3.2881 \(\int \frac{1}{(a+b (c+d x)^3)^3} \, dx\)

Optimal. Leaf size=198 \[ -\frac{5 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{8/3} \sqrt [3]{b} d}+\frac{5 (c+d x)}{18 a^2 d \left (a+b (c+d x)^3\right )}+\frac{5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{8/3} \sqrt [3]{b} d}-\frac{5 \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{8/3} \sqrt [3]{b} d}+\frac{c+d x}{6 a d \left (a+b (c+d x)^3\right )^2} \]

[Out]

(c + d*x)/(6*a*d*(a + b*(c + d*x)^3)^2) + (5*(c + d*x))/(18*a^2*d*(a + b*(c + d*x)^3)) - (5*ArcTan[(a^(1/3) -
2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(8/3)*b^(1/3)*d) + (5*Log[a^(1/3) + b^(1/3)*(c + d*x)])/
(27*a^(8/3)*b^(1/3)*d) - (5*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(54*a^(8/3)*b^(1/3
)*d)

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Rubi [A]  time = 0.156101, antiderivative size = 198, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.615, Rules used = {247, 199, 200, 31, 634, 617, 204, 628} \[ -\frac{5 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{8/3} \sqrt [3]{b} d}+\frac{5 (c+d x)}{18 a^2 d \left (a+b (c+d x)^3\right )}+\frac{5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{8/3} \sqrt [3]{b} d}-\frac{5 \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{9 \sqrt{3} a^{8/3} \sqrt [3]{b} d}+\frac{c+d x}{6 a d \left (a+b (c+d x)^3\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*(c + d*x)^3)^(-3),x]

[Out]

(c + d*x)/(6*a*d*(a + b*(c + d*x)^3)^2) + (5*(c + d*x))/(18*a^2*d*(a + b*(c + d*x)^3)) - (5*ArcTan[(a^(1/3) -
2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(8/3)*b^(1/3)*d) + (5*Log[a^(1/3) + b^(1/3)*(c + d*x)])/
(27*a^(8/3)*b^(1/3)*d) - (5*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/(54*a^(8/3)*b^(1/3
)*d)

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, 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{1}{\left (a+b (c+d x)^3\right )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (a+b x^3\right )^3} \, dx,x,c+d x\right )}{d}\\ &=\frac{c+d x}{6 a d \left (a+b (c+d x)^3\right )^2}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{\left (a+b x^3\right )^2} \, dx,x,c+d x\right )}{6 a d}\\ &=\frac{c+d x}{6 a d \left (a+b (c+d x)^3\right )^2}+\frac{5 (c+d x)}{18 a^2 d \left (a+b (c+d x)^3\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{a+b x^3} \, dx,x,c+d x\right )}{9 a^2 d}\\ &=\frac{c+d x}{6 a d \left (a+b (c+d x)^3\right )^2}+\frac{5 (c+d x)}{18 a^2 d \left (a+b (c+d x)^3\right )}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{27 a^{8/3} d}+\frac{5 \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^{8/3} d}\\ &=\frac{c+d x}{6 a d \left (a+b (c+d x)^3\right )^2}+\frac{5 (c+d x)}{18 a^2 d \left (a+b (c+d x)^3\right )}+\frac{5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{8/3} \sqrt [3]{b} d}+\frac{5 \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^{7/3} d}-\frac{5 \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^{8/3} \sqrt [3]{b} d}\\ &=\frac{c+d x}{6 a d \left (a+b (c+d x)^3\right )^2}+\frac{5 (c+d x)}{18 a^2 d \left (a+b (c+d x)^3\right )}+\frac{5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{8/3} \sqrt [3]{b} d}-\frac{5 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{8/3} \sqrt [3]{b} d}+\frac{5 \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^{8/3} \sqrt [3]{b} d}\\ &=\frac{c+d x}{6 a d \left (a+b (c+d x)^3\right )^2}+\frac{5 (c+d x)}{18 a^2 d \left (a+b (c+d x)^3\right )}-\frac{5 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{9 \sqrt{3} a^{8/3} \sqrt [3]{b} d}+\frac{5 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{27 a^{8/3} \sqrt [3]{b} d}-\frac{5 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{54 a^{8/3} \sqrt [3]{b} d}\\ \end{align*}

