3.2903 \(\int \frac{(c e+d e x)^3}{(a+b (c+d x)^3)^3} \, dx\)

Optimal. Leaf size=216 \[ \frac{e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{5/3} b^{4/3} d}-\frac{e^3 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{5/3} b^{4/3} d}-\frac{e^3 \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{5/3} b^{4/3} d}+\frac{e^3 (c+d x)}{18 a b d \left (a+b (c+d x)^3\right )}-\frac{e^3 (c+d x)}{6 b d \left (a+b (c+d x)^3\right )^2} \]

[Out]

-(e^3*(c + d*x))/(6*b*d*(a + b*(c + d*x)^3)^2) + (e^3*(c + d*x))/(18*a*b*d*(a + b*(c + d*x)^3)) - (e^3*ArcTan[
(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(5/3)*b^(4/3)*d) + (e^3*Log[a^(1/3) + b^(1/3)
*(c + d*x)])/(27*a^(5/3)*b^(4/3)*d) - (e^3*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(54
*a^(5/3)*b^(4/3)*d)

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Rubi [A]  time = 0.164838, antiderivative size = 216, 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, 288, 199, 200, 31, 634, 617, 204, 628} \[ \frac{e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{5/3} b^{4/3} d}-\frac{e^3 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{5/3} b^{4/3} d}-\frac{e^3 \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{5/3} b^{4/3} d}+\frac{e^3 (c+d x)}{18 a b d \left (a+b (c+d x)^3\right )}-\frac{e^3 (c+d x)}{6 b d \left (a+b (c+d x)^3\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(e^3*(c + d*x))/(6*b*d*(a + b*(c + d*x)^3)^2) + (e^3*(c + d*x))/(18*a*b*d*(a + b*(c + d*x)^3)) - (e^3*ArcTan[
(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(5/3)*b^(4/3)*d) + (e^3*Log[a^(1/3) + b^(1/3)
*(c + d*x)])/(27*a^(5/3)*b^(4/3)*d) - (e^3*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(54
*a^(5/3)*b^(4/3)*d)

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 288

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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{(c e+d e x)^3}{\left (a+b (c+d x)^3\right )^3} \, dx &=\frac{e^3 \operatorname{Subst}\left (\int \frac{x^3}{\left (a+b x^3\right )^3} \, dx,x,c+d x\right )}{d}\\ &=-\frac{e^3 (c+d x)}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac{e^3 \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{6 b d}\\ &=-\frac{e^3 (c+d x)}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac{e^3 (c+d x)}{18 a b d \left (a+b (c+d x)^3\right )}+\frac{e^3 \operatorname{Subst}\left (\int \frac{1}{a+b x^3} \, dx,x,c+d x\right )}{9 a b d}\\ &=-\frac{e^3 (c+d x)}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac{e^3 (c+d x)}{18 a b d \left (a+b (c+d x)^3\right )}+\frac{e^3 \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{27 a^{5/3} b d}+\frac{e^3 \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 )}{27 a^{5/3} b d}\\ &=-\frac{e^3 (c+d x)}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac{e^3 (c+d x)}{18 a b d \left (a+b (c+d x)^3\right )}+\frac{e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{5/3} b^{4/3} d}-\frac{e^3 \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 )}{54 a^{5/3} b^{4/3} d}+\frac{e^3 \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 )}{18 a^{4/3} b d}\\ &=-\frac{e^3 (c+d x)}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac{e^3 (c+d x)}{18 a b d \left (a+b (c+d x)^3\right )}+\frac{e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{5/3} b^{4/3} d}-\frac{e^3 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{5/3} b^{4/3} d}+\frac{e^3 \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}\right )}{9 a^{5/3} b^{4/3} d}\\ &=-\frac{e^3 (c+d x)}{6 b d \left (a+b (c+d x)^3\right )^2}+\frac{e^3 (c+d x)}{18 a b d \left (a+b (c+d x)^3\right )}-\frac{e^3 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{9 \sqrt{3} a^{5/3} b^{4/3} d}+\frac{e^3 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{5/3} b^{4/3} d}-\frac{e^3 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{5/3} b^{4/3} d}\\ \end{align*}

