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]

-1/6*e^3*(d*x+c)/b/d/(a+b*(d*x+c)^3)^2+1/18*e^3*(d*x+c)/a/b/d/(a+b*(d*x+c)^3)+1/27*e^3*ln(a^(1/3)+b^(1/3)*(d*x
+c))/a^(5/3)/b^(4/3)/d-1/54*e^3*ln(a^(2/3)-a^(1/3)*b^(1/3)*(d*x+c)+b^(2/3)*(d*x+c)^2)/a^(5/3)/b^(4/3)/d-1/27*e
^3*arctan(1/3*(a^(1/3)-2*b^(1/3)*(d*x+c))/a^(1/3)*3^(1/2))/a^(5/3)/b^(4/3)/d*3^(1/2)

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Rubi [A]  time = 0.16, antiderivative size = 216, normalized size of antiderivative = 1.00, 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 31

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

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*}

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Mathematica [A]  time = 0.02, 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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fricas [B]  time = 1.20, size = 1822, normalized size = 8.44 \[ \text {result too large to display} \]

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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giac [A]  time = 0.28, size = 288, normalized size = 1.33 \[ \frac {2 \, \sqrt {3} \left (\frac {1}{a^{2} b d^{3}}\right )^{\frac {1}{3}} \arctan \left (-\frac {b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}}}{\sqrt {3} b d x + \sqrt {3} b c - \sqrt {3} \left (a b^{2}\right )^{\frac {1}{3}}}\right ) e^{3} - \left (\frac {1}{a^{2} b d^{3}}\right )^{\frac {1}{3}} e^{3} \log \left (4 \, {\left (\sqrt {3} b d x + \sqrt {3} b c - \sqrt {3} \left (a b^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}}\right )}^{2}\right ) + 2 \, \left (\frac {1}{a^{2} b d^{3}}\right )^{\frac {1}{3}} e^{3} \log \left ({\left | b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}} \right |}\right )}{54 \, a b} + \frac {b d^{4} x^{4} e^{3} + 4 \, b c d^{3} x^{3} e^{3} + 6 \, b c^{2} d^{2} x^{2} e^{3} + 4 \, b c^{3} d x e^{3} + b c^{4} e^{3} - 2 \, a d x e^{3} - 2 \, a c e^{3}}{18 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )}^{2} a b d} \]

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]

1/54*(2*sqrt(3)*(1/(a^2*b*d^3))^(1/3)*arctan(-(b*d*x + b*c + (a*b^2)^(1/3))/(sqrt(3)*b*d*x + sqrt(3)*b*c - sqr
t(3)*(a*b^2)^(1/3)))*e^3 - (1/(a^2*b*d^3))^(1/3)*e^3*log(4*(sqrt(3)*b*d*x + sqrt(3)*b*c - sqrt(3)*(a*b^2)^(1/3
))^2 + 4*(b*d*x + b*c + (a*b^2)^(1/3))^2) + 2*(1/(a^2*b*d^3))^(1/3)*e^3*log(abs(b*d*x + b*c + (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)

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maple [C]  time = 0.02, size = 414, normalized size = 1.92 \[ \frac {d^{3} e^{3} x^{4}}{18 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right )^{2} a}+\frac {2 c \,d^{2} e^{3} x^{3}}{9 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right )^{2} a}+\frac {c^{2} d \,e^{3} x^{2}}{3 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right )^{2} a}+\frac {2 c^{3} e^{3} x}{9 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right )^{2} a}+\frac {c^{4} e^{3}}{18 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right )^{2} a d}-\frac {e^{3} x}{9 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right )^{2} b}-\frac {c \,e^{3}}{9 \left (b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a \right )^{2} b d}+\frac {e^{3} \ln \left (-\RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+x \right )}{27 a \,b^{2} d \left (d^{2} \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )^{2}+2 c d \RootOf \left (b \,d^{3} \textit {\_Z}^{3}+3 b \,d^{2} c \,\textit {\_Z}^{2}+3 b d \,c^{2} \textit {\_Z} +b \,c^{3}+a \right )+c^{2}\right )} \]

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/a*c^4/d-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*c/d+1/27*e^3/b^2/a/d*sum(1/(_R^2*d^2+2*_R*c*d+c^2)*ln(-_R+x),_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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {\frac {1}{6} \, e^{3} {\left (\frac {2 \, \sqrt {3} \left (\frac {1}{a^{2} b}\right )^{\frac {1}{3}} \arctan \left (-\frac {b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}}}{\sqrt {3} b d x + \sqrt {3} b c - \sqrt {3} \left (a b^{2}\right )^{\frac {1}{3}}}\right )}{d} - \frac {\left (\frac {1}{a^{2} b}\right )^{\frac {1}{3}} \log \left (4 \, {\left (\sqrt {3} b d x + \sqrt {3} b c - \sqrt {3} \left (a b^{2}\right )^{\frac {1}{3}}\right )}^{2} + 4 \, {\left (b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}}\right )}^{2}\right )}{d} + \frac {2 \, \left (\frac {1}{a^{2} b}\right )^{\frac {1}{3}} \log \left ({\left | b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}} \right |}\right )}{d}\right )}}{9 \, a b} + \frac {b d^{4} e^{3} x^{4} + 4 \, b c d^{3} e^{3} x^{3} + 6 \, b c^{2} d^{2} e^{3} x^{2} + 2 \, {\left (2 \, b c^{3} - a\right )} d e^{3} x + {\left (b c^{4} - 2 \, a c\right )} e^{3}}{18 \, {\left (a b^{3} d^{7} x^{6} + 6 \, a b^{3} c d^{6} x^{5} + 15 \, a b^{3} c^{2} d^{5} x^{4} + 2 \, {\left (10 \, a b^{3} c^{3} + a^{2} b^{2}\right )} d^{4} x^{3} + 3 \, {\left (5 \, a b^{3} c^{4} + 2 \, a^{2} b^{2} c\right )} d^{3} x^{2} + 6 \, {\left (a b^{3} c^{5} + a^{2} b^{2} c^{2}\right )} d^{2} x + {\left (a b^{3} c^{6} + 2 \, a^{2} b^{2} c^{3} + a^{3} b\right )} d\right )}} \]