Mathematica [A]  time = 0.0980796, size = 176, normalized size = 0.89 \[ \frac{-\frac{5 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{\sqrt [3]{b}}+\frac{9 a^{5/3} (c+d x)}{\left (a+b (c+d x)^3\right )^2}+\frac{15 a^{2/3} (c+d x)}{a+b (c+d x)^3}+\frac{10 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{\sqrt [3]{b}}+\frac{10 \sqrt{3} \tan ^{-1}\left (\frac{2 \sqrt [3]{b} (c+d x)-\sqrt [3]{a}}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt [3]{b}}}{54 a^{8/3} d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(c + d*x)^3)^(-3),x]

[Out]

((9*a^(5/3)*(c + d*x))/(a + b*(c + d*x)^3)^2 + (15*a^(2/3)*(c + d*x))/(a + b*(c + d*x)^3) + (10*Sqrt[3]*ArcTan
[(-a^(1/3) + 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/b^(1/3) + (10*Log[a^(1/3) + b^(1/3)*(c + d*x)])/b^(1/3)
- (5*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/3)*(c + d*x)^2])/b^(1/3))/(54*a^(8/3)*d)

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Maple [C]  time = 0.015, size = 185, normalized size = 0.9 \begin{align*}{\frac{1}{ \left ( b{d}^{3}{x}^{3}+3\,bc{d}^{2}{x}^{2}+3\,b{c}^{2}dx+b{c}^{3}+a \right ) ^{2}} \left ({\frac{5\,b{d}^{3}{x}^{4}}{18\,{a}^{2}}}+{\frac{10\,bc{d}^{2}{x}^{3}}{9\,{a}^{2}}}+{\frac{5\,b{c}^{2}d{x}^{2}}{3\,{a}^{2}}}+{\frac{ \left ( 10\,b{c}^{3}+4\,a \right ) x}{9\,{a}^{2}}}+{\frac{c \left ( 5\,b{c}^{3}+8\,a \right ) }{18\,{a}^{2}d}} \right ) }+{\frac{5}{27\,b{a}^{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(1/(a+b*(d*x+c)^3)^3,x)

[Out]

(5/18/a^2*b*d^3*x^4+10/9*c*d^2*b/a^2*x^3+5/3/a^2*b*c^2*d*x^2+2/9*(5*b*c^3+2*a)/a^2*x+1/18*c/d*(5*b*c^3+8*a)/a^
2)/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)^2+5/27/a^2/b/d*sum(1/(_R^2*d^2+2*_R*c*d+c^2)*ln(x-_R),_R=Root
Of(_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^{4} x^{4} + 20 \, b c d^{3} x^{3} + 30 \, b c^{2} d^{2} x^{2} + 5 \, b c^{4} + 4 \,{\left (5 \, b c^{3} + 2 \, a\right )} d x + 8 \, a c}{18 \,{\left (a^{2} b^{2} d^{7} x^{6} + 6 \, a^{2} b^{2} c d^{6} x^{5} + 15 \, a^{2} b^{2} c^{2} d^{5} x^{4} + 2 \,{\left (10 \, a^{2} b^{2} c^{3} + a^{3} b\right )} d^{4} x^{3} + 3 \,{\left (5 \, a^{2} b^{2} c^{4} + 2 \, a^{3} b c\right )} d^{3} x^{2} + 6 \,{\left (a^{2} b^{2} c^{5} + a^{3} b c^{2}\right )} d^{2} x +{\left (a^{2} b^{2} c^{6} + 2 \, a^{3} b c^{3} + a^{4}\right )} d\right )}} + \frac{5 \,{\left (\frac{1}{3} \, \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 ) - \frac{1}{6} \, \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 ) + \frac{1}{3} \, \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 )}}{9 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(5*b*d^4*x^4 + 20*b*c*d^3*x^3 + 30*b*c^2*d^2*x^2 + 5*b*c^4 + 4*(5*b*c^3 + 2*a)*d*x + 8*a*c)/(a^2*b^2*d^7*
x^6 + 6*a^2*b^2*c*d^6*x^5 + 15*a^2*b^2*c^2*d^5*x^4 + 2*(10*a^2*b^2*c^3 + a^3*b)*d^4*x^3 + 3*(5*a^2*b^2*c^4 + 2
*a^3*b*c)*d^3*x^2 + 6*(a^2*b^2*c^5 + a^3*b*c^2)*d^2*x + (a^2*b^2*c^6 + 2*a^3*b*c^3 + a^4)*d) + 5/9*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

