3.2899 \(\int \frac{1}{(c e+d e x)^2 (a+b (c+d x)^3)^2} \, dx\)

Optimal. Leaf size=204 \[ -\frac{2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d e^2}+\frac{4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d e^2}+\frac{4 \sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{7/3} d e^2}-\frac{4}{3 a^2 d e^2 (c+d x)}+\frac{1}{3 a d e^2 (c+d x) \left (a+b (c+d x)^3\right )} \]

[Out]

-4/(3*a^2*d*e^2*(c + d*x)) + 1/(3*a*d*e^2*(c + d*x)*(a + b*(c + d*x)^3)) + (4*b^(1/3)*ArcTan[(a^(1/3) - 2*b^(1
/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(7/3)*d*e^2) + (4*b^(1/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(9
*a^(7/3)*d*e^2) - (2*b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(9*a^(7/3)*d*e^2)

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Rubi [A]  time = 0.151237, 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, 292, 31, 634, 617, 204, 628} \[ -\frac{2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d e^2}+\frac{4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d e^2}+\frac{4 \sqrt [3]{b} \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{7/3} d e^2}-\frac{4}{3 a^2 d e^2 (c+d x)}+\frac{1}{3 a d e^2 (c+d x) \left (a+b (c+d x)^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-4/(3*a^2*d*e^2*(c + d*x)) + 1/(3*a*d*e^2*(c + d*x)*(a + b*(c + d*x)^3)) + (4*b^(1/3)*ArcTan[(a^(1/3) - 2*b^(1
/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(7/3)*d*e^2) + (4*b^(1/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(9
*a^(7/3)*d*e^2) - (2*b^(1/3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(9*a^(7/3)*d*e^2)

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 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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)^2 \left (a+b (c+d x)^3\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{d e^2}\\ &=\frac{1}{3 a d e^2 (c+d x) \left (a+b (c+d x)^3\right )}+\frac{4 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^3\right )} \, dx,x,c+d x\right )}{3 a d e^2}\\ &=-\frac{4}{3 a^2 d e^2 (c+d x)}+\frac{1}{3 a d e^2 (c+d x) \left (a+b (c+d x)^3\right )}-\frac{(4 b) \operatorname{Subst}\left (\int \frac{x}{a+b x^3} \, dx,x,c+d x\right )}{3 a^2 d e^2}\\ &=-\frac{4}{3 a^2 d e^2 (c+d x)}+\frac{1}{3 a d e^2 (c+d x) \left (a+b (c+d x)^3\right )}+\frac{\left (4 b^{2/3}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{9 a^{7/3} d e^2}-\frac{\left (4 b^{2/3}\right ) \operatorname{Subst}\left (\int \frac{\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^{7/3} d e^2}\\ &=-\frac{4}{3 a^2 d e^2 (c+d x)}+\frac{1}{3 a d e^2 (c+d x) \left (a+b (c+d x)^3\right )}+\frac{4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d e^2}-\frac{\left (2 \sqrt [3]{b}\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 )}{9 a^{7/3} d e^2}-\frac{\left (2 b^{2/3}\right ) \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 )}{3 a^2 d e^2}\\ &=-\frac{4}{3 a^2 d e^2 (c+d x)}+\frac{1}{3 a d e^2 (c+d x) \left (a+b (c+d x)^3\right )}+\frac{4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d e^2}-\frac{2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d e^2}-\frac{\left (4 \sqrt [3]{b}\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^{7/3} d e^2}\\ &=-\frac{4}{3 a^2 d e^2 (c+d x)}+\frac{1}{3 a d e^2 (c+d x) \left (a+b (c+d x)^3\right )}+\frac{4 \sqrt [3]{b} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{3 \sqrt{3} a^{7/3} d e^2}+\frac{4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{7/3} d e^2}-\frac{2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{7/3} d e^2}\\ \end{align*}

Mathematica [A]  time = 0.0940398, size = 171, normalized size = 0.84 \[ \frac{-2 \sqrt [3]{b} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )-\frac{3 \sqrt [3]{a} b (c+d x)^2}{a+b (c+d x)^3}+4 \sqrt [3]{b} \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )-4 \sqrt{3} \sqrt [3]{b} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )-\frac{9 \sqrt [3]{a}}{c+d x}}{9 a^{7/3} d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-9*a^(1/3))/(c + d*x) - (3*a^(1/3)*b*(c + d*x)^2)/(a + b*(c + d*x)^3) - 4*Sqrt[3]*b^(1/3)*ArcTan[(-a^(1/3) +
2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))] + 4*b^(1/3)*Log[a^(1/3) + b^(1/3)*(c + d*x)] - 2*b^(1/3)*Log[a^(2/3)
- a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(9*a^(7/3)*d*e^2)

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Maple [C]  time = 0.014, size = 242, normalized size = 1.2 \begin{align*} -{\frac{1}{{a}^{2}d{e}^{2} \left ( dx+c \right ) }}-{\frac{d{x}^{2}b}{3\,{e}^{2}{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{2\,bcx}{3\,{e}^{2}{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{b{c}^{2}}{3\,{e}^{2}{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{4}{9\,{a}^{2}d{e}^{2}}\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{ \left ({\it \_R}\,d+c \right ) \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)^2/(a+b*(d*x+c)^3)^2,x)

