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

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

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Rubi [A]  time = 0.16, antiderivative size = 198, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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 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 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 {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*}

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Mathematica [A]  time = 0.11, size = 176, normalized size = 0.89 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a+b (c+d x)^3\right )^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*(c + d*x)^3)^(-3),x]

[Out]

IntegrateAlgebraic[(a + b*(c + d*x)^3)^(-3), x]

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fricas [B]  time = 1.21, size = 1646, normalized size = 8.31

result too large to display

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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giac [A]  time = 0.19, size = 264, normalized size = 1.33 \begin {gather*} \frac {5 \, {\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 (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}} \log \left ({\left | b d x + b c + \left (a b^{2}\right )^{\frac {1}{3}} \right |}\right )\right )}}{54 \, a^{2}} + \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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/54*(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))) - (1/(a^2*b*d^3))^(1/3)*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)*log(abs(b*d*x + b*c + (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)

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maple [C]  time = 0.02, size = 185, normalized size = 0.93 \begin {gather*} \frac {5 \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^{2} b 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 )}+\frac {\frac {5 b \,d^{3} x^{4}}{18 a^{2}}+\frac {10 b c \,d^{2} x^{3}}{9 a^{2}}+\frac {5 b \,c^{2} d \,x^{2}}{3 a^{2}}+\frac {2 \left (5 b \,c^{3}+2 a \right ) x}{9 a^{2}}+\frac {\left (5 b \,c^{3}+8 a \right ) c}{18 a^{2} d}}{\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}} \end {gather*}

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*b*c*d^2/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(-_R+x),_R=Roo
tOf(_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 \begin {gather*} \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 {\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 )}{3 \, 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 )}{6 \, d} + \frac {\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 )}{3 \, d}\right )}}{9 \, a^{2}} \end {gather*}

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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mupad [B]  time = 1.61, size = 354, normalized size = 1.79 \begin {gather*} \frac {\frac {2\,x\,\left (5\,b\,c^3+2\,a\right )}{9\,a^2}+\frac {5\,b\,c^4+8\,a\,c}{18\,a^2\,d}+\frac {5\,b\,d^3\,x^4}{18\,a^2}+\frac {5\,b\,c^2\,d\,x^2}{3\,a^2}+\frac {10\,b\,c\,d^2\,x^3}{9\,a^2}}{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 {5\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{27\,a^{8/3}\,b^{1/3}\,d}+\frac {\ln \left (\frac {5\,b^2\,c\,d^5}{3\,a^2}+\frac {5\,b^2\,d^6\,x}{3\,a^2}+\frac {b^{5/3}\,d^5\,\left (-5+\sqrt {3}\,5{}\mathrm {i}\right )}{6\,a^{5/3}}\right )\,\left (-5+\sqrt {3}\,5{}\mathrm {i}\right )}{54\,a^{8/3}\,b^{1/3}\,d}-\frac {\ln \left (\frac {5\,b^2\,c\,d^5}{3\,a^2}+\frac {5\,b^2\,d^6\,x}{3\,a^2}-\frac {b^{5/3}\,d^5\,\left (5+\sqrt {3}\,5{}\mathrm {i}\right )}{6\,a^{5/3}}\right )\,\left (5+\sqrt {3}\,5{}\mathrm {i}\right )}{54\,a^{8/3}\,b^{1/3}\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*(c + d*x)^3)^3,x)

[Out]

((2*x*(2*a + 5*b*c^3))/(9*a^2) + (8*a*c + 5*b*c^4)/(18*a^2*d) + (5*b*d^3*x^4)/(18*a^2) + (5*b*c^2*d*x^2)/(3*a^
2) + (10*b*c*d^2*x^3)/(9*a^2))/(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) + (5
*log(b^(1/3)*c + a^(1/3) + b^(1/3)*d*x))/(27*a^(8/3)*b^(1/3)*d) + (log((5*b^2*c*d^5)/(3*a^2) + (5*b^2*d^6*x)/(
3*a^2) + (b^(5/3)*d^5*(3^(1/2)*5i - 5))/(6*a^(5/3)))*(3^(1/2)*5i - 5))/(54*a^(8/3)*b^(1/3)*d) - (log((5*b^2*c*
d^5)/(3*a^2) + (5*b^2*d^6*x)/(3*a^2) - (b^(5/3)*d^5*(3^(1/2)*5i + 5))/(6*a^(5/3)))*(3^(1/2)*5i + 5))/(54*a^(8/
3)*b^(1/3)*d)

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sympy [A]  time = 3.27, size = 267, normalized size = 1.35 \begin {gather*} \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 {gather*}

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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