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Fricas [B]  time = 1.88683, size = 3522, normalized size = 17.79 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/54*(15*a^2*b^2*d^4*x^4 + 60*a^2*b^2*c*d^3*x^3 + 90*a^2*b^2*c^2*d^2*x^2 + 15*a^2*b^2*c^4 + 24*a^3*b*c + 12*(
5*a^2*b^2*c^3 + 2*a^3*b)*d*x + 15*sqrt(1/3)*(a*b^3*d^6*x^6 + 6*a*b^3*c*d^5*x^5 + 15*a*b^3*c^2*d^4*x^4 + a*b^3*
c^6 + 2*a^2*b^2*c^3 + 2*(10*a*b^3*c^3 + a^2*b^2)*d^3*x^3 + 3*(5*a*b^3*c^4 + 2*a^2*b^2*c)*d^2*x^2 + a^3*b + 6*(
a*b^3*c^5 + a^2*b^2*c^2)*d*x)*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)) - 5*(b^2*d^6*x^6 + 6*b^2*c*d^5*x^5 + 15*b^2*c^2*d^4*x^4 + b^2*c^6 + 2*(10*b^2*c^3 + a*b)*d^3*x^3 + 2*a*b*
c^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*x + a^2)*(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) + 10*(b^2*d^6*x^6 + 6*b^2*c*d^5*x^5 + 15*b^2*c^2
*d^4*x^4 + b^2*c^6 + 2*(10*b^2*c^3 + a*b)*d^3*x^3 + 2*a*b*c^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*x^2 + 6*(b^2*c^5 +
a*b*c^2)*d*x + a^2)*(a^2*b)^(2/3)*log(a*b*d*x + a*b*c + (a^2*b)^(2/3)))/(a^4*b^3*d^7*x^6 + 6*a^4*b^3*c*d^6*x^
5 + 15*a^4*b^3*c^2*d^5*x^4 + 2*(10*a^4*b^3*c^3 + a^5*b^2)*d^4*x^3 + 3*(5*a^4*b^3*c^4 + 2*a^5*b^2*c)*d^3*x^2 +
6*(a^4*b^3*c^5 + a^5*b^2*c^2)*d^2*x + (a^4*b^3*c^6 + 2*a^5*b^2*c^3 + a^6*b)*d), 1/54*(15*a^2*b^2*d^4*x^4 + 60*
a^2*b^2*c*d^3*x^3 + 90*a^2*b^2*c^2*d^2*x^2 + 15*a^2*b^2*c^4 + 24*a^3*b*c + 12*(5*a^2*b^2*c^3 + 2*a^3*b)*d*x +
30*sqrt(1/3)*(a*b^3*d^6*x^6 + 6*a*b^3*c*d^5*x^5 + 15*a*b^3*c^2*d^4*x^4 + a*b^3*c^6 + 2*a^2*b^2*c^3 + 2*(10*a*b
^3*c^3 + a^2*b^2)*d^3*x^3 + 3*(5*a*b^3*c^4 + 2*a^2*b^2*c)*d^2*x^2 + a^3*b + 6*(a*b^3*c^5 + a^2*b^2*c^2)*d*x)*s
qrt((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)
- 5*(b^2*d^6*x^6 + 6*b^2*c*d^5*x^5 + 15*b^2*c^2*d^4*x^4 + b^2*c^6 + 2*(10*b^2*c^3 + a*b)*d^3*x^3 + 2*a*b*c^3
+ 3*(5*b^2*c^4 + 2*a*b*c)*d^2*x^2 + 6*(b^2*c^5 + a*b*c^2)*d*x + a^2)*(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) + 10*(b^2*d^6*x^6 + 6*b^2*c*d^5*x^5 + 15*b^2*c^2*d^4
*x^4 + b^2*c^6 + 2*(10*b^2*c^3 + a*b)*d^3*x^3 + 2*a*b*c^3 + 3*(5*b^2*c^4 + 2*a*b*c)*d^2*x^2 + 6*(b^2*c^5 + a*b
*c^2)*d*x + a^2)*(a^2*b)^(2/3)*log(a*b*d*x + a*b*c + (a^2*b)^(2/3)))/(a^4*b^3*d^7*x^6 + 6*a^4*b^3*c*d^6*x^5 +
15*a^4*b^3*c^2*d^5*x^4 + 2*(10*a^4*b^3*c^3 + a^5*b^2)*d^4*x^3 + 3*(5*a^4*b^3*c^4 + 2*a^5*b^2*c)*d^3*x^2 + 6*(a
^4*b^3*c^5 + a^5*b^2*c^2)*d^2*x + (a^4*b^3*c^6 + 2*a^5*b^2*c^3 + a^6*b)*d)]

