3.2.92 \(\int (d+e x)^2 \log (c (a+b x^3)^p) \, dx\) [192]

Optimal. Leaf size=250 \[ -3 d^2 p x-\frac {3}{2} d e p x^2-\frac {1}{3} e^2 p x^3-\frac {\sqrt {3} \sqrt [3]{a} d \left (\sqrt [3]{b} d+\sqrt [3]{a} e\right ) p \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{b^{2/3}}+\frac {\sqrt [3]{a} d \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) p \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}-\frac {\sqrt [3]{a} d \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) p \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 b^{2/3}}-\frac {\left (b d^3-a e^3\right ) p \log \left (a+b x^3\right )}{3 b e}+\frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e} \]

[Out]

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

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Rubi [A]
time = 0.32, antiderivative size = 250, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 11, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {2513, 1850, 1901, 1885, 1874, 31, 648, 631, 210, 642, 266} \begin {gather*} -\frac {\sqrt [3]{a} d p \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 p \text {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 p \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}+\frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}-\frac {p \left (b d^3-a e^3\right ) \log \left (a+b x^3\right )}{3 b e}-3 d^2 p x-\frac {3}{2} d e p x^2-\frac {1}{3} e^2 p x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2*Log[c*(a + b*x^3)^p],x]

[Out]

-3*d^2*p*x - (3*d*e*p*x^2)/2 - (e^2*p*x^3)/3 - (Sqrt[3]*a^(1/3)*d*(b^(1/3)*d + a^(1/3)*e)*p*ArcTan[(a^(1/3) -
2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/b^(2/3) + (a^(1/3)*d*(b^(1/3)*d - a^(1/3)*e)*p*Log[a^(1/3) + b^(1/3)*x])/b^(2
/3) - (a^(1/3)*d*(b^(1/3)*d - a^(1/3)*e)*p*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)*p*Log[a + b*x^3])/(3*b*e) + ((d + e*x)^3*Log[c*(a + b*x^3)^p])/(3*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*} \int (d+e x)^2 \log \left (c \left (a+b x^3\right )^p\right ) \, dx &=\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}{a+b x^3} \, dx}{e}\\ &=-\frac {1}{3} e^2 p x^3+\frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}-\frac {p \int \frac {x^2 \left (3 \left (b d^3-a e^3\right )+9 b d^2 e x+9 b d e^2 x^2\right )}{a+b x^3} \, dx}{3 e}\\ &=-\frac {1}{3} e^2 p x^3+\frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}-\frac {p \int \left (9 d^2 e+9 d e^2 x-\frac {3 \left (3 a d^2 e+3 a d e^2 x-\left (b d^3-a e^3\right ) x^2\right )}{a+b x^3}\right ) \, dx}{3 e}\\ &=-3 d^2 p x-\frac {3}{2} d e p x^2-\frac {1}{3} e^2 p x^3+\frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}+\frac {p \int \frac {3 a d^2 e+3 a d e^2 x-\left (b d^3-a e^3\right ) x^2}{a+b x^3} \, dx}{e}\\ &=-3 d^2 p x-\frac {3}{2} d e p x^2-\frac {1}{3} e^2 p x^3+\frac {(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}+\frac {p \int \frac {3 a d^2 e+3 a d e^2 x}{a+b x^3} \, dx}{e}-\frac {\left (\left (b d^3-a e^3\right ) p\right ) \int \frac {x^2}{a+b x^3} \, dx}{e}\\ &=-3 d^2 p x-\frac {3}{2} d e p x^2-\frac {1}{3} e^2 p x^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}+\frac {p \int \frac {\sqrt [3]{a} \left (6 a \sqrt [3]{b} d^2 e+3 a^{4/3} d e^2\right )+\sqrt [3]{b} \left (-3 a \sqrt [3]{b} d^2 e+3 a^{4/3} d e^2\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{3 a^{2/3} \sqrt [3]{b} e}-\frac {\left (\left (-3 a \sqrt [3]{b} d^2 e+3 a^{4/3} d e^2\right ) p\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a^{2/3} \sqrt [3]{b} e}\\ &=-3 d^2 p x-\frac {3}{2} d e p x^2-\frac {1}{3} e^2 p x^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 {\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}-\frac {\left (\sqrt [3]{a} d \left (\sqrt [3]{b} d-\sqrt [3]{a} e\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}{2 b^{2/3}}+\frac {1}{2} \left (3 a^{2/3} d \left (d+\frac {\sqrt [3]{a} e}{\sqrt [3]{b}}\right ) p\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, 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]{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}+\frac {\left (3 \sqrt [3]{a} d \left (\sqrt [3]{b} d+\sqrt [3]{a} e\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 )}{b^{2/3}}\\ &=-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 \tan ^{-1}\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}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 0.22, size = 218, normalized size = 0.87 \begin {gather*} \frac {-\frac {p \left (18 b d^2 e x+9 b d e^2 x^2+2 b e^3 x^3-9 b d e^2 x^2 \, _2F_1\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 \left (2 \sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )\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} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2*Log[c*(a + b*x^3)^p],x]

