3.2900 \(\int \frac{1}{(c e+d e x)^3 (a+b (c+d x)^3)^2} \, dx\)

Optimal. Leaf size=204 \[ -\frac{5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d e^3}+\frac{5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{8/3} d e^3}+\frac{5 b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{8/3} d e^3}-\frac{5}{6 a^2 d e^3 (c+d x)^2}+\frac{1}{3 a d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )} \]

[Out]

-5/(6*a^2*d*e^3*(c + d*x)^2) + 1/(3*a*d*e^3*(c + d*x)^2*(a + b*(c + d*x)^3)) + (5*b^(2/3)*ArcTan[(a^(1/3) - 2*
b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(8/3)*d*e^3) - (5*b^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)]
)/(9*a^(8/3)*d*e^3) + (5*b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(18*a^(8/3)*d
*e^3)

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Rubi [A]  time = 0.150436, antiderivative size = 204, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.375, Rules used = {372, 290, 325, 200, 31, 634, 617, 204, 628} \[ -\frac{5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d e^3}+\frac{5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{8/3} d e^3}+\frac{5 b^{2/3} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{8/3} d e^3}-\frac{5}{6 a^2 d e^3 (c+d x)^2}+\frac{1}{3 a d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*e + d*e*x)^3*(a + b*(c + d*x)^3)^2),x]

[Out]

-5/(6*a^2*d*e^3*(c + d*x)^2) + 1/(3*a*d*e^3*(c + d*x)^2*(a + b*(c + d*x)^3)) + (5*b^(2/3)*ArcTan[(a^(1/3) - 2*
b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(8/3)*d*e^3) - (5*b^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)]
)/(9*a^(8/3)*d*e^3) + (5*b^(2/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(18*a^(8/3)*d
*e^3)

Rule 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*Rt[a, 3] - Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3]^2*x^2), x], x]
/; FreeQ[{a, b}, x]

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]

Rubi steps

\begin{align*} \int \frac{1}{(c e+d e x)^3 \left (a+b (c+d x)^3\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^3 \left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{d e^3}\\ &=\frac{1}{3 a d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^3 \left (a+b x^3\right )} \, dx,x,c+d x\right )}{3 a d e^3}\\ &=-\frac{5}{6 a^2 d e^3 (c+d x)^2}+\frac{1}{3 a d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{a+b x^3} \, dx,x,c+d x\right )}{3 a^2 d e^3}\\ &=-\frac{5}{6 a^2 d e^3 (c+d x)^2}+\frac{1}{3 a d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{9 a^{8/3} d e^3}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{2 \sqrt [3]{a}-\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{9 a^{8/3} d e^3}\\ &=-\frac{5}{6 a^2 d e^3 (c+d x)^2}+\frac{1}{3 a d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )}-\frac{5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d e^3}+\frac{\left (5 b^{2/3}\right ) \operatorname{Subst}\left (\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,x,c+d x\right )}{18 a^{8/3} d e^3}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{6 a^{7/3} d e^3}\\ &=-\frac{5}{6 a^2 d e^3 (c+d x)^2}+\frac{1}{3 a d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )}-\frac{5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d e^3}+\frac{5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{8/3} d e^3}-\frac{\left (5 b^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{3 a^{8/3} d e^3}\\ &=-\frac{5}{6 a^2 d e^3 (c+d x)^2}+\frac{1}{3 a d e^3 (c+d x)^2 \left (a+b (c+d x)^3\right )}+\frac{5 b^{2/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{3 \sqrt{3} a^{8/3} d e^3}-\frac{5 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{8/3} d e^3}+\frac{5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{18 a^{8/3} d e^3}\\ \end{align*}

Mathematica [A]  time = 0.0796293, size = 169, normalized size = 0.83 \[ \frac{5 b^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )-\frac{6 a^{2/3} b (c+d x)}{a+b (c+d x)^3}-\frac{9 a^{2/3}}{(c+d x)^2}-10 b^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-10 \sqrt{3} b^{2/3} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{18 a^{8/3} d e^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*e + d*e*x)^3*(a + b*(c + d*x)^3)^2),x]

[Out]