[Out]

-1/a^2/d/e^2/(d*x+c)-1/3/e^2*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^2*d-2/3/e^2*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*c-1/3/e^2*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^2/d-4/9
/e^2/a^2/d*sum((_R*d+c)/(_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{4 \, b d^{3} x^{3} + 12 \, b c d^{2} x^{2} + 12 \, b c^{2} d x + 4 \, b c^{3} + 3 \, a}{3 \,{\left (a^{2} b d^{5} e^{2} x^{4} + 4 \, a^{2} b c d^{4} e^{2} x^{3} + 6 \, a^{2} b c^{2} d^{3} e^{2} x^{2} +{\left (4 \, a^{2} b c^{3} + a^{3}\right )} d^{2} e^{2} x +{\left (a^{2} b c^{4} + a^{3} c\right )} d e^{2}\right )}} - \frac{-\frac{2}{3} \,{\left (2 \, \sqrt{3} \left (-\frac{1}{a b^{2} d^{3}}\right )^{\frac{1}{3}} \arctan \left (\frac{\sqrt{3}{\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac{2}{3}}\right )}}{3 \, \left (-a^{2} b\right )^{\frac{2}{3}}}\right ) + \left (-\frac{1}{a b^{2} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left (2 \, a b d x + 2 \, a b c - \left (-a^{2} b\right )^{\frac{2}{3}}\right )}^{2} + 3 \, \left (-a^{2} b\right )^{\frac{4}{3}}\right ) - 2 \, \left (-\frac{1}{a b^{2} d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | a b d x + a b c + \left (-a^{2} b\right )^{\frac{2}{3}} \right |}\right )\right )} b}{3 \, a^{2} e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(4*b*d^3*x^3 + 12*b*c*d^2*x^2 + 12*b*c^2*d*x + 4*b*c^3 + 3*a)/(a^2*b*d^5*e^2*x^4 + 4*a^2*b*c*d^4*e^2*x^3
+ 6*a^2*b*c^2*d^3*e^2*x^2 + (4*a^2*b*c^3 + a^3)*d^2*e^2*x + (a^2*b*c^4 + a^3*c)*d*e^2) - 4/3*b*integrate((d*x
+ c)/(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^2)

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Fricas [B]  time = 1.69315, size = 896, normalized size = 4.39 \begin{align*} -\frac{12 \, b d^{3} x^{3} + 36 \, b c d^{2} x^{2} + 36 \, b c^{2} d x + 12 \, b c^{3} + 4 \, \sqrt{3}{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} +{\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \arctan \left (\frac{2}{3} \, \sqrt{3}{\left (d x + c\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} - \frac{1}{3} \, \sqrt{3}\right ) + 2 \,{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} +{\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \log \left (b d^{2} x^{2} + 2 \, b c d x + b c^{2} -{\left (a d x + a c\right )} \left (\frac{b}{a}\right )^{\frac{2}{3}} + a \left (\frac{b}{a}\right )^{\frac{1}{3}}\right ) - 4 \,{\left (b d^{4} x^{4} + 4 \, b c d^{3} x^{3} + 6 \, b c^{2} d^{2} x^{2} + b c^{4} +{\left (4 \, b c^{3} + a\right )} d x + a c\right )} \left (\frac{b}{a}\right )^{\frac{1}{3}} \log \left (b d x + b c + a \left (\frac{b}{a}\right )^{\frac{2}{3}}\right ) + 9 \, a}{9 \,{\left (a^{2} b d^{5} e^{2} x^{4} + 4 \, a^{2} b c d^{4} e^{2} x^{3} + 6 \, a^{2} b c^{2} d^{3} e^{2} x^{2} +{\left (4 \, a^{2} b c^{3} + a^{3}\right )} d^{2} e^{2} x +{\left (a^{2} b c^{4} + a^{3} c\right )} d e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(12*b*d^3*x^3 + 36*b*c*d^2*x^2 + 36*b*c^2*d*x + 12*b*c^3 + 4*sqrt(3)*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2
*d^2*x^2 + b*c^4 + (4*b*c^3 + a)*d*x + a*c)*(b/a)^(1/3)*arctan(2/3*sqrt(3)*(d*x + c)*(b/a)^(1/3) - 1/3*sqrt(3)
) + 2*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^2*x^2 + b*c^4 + (4*b*c^3 + a)*d*x + a*c)*(b/a)^(1/3)*log(b*d^2*x^
2 + 2*b*c*d*x + b*c^2 - (a*d*x + a*c)*(b/a)^(2/3) + a*(b/a)^(1/3)) - 4*(b*d^4*x^4 + 4*b*c*d^3*x^3 + 6*b*c^2*d^
2*x^2 + b*c^4 + (4*b*c^3 + a)*d*x + a*c)*(b/a)^(1/3)*log(b*d*x + b*c + a*(b/a)^(2/3)) + 9*a)/(a^2*b*d^5*e^2*x^
4 + 4*a^2*b*c*d^4*e^2*x^3 + 6*a^2*b*c^2*d^3*e^2*x^2 + (4*a^2*b*c^3 + a^3)*d^2*e^2*x + (a^2*b*c^4 + a^3*c)*d*e^
2)