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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mupad [B]  time = 1.67, size = 349, normalized size = 1.62 \[ \frac {\frac {b\,c^4\,e^3-2\,a\,c\,e^3}{18\,a\,b\,d}+\frac {d^3\,e^3\,x^4}{18\,a}-\frac {e^3\,x\,\left (a-2\,b\,c^3\right )}{9\,a\,b}+\frac {c^2\,d\,e^3\,x^2}{3\,a}+\frac {2\,c\,d^2\,e^3\,x^3}{9\,a}}{x^3\,\left (20\,b^2\,c^3\,d^3+2\,a\,b\,d^3\right )+x^2\,\left (15\,b^2\,c^4\,d^2+6\,a\,b\,c\,d^2\right )+a^2+x\,\left (6\,d\,b^2\,c^5+6\,a\,d\,b\,c^2\right )+b^2\,c^6+b^2\,d^6\,x^6+2\,a\,b\,c^3+6\,b^2\,c\,d^5\,x^5+15\,b^2\,c^2\,d^4\,x^4}+\frac {e^3\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{27\,a^{5/3}\,b^{4/3}\,d}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,c-2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (e^3+\sqrt {3}\,e^3\,1{}\mathrm {i}\right )}{54\,a^{5/3}\,b^{4/3}\,d}+\frac {e^3\,\ln \left (2\,b^{1/3}\,c-a^{1/3}+2\,b^{1/3}\,d\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )}{27\,a^{5/3}\,b^{4/3}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*c^4*e^3 - 2*a*c*e^3)/(18*a*b*d) + (d^3*e^3*x^4)/(18*a) - (e^3*x*(a - 2*b*c^3))/(9*a*b) + (c^2*d*e^3*x^2)/(
3*a) + (2*c*d^2*e^3*x^3)/(9*a))/(x^3*(20*b^2*c^3*d^3 + 2*a*b*d^3) + x^2*(15*b^2*c^4*d^2 + 6*a*b*c*d^2) + a^2 +
x*(6*b^2*c^5*d + 6*a*b*c^2*d) + b^2*c^6 + b^2*d^6*x^6 + 2*a*b*c^3 + 6*b^2*c*d^5*x^5 + 15*b^2*c^2*d^4*x^4) + (
e^3*log(b^(1/3)*c + a^(1/3) + b^(1/3)*d*x))/(27*a^(5/3)*b^(4/3)*d) - (log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*c + a
^(1/3) - 2*b^(1/3)*d*x)*(3^(1/2)*e^3*1i + e^3))/(54*a^(5/3)*b^(4/3)*d) + (e^3*log(3^(1/2)*a^(1/3)*1i + 2*b^(1/
3)*c - a^(1/3) + 2*b^(1/3)*d*x)*((3^(1/2)*1i)/2 - 1/2))/(27*a^(5/3)*b^(4/3)*d)

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sympy [A]  time = 3.66, size = 298, normalized size = 1.38 \[ \frac {- 2 a c e^{3} + b c^{4} e^{3} + 6 b c^{2} d^{2} e^{3} x^{2} + 4 b c d^{3} e^{3} x^{3} + b d^{4} e^{3} x^{4} + x \left (- 2 a d e^{3} + 4 b c^{3} d e^{3}\right )}{18 a^{3} b d + 36 a^{2} b^{2} c^{3} d + 18 a b^{3} c^{6} d + 270 a b^{3} c^{2} d^{5} x^{4} + 108 a b^{3} c d^{6} x^{5} + 18 a b^{3} d^{7} x^{6} + x^{3} \left (36 a^{2} b^{2} d^{4} + 360 a b^{3} c^{3} d^{4}\right ) + x^{2} \left (108 a^{2} b^{2} c d^{3} + 270 a b^{3} c^{4} d^{3}\right ) + x \left (108 a^{2} b^{2} c^{2} d^{2} + 108 a b^{3} c^{5} d^{2}\right )} + \frac {e^{3} \operatorname {RootSum} {\left (19683 t^{3} a^{5} b^{4} - 1, \left (t \mapsto t \log {\left (x + \frac {27 t a^{2} b e^{3} + c e^{3}}{d e^{3}} \right )} \right )\right )}}{d} \]

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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