\(\int (d+e x)^3 \log (c (a+b x^3)^p) \, dx\) [191]

Optimal result

Integrand size = 20, antiderivative size = 320 \[ \int (d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {3 \left (4 b d^3-a e^3\right ) p x}{4 b}-\frac {9}{4} d^2 e p x^2-d e^2 p x^3-\frac {3}{16} e^3 p x^4-\frac {\sqrt {3} \sqrt [3]{a} \left (4 b d^3+6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{4 b^{4/3}}+\frac {\sqrt [3]{a} \left (4 b d^3-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}-\frac {\sqrt [3]{a} \left (4 b d^3-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 b^{4/3}}-\frac {d \left (b d^3-4 a e^3\right ) p \log \left (a+b x^3\right )}{4 b e}+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e} \]

[Out]

-3/4*(-a*e^3+4*b*d^3)*p*x/b-9/4*d^2*e*p*x^2-d*e^2*p*x^3-3/16*e^3*p*x^4+1/4*a^(1/3)*(4*b*d^3-6*a^(1/3)*b^(2/3)*
d^2*e-a*e^3)*p*ln(a^(1/3)+b^(1/3)*x)/b^(4/3)-1/8*a^(1/3)*(4*b*d^3-6*a^(1/3)*b^(2/3)*d^2*e-a*e^3)*p*ln(a^(2/3)-
a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(4/3)-1/4*d*(-4*a*e^3+b*d^3)*p*ln(b*x^3+a)/b/e+1/4*(e*x+d)^4*ln(c*(b*x^3+a)^p
)/e-1/4*a^(1/3)*(4*b*d^3+6*a^(1/3)*b^(2/3)*d^2*e-a*e^3)*p*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))*3^
(1/2)/b^(4/3)

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {2513, 1850, 1901, 1885, 1874, 31, 648, 631, 210, 642, 266} \[ \int (d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {\sqrt [3]{a} p \left (-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3+4 b d^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 b^{4/3}}-\frac {\sqrt {3} \sqrt [3]{a} p \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right ) \left (6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3+4 b d^3\right )}{4 b^{4/3}}+\frac {\sqrt [3]{a} p \left (-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3+4 b d^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e}-\frac {d p \left (b d^3-4 a e^3\right ) \log \left (a+b x^3\right )}{4 b e}-\frac {3 p x \left (4 b d^3-a e^3\right )}{4 b}-\frac {9}{4} d^2 e p x^2-d e^2 p x^3-\frac {3}{16} e^3 p x^4 \]

[In]

Int[(d + e*x)^3*Log[c*(a + b*x^3)^p],x]

[Out]

(-3*(4*b*d^3 - a*e^3)*p*x)/(4*b) - (9*d^2*e*p*x^2)/4 - d*e^2*p*x^3 - (3*e^3*p*x^4)/16 - (Sqrt[3]*a^(1/3)*(4*b*
d^3 + 6*a^(1/3)*b^(2/3)*d^2*e - a*e^3)*p*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(4*b^(4/3)) + (a^(
1/3)*(4*b*d^3 - 6*a^(1/3)*b^(2/3)*d^2*e - a*e^3)*p*Log[a^(1/3) + b^(1/3)*x])/(4*b^(4/3)) - (a^(1/3)*(4*b*d^3 -
6*a^(1/3)*b^(2/3)*d^2*e - a*e^3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(8*b^(4/3)) - (d*(b*d^3 -
4*a*e^3)*p*Log[a + b*x^3])/(4*b*e) + ((d + e*x)^4*Log[c*(a + b*x^3)^p])/(4*e)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 210

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])

Rule 266

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 631

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[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])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 642

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[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]

Rule 648

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

Rule 1850

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{q = Expon[Pq, x]}, With[{Pqq =
Coeff[Pq, x, q]}, Dist[1/(b*(m + q + n*p + 1)), Int[(c*x)^m*ExpandToSum[b*(m + q + n*p + 1)*(Pq - Pqq*x^q) - a
*Pqq*(m + q - n + 1)*x^(q - n), x]*(a + b*x^n)^p, x], x] + 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]] /; NeQ[m + q + n*p + 1, 0] && q - n >= 0 && (IntegerQ[2*p] || Integ
erQ[p + (q + 1)/(2*n)])] /; FreeQ[{a, b, c, m, p}, x] && PolyQ[Pq, x] && IGtQ[n, 0]