Mathematica [A]  time = 0.110703, size = 182, normalized size = 0.84 \[ \frac{e^3 \left (-\frac{\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{a^{5/3}}+\frac{2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{a^{5/3}}+\frac{2 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{a^{5/3}}+\frac{3 \sqrt [3]{b} (c+d x)}{a \left (a+b (c+d x)^3\right )}-\frac{9 \sqrt [3]{b} (c+d x)}{\left (a+b (c+d x)^3\right )^2}\right )}{54 b^{4/3} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e^3*((-9*b^(1/3)*(c + d*x))/(a + b*(c + d*x)^3)^2 + (3*b^(1/3)*(c + d*x))/(a*(a + b*(c + d*x)^3)) + (2*Sqrt[3
]*ArcTan[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/a^(5/3) + (2*Log[a^(1/3) + b^(1/3)*(c + d*x)])/a
^(5/3) - Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2]/a^(5/3)))/(54*b^(4/3)*d)

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Maple [C]  time = 0.018, size = 414, normalized size = 1.9 \begin{align*}{\frac{{e}^{3}{d}^{3}{x}^{4}}{18\, \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}a}}+{\frac{2\,{e}^{3}c{d}^{2}{x}^{3}}{9\, \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}a}}+{\frac{{e}^{3}{c}^{2}d{x}^{2}}{3\, \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}a}}+{\frac{2\,{e}^{3}x{c}^{3}}{9\, \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}a}}-{\frac{{e}^{3}x}{9\, \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}b}}+{\frac{{e}^{3}{c}^{4}}{18\, \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}da}}-{\frac{{e}^{3}c}{9\, \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}db}}+{\frac{{e}^{3}}{27\,a{b}^{2}d}\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((d*e*x+c*e)^3/(a+b*(d*x+c)^3)^3,x)

[Out]

1/18*e^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*d^3/a*x^4+2/9*e^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*
x+b*c^3+a)^2*c*d^2/a*x^3+1/3*e^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2/a*c^2*d*x^2+2/9*e^3/(b*d^3*x^
3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2/a*x*c^3-1/9*e^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2/b*x+1/1
8*e^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2*c^4/d/a-1/9*e^3/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c
^3+a)^2*c/d/b+1/27*e^3/b^2/a/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{\frac{1}{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 )} e^{3}}{9 \, a b} + \frac{b d^{4} e^{3} x^{4} + 4 \, b c d^{3} e^{3} x^{3} + 6 \, b c^{2} d^{2} e^{3} x^{2} + 2 \,{\left (2 \, b c^{3} - a\right )} d e^{3} x +{\left (b c^{4} - 2 \, a c\right )} e^{3}}{18 \,{\left (a b^{3} d^{7} x^{6} + 6 \, a b^{3} c d^{6} x^{5} + 15 \, a b^{3} c^{2} d^{5} x^{4} + 2 \,{\left (10 \, a b^{3} c^{3} + a^{2} b^{2}\right )} d^{4} x^{3} + 3 \,{\left (5 \, a b^{3} c^{4} + 2 \, a^{2} b^{2} c\right )} d^{3} x^{2} + 6 \,{\left (a b^{3} c^{5} + a^{2} b^{2} c^{2}\right )} d^{2} x +{\left (a b^{3} c^{6} + 2 \, a^{2} b^{2} c^{3} + a^{3} b\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9*e^3*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*b) + 1/18*(b*d^4*e^3*x^4 + 4*
b*c*d^3*e^3*x^3 + 6*b*c^2*d^2*e^3*x^2 + 2*(2*b*c^3 - a)*d*e^3*x + (b*c^4 - 2*a*c)*e^3)/(a*b^3*d^7*x^6 + 6*a*b^
3*c*d^6*x^5 + 15*a*b^3*c^2*d^5*x^4 + 2*(10*a*b^3*c^3 + a^2*b^2)*d^4*x^3 + 3*(5*a*b^3*c^4 + 2*a^2*b^2*c)*d^3*x^
2 + 6*(a*b^3*c^5 + a^2*b^2*c^2)*d^2*x + (a*b^3*c^6 + 2*a^2*b^2*c^3 + a^3*b)*d)