((-9*a^(2/3))/(c + d*x)^2 - (6*a^(2/3)*b*(c + d*x))/(a + b*(c + d*x)^3) - 10*Sqrt[3]*b^(2/3)*ArcTan[(-a^(1/3)
+ 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))] - 10*b^(2/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)] + 5*b^(2/3)*Log[a^(2/3
) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(18*a^(8/3)*d*e^3)

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Maple [C]  time = 0.013, size = 186, normalized size = 0.9 \begin{align*} -{\frac{1}{2\,{a}^{2}d{e}^{3} \left ( dx+c \right ) ^{2}}}-{\frac{bx}{3\,{e}^{3}{a}^{2} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) }}-{\frac{bc}{3\,{e}^{3}{a}^{2} \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) d}}-{\frac{5}{9\,{a}^{2}d{e}^{3}}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{3}b{d}^{3}+3\,{{\it \_Z}}^{2}bc{d}^{2}+3\,{\it \_Z}\,b{c}^{2}d+b{c}^{3}+a \right ) }{\frac{\ln \left ( x-{\it \_R} \right ) }{{d}^{2}{{\it \_R}}^{2}+2\,cd{\it \_R}+{c}^{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(d*e*x+c*e)^3/(a+b*(d*x+c)^3)^2,x)

[Out]

-1/2/a^2/d/e^3/(d*x+c)^2-1/3/e^3*b/a^2/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)*x-1/3/e^3*b/a^2/(b*d^3*x^
3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)*c/d-5/9/e^3/a^2/d*sum(1/(_R^2*d^2+2*_R*c*d+c^2)*ln(x-_R),_R=RootOf(_Z^3*b
*d^3+3*_Z^2*b*c*d^2+3*_Z*b*c^2*d+b*c^3+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{5 \, b d^{3} x^{3} + 15 \, b c d^{2} x^{2} + 15 \, b c^{2} d x + 5 \, b c^{3} + 3 \, a}{6 \,{\left (a^{2} b d^{6} e^{3} x^{5} + 5 \, a^{2} b c d^{5} e^{3} x^{4} + 10 \, a^{2} b c^{2} d^{4} e^{3} x^{3} +{\left (10 \, a^{2} b c^{3} + a^{3}\right )} d^{3} e^{3} x^{2} +{\left (5 \, a^{2} b c^{4} + 2 \, a^{3} c\right )} d^{2} e^{3} x +{\left (a^{2} b c^{5} + a^{3} c^{2}\right )} d e^{3}\right )}} - \frac{\frac{5}{6} \,{\left (2 \, \sqrt{3} \left (\frac{1}{a^{2} b d^{3}}\right )^{\frac{1}{3}} \arctan \left (-\frac{b d x + b c + \left (a b^{2}\right )^{\frac{1}{3}}}{\sqrt{3} b d x + \sqrt{3} b c - \sqrt{3} \left (a b^{2}\right )^{\frac{1}{3}}}\right ) - \left (\frac{1}{a^{2} b d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (\sqrt{3} b d x + \sqrt{3} b c - \sqrt{3} \left (a b^{2}\right )^{\frac{1}{3}}\right )}^{2} +{\left (b d x + b c + \left (a b^{2}\right )^{\frac{1}{3}}\right )}^{2}\right ) + 2 \, \left (\frac{1}{a^{2} b d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | b d x + b c + \left (a b^{2}\right )^{\frac{1}{3}} \right |}\right )\right )} b}{3 \, a^{2} e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^3/(a+b*(d*x+c)^3)^2,x, algorithm="maxima")

[Out]

-1/6*(5*b*d^3*x^3 + 15*b*c*d^2*x^2 + 15*b*c^2*d*x + 5*b*c^3 + 3*a)/(a^2*b*d^6*e^3*x^5 + 5*a^2*b*c*d^5*e^3*x^4
+ 10*a^2*b*c^2*d^4*e^3*x^3 + (10*a^2*b*c^3 + a^3)*d^3*e^3*x^2 + (5*a^2*b*c^4 + 2*a^3*c)*d^2*e^3*x + (a^2*b*c^5
+ a^3*c^2)*d*e^3) - 5/3*b*integrate(1/(b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a), x)/(a^2*e^3)