Rule 1874

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R
t[a/b, 3]]}, Dist[(-r)*((B*r - A*s)/(3*a*s)), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) +
s*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[
a/b]

Rule 1885

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,
2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] ||  !Ration
alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 1901

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]

Rule 2513

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e}-\frac {(3 b p) \int \frac {x^2 (d+e x)^4}{a+b x^3} \, dx}{4 e} \\ & = -\frac {3}{16} e^3 p x^4+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e}-\frac {(3 p) \int \frac {x^2 \left (4 b d^4+4 e \left (4 b d^3-a e^3\right ) x+24 b d^2 e^2 x^2+16 b d e^3 x^3\right )}{a+b x^3} \, dx}{16 e} \\ & = -d e^2 p x^3-\frac {3}{16} e^3 p x^4+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e}-\frac {p \int \frac {x^2 \left (12 b d \left (b d^3-4 a e^3\right )+12 b e \left (4 b d^3-a e^3\right ) x+72 b^2 d^2 e^2 x^2\right )}{a+b x^3} \, dx}{16 b e} \\ & = -d e^2 p x^3-\frac {3}{16} e^3 p x^4+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e}-\frac {p \int \left (12 e \left (4 b d^3-a e^3\right )+72 b d^2 e^2 x-\frac {12 \left (a e \left (4 b d^3-a e^3\right )+6 a b d^2 e^2 x-b d \left (b d^3-4 a e^3\right ) x^2\right )}{a+b x^3}\right ) \, dx}{16 b e} \\ & = -\frac {3 \left (4 b d^3-a e^3\right ) p x}{4 b}-\frac {9}{4} d^2 e p x^2-d e^2 p x^3-\frac {3}{16} e^3 p x^4+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e}+\frac {(3 p) \int \frac {a e \left (4 b d^3-a e^3\right )+6 a b d^2 e^2 x-b d \left (b d^3-4 a e^3\right ) x^2}{a+b x^3} \, dx}{4 b e} \\ & = -\frac {3 \left (4 b d^3-a e^3\right ) p x}{4 b}-\frac {9}{4} d^2 e p x^2-d e^2 p x^3-\frac {3}{16} e^3 p x^4+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e}+\frac {(3 p) \int \frac {a e \left (4 b d^3-a e^3\right )+6 a b d^2 e^2 x}{a+b x^3} \, dx}{4 b e}-\frac {\left (3 d \left (b d^3-4 a e^3\right ) p\right ) \int \frac {x^2}{a+b x^3} \, dx}{4 e} \\ & = -\frac {3 \left (4 b d^3-a e^3\right ) p x}{4 b}-\frac {9}{4} d^2 e p x^2-d e^2 p x^3-\frac {3}{16} e^3 p x^4-\frac {d \left (b d^3-4 a e^3\right ) p \log \left (a+b x^3\right )}{4 b e}+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e}+\frac {p \int \frac {\sqrt [3]{a} \left (6 a^{4/3} b d^2 e^2+2 a \sqrt [3]{b} e \left (4 b d^3-a e^3\right )\right )+\sqrt [3]{b} \left (6 a^{4/3} b d^2 e^2-a \sqrt [3]{b} e \left (4 b d^3-a e^3\right )\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{4 a^{2/3} b^{4/3} e}+\frac {\left (\sqrt [3]{a} \left (4 b