[Out]

(-1/2*(p*(18*b*d^2*e*x + 9*b*d*e^2*x^2 + 2*b*e^3*x^3 - 9*b*d*e^2*x^2*Hypergeometric2F1[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*(2*Sqrt[3]*ArcTan[(1 - (2*b^(
1/3)*x)/a^(1/3))/Sqrt[3]] + 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*Log[c*(a + b*x^3)^p])/(3*e)

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 3.
time = 0.91, size = 537, normalized size = 2.15

method	result	size
risch	\(\frac {\left (e x +d \right )^{3} \ln \left (\left (x^{3} b +a \right )^{p}\right )}{3 e}-\frac {i e^{2} \pi \,x^{3} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{6}+\frac {i \pi \,d^{2} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2} x}{2}-\frac {i e^{2} \pi \,x^{3} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{3}}{6}+\frac {i e \pi d \,x^{2} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2}}{2}-\frac {i e \pi d \,x^{2} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right )}{2}-\frac {i \pi \,d^{2} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \right ) x}{2}-\frac {i \pi \,d^{2} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{3} x}{2}+\frac {i e^{2} \pi \,x^{3} \mathrm {csgn}\left (i \left (x^{3} b +a \right )^{p}\right ) \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2}}{6}+\frac {i e \pi d \,x^{2} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{2}+\frac {i \pi \,d^{2} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right ) x}{2}+\frac {i e^{2} \pi \,x^{3} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{2} \mathrm {csgn}\left (i c \right )}{6}-\frac {i e \pi d \,x^{2} \mathrm {csgn}\left (i c \left (x^{3} b +a \right )^{p}\right )^{3}}{2}+\frac {e^{2} \ln \left (c \right ) 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}+\ln \left (c \right ) d^{2} x -3 d^{2} p x +\frac {p \left (\munderset {\textit {\_R} =\RootOf \left (\textit {\_Z}^{3} b +a \right )}{\sum }\frac {\left (\left (e^{3} a -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}\)	\(537\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2*ln(c*(b*x^3+a)^p),x,method=_RETURNVERBOSE)

[Out]