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Fricas [B]  time = 2.02395, size = 3803, normalized size = 17.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/54*(3*a^2*b^2*d^4*e^3*x^4 + 12*a^2*b^2*c*d^3*e^3*x^3 + 18*a^2*b^2*c^2*d^2*e^3*x^2 + 6*(2*a^2*b^2*c^3 - a^3*
b)*d*e^3*x + 3*(a^2*b^2*c^4 - 2*a^3*b*c)*e^3 + 3*sqrt(1/3)*(a*b^3*d^6*e^3*x^6 + 6*a*b^3*c*d^5*e^3*x^5 + 15*a*b
^3*c^2*d^4*e^3*x^4 + 2*(10*a*b^3*c^3 + a^2*b^2)*d^3*e^3*x^3 + 3*(5*a*b^3*c^4 + 2*a^2*b^2*c)*d^2*e^3*x^2 + 6*(a
*b^3*c^5 + a^2*b^2*c^2)*d*e^3*x + (a*b^3*c^6 + 2*a^2*b^2*c^3 + a^3*b)*e^3)*sqrt(-(a^2*b)^(1/3)/b)*log((2*a*b*d
^3*x^3 + 6*a*b*c*d^2*x^2 + 6*a*b*c^2*d*x + 2*a*b*c^3 - a^2 + 3*sqrt(1/3)*(2*a*b*d^2*x^2 + 4*a*b*c*d*x + 2*a*b*
c^2 + (a^2*b)^(2/3)*(d*x + c) - (a^2*b)^(1/3)*a)*sqrt(-(a^2*b)^(1/3)/b) - 3*(a^2*b)^(1/3)*(a*d*x + a*c))/(b*d^
3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)) - (b^2*d^6*e^3*x^6 + 6*b^2*c*d^5*e^3*x^5 + 15*b^2*c^2*d^4*e^
3*x^4 + 2*(10*b^2*c^3 + a*b)*d^3*e^3*x^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*e^3*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*e^3*x
+ (b^2*c^6 + 2*a*b*c^3 + a^2)*e^3)*(a^2*b)^(2/3)*log(a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^2 - (a^2*b)^(2/3)*(d*x
+ c) + (a^2*b)^(1/3)*a) + 2*(b^2*d^6*e^3*x^6 + 6*b^2*c*d^5*e^3*x^5 + 15*b^2*c^2*d^4*e^3*x^4 + 2*(10*b^2*c^3 +
a*b)*d^3*e^3*x^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*e^3*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*e^3*x + (b^2*c^6 + 2*a*b*c^3
+ a^2)*e^3)*(a^2*b)^(2/3)*log(a*b*d*x + a*b*c + (a^2*b)^(2/3)))/(a^3*b^4*d^7*x^6 + 6*a^3*b^4*c*d^6*x^5 + 15*a
^3*b^4*c^2*d^5*x^4 + 2*(10*a^3*b^4*c^3 + a^4*b^3)*d^4*x^3 + 3*(5*a^3*b^4*c^4 + 2*a^4*b^3*c)*d^3*x^2 + 6*(a^3*b
^4*c^5 + a^4*b^3*c^2)*d^2*x + (a^3*b^4*c^6 + 2*a^4*b^3*c^3 + a^5*b^2)*d), 1/54*(3*a^2*b^2*d^4*e^3*x^4 + 12*a^2
*b^2*c*d^3*e^3*x^3 + 18*a^2*b^2*c^2*d^2*e^3*x^2 + 6*(2*a^2*b^2*c^3 - a^3*b)*d*e^3*x + 3*(a^2*b^2*c^4 - 2*a^3*b
*c)*e^3 + 6*sqrt(1/3)*(a*b^3*d^6*e^3*x^6 + 6*a*b^3*c*d^5*e^3*x^5 + 15*a*b^3*c^2*d^4*e^3*x^4 + 2*(10*a*b^3*c^3
+ a^2*b^2)*d^3*e^3*x^3 + 3*(5*a*b^3*c^4 + 2*a^2*b^2*c)*d^2*e^3*x^2 + 6*(a*b^3*c^5 + a^2*b^2*c^2)*d*e^3*x + (a*
b^3*c^6 + 2*a^2*b^2*c^3 + a^3*b)*e^3)*sqrt((a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*(2*(a^2*b)^(2/3)*(d*x + c) - (a^2
*b)^(1/3)*a)*sqrt((a^2*b)^(1/3)/b)/a^2) - (b^2*d^6*e^3*x^6 + 6*b^2*c*d^5*e^3*x^5 + 15*b^2*c^2*d^4*e^3*x^4 + 2*
(10*b^2*c^3 + a*b)*d^3*e^3*x^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*e^3*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*e^3*x + (b^2*c^
6 + 2*a*b*c^3 + a^2)*e^3)*(a^2*b)^(2/3)*log(a*b*d^2*x^2 + 2*a*b*c*d*x + a*b*c^2 - (a^2*b)^(2/3)*(d*x + c) + (a
^2*b)^(1/3)*a) + 2*(b^2*d^6*e^3*x^6 + 6*b^2*c*d^5*e^3*x^5 + 15*b^2*c^2*d^4*e^3*x^4 + 2*(10*b^2*c^3 + a*b)*d^3*
e^3*x^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*e^3*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*e^3*x + (b^2*c^6 + 2*a*b*c^3 + a^2)*e^
3)*(a^2*b)^(2/3)*log(a*b*d*x + a*b*c + (a^2*b)^(2/3)))/(a^3*b^4*d^7*x^6 + 6*a^3*b^4*c*d^6*x^5 + 15*a^3*b^4*c^2
*d^5*x^4 + 2*(10*a^3*b^4*c^3 + a^4*b^3)*d^4*x^3 + 3*(5*a^3*b^4*c^4 + 2*a^4*b^3*c)*d^3*x^2 + 6*(a^3*b^4*c^5 + a
^4*b^3*c^2)*d^2*x + (a^3*b^4*c^6 + 2*a^4*b^3*c^3 + a^5*b^2)*d)]