d^3-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{4 b} \\ & = -\frac {3 \left (4 b d^3-a e^3\right ) p x}{4 b}-\frac {9}{4} d^2 e p x^2-d e^2 p x^3-\frac {3}{16} e^3 p x^4+\frac {\sqrt [3]{a} \left (4 b d^3-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}-\frac {d \left (b d^3-4 a e^3\right ) p \log \left (a+b x^3\right )}{4 b e}+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e}-\frac {\left (\sqrt [3]{a} \left (4 b d^3-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p\right ) \int \frac {-\sqrt [3]{a} \sqrt [3]{b}+2 b^{2/3} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{8 b^{4/3}}+\frac {\left (3 a^{2/3} \left (4 b d^3+6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{8 b} \\ & = -\frac {3 \left (4 b d^3-a e^3\right ) p x}{4 b}-\frac {9}{4} d^2 e p x^2-d e^2 p x^3-\frac {3}{16} e^3 p x^4+\frac {\sqrt [3]{a} \left (4 b d^3-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}-\frac {\sqrt [3]{a} \left (4 b d^3-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 b^{4/3}}-\frac {d \left (b d^3-4 a e^3\right ) p \log \left (a+b x^3\right )}{4 b e}+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e}+\frac {\left (3 \sqrt [3]{a} \left (4 b d^3+6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{4 b^{4/3}} \\ & = -\frac {3 \left (4 b d^3-a e^3\right ) p x}{4 b}-\frac {9}{4} d^2 e p x^2-d e^2 p x^3-\frac {3}{16} e^3 p x^4-\frac {\sqrt {3} \sqrt [3]{a} \left (4 b d^3+6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{4 b^{4/3}}+\frac {\sqrt [3]{a} \left (4 b d^3-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{4 b^{4/3}}-\frac {\sqrt [3]{a} \left (4 b d^3-6 \sqrt [3]{a} b^{2/3} d^2 e-a e^3\right ) p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{8 b^{4/3}}-\frac {d \left (b d^3-4 a e^3\right ) p \log \left (a+b x^3\right )}{4 b e}+\frac {(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.32 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.89 \[ \int (d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {\frac {3 e \left (-4 b d^3+a e^3\right ) p x}{b}-9 d^2 e^2 p x^2-4 d e^3 p x^3-\frac {3}{4} e^4 p x^4+\frac {\sqrt {3} \sqrt [3]{a} e \left (-4 b d^3+a e^3\right ) p \arctan \left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{4/3}}+9 d^2 e^2 p x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {b x^3}{a}\right )+\frac {\sqrt [3]{a} e \left (4 b d^3-a e^3\right ) p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{4/3}}+\frac {\sqrt [3]{a} e \left (-4 b d^3+a e^3\right ) p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 b^{4/3}}-\frac {d \left (b d^3-4 a e^3\right ) p \log \left (a+b x^3\right )}{b}+(d+e x)^4 \log \left (c \left (a+b x^3\right )^p\right )}{4 e} \]