1/3*(e*x+d)^3/e*ln((b*x^3+a)^p)-1/6*I*e^2*Pi*x^3*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)*csgn(I*c)+1/2*I*Pi*
d^2*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)^2*x-1/6*I*e^2*Pi*x^3*csgn(I*c*(b*x^3+a)^p)^3+1/2*I*e*Pi*d*x^2*cs
gn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)^2-1/2*I*e*Pi*d*x^2*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)*csgn(I*c)
-1/2*I*Pi*d^2*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)*csgn(I*c)*x-1/2*I*Pi*d^2*csgn(I*c*(b*x^3+a)^p)^3*x+1/6
*I*e^2*Pi*x^3*csgn(I*(b*x^3+a)^p)*csgn(I*c*(b*x^3+a)^p)^2+1/2*I*e*Pi*d*x^2*csgn(I*c*(b*x^3+a)^p)^2*csgn(I*c)+1
/2*I*Pi*d^2*csgn(I*c*(b*x^3+a)^p)^2*csgn(I*c)*x+1/6*I*e^2*Pi*x^3*csgn(I*c*(b*x^3+a)^p)^2*csgn(I*c)-1/2*I*e*Pi*
d*x^2*csgn(I*c*(b*x^3+a)^p)^3+1/3*e^2*ln(c)*x^3-1/3*e^2*p*x^3+ln(c)*d*e*x^2-3/2*d*e*p*x^2+ln(c)*d^2*x-3*d^2*p*
x+1/3*p/b/e*sum(((a*e^3-b*d^3)*_R^2+3*e^2*d*a*_R+3*e*d^2*a)/_R^2*ln(x-_R),_R=RootOf(_Z^3*b+a))

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Maxima [A]
time = 0.59, size = 251, normalized size = 1.00 \begin {gather*} -\frac {1}{6} \, b p {\left (\frac {2 \, x^{3} e^{2} + 9 \, d x^{2} e + 18 \, d^{2} x}{b} - \frac {6 \, \sqrt {3} {\left (a b d \left (\frac {a}{b}\right )^{\frac {2}{3}} e + 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 (3 \, a d \left (\frac {a}{b}\right )^{\frac {1}{3}} e - 3 \, a d^{2} + 2 \, a \left (\frac {a}{b}\right )^{\frac {2}{3}} e^{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 (3 \, a d \left (\frac {a}{b}\right )^{\frac {1}{3}} e - 3 \, a d^{2} - a \left (\frac {a}{b}\right )^{\frac {2}{3}} e^{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 (x^{3} e^{2} + 3 \, d x^{2} e + 3 \, d^{2} x\right )} \log \left ({\left (b x^{3} + a\right )}^{p} c\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*log(c*(b*x^3+a)^p),x, algorithm="maxima")

[Out]

-1/6*b*p*((2*x^3*e^2 + 9*d*x^2*e + 18*d^2*x)/b - 6*sqrt(3)*(a*b*d*(a/b)^(2/3)*e + 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) - (3*a*d*(a/b)^(1/3)*e - 3*a*d^2 + 2*a*(a/b)^(2/3)*e^2)*l
og(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^2*(a/b)^(2/3)) + 2*(3*a*d*(a/b)^(1/3)*e - 3*a*d^2 - a*(a/b)^(2/3)*e^2
)*log(x + (a/b)^(1/3))/(b^2*(a/b)^(2/3))) + 1/3*(x^3*e^2 + 3*d*x^2*e + 3*d^2*x)*log((b*x^3 + a)^p*c)

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Fricas [C] Result contains complex when optimal does not.
time = 2.81, size = 5590, normalized size = 22.36 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*log(c*(b*x^3+a)^p),x, algorithm="fricas")

[Out]