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Sympy [A]  time = 23.5682, size = 298, normalized size = 1.38 \begin{align*} \frac{- 2 a c e^{3} + b c^{4} e^{3} + 6 b c^{2} d^{2} e^{3} x^{2} + 4 b c d^{3} e^{3} x^{3} + b d^{4} e^{3} x^{4} + x \left (- 2 a d e^{3} + 4 b c^{3} d e^{3}\right )}{18 a^{3} b d + 36 a^{2} b^{2} c^{3} d + 18 a b^{3} c^{6} d + 270 a b^{3} c^{2} d^{5} x^{4} + 108 a b^{3} c d^{6} x^{5} + 18 a b^{3} d^{7} x^{6} + x^{3} \left (36 a^{2} b^{2} d^{4} + 360 a b^{3} c^{3} d^{4}\right ) + x^{2} \left (108 a^{2} b^{2} c d^{3} + 270 a b^{3} c^{4} d^{3}\right ) + x \left (108 a^{2} b^{2} c^{2} d^{2} + 108 a b^{3} c^{5} d^{2}\right )} + \frac{e^{3} \operatorname{RootSum}{\left (19683 t^{3} a^{5} b^{4} - 1, \left ( t \mapsto t \log{\left (x + \frac{27 t a^{2} b e^{3} + c e^{3}}{d e^{3}} \right )} \right )\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*a*c*e**3 + b*c**4*e**3 + 6*b*c**2*d**2*e**3*x**2 + 4*b*c*d**3*e**3*x**3 + b*d**4*e**3*x**4 + x*(-2*a*d*e**
3 + 4*b*c**3*d*e**3))/(18*a**3*b*d + 36*a**2*b**2*c**3*d + 18*a*b**3*c**6*d + 270*a*b**3*c**2*d**5*x**4 + 108*
a*b**3*c*d**6*x**5 + 18*a*b**3*d**7*x**6 + x**3*(36*a**2*b**2*d**4 + 360*a*b**3*c**3*d**4) + x**2*(108*a**2*b*
*2*c*d**3 + 270*a*b**3*c**4*d**3) + x*(108*a**2*b**2*c**2*d**2 + 108*a*b**3*c**5*d**2)) + e**3*RootSum(19683*_
t**3*a**5*b**4 - 1, Lambda(_t, _t*log(x + (27*_t*a**2*b*e**3 + c*e**3)/(d*e**3))))/d

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Giac [A]  time = 1.21538, size = 387, normalized size = 1.79 \begin{align*} \frac{1}{27} \, \sqrt{3} \left (\frac{1}{a^{5} b^{4} 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 ) e^{3} - \frac{1}{54} \, \left (\frac{1}{a^{5} b^{4} d^{3}}\right )^{\frac{1}{3}} e^{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{1}{27} \, \left (\frac{1}{a^{5} b^{4} d^{3}}\right )^{\frac{1}{3}} e^{3} \log \left ({\left | 9 \, a b^{2} d x + 9 \, a b^{2} c + 9 \, \left (a b^{2}\right )^{\frac{1}{3}} a b \right |}\right ) + \frac{b d^{4} x^{4} e^{3} + 4 \, b c d^{3} x^{3} e^{3} + 6 \, b c^{2} d^{2} x^{2} e^{3} + 4 \, b c^{3} d x e^{3} + b c^{4} e^{3} - 2 \, a d x e^{3} - 2 \, a c e^{3}}{18 \,{\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )}^{2} a b d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/27*sqrt(3)*(1/(a^5*b^4*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)))*e^3 - 1/54*(1/(a^5*b^4*d^3))^(1/3)*e^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) + 1/27*(1/(a^5*b^4*d^3))^(1/3)*e^3*log(abs(9*a*b^2*d*x + 9*a*b^2*c
+ 9*(a*b^2)^(1/3)*a*b)) + 1/18*(b*d^4*x^4*e^3 + 4*b*c*d^3*x^3*e^3 + 6*b*c^2*d^2*x^2*e^3 + 4*b*c^3*d*x*e^3 + b
*c^4*e^3 - 2*a*d*x*e^3 - 2*a*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)^2*a*b*d)