[In]

Integrate[(d + e*x)^3*Log[c*(a + b*x^3)^p],x]

[Out]

((3*e*(-4*b*d^3 + a*e^3)*p*x)/b - 9*d^2*e^2*p*x^2 - 4*d*e^3*p*x^3 - (3*e^4*p*x^4)/4 + (Sqrt[3]*a^(1/3)*e*(-4*b
*d^3 + a*e^3)*p*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(4/3) + 9*d^2*e^2*p*x^2*Hypergeometric2F1[2/3,
1, 5/3, -((b*x^3)/a)] + (a^(1/3)*e*(4*b*d^3 - a*e^3)*p*Log[a^(1/3) + b^(1/3)*x])/b^(4/3) + (a^(1/3)*e*(-4*b*d^
3 + a*e^3)*p*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(2*b^(4/3)) - (d*(b*d^3 - 4*a*e^3)*p*Log[a + b*x^
3])/b + (d + e*x)^4*Log[c*(a + b*x^3)^p])/(4*e)

Maple [A] (verified)

Time = 2.32 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.24

method	result	size
parts	\(\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) e^{3} x^{4}}{4}+\ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) e^{2} d \,x^{3}+\frac {3 \ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) e \,d^{2} x^{2}}{2}+d^{3} \ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) x +\frac {\ln \left (c \left (b \,x^{3}+a \right )^{p}\right ) d^{4}}{4 e}-\frac {3 p b \left (-\frac {e \left (-\frac {1}{4} x^{4} b \,e^{3}-\frac {4}{3} x^{3} b d \,e^{2}-3 e \,d^{2} b \,x^{2}+x a \,e^{3}-4 b \,d^{3} x \right )}{b^{2}}+\frac {\left (a^{2} e^{4}-4 a b \,d^{3} e \right ) \left (\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}-\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )-6 a b \,d^{2} e^{2} \left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )+\frac {\left (-4 a b d \,e^{3}+b^{2} d^{4}\right ) \ln \left (b \,x^{3}+a \right )}{3 b}}{b^{2}}\right )}{4 e}\)	\(398\)
risch	\(-\frac {i e^{3} \pi \,x^{4} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{8}-\frac {i \pi \,d^{3} x {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{2}+\frac {\left (e x +d \right )^{4} \ln \left (\left (b \,x^{3}+a \right )^{p}\right )}{4 e}-3 d^{3} p x +\frac {3 e^{3} a p x}{4 b}+\frac {e^{3} \ln \left (c \right ) x^{4}}{4}-\frac {i e^{2} \pi d \,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 )}{2}+\ln \left (c \right ) d \,e^{2} x^{3}+\frac {3 \ln \left (c \right ) d^{2} e \,x^{2}}{2}+x \ln \left (c \right ) d^{3}-\frac {3 i e \pi \,d^{2} 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 e^{3} \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 {p \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (b d \left (4 a \,e^{3}-b \,d^{3}\right ) \textit {\_R}^{2}+6 a b \,d^{2} e^{2} \textit {\_R} -a^{2} e^{4}+4 a b \,d^{3} e \right ) \ln \left (x -\textit {\_R} \right )}{\textit {\_R}^{2}}\right )}{4 b^{2} e}-\frac {i e^{2} \pi d \,x^{3} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{2}-\frac {3 i e \pi \,d^{2} x^{2} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{3}}{4}+\frac {i \pi \,d^{3} x \,\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 \,d^{3} x {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {i e^{3} \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 \,d^{3} x \,\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 e^{3} \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 e^{2} \pi d \,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}}{2}+\frac {i e^{2} \pi d \,x^{3} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\frac {3 i e \pi \,d^{2} 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}}{4}+\frac {3 i e \pi \,d^{2} x^{2} {\operatorname {csgn}\left (i c \left (b \,x^{3}+a \right )^{p}\right )}^{2} \operatorname {csgn}\left (i c \right )}{4}-d \,e^{2} p \,x^{3}-\frac {9 d^{2} e p \,x^{2}}{4}-\frac {3 e^{3} p \,x^{4}}{16}\)	\(738\)

[In]

int((e*x+d)^3*ln(c*(b*x^3+a)^p),x,method=_RETURNVERBOSE)

[Out]

1/4*ln(c*(b*x^3+a)^p)*e^3*x^4+ln(c*(b*x^3+a)^p)*e^2*d*x^3+3/2*ln(c*(b*x^3+a)^p)*e*d^2*x^2+d^3*ln(c*(b*x^3+a)^p
)*x+1/4*ln(c*(b*x^3+a)^p)/e*d^4-3/4*p*b/e*(-e/b^2*(-1/4*x^4*b*e^3-4/3*x^3*b*d*e^2-3*e*d^2*b*x^2+x*a*e^3-4*b*d^
3*x)+((a^2*e^4-4*a*b*d^3*e)*(1/3/b/(a/b)^(2/3)*ln(x+(a/b)^(1/3))-1/6/b/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^
(2/3))+1/3/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1)))-6*a*b*d^2*e^2*(-1/3/b/(a/b)^(1/3)*ln
(x+(a/b)^(1/3))+1/6/b/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+1/3*3^(1/2)/b/(a/b)^(1/3)*arctan(1/3*3^(1/
2)*(2/(a/b)^(1/3)*x-1)))+1/3*(-4*a*b*d*e^3+b^2*d^4)/b*ln(b*x^3+a))/b^2)

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 9.18 (sec) , antiderivative size = 8840, normalized size of antiderivative = 27.62 \[ \int (d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\text {Too large to display} \]

[In]

integrate((e*x+d)^3*log(c*(b*x^3+a)^p),x, algorithm="fricas")

[Out]

Too large to include

Sympy [A] (verification not implemented)