-1/12*(4*b*p*x^3*e^2 + 18*b*d*p*x^2*e + 36*b*d^2*p*x + 2*(2*(1/2)^(2/3)*(a^2*p^2*e^4/b^2 - (9*a*b*d^3*p^2*e +
a^2*p^2*e^4)/b^2)*(-I*sqrt(3) + 1)/(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e
+ a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3) - 2*a*p*e^2/b + (1/2)^(1/3)*(27*(b*d^
3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p
^3 + a^3*p^3*e^6)/b^3)^(1/3)*(I*sqrt(3) + 1))*b*log(9*b^2*d^5*p^2*x + 9*a*b*d^2*p^2*x*e^3 + 15*a*b*d^3*p^2*e^2
+ a^2*p^2*e^5 + 1/4*(2*(1/2)^(2/3)*(a^2*p^2*e^4/b^2 - (9*a*b*d^3*p^2*e + a^2*p^2*e^4)/b^2)*(-I*sqrt(3) + 1)/(
27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b
^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3) - 2*a*p*e^2/b + (1/2)^(1/3)*(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3
*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3)*(I*sqrt
(3) + 1))^2*b^2*e - 1/2*(3*b^2*d^3*p - 2*a*b*p*e^3)*(2*(1/2)^(2/3)*(a^2*p^2*e^4/b^2 - (9*a*b*d^3*p^2*e + a^2*p
^2*e^4)/b^2)*(-I*sqrt(3) + 1)/(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2
*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3) - 2*a*p*e^2/b + (1/2)^(1/3)*(27*(b*d^3 + a
*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 +
a^3*p^3*e^6)/b^3)^(1/3)*(I*sqrt(3) + 1))) - (6*a*p*e^2 + (2*(1/2)^(2/3)*(a^2*p^2*e^4/b^2 - (9*a*b*d^3*p^2*e +
a^2*p^2*e^4)/b^2)*(-I*sqrt(3) + 1)/(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e
+ a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3) - 2*a*p*e^2/b + (1/2)^(1/3)*(27*(b*d^
3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p
^3 + a^3*p^3*e^6)/b^3)^(1/3)*(I*sqrt(3) + 1))*b + 3*sqrt(1/3)*b*sqrt(-(144*a*b*d^3*p^2*e + 4*a^2*p^2*e^4 + 4*(
2*(1/2)^(2/3)*(a^2*p^2*e^4/b^2 - (9*a*b*d^3*p^2*e + a^2*p^2*e^4)/b^2)*(-I*sqrt(3) + 1)/(27*(b*d^3 + a*e^3)*a*d
^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e
^6)/b^3)^(1/3) - 2*a*p*e^2/b + (1/2)^(1/3)*(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^
3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3)*(I*sqrt(3) + 1))*a*b*p*e^2 +
(2*(1/2)^(2/3)*(a^2*p^2*e^4/b^2 - (9*a*b*d^3*p^2*e + a^2*p^2*e^4)/b^2)*(-I*sqrt(3) + 1)/(27*(b*d^3 + a*e^3)*a*
d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*
e^6)/b^3)^(1/3) - 2*a*p*e^2/b + (1/2)^(1/3)*(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d
^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3)*(I*sqrt(3) + 1))^2*b^2)/b^2)
)*log(18*b^2*d^5*p^2*x + 18*a*b*d^2*p^2*x*e^3 - 15*a*b*d^3*p^2*e^2 - a^2*p^2*e^5 - 1/4*(2*(1/2)^(2/3)*(a^2*p^2
*e^4/b^2 - (9*a*b*d^3*p^2*e + a^2*p^2*e^4)/b^2)*(-I*sqrt(3) + 1)/(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3
*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3) - 2*a*p
*e^2/b + (1/2)^(1/3)*(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)
*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3)*(I*sqrt(3) + 1))^2*b^2*e + 1/2*(3*b^2*d^3*p - 2*a*b
*p*e^3)*(2*(1/2)^(2/3)*(a^2*p^2*e^4/b^2 - (9*a*b*d^3*p^2*e + a^2*p^2*e^4)/b^2)*(-I*sqrt(3) + 1)/(27*(b*d^3 + a
*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 +
a^3*p^3*e^6)/b^3)^(1/3) - 2*a*p*e^2/b + (1/2)^(1/3)*(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*
(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3)*(I*sqrt(3) + 1)) + 3
/4*sqrt(1/3)*(6*b^2*d^3*p + 2*a*b*p*e^3 + (2*(1/2)^(2/3)*(a^2*p^2*e^4/b^2 - (9*a*b*d^3*p^2*e + a^2*p^2*e^4)/b^
2)*(-I*sqrt(3) + 1)/(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*
a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3) - 2*a*p*e^2/b + (1/2)^(1/3)*(27*(b*d^3 + a*e^3)*a*d^
3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^
6)/b^3)^(1/3)*(I*sqrt(3) + 1))*b^2*e)*sqrt(-(144*a*b*d^3*p^2*e + 4*a^2*p^2*e^4 + 4*(2*(1/2)^(2/3)*(a^2*p^2*e^4
/b^2 - (9*a*b*d^3*p^2*e + a^2*p^2*e^4)/b^2)*(-I*sqrt(3) + 1)/(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6
/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3) - 2*a*p*e^2
/b + (1/2)^(1/3)*(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p
*e^2/b^3 + (27*a*b^2*d^6*p^3 + a^3*p^3*e^6)/b^3)^(1/3)*(I*sqrt(3) + 1))*a*b*p*e^2 + (2*(1/2)^(2/3)*(a^2*p^2*e^
4/b^2 - (9*a*b*d^3*p^2*e + a^2*p^2*e^4)/b^2)*(-I*sqrt(3) + 1)/(27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 + 2*a^3*p^3*e^
6/b^3 - 3*(9*a*b*d^3*p^2*e + a^2*p^2*e^4)*a*p*e...

