∫ 1_____ x3(a+b(c+dx)3)dx

3.109 $\int \frac{1}{x^3 (a+b (c+d x)^3)} \, dx$

Optimal. Leaf size=393 \[ -\frac{b^{2/3} d^2 \left (6 a^{4/3} b^{2/3} c^2+a^2-3 \sqrt [3]{a} b^{5/3} c^5-7 a b c^3+b^2 c^6\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \left (a+b c^3\right )^3}+\frac{b^{2/3} d^2 \left (6 a^{4/3} b^{2/3} c^2+a^2-3 \sqrt [3]{a} b^{5/3} c^5-7 a b c^3+b^2 c^6\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{2/3} \left (a+b c^3\right )^3}+\frac{b^{2/3} d^2 \left (-3 a^{2/3} \sqrt [3]{b} c+a+b c^3\right ) \left (\sqrt [3]{a}+\sqrt [3]{b} c\right )^3 \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{2/3} \left (a+b c^3\right )^3}-\frac{3 b c d^2 \log (x) \left (a-2 b c^3\right )}{\left (a+b c^3\right )^3}+\frac{b c d^2 \left (a-2 b c^3\right ) \log \left (a+b (c+d x)^3\right )}{\left (a+b c^3\right )^3}+\frac{3 b c^2 d}{x \left (a+b c^3\right )^2}-\frac{1}{2 x^2 \left (a+b c^3\right )} \]

[Out]

-1/(2*(a + b*c^3)*x^2) + (3*b*c^2*d)/((a + b*c^3)^2*x) + (b^(2/3)*(a^(1/3) + b^(1/3)*c)^3*(a - 3*a^(2/3)*b^(1/

3)*c + b*c^3)*d^2*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(2/3)*(a + b*c^3)^3) -

(3*b*c*(a - 2*b*c^3)*d^2*Log[x])/(a + b*c^3)^3 - (b^(2/3)*(a^2 + 6*a^(4/3)*b^(2/3)*c^2 - 7*a*b*c^3 - 3*a^(1/3

)*b^(5/3)*c^5 + b^2*c^6)*d^2*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(3*a^(2/3)*(a + b*c^3)^3) + (b^(2/3)*(a^2 + 6*a

^(4/3)*b^(2/3)*c^2 - 7*a*b*c^3 - 3*a^(1/3)*b^(5/3)*c^5 + b^2*c^6)*d^2*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x)

+ b^(2/3)*(c + d*x)^2])/(6*a^(2/3)*(a + b*c^3)^3) + (b*c*(a - 2*b*c^3)*d^2*Log[a + b*(c + d*x)^3])/(a + b*c^3)

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Rubi [A] time = 0.603958, antiderivative size = 393, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 10, integrand size = 17, $\frac{\text{number of rules}}{\text{integrand size}}$ = 0.588, Rules used = {371, 6725, 1871, 1860, 31, 634, 617, 204, 628, 260} \[ -\frac{b^{2/3} d^2 \left (6 a^{4/3} b^{2/3} c^2+a^2-3 \sqrt [3]{a} b^{5/3} c^5-7 a b c^3+b^2 c^6\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \left (a+b c^3\right )^3}+\frac{b^{2/3} d^2 \left (6 a^{4/3} b^{2/3} c^2+a^2-3 \sqrt [3]{a} b^{5/3} c^5-7 a b c^3+b^2 c^6\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{2/3} \left (a+b c^3\right )^3}+\frac{b^{2/3} d^2 \left (-3 a^{2/3} \sqrt [3]{b} c+a+b c^3\right ) \left (\sqrt [3]{a}+\sqrt [3]{b} c\right )^3 \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{2/3} \left (a+b c^3\right )^3}-\frac{3 b c d^2 \log (x) \left (a-2 b c^3\right )}{\left (a+b c^3\right )^3}+\frac{b c d^2 \left (a-2 b c^3\right ) \log \left (a+b (c+d x)^3\right )}{\left (a+b c^3\right )^3}+\frac{3 b c^2 d}{x \left (a+b c^3\right )^2}-\frac{1}{2 x^2 \left (a+b c^3\right )} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-1/(2*(a + b*c^3)*x^2) + (3*b*c^2*d)/((a + b*c^3)^2*x) + (b^(2/3)*(a^(1/3) + b^(1/3)*c)^3*(a - 3*a^(2/3)*b^(1/