Time = 23.69 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.83 \[ \int (d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=- \frac {3 a^{2} e^{3} p \operatorname {RootSum} {\left (27 t^{3} a^{2} b - 1, \left ( t \mapsto t \log {\left (3 t a + x \right )} \right )\right )}}{4 b} + 3 a d^{3} p \operatorname {RootSum} {\left (27 t^{3} a^{2} b - 1, \left ( t \mapsto t \log {\left (3 t a + x \right )} \right )\right )} + \frac {9 a d^{2} e p \operatorname {RootSum} {\left (27 t^{3} a b^{2} + 1, \left ( t \mapsto t \log {\left (9 t^{2} a b + x \right )} \right )\right )}}{2} + a d e^{2} p \left (\begin {cases} \frac {x^{3}}{a} & \text {for}\: b = 0 \\\frac {\log {\left (a + b x^{3} \right )}}{b} & \text {otherwise} \end {cases}\right ) + \frac {3 a e^{3} p x}{4 b} - 3 d^{3} p x + d^{3} x \log {\left (c \left (a + b x^{3}\right )^{p} \right )} - \frac {9 d^{2} e p x^{2}}{4} + \frac {3 d^{2} e x^{2} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{2} - d e^{2} p x^{3} + d e^{2} x^{3} \log {\left (c \left (a + b x^{3}\right )^{p} \right )} - \frac {3 e^{3} p x^{4}}{16} + \frac {e^{3} x^{4} \log {\left (c \left (a + b x^{3}\right )^{p} \right )}}{4} \]

[In]

integrate((e*x+d)**3*ln(c*(b*x**3+a)**p),x)

[Out]

-3*a**2*e**3*p*RootSum(27*_t**3*a**2*b - 1, Lambda(_t, _t*log(3*_t*a + x)))/(4*b) + 3*a*d**3*p*RootSum(27*_t**
3*a**2*b - 1, Lambda(_t, _t*log(3*_t*a + x))) + 9*a*d**2*e*p*RootSum(27*_t**3*a*b**2 + 1, Lambda(_t, _t*log(9*
_t**2*a*b + x)))/2 + a*d*e**2*p*Piecewise((x**3/a, Eq(b, 0)), (log(a + b*x**3)/b, True)) + 3*a*e**3*p*x/(4*b)
- 3*d**3*p*x + d**3*x*log(c*(a + b*x**3)**p) - 9*d**2*e*p*x**2/4 + 3*d**2*e*x**2*log(c*(a + b*x**3)**p)/2 - d*
e**2*p*x**3 + d*e**2*x**3*log(c*(a + b*x**3)**p) - 3*e**3*p*x**4/16 + e**3*x**4*log(c*(a + b*x**3)**p)/4

Maxima [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 332, normalized size of antiderivative = 1.04 \[ \int (d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\frac {1}{16} \, b p {\left (\frac {4 \, \sqrt {3} {\left (6 \, a b d^{2} e \left (\frac {a}{b}\right )^{\frac {2}{3}} + 4 \, a b d^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}} - a^{2} e^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{a b^{2}} - \frac {3 \, b e^{3} x^{4} + 16 \, b d e^{2} x^{3} + 36 \, b d^{2} e x^{2} + 12 \, {\left (4 \, b d^{3} - a e^{3}\right )} x}{b^{2}} + \frac {2 \, {\left (8 \, a b d e^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} + 6 \, a b d^{2} e \left (\frac {a}{b}\right )^{\frac {1}{3}} - 4 \, a b d^{3} + a^{2} e^{3}\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {4 \, {\left (4 \, a b d e^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}} - 6 \, a b d^{2} e \left (\frac {a}{b}\right )^{\frac {1}{3}} + 4 \, a b d^{3} - a^{2} e^{3}\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}}\right )} + \frac {1}{4} \, {\left (e^{3} x^{4} + 4 \, d e^{2} x^{3} + 6 \, d^{2} e x^{2} + 4 \, d^{3} x\right )} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) \]

[In]

integrate((e*x+d)^3*log(c*(b*x^3+a)^p),x, algorithm="maxima")

[Out]

1/16*b*p*(4*sqrt(3)*(6*a*b*d^2*e*(a/b)^(2/3) + 4*a*b*d^3*(a/b)^(1/3) - a^2*e^3*(a/b)^(1/3))*arctan(1/3*sqrt(3)
*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a*b^2) - (3*b*e^3*x^4 + 16*b*d*e^2*x^3 + 36*b*d^2*e*x^2 + 12*(4*b*d^3 - a*e
^3)*x)/b^2 + 2*(8*a*b*d*e^2*(a/b)^(2/3) + 6*a*b*d^2*e*(a/b)^(1/3) - 4*a*b*d^3 + a^2*e^3)*log(x^2 - x*(a/b)^(1/
3) + (a/b)^(2/3))/(b^3*(a/b)^(2/3)) + 4*(4*a*b*d*e^2*(a/b)^(2/3) - 6*a*b*d^2*e*(a/b)^(1/3) + 4*a*b*d^3 - a^2*e
^3)*log(x + (a/b)^(1/3))/(b^3*(a/b)^(2/3))) + 1/4*(e^3*x^4 + 4*d*e^2*x^3 + 6*d^2*e*x^2 + 4*d^3*x)*log((b*x^3 +
a)^p*c)