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Fricas [B]  time = 1.72608, size = 1161, normalized size = 5.69 \begin{align*} -\frac{15 \, b d^{3} x^{3} + 45 \, b c d^{2} x^{2} + 45 \, b c^{2} d x + 15 \, b c^{3} - 10 \, \sqrt{3}{\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} +{\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} +{\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \arctan \left (\frac{2 \, \sqrt{3}{\left (a d x + a c\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{2}{3}} - \sqrt{3} b}{3 \, b}\right ) + 5 \,{\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} +{\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} +{\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \log \left (b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2} + a^{2} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{2}{3}} +{\left (a b d x + a b c\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}}\right ) - 10 \,{\left (b d^{5} x^{5} + 5 \, b c d^{4} x^{4} + 10 \, b c^{2} d^{3} x^{3} + b c^{5} +{\left (10 \, b c^{3} + a\right )} d^{2} x^{2} + a c^{2} +{\left (5 \, b c^{4} + 2 \, a c\right )} d x\right )} \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}} \log \left (b d x + b c - a \left (-\frac{b^{2}}{a^{2}}\right )^{\frac{1}{3}}\right ) + 9 \, a}{18 \,{\left (a^{2} b d^{6} e^{3} x^{5} + 5 \, a^{2} b c d^{5} e^{3} x^{4} + 10 \, a^{2} b c^{2} d^{4} e^{3} x^{3} +{\left (10 \, a^{2} b c^{3} + a^{3}\right )} d^{3} e^{3} x^{2} +{\left (5 \, a^{2} b c^{4} + 2 \, a^{3} c\right )} d^{2} e^{3} x +{\left (a^{2} b c^{5} + a^{3} c^{2}\right )} d e^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^3/(a+b*(d*x+c)^3)^2,x, algorithm="fricas")

[Out]

-1/18*(15*b*d^3*x^3 + 45*b*c*d^2*x^2 + 45*b*c^2*d*x + 15*b*c^3 - 10*sqrt(3)*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*
c^2*d^3*x^3 + b*c^5 + (10*b*c^3 + a)*d^2*x^2 + a*c^2 + (5*b*c^4 + 2*a*c)*d*x)*(-b^2/a^2)^(1/3)*arctan(1/3*(2*s
qrt(3)*(a*d*x + a*c)*(-b^2/a^2)^(2/3) - sqrt(3)*b)/b) + 5*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2*d^3*x^3 + b*c^
5 + (10*b*c^3 + a)*d^2*x^2 + a*c^2 + (5*b*c^4 + 2*a*c)*d*x)*(-b^2/a^2)^(1/3)*log(b^2*d^2*x^2 + 2*b^2*c*d*x + b
^2*c^2 + a^2*(-b^2/a^2)^(2/3) + (a*b*d*x + a*b*c)*(-b^2/a^2)^(1/3)) - 10*(b*d^5*x^5 + 5*b*c*d^4*x^4 + 10*b*c^2
*d^3*x^3 + b*c^5 + (10*b*c^3 + a)*d^2*x^2 + a*c^2 + (5*b*c^4 + 2*a*c)*d*x)*(-b^2/a^2)^(1/3)*log(b*d*x + b*c -
a*(-b^2/a^2)^(1/3)) + 9*a)/(a^2*b*d^6*e^3*x^5 + 5*a^2*b*c*d^5*e^3*x^4 + 10*a^2*b*c^2*d^4*e^3*x^3 + (10*a^2*b*c
^3 + a^3)*d^3*e^3*x^2 + (5*a^2*b*c^4 + 2*a^3*c)*d^2*e^3*x + (a^2*b*c^5 + a^3*c^2)*d*e^3)

