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

Optimal. Leaf size=215 \[ -\frac {\left (5 \sqrt [3]{b} c-2 \sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{8/3} b^{2/3}}+\frac {\left (5 \sqrt [3]{b} c-2 \sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{2/3}}-\frac {\left (2 \sqrt [3]{a} d+5 \sqrt [3]{b} c\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{8/3} b^{2/3}}+\frac {x (5 c+4 d x)}{18 a^2 \left (a+b x^3\right )}+\frac {x (c+d x)}{6 a \left (a+b x^3\right )^2} \]

[Out]

1/6*x*(d*x+c)/a/(b*x^3+a)^2+1/18*x*(4*d*x+5*c)/a^2/(b*x^3+a)+1/27*(5*b^(1/3)*c-2*a^(1/3)*d)*ln(a^(1/3)+b^(1/3)
*x)/a^(8/3)/b^(2/3)-1/54*(5*b^(1/3)*c-2*a^(1/3)*d)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(8/3)/b^(2/3)-1
/27*(5*b^(1/3)*c+2*a^(1/3)*d)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/a^(8/3)/b^(2/3)*3^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {1855, 1860, 31, 634, 617, 204, 628} \[ -\frac {\left (5 \sqrt [3]{b} c-2 \sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{8/3} b^{2/3}}+\frac {\left (5 \sqrt [3]{b} c-2 \sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{2/3}}-\frac {\left (2 \sqrt [3]{a} d+5 \sqrt [3]{b} c\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{8/3} b^{2/3}}+\frac {x (5 c+4 d x)}{18 a^2 \left (a+b x^3\right )}+\frac {x (c+d x)}{6 a \left (a+b x^3\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*x^3)^3,x]

[Out]

(x*(c + d*x))/(6*a*(a + b*x^3)^2) + (x*(5*c + 4*d*x))/(18*a^2*(a + b*x^3)) - ((5*b^(1/3)*c + 2*a^(1/3)*d)*ArcT
an[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(8/3)*b^(2/3)) + ((5*b^(1/3)*c - 2*a^(1/3)*d)*Log[
a^(1/3) + b^(1/3)*x])/(27*a^(8/3)*b^(2/3)) - ((5*b^(1/3)*c - 2*a^(1/3)*d)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^
(2/3)*x^2])/(54*a^(8/3)*b^(2/3))

Rule 31

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

Rule 1855

Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> -Simp[(x*Pq*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Di
st[1/(a*n*(p + 1)), Int[ExpandToSum[n*(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b},
x] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]

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/
b]

