3.24.16 \(\int \frac {1}{(a+b (c+d x)^3)^2} \, dx\)

Optimal. Leaf size=170 \[ -\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{5/3} \sqrt [3]{b} d}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} \sqrt [3]{b} d}+\frac {c+d x}{3 a d \left (a+b (c+d x)^3\right )} \]

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Rubi [A]  time = 0.13, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 8, 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 {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} (c+d x)+b^{2/3} (c+d x)^2\right )}{9 a^{5/3} \sqrt [3]{b} d}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{9 a^{5/3} \sqrt [3]{b} d}-\frac {2 \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{5/3} \sqrt [3]{b} d}+\frac {c+d x}{3 a d \left (a+b (c+d x)^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*(c + d*x)^3)^(-2),x]

[Out]

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

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Mathematica [A]  time = 0.06, size = 151, normalized size = 0.89 \begin {gather*} \frac {-\frac {\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 {3 a^{2/3} (c+d x)}{a+b (c+d x)^3}+\frac {2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} (c+d x)\right )}{\sqrt [3]{b}}+\frac {2 \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}}}{9 a^{5/3} d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*(c + d*x)^3)^(-2),x]

[Out]

((3*a^(2/3)*(c + d*x))/(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))])/b^(1/3) + (2*Log[a^(1/3) + b^(1/3)*(c + d*x)])/b^(1/3) - Log[a^(2/3) - a^(1/3)*b^(1/3)*(c + d*x) + b^(2/
3)*(c + d*x)^2]/b^(1/3))/(9*a^(5/3)*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 )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*(c + d*x)^3)^(-2),x]

[Out]

IntegrateAlgebraic[(a + b*(c + d*x)^3)^(-2), x]

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fricas [B]  time = 1.19, size = 808, normalized size = 4.75 \begin {gather*} \left [\frac {3 \, a^{2} b d x + 3 \, a^{2} b c + 3 \, \sqrt {\frac {1}{3}} {\left (a b^{2} d^{3} x^{3} + 3 \, a b^{2} c d^{2} x^{2} + 3 \, a b^{2} c^{2} d x + a b^{2} c^{3} + a^{2} b\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {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 {\frac {1}{3}} {\left (2 \, a b d^{2} x^{2} + 4 \, a b c d x + 2 \, a b c^{2} + \left (a^{2} b\right )^{\frac {2}{3}} {\left (d x + c\right )} - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} {\left (a d x + a c\right )}}{b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a}\right ) - {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b d^{2} x^{2} + 2 \, a b c d x + a b c^{2} - \left (a^{2} b\right )^{\frac {2}{3}} {\left (d x + c\right )} + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) + 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b d x + a b c + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{9 \, {\left (a^{3} b^{2} d^{4} x^{3} + 3 \, a^{3} b^{2} c d^{3} x^{2} + 3 \, a^{3} b^{2} c^{2} d^{2} x + {\left (a^{3} b^{2} c^{3} + a^{4} b\right )} d\right )}}, \frac {3 \, a^{2} b d x + 3 \, a^{2} b c + 6 \, \sqrt {\frac {1}{3}} {\left (a b^{2} d^{3} x^{3} + 3 \, a b^{2} c d^{2} x^{2} + 3 \, a b^{2} c^{2} d x + a b^{2} c^{3} + a^{2} b\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (a^{2} b\right )^{\frac {2}{3}} {\left (d x + c\right )} - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{a^{2}}\right ) - {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b d^{2} x^{2} + 2 \, a b c d x + a b c^{2} - \left (a^{2} b\right )^{\frac {2}{3}} {\left (d x + c\right )} + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) + 2 \, {\left (b d^{3} x^{3} + 3 \, b c d^{2} x^{2} + 3 \, b c^{2} d x + b c^{3} + a\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b d x + a b c + \left (a^{2} b\right )^{\frac {2}{3}}\right )}{9 \, {\left (a^{3} b^{2} d^{4} x^{3} + 3 \, a^{3} b^{2} c d^{3} x^{2} + 3 \, a^{3} b^{2} c^{2} d^{2} x + {\left (a^{3} b^{2} c^{3} + a^{4} b\right )} d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(d*x+c)^3)^2,x, algorithm="fricas")

[Out]

[1/9*(3*a^2*b*d*x + 3*a^2*b*c + 3*sqrt(1/3)*(a*b^2*d^3*x^3 + 3*a*b^2*c*d^2*x^2 + 3*a*b^2*c^2*d*x + a*b^2*c^3 +
a^2*b)*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*d^3*x^3 + 3*
b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*(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*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*(a^2*b)^(2/3)*log(a*b*d*
x + a*b*c + (a^2*b)^(2/3)))/(a^3*b^2*d^4*x^3 + 3*a^3*b^2*c*d^3*x^2 + 3*a^3*b^2*c^2*d^2*x + (a^3*b^2*c^3 + a^4*
b)*d), 1/9*(3*a^2*b*d*x + 3*a^2*b*c + 6*sqrt(1/3)*(a*b^2*d^3*x^3 + 3*a*b^2*c*d^2*x^2 + 3*a*b^2*c^2*d*x + a*b^2
*c^3 + a^2*b)*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*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*c^3 + a)*(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*d^3*x^3 + 3*b*c*d^2*x^2 + 3*b*c^2*d*x + b*
c^3 + a)*(a^2*b)^(2/3)*log(a*b*d*x + a*b*c + (a^2*b)^(2/3)))/(a^3*b^2*d^4*x^3 + 3*a^3*b^2*c*d^3*x^2 + 3*a^3*b^
2*c^2*d^2*x + (a^3*b^2*c^3 + a^4*b)*d)]