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Sympy [A]  time = 12.8756, size = 267, normalized size = 1.35 \begin{align*} \frac{8 a c + 5 b c^{4} + 30 b c^{2} d^{2} x^{2} + 20 b c d^{3} x^{3} + 5 b d^{4} x^{4} + x \left (8 a d + 20 b c^{3} d\right )}{18 a^{4} d + 36 a^{3} b c^{3} d + 18 a^{2} b^{2} c^{6} d + 270 a^{2} b^{2} c^{2} d^{5} x^{4} + 108 a^{2} b^{2} c d^{6} x^{5} + 18 a^{2} b^{2} d^{7} x^{6} + x^{3} \left (36 a^{3} b d^{4} + 360 a^{2} b^{2} c^{3} d^{4}\right ) + x^{2} \left (108 a^{3} b c d^{3} + 270 a^{2} b^{2} c^{4} d^{3}\right ) + x \left (108 a^{3} b c^{2} d^{2} + 108 a^{2} b^{2} c^{5} d^{2}\right )} + \frac{\operatorname{RootSum}{\left (19683 t^{3} a^{8} b - 125, \left ( t \mapsto t \log{\left (x + \frac{27 t a^{3} + 5 c}{5 d} \right )} \right )\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(8*a*c + 5*b*c**4 + 30*b*c**2*d**2*x**2 + 20*b*c*d**3*x**3 + 5*b*d**4*x**4 + x*(8*a*d + 20*b*c**3*d))/(18*a**4
*d + 36*a**3*b*c**3*d + 18*a**2*b**2*c**6*d + 270*a**2*b**2*c**2*d**5*x**4 + 108*a**2*b**2*c*d**6*x**5 + 18*a*
*2*b**2*d**7*x**6 + x**3*(36*a**3*b*d**4 + 360*a**2*b**2*c**3*d**4) + x**2*(108*a**3*b*c*d**3 + 270*a**2*b**2*
c**4*d**3) + x*(108*a**3*b*c**2*d**2 + 108*a**2*b**2*c**5*d**2)) + RootSum(19683*_t**3*a**8*b - 125, Lambda(_t
, _t*log(x + (27*_t*a**3 + 5*c)/(5*d))))/d

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Giac [A]  time = 1.14475, size = 360, normalized size = 1.82 \begin{align*} \frac{5}{27} \, \sqrt{3} \left (\frac{1}{a^{8} 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 ) - \frac{5}{54} \, \left (\frac{1}{a^{8} 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 ) + \frac{5}{27} \, \left (\frac{1}{a^{8} b d^{3}}\right )^{\frac{1}{3}} \log \left ({\left | 9 \, a^{2} b d x + 9 \, a^{2} b c + 9 \, \left (a b^{2}\right )^{\frac{1}{3}} a^{2} \right |}\right ) + \frac{5 \, b d^{4} x^{4} + 20 \, b c d^{3} x^{3} + 30 \, b c^{2} d^{2} x^{2} + 20 \, b c^{3} d x + 5 \, b c^{4} + 8 \, a d x + 8 \, a c}{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^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/27*sqrt(3)*(1/(a^8*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 - sqrt(3
)*(a*b^2)^(1/3))) - 5/54*(1/(a^8*b*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) + 5/27*(1/(a^8*b*d^3))^(1/3)*log(abs(9*a^2*b*d*x + 9*a^2*b*c + 9*(a*b^2)^(1/3)
*a^2)) + 1/18*(5*b*d^4*x^4 + 20*b*c*d^3*x^3 + 30*b*c^2*d^2*x^2 + 20*b*c^3*d*x + 5*b*c^4 + 8*a*d*x + 8*a*c)/((b
*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)^2*a^2*d)