+81\,a\,b^2\,c\,d^3\,e\,p^2+9\,a^2\,b\,c\,e^4\,p^2-27\,a\,b^2\,d^6\,p^3-a^3\,e^6\,p^3,c,k\right )\,\left (\mathrm {root}\left (27\,b^3\,c^3-27\,a\,b^2\,c^2\,e^2\,p+81\,a\,b^2\,c\,d^3\,e\,p^2+9\,a^2\,b\,c\,e^4\,p^2-27\,a\,b^2\,d^6\,p^3-a^3\,e^6\,p^3,c,k\right )\,a\,b^2\,9-6\,a^2\,b\,e^2\,p+9\,a\,b^2\,d^2\,p\,x\right )+a^3\,e^4\,p^2+9\,a^2\,b\,d^3\,e\,p^2+6\,a^2\,b\,d^2\,e^2\,p^2\,x\right )\,\mathrm {root}\left (27\,b^3\,c^3-27\,a\,b^2\,c^2\,e^2\,p+81\,a\,b^2\,c\,d^3\,e\,p^2+9\,a^2\,b\,c\,e^4\,p^2-27\,a\,b^2\,d^6\,p^3-a^3\,e^6\,p^3,c,k\right )\right )+\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )\,\left (d^2\,x+d\,e\,x^2+\frac {e^2\,x^3}{3}\right )-3\,d^2\,p\,x-\frac {e^2\,p\,x^3}{3}-\frac {3\,d\,e\,p\,x^2}{2} \] Input:
int(log(c*(a + b*x^3)^p)*(d + e*x)^2,x)
Output:
symsum(log(root(27*b^3*c^3 - 27*a*b^2*c^2*e^2*p + 81*a*b^2*c*d^3*e*p^2 + 9 *a^2*b*c*e^4*p^2 - 27*a*b^2*d^6*p^3 - a^3*e^6*p^3, c, k)*(9*root(27*b^3*c^ 3 - 27*a*b^2*c^2*e^2*p + 81*a*b^2*c*d^3*e*p^2 + 9*a^2*b*c*e^4*p^2 - 27*a*b ^2*d^6*p^3 - a^3*e^6*p^3, c, k)*a*b^2 - 6*a^2*b*e^2*p + 9*a*b^2*d^2*p*x) + a^3*e^4*p^2 + 9*a^2*b*d^3*e*p^2 + 6*a^2*b*d^2*e^2*p^2*x)*root(27*b^3*c^3 - 27*a*b^2*c^2*e^2*p + 81*a*b^2*c*d^3*e*p^2 + 9*a^2*b*c*e^4*p^2 - 27*a*b^2 *d^6*p^3 - a^3*e^6*p^3, c, k), k, 1, 3) + log(c*(a + b*x^3)^p)*(d^2*x + (e ^2*x^3)/3 + d*e*x^2) - 3*d^2*p*x - (e^2*p*x^3)/3 - (3*d*e*p*x^2)/2
Time = 0.19 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.16 \[ \int (d+e x)^2 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {-6 b^{\frac {4}{3}} a^{\frac {2}{3}} \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) d^{2} p -6 \sqrt {3}\, \mathit {atan} \left (\frac {a^{\frac {1}{3}}-2 b^{\frac {1}{3}} x}{a^{\frac {1}{3}} \sqrt {3}}\right ) a b d e p +9 b^{\frac {4}{3}} a^{\frac {2}{3}} \mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) d^{2} p -3 b^{\frac {4}{3}} a^{\frac {2}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) d^{2}+2 b^{\frac {2}{3}} a^{\frac {4}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) e^{2}+6 b^{\frac {5}{3}} a^{\frac {1}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) d^{2} x +6 b^{\frac {5}{3}} a^{\frac {1}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) d e \,x^{2}+2 b^{\frac {5}{3}} a^{\frac {1}{3}} \mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) e^{2} x^{3}-18 b^{\frac {5}{3}} a^{\frac {1}{3}} d^{2} p x -9 b^{\frac {5}{3}} a^{\frac {1}{3}} d e p \,x^{2}-2 b^{\frac {5}{3}} a^{\frac {1}{3}} e^{2} p \,x^{3}-9 \,\mathrm {log}\left (a^{\frac {1}{3}}+b^{\frac {1}{3}} x \right ) a b d e p +3 \,\mathrm {log}\left (\left (b \,x^{3}+a \right )^{p} c \right ) a b d e}{6 b^{\frac {5}{3}} a^{\frac {1}{3}}} \] Input:
int((e*x+d)^2*log(c*(b*x^3+a)^p),x)
Output:
( - 6*b**(1/3)*a**(2/3)*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*s qrt(3)))*b*d**2*p - 6*sqrt(3)*atan((a**(1/3) - 2*b**(1/3)*x)/(a**(1/3)*sqr t(3)))*a*b*d*e*p + 9*b**(1/3)*a**(2/3)*log(a**(1/3) + b**(1/3)*x)*b*d**2*p - 3*b**(1/3)*a**(2/3)*log((a + b*x**3)**p*c)*b*d**2 + 2*b**(2/3)*a**(1/3) *log((a + b*x**3)**p*c)*a*e**2 + 6*b**(2/3)*a**(1/3)*log((a + b*x**3)**p*c )*b*d**2*x + 6*b**(2/3)*a**(1/3)*log((a + b*x**3)**p*c)*b*d*e*x**2 + 2*b** (2/3)*a**(1/3)*log((a + b*x**3)**p*c)*b*e**2*x**3 - 18*b**(2/3)*a**(1/3)*b *d**2*p*x - 9*b**(2/3)*a**(1/3)*b*d*e*p*x**2 - 2*b**(2/3)*a**(1/3)*b*e**2* p*x**3 - 9*log(a**(1/3) + b**(1/3)*x)*a*b*d*e*p + 3*log((a + b*x**3)**p*c) *a*b*d*e)/(6*b**(2/3)*a**(1/3)*b)