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Sympy [A]  time = 41.2039, size = 231, normalized size = 1.13 \begin{align*} - \frac{3 a + 5 b c^{3} + 15 b c^{2} d x + 15 b c d^{2} x^{2} + 5 b d^{3} x^{3}}{6 a^{3} c^{2} d e^{3} + 6 a^{2} b c^{5} d e^{3} + 60 a^{2} b c^{2} d^{4} e^{3} x^{3} + 30 a^{2} b c d^{5} e^{3} x^{4} + 6 a^{2} b d^{6} e^{3} x^{5} + x^{2} \left (6 a^{3} d^{3} e^{3} + 60 a^{2} b c^{3} d^{3} e^{3}\right ) + x \left (12 a^{3} c d^{2} e^{3} + 30 a^{2} b c^{4} d^{2} e^{3}\right )} + \frac{\operatorname{RootSum}{\left (729 t^{3} a^{8} + 125 b^{2}, \left ( t \mapsto t \log{\left (x + \frac{- 9 t a^{3} + 5 b c}{5 b d} \right )} \right )\right )}}{d e^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)**3/(a+b*(d*x+c)**3)**2,x)

[Out]

-(3*a + 5*b*c**3 + 15*b*c**2*d*x + 15*b*c*d**2*x**2 + 5*b*d**3*x**3)/(6*a**3*c**2*d*e**3 + 6*a**2*b*c**5*d*e**
3 + 60*a**2*b*c**2*d**4*e**3*x**3 + 30*a**2*b*c*d**5*e**3*x**4 + 6*a**2*b*d**6*e**3*x**5 + x**2*(6*a**3*d**3*e
**3 + 60*a**2*b*c**3*d**3*e**3) + x*(12*a**3*c*d**2*e**3 + 30*a**2*b*c**4*d**2*e**3)) + RootSum(729*_t**3*a**8
+ 125*b**2, Lambda(_t, _t*log(x + (-9*_t*a**3 + 5*b*c)/(5*b*d))))/(d*e**3)

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Giac [A]  time = 1.18825, size = 352, normalized size = 1.73 \begin{align*} \frac{5}{9} \, \sqrt{3} \left (-\frac{b^{2} e^{\left (-9\right )}}{a^{8} d^{3}}\right )^{\frac{1}{3}} \arctan \left (-\frac{b d x + b c - \left (-a b^{2}\right )^{\frac{1}{3}}}{\sqrt{3} b d x + \sqrt{3} b c + \sqrt{3} \left (-a b^{2}\right )^{\frac{1}{3}}}\right ) - \frac{5}{18} \, \left (-\frac{b^{2} e^{\left (-9\right )}}{a^{8} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (\sqrt{3} b d x + \sqrt{3} b c + \sqrt{3} \left (-a b^{2}\right )^{\frac{1}{3}}\right )}^{2} +{\left (b d x + b c - \left (-a b^{2}\right )^{\frac{1}{3}}\right )}^{2}\right ) + \frac{5}{9} \, \left (-\frac{b^{2} e^{\left (-9\right )}}{a^{8} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | 3 \, a^{2} b d x e^{3} + 3 \, a^{2} b c e^{3} - 3 \, \left (-a b^{2}\right )^{\frac{1}{3}} a^{2} e^{3} \right |}\right ) - \frac{{\left (b d x + b c\right )} e^{\left (-3\right )}}{3 \,{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} a^{2} d} - \frac{e^{\left (-3\right )}}{2 \,{\left (d x + c\right )}^{2} a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(d*e*x+c*e)^3/(a+b*(d*x+c)^3)^2,x, algorithm="giac")

[Out]

5/9*sqrt(3)*(-b^2*e^(-9)/(a^8*d^3))^(1/3)*arctan(-(b*d*x + b*c - (-a*b^2)^(1/3))/(sqrt(3)*b*d*x + sqrt(3)*b*c
+ sqrt(3)*(-a*b^2)^(1/3))) - 5/18*(-b^2*e^(-9)/(a^8*d^3))^(1/3)*log((sqrt(3)*b*d*x + sqrt(3)*b*c + sqrt(3)*(-a
*b^2)^(1/3))^2 + (b*d*x + b*c - (-a*b^2)^(1/3))^2) + 5/9*(-b^2*e^(-9)/(a^8*d^3))^(1/3)*log(abs(3*a^2*b*d*x*e^3
+ 3*a^2*b*c*e^3 - 3*(-a*b^2)^(1/3)*a^2*e^3)) - 1/3*(b*d*x + b*c)*e^(-3)/((b*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2
*d*x + b*c^3 + a)*a^2*d) - 1/2*e^(-3)/((d*x + c)^2*a^2*d)