Rubi steps

\begin {align*} \int \frac {c+d x}{\left (a+b x^3\right )^3} \, dx &=\frac {x (c+d x)}{6 a \left (a+b x^3\right )^2}-\frac {\int \frac {-5 c-4 d x}{\left (a+b x^3\right )^2} \, dx}{6 a}\\ &=\frac {x (c+d x)}{6 a \left (a+b x^3\right )^2}+\frac {x (5 c+4 d x)}{18 a^2 \left (a+b x^3\right )}+\frac {\int \frac {10 c+4 d x}{a+b x^3} \, dx}{18 a^2}\\ &=\frac {x (c+d x)}{6 a \left (a+b x^3\right )^2}+\frac {x (5 c+4 d x)}{18 a^2 \left (a+b x^3\right )}+\frac {\int \frac {\sqrt [3]{a} \left (20 \sqrt [3]{b} c+4 \sqrt [3]{a} d\right )+\sqrt [3]{b} \left (-10 \sqrt [3]{b} c+4 \sqrt [3]{a} d\right ) x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{54 a^{8/3} \sqrt [3]{b}}+\frac {\left (5 c-\frac {2 \sqrt [3]{a} d}{\sqrt [3]{b}}\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{27 a^{8/3}}\\ &=\frac {x (c+d x)}{6 a \left (a+b x^3\right )^2}+\frac {x (5 c+4 d x)}{18 a^2 \left (a+b x^3\right )}+\frac {\left (5 \sqrt [3]{b} c-2 \sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{2/3}}-\frac {\left (5 \sqrt [3]{b} c-2 \sqrt [3]{a} d\right ) \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}{54 a^{8/3} b^{2/3}}+\frac {\left (5 \sqrt [3]{b} c+2 \sqrt [3]{a} d\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{18 a^{7/3} \sqrt [3]{b}}\\ &=\frac {x (c+d x)}{6 a \left (a+b x^3\right )^2}+\frac {x (5 c+4 d x)}{18 a^2 \left (a+b x^3\right )}+\frac {\left (5 \sqrt [3]{b} c-2 \sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{2/3}}-\frac {\left (5 \sqrt [3]{b} c-2 \sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{8/3} b^{2/3}}+\frac {\left (5 \sqrt [3]{b} c+2 \sqrt [3]{a} d\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{9 a^{8/3} b^{2/3}}\\ &=\frac {x (c+d x)}{6 a \left (a+b x^3\right )^2}+\frac {x (5 c+4 d x)}{18 a^2 \left (a+b x^3\right )}-\frac {\left (5 \sqrt [3]{b} c+2 \sqrt [3]{a} d\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{8/3} b^{2/3}}+\frac {\left (5 \sqrt [3]{b} c-2 \sqrt [3]{a} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{8/3} b^{2/3}}-\frac {\left (5 \sqrt [3]{b} c-2 \sqrt [3]{a} d\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{8/3} b^{2/3}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 205, normalized size = 0.95 \[ \frac {\frac {\left (2 a^{2/3} d-5 \sqrt [3]{a} \sqrt [3]{b} c\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{2/3}}+\frac {2 \left (5 \sqrt [3]{a} \sqrt [3]{b} c-2 a^{2/3} d\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}+\frac {9 a^2 x (c+d x)}{\left (a+b x^3\right )^2}-\frac {2 \sqrt {3} \sqrt [3]{a} \left (2 \sqrt [3]{a} d+5 \sqrt [3]{b} c\right ) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{b^{2/3}}+\frac {3 a x (5 c+4 d x)}{a+b x^3}}{54 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(a + b*x^3)^3,x]

[Out]

((9*a^2*x*(c + d*x))/(a + b*x^3)^2 + (3*a*x*(5*c + 4*d*x))/(a + b*x^3) - (2*Sqrt[3]*a^(1/3)*(5*b^(1/3)*c + 2*a
^(1/3)*d)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(2/3) + (2*(5*a^(1/3)*b^(1/3)*c - 2*a^(2/3)*d)*Log[a^
(1/3) + b^(1/3)*x])/b^(2/3) + ((-5*a^(1/3)*b^(1/3)*c + 2*a^(2/3)*d)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*
x^2])/b^(2/3))/(54*a^3)