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giac [A]  time = 0.23, size = 214, normalized size = 1.26 \begin {gather*} \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 ) - \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 )}{9 \, a} + \frac {d x + c}{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 )} a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(d*x+c)^3)^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.01, size = 127, normalized size = 0.75 \begin {gather*} \frac {2 \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 )}{9 a 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 {x}{3 a}+\frac {c}{3 a d}}{b \,d^{3} x^{3}+3 b c \,d^{2} x^{2}+3 b \,c^{2} d x +b \,c^{3}+a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/3/a*x+1/3*c/d/a)/(b*d^3*x^3+3*b*c*d^2*x^2+3*b*c^2*d*x+b*c^3+a)+2/9/a/b/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 \begin {gather*} \frac {d x + c}{3 \, {\left (a b d^{4} x^{3} + 3 \, a b c d^{3} x^{2} + 3 \, a b c^{2} d^{2} x + {\left (a b c^{3} + a^{2}\right )} d\right )}} + \frac {2 \, {\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 )}}{3 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*(d*x+c)^3)^2,x, algorithm="maxima")

[Out]

1/3*(d*x + c)/(a*b*d^4*x^3 + 3*a*b*c*d^3*x^2 + 3*a*b*c^2*d^2*x + (a*b*c^3 + a^2)*d) + 2/3*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

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mupad [B]  time = 1.46, size = 207, normalized size = 1.22 \begin {gather*} \frac {\frac {x}{3\,a}+\frac {c}{3\,a\,d}}{b\,c^3+3\,b\,c^2\,d\,x+3\,b\,c\,d^2\,x^2+b\,d^3\,x^3+a}+\frac {2\,\ln \left (b^{1/3}\,c+a^{1/3}+b^{1/3}\,d\,x\right )}{9\,a^{5/3}\,b^{1/3}\,d}+\frac {\ln \left (\frac {2\,b^2\,c\,d^5}{a}+\frac {2\,b^2\,d^6\,x}{a}+\frac {b^{5/3}\,d^5\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{a^{2/3}}\right )\,\left (-1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,a^{5/3}\,b^{1/3}\,d}-\frac {\ln \left (\frac {2\,b^2\,c\,d^5}{a}+\frac {2\,b^2\,d^6\,x}{a}-\frac {b^{5/3}\,d^5\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{a^{2/3}}\right )\,\left (1+\sqrt {3}\,1{}\mathrm {i}\right )}{9\,a^{5/3}\,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)^2,x)

[Out]

(x/(3*a) + c/(3*a*d))/(a + b*c^3 + b*d^3*x^3 + 3*b*c^2*d*x + 3*b*c*d^2*x^2) + (2*log(b^(1/3)*c + a^(1/3) + b^(
1/3)*d*x))/(9*a^(5/3)*b^(1/3)*d) + (log((2*b^2*c*d^5)/a + (2*b^2*d^6*x)/a + (b^(5/3)*d^5*(3^(1/2)*1i - 1))/a^(
2/3))*(3^(1/2)*1i - 1))/(9*a^(5/3)*b^(1/3)*d) - (log((2*b^2*c*d^5)/a + (2*b^2*d^6*x)/a - (b^(5/3)*d^5*(3^(1/2)
*1i + 1))/a^(2/3))*(3^(1/2)*1i + 1))/(9*a^(5/3)*b^(1/3)*d)

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sympy [A]  time = 0.94, size = 92, normalized size = 0.54 \begin {gather*} \frac {c + d x}{3 a^{2} d + 3 a b c^{3} d + 9 a b c^{2} d^{2} x + 9 a b c d^{3} x^{2} + 3 a b d^{4} x^{3}} + \frac {\operatorname {RootSum} {\left (729 t^{3} a^{5} b - 8, \left (t \mapsto t \log {\left (x + \frac {9 t a^{2} + 2 c}{2 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)**2,x)

[Out]

(c + d*x)/(3*a**2*d + 3*a*b*c**3*d + 9*a*b*c**2*d**2*x + 9*a*b*c*d**3*x**2 + 3*a*b*d**4*x**3) + RootSum(729*_t
**3*a**5*b - 8, Lambda(_t, _t*log(x + (9*_t*a**2 + 2*c)/(2*d))))/d

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