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Sympy [A]
time = 18.54, size = 173, normalized size = 0.69 \begin {gather*} 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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2*ln(c*(b*x**3+a)**p),x)

[Out]

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

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Giac [A]
time = 5.00, size = 298, normalized size = 1.19 \begin {gather*} -\frac {{\left (a d p \left (-\frac {a}{b}\right )^{\frac {1}{3}} e + a 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} + \frac {2 \, b p x^{3} e^{2} \log \left (b x^{3} + a\right ) + 6 \, b d p x^{2} e \log \left (b x^{3} + a\right ) - 2 \, b p x^{3} e^{2} - 9 \, b d p x^{2} e + 6 \, b d^{2} p x \log \left (b x^{3} + a\right ) + 2 \, b x^{3} e^{2} \log \left (c\right ) + 6 \, b d x^{2} e \log \left (c\right ) - 18 \, b d^{2} p x + 6 \, b d^{2} x \log \left (c\right ) + 2 \, a p e^{2} \log \left (b x^{3} + a\right )}{6 \, b} + \frac {{\left (\sqrt {3} \left (-a b^{2}\right )^{\frac {1}{3}} b d^{2} p - \sqrt {3} \left (-a b^{2}\right )^{\frac {2}{3}} d p e\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 (\left (-a b^{2}\right )^{\frac {1}{3}} b d^{2} p + \left (-a b^{2}\right )^{\frac {2}{3}} d p e\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*log(c*(b*x^3+a)^p),x, algorithm="giac")

[Out]

-(a*d*p*(-a/b)^(1/3)*e + a*d^2*p)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a + 1/6*(2*b*p*x^3*e^2*log(b*x^3 + a
) + 6*b*d*p*x^2*e*log(b*x^3 + a) - 2*b*p*x^3*e^2 - 9*b*d*p*x^2*e + 6*b*d^2*p*x*log(b*x^3 + a) + 2*b*x^3*e^2*lo
g(c) + 6*b*d*x^2*e*log(c) - 18*b*d^2*p*x + 6*b*d^2*x*log(c) + 2*a*p*e^2*log(b*x^3 + a))/b + (sqrt(3)*(-a*b^2)^
(1/3)*b*d^2*p - sqrt(3)*(-a*b^2)^(2/3)*d*p*e)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/b^2 + 1/2*
((-a*b^2)^(1/3)*b*d^2*p + (-a*b^2)^(2/3)*d*p*e)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/b^2

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Mupad [B]
time = 0.32, size = 358, normalized size = 1.43 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*(a + b*x^3)^p)*(d + e*x)^2,x)

[Out]

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

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