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fricas [C]  time = 2.36, size = 2215, normalized size = 10.30 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/108*(24*b*d*x^5 + 30*b*c*x^4 + 42*a*d*x^2 + 48*a*c*x - 2*(a^2*b^2*x^6 + 2*a^3*b*x^3 + a^4)*((1/2)^(1/3)*(I*s
qrt(3) + 1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)^(2/3)*c*d*(-I
*sqrt(3) + 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3)))*log(1/2*((1/2
)^(1/3)*(I*sqrt(3) + 1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)^(
2/3)*c*d*(-I*sqrt(3) + 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3)))^2
*a^6*b*d - 25/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2
))^(1/3) - 20*(1/2)^(2/3)*c*d*(-I*sqrt(3) + 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)
/(a^8*b^2))^(1/3)))*a^3*b*c^2 + 40*a*c*d^2 + (125*b*c^3 + 8*a*d^3)*x) + ((a^2*b^2*x^6 + 2*a^3*b*x^3 + a^4)*((1
/2)^(1/3)*(I*sqrt(3) + 1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)
^(2/3)*c*d*(-I*sqrt(3) + 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3)))
+ 3*sqrt(1/3)*(a^2*b^2*x^6 + 2*a^3*b*x^3 + a^4)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((125*b*c^3 + 8*a*d^3)/(a
^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)^(2/3)*c*d*(-I*sqrt(3) + 1)/(a^5*b*((125*b*c^3 + 8*
a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3)))^2*a^5*b + 160*c*d)/(a^5*b)))*log(-1/2*((1/2)^(1/3)
*(I*sqrt(3) + 1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)^(2/3)*c*
d*(-I*sqrt(3) + 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3)))^2*a^6*b*
d + 25/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3
) - 20*(1/2)^(2/3)*c*d*(-I*sqrt(3) + 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b
^2))^(1/3)))*a^3*b*c^2 - 40*a*c*d^2 + 2*(125*b*c^3 + 8*a*d^3)*x + 3/2*sqrt(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*
((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)^(2/3)*c*d*(-I*sqrt(3) + 1
)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3)))*a^6*b*d + 25*a^3*b*c^2)*s
qrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) -
20*(1/2)^(2/3)*c*d*(-I*sqrt(3) + 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2)
)^(1/3)))^2*a^5*b + 160*c*d)/(a^5*b))) + ((a^2*b^2*x^6 + 2*a^3*b*x^3 + a^4)*((1/2)^(1/3)*(I*sqrt(3) + 1)*((125
*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)^(2/3)*c*d*(-I*sqrt(3) + 1)/(a^
5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3))) - 3*sqrt(1/3)*(a^2*b^2*x^6 + 2
*a^3*b*x^3 + a^4)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) + 1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)
/(a^8*b^2))^(1/3) - 20*(1/2)^(2/3)*c*d*(-I*sqrt(3) + 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 -
8*a*d^3)/(a^8*b^2))^(1/3)))^2*a^5*b + 160*c*d)/(a^5*b)))*log(-1/2*((1/2)^(1/3)*(I*sqrt(3) + 1)*((125*b*c^3 +
8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)^(2/3)*c*d*(-I*sqrt(3) + 1)/(a^5*b*((125
*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3)))^2*a^6*b*d + 25/2*((1/2)^(1/3)*(I*sqrt(3
) + 1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)^(2/3)*c*d*(-I*sqrt
(3) + 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3)))*a^3*b*c^2 - 40*a*c
*d^2 + 2*(125*b*c^3 + 8*a*d^3)*x - 3/2*sqrt(1/3)*(((1/2)^(1/3)*(I*sqrt(3) + 1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2
) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)^(2/3)*c*d*(-I*sqrt(3) + 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)
/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3)))*a^6*b*d + 25*a^3*b*c^2)*sqrt(-(((1/2)^(1/3)*(I*sqrt(3) +
1)*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3) - 20*(1/2)^(2/3)*c*d*(-I*sqrt(3)
+ 1)/(a^5*b*((125*b*c^3 + 8*a*d^3)/(a^8*b^2) + (125*b*c^3 - 8*a*d^3)/(a^8*b^2))^(1/3)))^2*a^5*b + 160*c*d)/(a
^5*b))))/(a^2*b^2*x^6 + 2*a^3*b*x^3 + a^4)

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giac [A]  time = 0.23, size = 194, normalized size = 0.90 \[ -\frac {\sqrt {3} {\left (5 \, b c - 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} d\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{2}} - \frac {{\left (5 \, b c + 2 \, \left (-a b^{2}\right )^{\frac {1}{3}} d\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, \left (-a b^{2}\right )^{\frac {2}{3}} a^{2}} - \frac {{\left (2 \, d \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 5 \, c\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{27 \, a^{3}} + \frac {4 \, b d x^{5} + 5 \, b c x^{4} + 7 \, a d x^{2} + 8 \, a c x}{18 \, {\left (b x^{3} + a\right )}^{2} a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/27*sqrt(3)*(5*b*c - 2*(-a*b^2)^(1/3)*d)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(2/
3)*a^2) - 1/54*(5*b*c + 2*(-a*b^2)^(1/3)*d)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*a^2) - 1/
27*(2*d*(-a/b)^(1/3) + 5*c)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^3 + 1/18*(4*b*d*x^5 + 5*b*c*x^4 + 7*a*d*
x^2 + 8*a*c*x)/((b*x^3 + a)^2*a^2)