Giac [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 381, normalized size of antiderivative = 1.19 \[ \int (d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=-\frac {1}{16} \, {\left (3 \, e^{3} p - 4 \, e^{3} \log \left (c\right )\right )} x^{4} + \frac {a d e^{2} p \log \left ({\left | b x^{3} + a \right |}\right )}{b} - {\left (d e^{2} p - d e^{2} \log \left (c\right )\right )} x^{3} - \frac {3}{4} \, {\left (3 \, d^{2} e p - 2 \, d^{2} e \log \left (c\right )\right )} x^{2} + \frac {1}{4} \, {\left (e^{3} p x^{4} + 4 \, d e^{2} p x^{3} + 6 \, d^{2} e p x^{2} + 4 \, d^{3} p x\right )} \log \left (b x^{3} + a\right ) - \frac {{\left (12 \, b d^{3} p - 3 \, a e^{3} p - 4 \, b d^{3} \log \left (c\right )\right )} x}{4 \, b} + \frac {\sqrt {3} {\left (4 \, \left (-a b^{2}\right )^{\frac {1}{3}} b d^{3} p - \left (-a b^{2}\right )^{\frac {1}{3}} a e^{3} p - 6 \, \left (-a b^{2}\right )^{\frac {2}{3}} d^{2} e p\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{4 \, b^{2}} + \frac {{\left (4 \, \left (-a b^{2}\right )^{\frac {1}{3}} b d^{3} p - \left (-a b^{2}\right )^{\frac {1}{3}} a e^{3} p + 6 \, \left (-a b^{2}\right )^{\frac {2}{3}} d^{2} e p\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{8 \, b^{2}} - \frac {{\left (6 \, a b^{3} d^{2} e p \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 4 \, a b^{3} d^{3} p - a^{2} b^{2} e^{3} p\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{4 \, a b^{3}} \]

[In]

integrate((e*x+d)^3*log(c*(b*x^3+a)^p),x, algorithm="giac")

[Out]

-1/16*(3*e^3*p - 4*e^3*log(c))*x^4 + a*d*e^2*p*log(abs(b*x^3 + a))/b - (d*e^2*p - d*e^2*log(c))*x^3 - 3/4*(3*d
^2*e*p - 2*d^2*e*log(c))*x^2 + 1/4*(e^3*p*x^4 + 4*d*e^2*p*x^3 + 6*d^2*e*p*x^2 + 4*d^3*p*x)*log(b*x^3 + a) - 1/
4*(12*b*d^3*p - 3*a*e^3*p - 4*b*d^3*log(c))*x/b + 1/4*sqrt(3)*(4*(-a*b^2)^(1/3)*b*d^3*p - (-a*b^2)^(1/3)*a*e^3
*p - 6*(-a*b^2)^(2/3)*d^2*e*p)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/b^2 + 1/8*(4*(-a*b^2)^(1/
3)*b*d^3*p - (-a*b^2)^(1/3)*a*e^3*p + 6*(-a*b^2)^(2/3)*d^2*e*p)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/b^2 -
1/4*(6*a*b^3*d^2*e*p*(-a/b)^(1/3) + 4*a*b^3*d^3*p - a^2*b^2*e^3*p)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a
*b^3)

Mupad [B] (verification not implemented)