3)*c + b*c^3)*d^2*ArcTan[(a^(1/3) - 2*b^(1/3)*(c + d*x))/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(2/3)*(a + b*c^3)^3) -

(3*b*c*(a - 2*b*c^3)*d^2*Log[x])/(a + b*c^3)^3 - (b^(2/3)*(a^2 + 6*a^(4/3)*b^(2/3)*c^2 - 7*a*b*c^3 - 3*a^(1/3

)*b^(5/3)*c^5 + b^2*c^6)*d^2*Log[a^(1/3) + b^(1/3)*(c + d*x)])/(3*a^(2/3)*(a + b*c^3)^3) + (b^(2/3)*(a^2 + 6*a

^(4/3)*b^(2/3)*c^2 - 7*a*b*c^3 - 3*a^(1/3)*b^(5/3)*c^5 + b^2*c^6)*d^2*Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x)

+ b^(2/3)*(c + d*x)^2])/(6*a^(2/3)*(a + b*c^3)^3) + (b*c*(a - 2*b*c^3)*d^2*Log[a + b*(c + d*x)^3])/(a + b*c^3)

Rule 371

Int[((a_) + (b_.)*(v_)^(n_))^(p_.)*(x_)^(m_.), x_Symbol] :> With[{c = Coefficient[v, x, 0], d = Coefficient[v,

x, 1]}, Dist[1/d^(m + 1), Subst[Int[SimplifyIntegrand[(x - c)^m*(a + b*x^n)^p, x], x], x, v], x] /; NeQ[c, 0]

] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && IntegerQ[m]

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]

/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rule 1871

Int[(P2_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{A = Coeff[P2, x, 0], B = Coeff[P2, x, 1], C = Coeff[P2, x,

2]}, Int[(A + B*x)/(a + b*x^3), x] + Dist[C, Int[x^2/(a + b*x^3), x], x] /; EqQ[a*B^3 - b*A^3, 0] || !Ration

alQ[a/b]] /; FreeQ[{a, b}, x] && PolyQ[P2, x, 2]

Rule 1860

Int[((A_) + (B_.)*(x_))/((a_) + (b_.)*(x_)^3), x_Symbol] :> With[{r = Numerator[Rt[a/b, 3]], s = Denominator[R

t[a/b, 3]]}, -Dist[(r*(B*r - A*s))/(3*a*s), Int[1/(r + s*x), x], x] + Dist[r/(3*a*s), Int[(r*(B*r + 2*A*s) + s

*(B*r - A*s)*x)/(r^2 - r*s*x + s^2*x^2), x], x]] /; FreeQ[{a, b, A, B}, x] && NeQ[a*B^3 - b*A^3, 0] && PosQ[a/

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]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ

[{a, b, m, n}, x] && EqQ[m, n - 1]

