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

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

[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)

________________________________________________________________________________________

Rubi [A]  time = 0.482857, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 11, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.55, Rules used = {2463, 1836, 1887, 1871, 1860, 31, 634, 617, 204, 628, 260} \[ -\frac{\sqrt [3]{a} d p \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{2 b^{2/3}}+\frac{\sqrt [3]{a} d p \left (\sqrt [3]{b} d-\sqrt [3]{a} e\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}-\frac{\sqrt{3} \sqrt [3]{a} d p \left (\sqrt [3]{a} e+\sqrt [3]{b} d\right ) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{b^{2/3}}+\frac{(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )}{3 e}-\frac{p \left (b d^3-a e^3\right ) \log \left (a+b x^3\right )}{3 b e}-3 d^2 p x-\frac{3}{2} d e p x^2-\frac{1}{3} e^2 p x^3 \]

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 2463

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]

Rule 1836

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 1887

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 1871

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 1860

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 31

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

Rule 634

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 617

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 628

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 260

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]

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 ) \operatorname{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*}

Mathematica [C]  time = 0.304001, size = 218, normalized size = 0.87 \[ \frac{(d+e x)^3 \log \left (c \left (a+b x^3\right )^p\right )-\frac{p \left (3 \sqrt [3]{a} b^{2/3} d^2 e \left (\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+2 \sqrt{3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )\right )-6 \sqrt [3]{a} b^{2/3} d^2 e \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 \left (b d^3-a e^3\right ) \log \left (a+b x^3\right )-9 b d e^2 x^2 \, _2F_1\left (\frac{2}{3},1;\frac{5}{3};-\frac{b x^3}{a}\right )+18 b d^2 e x+9 b d e^2 x^2+2 b e^3 x^3\right )}{2 b}}{3 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.733, size = 537, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [C]  time = 44.5933, size = 12299, normalized size = 49.2 \begin{align*} \text{result too large to display} \end{align*}

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*e^2*p*x^3 + 18*b*d*e*p*x^2 + 36*b*d^2*p*x - 2*(2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*
b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*
a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^
6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p
^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))*b*log(1/4*(2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b
*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a
*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6
*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^
3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))^2*b^2*e + 9*(b^2*d^5 + a*b*d^2*e^3)*p^2*x + 1/2*(3*b^2*d^3 - 2*a*
b*e^3)*(2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(
2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b
^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(
9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))*p +
(15*a*b*d^3*e^2 + a^2*e^5)*p^2) - (6*a*e^2*p - (2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^
2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*
p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3
+ 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e
^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))*b - 3*sqrt(1/3)*b*sqrt((4*(2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 -
(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 -
3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a
^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*
d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))*a*b*e^2*p - (2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 -
(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 -
3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*
a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2
*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))^2*b^2 - 4*(36*a*b*d^3*e + a^2*e^4)*p^2)/b^2))*log(-1/4*(2*
a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*
p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3
+ a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*
e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))^2*b^2*e + 18*(
b^2*d^5 + a*b*d^2*e^3)*p^2*x - 1/2*(3*b^2*d^3 - 2*a*b*e^3)*(2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*
a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(
9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*
e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6
*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))*p - (15*a*b*d^3*e^2 + a^2*e^5)*p^2 + 3/4*sqrt(1/3)*((2*a*e^2*p
/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3
+ 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*
e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 +
a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))*b^2*e - 2*(3*b^2*d^3
+ a*b*e^3)*p)*sqrt((4*(2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I
*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*
p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^
3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqr
t(3) + 1))*a*b*e^2*p - (2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-
I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2
*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d
^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sq
rt(3) + 1))^2*b^2 - 4*(36*a*b*d^3*e + a^2*e^4)*p^2)/b^2)) - (6*a*e^2*p - (2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4
*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3
*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)
^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 +
(27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))*b + 3*sqrt(1/3)*b*sqrt((4*(2*a*e^2*p/b - 2*(1/2)
^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3
+ a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3
)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2
)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))*a*b*e^2*p - (2*a*e^2*p/b - 2*(1/2
)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^
3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^
3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^
2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))^2*b^2 - 4*(36*a*b*d^3*e + a^2*e^
4)*p^2)/b^2))*log(-1/4*(2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-
I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2
*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d
^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sq
rt(3) + 1))^2*b^2*e + 18*(b^2*d^5 + a*b*d^2*e^3)*p^2*x - 1/2*(3*b^2*d^3 - 2*a*b*e^3)*(2*a*e^2*p/b - 2*(1/2)^(2
/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 +
a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(
1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a
*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))*p - (15*a*b*d^3*e^2 + a^2*e^5)*p^2 -
3/4*sqrt(1/3)*((2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(
3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3
+ (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/
b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) +
1))*b^2*e - 2*(3*b^2*d^3 + a*b*e^3)*p)*sqrt((4*(2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*p
^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e
*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^3
+ 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*
e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))*a*b*e^2*p - (2*a*e^2*p/b - 2*(1/2)^(2/3)*(a^2*e^4*p^2/b^2 - (9*a*b*d^3*e*
p^2 + a^2*e^4*p^2)/b^2)*(-I*sqrt(3) + 1)/(2*a^3*e^6*p^3/b^3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*
e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3*e^6*p^3)/b^3)^(1/3) - (1/2)^(1/3)*(2*a^3*e^6*p^3/b^
3 + 27*(b*d^3 + a*e^3)*a*d^3*p^3/b^2 - 3*(9*a*b*d^3*e*p^2 + a^2*e^4*p^2)*a*e^2*p/b^3 + (27*a*b^2*d^6*p^3 + a^3
*e^6*p^3)/b^3)^(1/3)*(I*sqrt(3) + 1))^2*b^2 - 4*(36*a*b*d^3*e + a^2*e^4)*p^2)/b^2)) - 4*(b*e^2*p*x^3 + 3*b*d*e
*p*x^2 + 3*b*d^2*p*x)*log(b*x^3 + a) - 4*(b*e^2*x^3 + 3*b*d*e*x^2 + 3*b*d^2*x)*log(c))/b

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.22488, size = 509, normalized size = 2.04 \begin{align*} -\frac{1}{2} \, a b^{2} d 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 )} e - \frac{1}{2} \, a b d^{2} 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} - \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^{2}} - \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^{2}}\right )} + \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} \end{align*}

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]

-1/2*a*b^2*d*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)
)*e - 1/2*a*b*d^2*p*(2*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b) - 2*sqrt(3)*(-a*b^2)^(1/3)*arctan(1/3*sqr
t(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a*b^2) - (-a*b^2)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a*b
^2)) + 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*log(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