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Sympy [A]  time = 7.75428, size = 196, normalized size = 0.96 \begin{align*} - \frac{3 a + 4 b c^{3} + 12 b c^{2} d x + 12 b c d^{2} x^{2} + 4 b d^{3} x^{3}}{3 a^{3} c d e^{2} + 3 a^{2} b c^{4} d e^{2} + 18 a^{2} b c^{2} d^{3} e^{2} x^{2} + 12 a^{2} b c d^{4} e^{2} x^{3} + 3 a^{2} b d^{5} e^{2} x^{4} + x \left (3 a^{3} d^{2} e^{2} + 12 a^{2} b c^{3} d^{2} e^{2}\right )} + \frac{\operatorname{RootSum}{\left (729 t^{3} a^{7} - 64 b, \left ( t \mapsto t \log{\left (x + \frac{81 t^{2} a^{5} + 16 b c}{16 b d} \right )} \right )\right )}}{d e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(3*a + 4*b*c**3 + 12*b*c**2*d*x + 12*b*c*d**2*x**2 + 4*b*d**3*x**3)/(3*a**3*c*d*e**2 + 3*a**2*b*c**4*d*e**2 +
18*a**2*b*c**2*d**3*e**2*x**2 + 12*a**2*b*c*d**4*e**2*x**3 + 3*a**2*b*d**5*e**2*x**4 + x*(3*a**3*d**2*e**2 +
12*a**2*b*c**3*d**2*e**2)) + RootSum(729*_t**3*a**7 - 64*b, Lambda(_t, _t*log(x + (81*_t**2*a**5 + 16*b*c)/(16
*b*d))))/(d*e**2)

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Giac [A]  time = 1.1818, size = 363, normalized size = 1.78 \begin{align*} \frac{4 \, \left (\frac{b}{a d^{3}}\right )^{\frac{1}{3}} e^{\left (-2\right )} \log \left ({\left | -\left (\frac{b}{a d^{3}}\right )^{\frac{1}{3}} e^{\left (-2\right )} - \frac{e^{\left (-1\right )}}{{\left (d x e + c e\right )} d} \right |}\right )}{9 \, a^{2}} - \frac{4 \, \sqrt{3} \left (a^{2} b\right )^{\frac{1}{3}} \arctan \left (\frac{\sqrt{3}{\left (\left (\frac{b}{a d^{3}}\right )^{\frac{1}{3}} e^{\left (-2\right )} - \frac{2 \, e^{\left (-1\right )}}{{\left (d x e + c e\right )} d}\right )} e^{2}}{3 \, \left (\frac{b}{a d^{3}}\right )^{\frac{1}{3}}}\right ) e^{\left (-2\right )}}{9 \, a^{3} d} - \frac{2 \, \left (a^{2} b\right )^{\frac{1}{3}} e^{\left (-2\right )} \log \left (\left (\frac{b}{a d^{3}}\right )^{\frac{2}{3}} e^{\left (-4\right )} - \frac{\left (\frac{b}{a d^{3}}\right )^{\frac{1}{3}} e^{\left (-3\right )}}{{\left (d x e + c e\right )} d} + \frac{e^{\left (-2\right )}}{{\left (d x e + c e\right )}^{2} d^{2}}\right )}{9 \, a^{3} d} - \frac{e^{\left (-1\right )}}{{\left (d x e + c e\right )} a^{2} d} - \frac{b e^{\left (-1\right )}}{3 \,{\left (d x e + c e\right )} a^{2}{\left (b + \frac{a e^{3}}{{\left (d x e + c e\right )}^{3}}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/9*(b/(a*d^3))^(1/3)*e^(-2)*log(abs(-(b/(a*d^3))^(1/3)*e^(-2) - e^(-1)/((d*x*e + c*e)*d)))/a^2 - 4/9*sqrt(3)*
(a^2*b)^(1/3)*arctan(1/3*sqrt(3)*((b/(a*d^3))^(1/3)*e^(-2) - 2*e^(-1)/((d*x*e + c*e)*d))*e^2/(b/(a*d^3))^(1/3)
)*e^(-2)/(a^3*d) - 2/9*(a^2*b)^(1/3)*e^(-2)*log((b/(a*d^3))^(2/3)*e^(-4) - (b/(a*d^3))^(1/3)*e^(-3)/((d*x*e +
c*e)*d) + e^(-2)/((d*x*e + c*e)^2*d^2))/(a^3*d) - e^(-1)/((d*x*e + c*e)*a^2*d) - 1/3*b*e^(-1)/((d*x*e + c*e)*a
^2*(b + a*e^3/(d*x*e + c*e)^3)*d)