Rubi steps

\begin{align*} \int \frac{1}{x^3 \left (a+b (c+d x)^3\right )} \, dx &=d^2 \operatorname{Subst}\left (\int \frac{1}{(-c+x)^3 \left (a+b x^3\right )} \, dx,x,c+d x\right )\\ &=d^2 \operatorname{Subst}\left (\int \left (-\frac{1}{\left (a+b c^3\right ) (c-x)^3}-\frac{3 b c^2}{\left (a+b c^3\right )^2 (c-x)^2}-\frac{3 b c \left (-a+2 b c^3\right )}{\left (a+b c^3\right )^3 (c-x)}+\frac{b \left (-a^2+7 a b c^3-b^2 c^6+3 b c^2 \left (2 a-b c^3\right ) x+3 b c \left (a-2 b c^3\right ) x^2\right )}{\left (a+b c^3\right )^3 \left (a+b x^3\right )}\right ) \, dx,x,c+d x\right )\\ &=-\frac{1}{2 \left (a+b c^3\right ) x^2}+\frac{3 b c^2 d}{\left (a+b c^3\right )^2 x}-\frac{3 b c \left (a-2 b c^3\right ) d^2 \log (x)}{\left (a+b c^3\right )^3}+\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{-a^2+7 a b c^3-b^2 c^6+3 b c^2 \left (2 a-b c^3\right ) x+3 b c \left (a-2 b c^3\right ) x^2}{a+b x^3} \, dx,x,c+d x\right )}{\left (a+b c^3\right )^3}\\ &=-\frac{1}{2 \left (a+b c^3\right ) x^2}+\frac{3 b c^2 d}{\left (a+b c^3\right )^2 x}-\frac{3 b c \left (a-2 b c^3\right ) d^2 \log (x)}{\left (a+b c^3\right )^3}+\frac{\left (b d^2\right ) \operatorname{Subst}\left (\int \frac{-a^2+7 a b c^3-b^2 c^6+3 b c^2 \left (2 a-b c^3\right ) x}{a+b x^3} \, dx,x,c+d x\right )}{\left (a+b c^3\right )^3}+\frac{\left (3 b^2 c \left (a-2 b c^3\right ) d^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{a+b x^3} \, dx,x,c+d x\right )}{\left (a+b c^3\right )^3}\\ &=-\frac{1}{2 \left (a+b c^3\right ) x^2}+\frac{3 b c^2 d}{\left (a+b c^3\right )^2 x}-\frac{3 b c \left (a-2 b c^3\right ) d^2 \log (x)}{\left (a+b c^3\right )^3}+\frac{b c \left (a-2 b c^3\right ) d^2 \log \left (a+b (c+d x)^3\right )}{\left (a+b c^3\right )^3}+\frac{\left (b^{2/3} d^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt [3]{a} \left (3 \sqrt [3]{a} b c^2 \left (2 a-b c^3\right )+2 \sqrt [3]{b} \left (-a^2+7 a b c^3-b^2 c^6\right )\right )+\sqrt [3]{b} \left (3 \sqrt [3]{a} b c^2 \left (2 a-b c^3\right )-\sqrt [3]{b} \left (-a^2+7 a b c^3-b^2 c^6\right )\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx,x,c+d x\right )}{3 a^{2/3} \left (a+b c^3\right )^3}-\frac{\left (b \left (a^2+6 a^{4/3} b^{2/3} c^2-7 a b c^3-3 \sqrt [3]{a} b^{5/3} c^5+b^2 c^6\right ) d^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx,x,c+d x\right )}{3 a^{2/3} \left (a+b c^3\right )^3}\\ &=-\frac{1}{2 \left (a+b c^3\right ) x^2}+\frac{3 b c^2 d}{\left (a+b c^3\right )^2 x}-\frac{3 b c \left (a-2 b c^3\right ) d^2 \log (x)}{\left (a+b c^3\right )^3}-\frac{b^{2/3} \left (a^2+6 a^{4/3} b^{2/3} c^2-7 a b c^3-3 \sqrt [3]{a} b^{5/3} c^5+b^2 c^6\right ) d^2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \left (a+b c^3\right )^3}+\frac{b c \left (a-2 b c^3\right ) d^2 \log \left (a+b (c+d x)^3\right )}{\left (a+b c^3\right )^3}-\frac{\left (b \left (\sqrt [3]{a}+\sqrt [3]{b} c\right )^3 \left (a-3 a^{2/3} \sqrt [3]{b} c+b c^3\right ) d^2\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 )}{2 \sqrt [3]{a} \left (a+b c^3\right )^3}+\frac{\left (b^{2/3} \left (a^2+6 a^{4/3} b^{2/3} c^2-7 a b c^3-3 \sqrt [3]{a} b^{5/3} c^5+b^2 c^6\right ) d^2\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 )}{6 a^{2/3} \left (a+b c^3\right )^3}\\ &=-\frac{1}{2 \left (a+b c^3\right ) x^2}+\frac{3 b c^2 d}{\left (a+b c^3\right )^2 x}-\frac{3 b c \left (a-2 b c^3\right ) d^2 \log (x)}{\left (a+b c^3\right )^3}-\frac{b^{2/3} \left (a^2+6 a^{4/3} b^{2/3} c^2-7 a b c^3-3 \sqrt [3]{a} b^{5/3} c^5+b^2 c^6\right ) d^2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \left (a+b c^3\right )^3}+\frac{b^{2/3} \left (a^2+6 a^{4/3} b^{2/3} c^2-7 a b c^3-3 \sqrt [3]{a} b^{5/3} c^5+b^2 c^6\right ) d^2 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{2/3} \left (a+b c^3\right )^3}+\frac{b c \left (a-2 b c^3\right ) d^2 \log \left (a+b (c+d x)^3\right )}{\left (a+b c^3\right )^3}-\frac{\left (b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} c\right )^3 \left (a-3 a^{2/3} \sqrt [3]{b} c+b c^3\right ) d^2\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 )}{a^{2/3} \left (a+b c^3\right )^3}\\ &=-\frac{1}{2 \left (a+b c^3\right ) x^2}+\frac{3 b c^2 d}{\left (a+b c^3\right )^2 x}+\frac{b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} c\right )^3 \left (a-3 a^{2/3} \sqrt [3]{b} c+b c^3\right ) d^2 \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} (c+d x)}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt{3} a^{2/3} \left (a+b c^3\right )^3}-\frac{3 b c \left (a-2 b c^3\right ) d^2 \log (x)}{\left (a+b c^3\right )^3}-\frac{b^{2/3} \left (a^2+6 a^{4/3} b^{2/3} c^2-7 a b c^3-3 \sqrt [3]{a} b^{5/3} c^5+b^2 c^6\right ) d^2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{3 a^{2/3} \left (a+b c^3\right )^3}+\frac{b^{2/3} \left (a^2+6 a^{4/3} b^{2/3} c^2-7 a b c^3-3 \sqrt [3]{a} b^{5/3} c^5+b^2 c^6\right ) d^2 \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{6 a^{2/3} \left (a+b c^3\right )^3}+\frac{b c \left (a-2 b c^3\right ) d^2 \log \left (a+b (c+d x)^3\right )}{\left (a+b c^3\right )^3}\\ \end{align*}