Time = 2.05 (sec) , antiderivative size = 536, normalized size of antiderivative = 1.68 \[ \int (d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right ) \, dx=\ln \left (c\,{\left (b\,x^3+a\right )}^p\right )\,\left (d^3\,x+\frac {3\,d^2\,e\,x^2}{2}+d\,e^2\,x^3+\frac {e^3\,x^4}{4}\right )-x\,\left (3\,d^3\,p-\frac {3\,a\,e^3\,p}{4\,b}\right )+\left (\sum _{k=1}^3\ln \left (x\,\left (\frac {9\,a^3\,d\,e^5\,p^2}{4}+\frac {45\,b\,a^2\,d^4\,e^2\,p^2}{4}\right )+\mathrm {root}\left (64\,b^4\,c^3-192\,a\,b^3\,c^2\,d\,e^2\,p+288\,a\,b^3\,c\,d^5\,e\,p^2+120\,a^2\,b^2\,c\,d^2\,e^4\,p^2-4\,a^3\,b\,d^3\,e^6\,p^3-24\,a^2\,b^2\,d^6\,e^3\,p^3-64\,a\,b^3\,d^9\,p^3+a^4\,e^9\,p^3,c,k\right )\,\left (x\,\left (9\,a\,b^2\,d^3\,p-\frac {9\,a^2\,b\,e^3\,p}{4}\right )+\mathrm {root}\left (64\,b^4\,c^3-192\,a\,b^3\,c^2\,d\,e^2\,p+288\,a\,b^3\,c\,d^5\,e\,p^2+120\,a^2\,b^2\,c\,d^2\,e^4\,p^2-4\,a^3\,b\,d^3\,e^6\,p^3-24\,a^2\,b^2\,d^6\,e^3\,p^3-64\,a\,b^3\,d^9\,p^3+a^4\,e^9\,p^3,c,k\right )\,a\,b^2\,9-18\,a^2\,b\,d\,e^2\,p\right )+\frac {45\,a^3\,d^2\,e^4\,p^2}{8}+\frac {27\,a^2\,b\,d^5\,e\,p^2}{2}\right )\,\mathrm {root}\left (64\,b^4\,c^3-192\,a\,b^3\,c^2\,d\,e^2\,p+288\,a\,b^3\,c\,d^5\,e\,p^2+120\,a^2\,b^2\,c\,d^2\,e^4\,p^2-4\,a^3\,b\,d^3\,e^6\,p^3-24\,a^2\,b^2\,d^6\,e^3\,p^3-64\,a\,b^3\,d^9\,p^3+a^4\,e^9\,p^3,c,k\right )\right )-\frac {3\,e^3\,p\,x^4}{16}-\frac {9\,d^2\,e\,p\,x^2}{4}-d\,e^2\,p\,x^3 \]

[In]

int(log(c*(a + b*x^3)^p)*(d + e*x)^3,x)

[Out]

log(c*(a + b*x^3)^p)*(d^3*x + (e^3*x^4)/4 + (3*d^2*e*x^2)/2 + d*e^2*x^3) - x*(3*d^3*p - (3*a*e^3*p)/(4*b)) + s
ymsum(log(x*((9*a^3*d*e^5*p^2)/4 + (45*a^2*b*d^4*e^2*p^2)/4) + root(64*b^4*c^3 - 192*a*b^3*c^2*d*e^2*p + 288*a
*b^3*c*d^5*e*p^2 + 120*a^2*b^2*c*d^2*e^4*p^2 - 4*a^3*b*d^3*e^6*p^3 - 24*a^2*b^2*d^6*e^3*p^3 - 64*a*b^3*d^9*p^3
+ a^4*e^9*p^3, c, k)*(x*(9*a*b^2*d^3*p - (9*a^2*b*e^3*p)/4) + 9*root(64*b^4*c^3 - 192*a*b^3*c^2*d*e^2*p + 288
*a*b^3*c*d^5*e*p^2 + 120*a^2*b^2*c*d^2*e^4*p^2 - 4*a^3*b*d^3*e^6*p^3 - 24*a^2*b^2*d^6*e^3*p^3 - 64*a*b^3*d^9*p
^3 + a^4*e^9*p^3, c, k)*a*b^2 - 18*a^2*b*d*e^2*p) + (45*a^3*d^2*e^4*p^2)/8 + (27*a^2*b*d^5*e*p^2)/2)*root(64*b
^4*c^3 - 192*a*b^3*c^2*d*e^2*p + 288*a*b^3*c*d^5*e*p^2 + 120*a^2*b^2*c*d^2*e^4*p^2 - 4*a^3*b*d^3*e^6*p^3 - 24*
a^2*b^2*d^6*e^3*p^3 - 64*a*b^3*d^9*p^3 + a^4*e^9*p^3, c, k), k, 1, 3) - (3*e^3*p*x^4)/16 - (9*d^2*e*p*x^2)/4 -
d*e^2*p*x^3