Mathematica [C] time = 0.143322, size = 244, normalized size = 0.62 \[ -\frac{2 d^2 x^2 \text{RootSum}\left [3 \text{$\#$1}^2 b c d^2+\text{$\#$1}^3 b d^3+3 \text{$\#$1} b c^2 d+a+b c^3\& ,\frac{-3 \text{$\#$1}^2 a b c d^2 \log (x-\text{$\#$1})+6 \text{$\#$1}^2 b^2 c^4 d^2 \log (x-\text{$\#$1})+a^2 \log (x-\text{$\#$1})-12 \text{$\#$1} a b c^2 d \log (x-\text{$\#$1})-16 a b c^3 \log (x-\text{$\#$1})+15 \text{$\#$1} b^2 c^5 d \log (x-\text{$\#$1})+10 b^2 c^6 \log (x-\text{$\#$1})}{\text{$\#$1}^2 d^2+2 \text{$\#$1} c d+c^2}\& \right ]+18 b c d^2 x^2 \log (x) \left (a-2 b c^3\right )+3 \left (a+b c^3\right ) \left (a+b c^2 (c-6 d x)\right )}{6 x^2 \left (a+b c^3\right )^3} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(3*(a + b*c^3)*(a + b*c^2*(c - 6*d*x)) + 18*b*c*(a - 2*b*c^3)*d^2*x^2*Log[x] + 2*d^2*x^2*RootSum[a + b*c^3 +

3*b*c^2*d*#1 + 3*b*c*d^2*#1^2 + b*d^3*#1^3 & , (a^2*Log[x - #1] - 16*a*b*c^3*Log[x - #1] + 10*b^2*c^6*Log[x -

#1] - 12*a*b*c^2*d*Log[x - #1]*#1 + 15*b^2*c^5*d*Log[x - #1]*#1 - 3*a*b*c*d^2*Log[x - #1]*#1^2 + 6*b^2*c^4*d^2

*Log[x - #1]*#1^2)/(c^2 + 2*c*d*#1 + d^2*#1^2) & ])/(6*(a + b*c^3)^3*x^2)

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Maple [C] time = 0.01, size = 217, normalized size = 0.6 \begin{align*}{\frac{{d}^{2}}{3\, \left ( b{c}^{3}+a \right ) ^{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{ \left ( -6\,{{\it \_R}}^{2}{b}^{2}{c}^{4}{d}^{2}-15\,{\it \_R}\,{b}^{2}{c}^{5}d-10\,{b}^{2}{c}^{6}+3\,{{\it \_R}}^{2}abc{d}^{2}+12\,{\it \_R}\,ab{c}^{2}d+16\,ab{c}^{3}-{a}^{2} \right ) \ln \left ( x-{\it \_R} \right ) }{{d}^{2}{{\it \_R}}^{2}+2\,cd{\it \_R}+{c}^{2}}}}-{\frac{1}{ \left ( 2\,b{c}^{3}+2\,a \right ){x}^{2}}}+3\,{\frac{b{c}^{2}d}{ \left ( b{c}^{3}+a \right ) ^{2}x}}+6\,{\frac{{b}^{2}{c}^{4}{d}^{2}\ln \left ( x \right ) }{ \left ( b{c}^{3}+a \right ) ^{3}}}-3\,{\frac{bc{d}^{2}\ln \left ( x \right ) a}{ \left ( b{c}^{3}+a \right ) ^{3}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*d^2*sum((-6*_R^2*b^2*c^4*d^2-15*_R*b^2*c^5*d-10*b^2*c^6+3*_R^2*a*b*c*d^2+12*_R*a*b*c^2*d+16*a*b*c^3-a^2)/(

_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))/(b*c^3+a)^3-1/2/(b*

c^3+a)/x^2+3*b*c^2*d/(b*c^3+a)^2/x+6*b^2*c^4*d^2/(b*c^3+a)^3*ln(x)-3*b*c*d^2/(b*c^3+a)^3*ln(x)*a

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Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [C] time = 53.0366, size = 29965, normalized size = 76.25 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

+ a^3)*(6*(1/2)^(2/3)*(b^2*c^2*d^4/(a*b^3*c^9 + 3*a^2*b^2*c^6 + 3*a^3*b*c^3 + a^4) - 3*(2*b^2*c^4*d^2 - a*b*c

*d^2)^2/(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3)^2)*(-I*sqrt(3) + 1)/(27*(2*b^2*c^4*d^2 - a*b*c*d^2)*b^2*c^

2*d^4/((a*b^3*c^9 + 3*a^2*b^2*c^6 + 3*a^3*b*c^3 + a^4)*(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3)) - b^2*d^6/

(a^2*b^3*c^9 + 3*a^3*b^2*c^6 + 3*a^4*b*c^3 + a^5) + (b^3*c^9 + 3*a*b^2*c^6 - 24*a^2*b*c^3 + a^3)*b^2*d^6/((b*c

^3 + a)^6*a^2) - 54*(2*b^2*c^4*d^2 - a*b*c*d^2)^3/(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3)^3)^(1/3) - (1/2)

^(1/3)*(27*(2*b^2*c^4*d^2 - a*b*c*d^2)*b^2*c^2*d^4/((a*b^3*c^9 + 3*a^2*b^2*c^6 + 3*a^3*b*c^3 + a^4)*(b^3*c^9 +

3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3)) - b^2*d^6/(a^2*b^3*c^9 + 3*a^3*b^2*c^6 + 3*a^4*b*c^3 + a^5) + (b^3*c^9 + 3*

a*b^2*c^6 - 24*a^2*b*c^3 + a^3)*b^2*d^6/((b*c^3 + a)^6*a^2) - 54*(2*b^2*c^4*d^2 - a*b*c*d^2)^3/(b^3*c^9 + 3*a*

b^2*c^6 + 3*a^2*b*c^3 + a^3)^3)^(1/3)*(I*sqrt(3) + 1) - 6*(2*b^2*c^4*d^2 - a*b*c*d^2)/(b^3*c^9 + 3*a*b^2*c^6 +

3*a^2*b*c^3 + a^3))*x^2*log((b^4*c^9 + 3*a*b^3*c^6 - 24*a^2*b^2*c^3 + a^3*b)*d^5*x + (b^4*c^10 + 15*a*b^3*c^7

- 63*a^2*b^2*c^4 + 4*a^3*b*c)*d^4 - 1/2*(a*b^4*c^12 - 50*a^2*b^3*c^9 + 141*a^3*b^2*c^6 - 50*a^4*b*c^3 + a^5)*

(6*(1/2)^(2/3)*(b^2*c^2*d^4/(a*b^3*c^9 + 3*a^2*b^2*c^6 + 3*a^3*b*c^3 + a^4) - 3*(2*b^2*c^4*d^2 - a*b*c*d^2)^2/

(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3)^2)*(-I*sqrt(3) + 1)/(27*(2*b^2*c^4*d^2 - a*b*c*d^2)*b^2*c^2*d^4/((

a*b^3*c^9 + 3*a^2*b^2*c^6 + 3*a^3*b*c^3 + a^4)*(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3)) - b^2*d^6/(a^2*b^3

*c^9 + 3*a^3*b^2*c^6 + 3*a^4*b*c^3 + a^5) + (b^3*c^9 + 3*a*b^2*c^6 - 24*a^2*b*c^3 + a^3)*b^2*d^6/((b*c^3 + a)^

6*a^2) - 54*(2*b^2*c^4*d^2 - a*b*c*d^2)^3/(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3)^3)^(1/3) - (1/2)^(1/3)*(

27*(2*b^2*c^4*d^2 - a*b*c*d^2)*b^2*c^2*d^4/((a*b^3*c^9 + 3*a^2*b^2*c^6 + 3*a^3*b*c^3 + a^4)*(b^3*c^9 + 3*a*b^2

*c^6 + 3*a^2*b*c^3 + a^3)) - b^2*d^6/(a^2*b^3*c^9 + 3*a^3*b^2*c^6 + 3*a^4*b*c^3 + a^5) + (b^3*c^9 + 3*a*b^2*c^

6 - 24*a^2*b*c^3 + a^3)*b^2*d^6/((b*c^3 + a)^6*a^2) - 54*(2*b^2*c^4*d^2 - a*b*c*d^2)^3/(b^3*c^9 + 3*a*b^2*c^6

+ 3*a^2*b*c^3 + a^3)^3)^(1/3)*(I*sqrt(3) + 1) - 6*(2*b^2*c^4*d^2 - a*b*c*d^2)/(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b

*c^3 + a^3))*d^2 + 3/4*(a^2*b^4*c^14 + a^3*b^3*c^11 - 3*a^4*b^2*c^8 - 5*a^5*b*c^5 - 2*a^6*c^2)*(6*(1/2)^(2/3)*

(b^2*c^2*d^4/(a*b^3*c^9 + 3*a^2*b^2*c^6 + 3*a^3*b*c^3 + a^4) - 3*(2*b^2*c^4*d^2 - a*b*c*d^2)^2/(b^3*c^9 + 3*a*

b^2*c^6 + 3*a^2*b*c^3 + a^3)^2)*(-I*sqrt(3) + 1)/(27*(2*b^2*c^4*d^2 - a*b*c*d^2)*b^2*c^2*d^4/((a*b^3*c^9 + 3*a

^2*b^2*c^6 + 3*a^3*b*c^3 + a^4)*(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3)) - b^2*d^6/(a^2*b^3*c^9 + 3*a^3*b^

2*c^6 + 3*a^4*b*c^3 + a^5) + (b^3*c^9 + 3*a*b^2*c^6 - 24*a^2*b*c^3 + a^3)*b^2*d^6/((b*c^3 + a)^6*a^2) - 54*(2*

b^2*c^4*d^2 - a*b*c*d^2)^3/(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3)^3)^(1/3) - (1/2)^(1/3)*(27*(2*b^2*c^4*d

^2 - a*b*c*d^2)*b^2*c^2*d^4/((a*b^3*c^9 + 3*a^2*b^2*c^6 + 3*a^3*b*c^3 + a^4)*(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*

c^3 + a^3)) - b^2*d^6/(a^2*b^3*c^9 + 3*a^3*b^2*c^6 + 3*a^4*b*c^3 + a^5) + (b^3*c^9 + 3*a*b^2*c^6 - 24*a^2*b*c^

3 + a^3)*b^2*d^6/((b*c^3 + a)^6*a^2) - 54*(2*b^2*c^4*d^2 - a*b*c*d^2)^3/(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 +

a^3)^3)^(1/3)*(I*sqrt(3) + 1) - 6*(2*b^2*c^4*d^2 - a*b*c*d^2)/(b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3 + a^3))^2)

- 36*(b^2*c^5 + a*b*c^2)*d*x + 6*a^2 + (18*(2*b^2*c^4 - a*b*c)*d^2*x^2 + (b^3*c^9 + 3*a*b^2*c^6 + 3*a^2*b*c^3