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maple [A]  time = 0.06, size = 272, normalized size = 1.27 \[ \frac {d \,x^{2}}{6 \left (b \,x^{3}+a \right )^{2} a}+\frac {c x}{6 \left (b \,x^{3}+a \right )^{2} a}+\frac {2 d \,x^{2}}{9 \left (b \,x^{3}+a \right ) a^{2}}+\frac {5 c x}{18 \left (b \,x^{3}+a \right ) a^{2}}+\frac {5 \sqrt {3}\, c \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{2} b}+\frac {5 c \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{2} b}-\frac {5 c \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{2} b}+\frac {2 \sqrt {3}\, d \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{27 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{2} b}-\frac {2 d \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{2} b}+\frac {d \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*c/a*x/(b*x^3+a)^2+5/18*c/a^2*x/(b*x^3+a)+5/27*c/a^2/b/(a/b)^(2/3)*ln(x+(a/b)^(1/3))-5/54*c/a^2/b/(a/b)^(2/
3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+5/27*c/a^2/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))+
1/6*d/a*x^2/(b*x^3+a)^2+2/9*d/a^2*x^2/(b*x^3+a)-2/27*d/a^2/b/(a/b)^(1/3)*ln(x+(a/b)^(1/3))+1/27*d/a^2/b/(a/b)^
(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+2/27*d/a^2*3^(1/2)/b/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1
))

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maxima [A]  time = 1.96, size = 203, normalized size = 0.94 \[ \frac {4 \, b d x^{5} + 5 \, b c x^{4} + 7 \, a d x^{2} + 8 \, a c x}{18 \, {\left (a^{2} b^{2} x^{6} + 2 \, a^{3} b x^{3} + a^{4}\right )}} + \frac {\sqrt {3} {\left (2 \, d \left (\frac {a}{b}\right )^{\frac {1}{3}} + 5 \, c\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (2 \, d \left (\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, c\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{54 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (2 \, d \left (\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, c\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, a^{2} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(4*b*d*x^5 + 5*b*c*x^4 + 7*a*d*x^2 + 8*a*c*x)/(a^2*b^2*x^6 + 2*a^3*b*x^3 + a^4) + 1/27*sqrt(3)*(2*d*(a/b)
^(1/3) + 5*c)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a^2*b*(a/b)^(2/3)) + 1/54*(2*d*(a/b)^(1/3)
- 5*c)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*b*(a/b)^(2/3)) - 1/27*(2*d*(a/b)^(1/3) - 5*c)*log(x + (a/b)
^(1/3))/(a^2*b*(a/b)^(2/3))

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mupad [B]  time = 0.27, size = 206, normalized size = 0.96 \[ \frac {\frac {7\,d\,x^2}{18\,a}+\frac {4\,c\,x}{9\,a}+\frac {5\,b\,c\,x^4}{18\,a^2}+\frac {2\,b\,d\,x^5}{9\,a^2}}{a^2+2\,a\,b\,x^3+b^2\,x^6}+\left (\sum _{k=1}^3\ln \left (\frac {b\,\left (10\,c\,d+4\,d^2\,x+{\mathrm {root}\left (19683\,a^8\,b^2\,z^3+810\,a^3\,b\,c\,d\,z-125\,b\,c^3+8\,a\,d^3,z,k\right )}^2\,a^5\,b\,729+\mathrm {root}\left (19683\,a^8\,b^2\,z^3+810\,a^3\,b\,c\,d\,z-125\,b\,c^3+8\,a\,d^3,z,k\right )\,a^2\,b\,c\,x\,135\right )}{a^4\,81}\right )\,\mathrm {root}\left (19683\,a^8\,b^2\,z^3+810\,a^3\,b\,c\,d\,z-125\,b\,c^3+8\,a\,d^3,z,k\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)/(a + b*x^3)^3,x)

[Out]

((7*d*x^2)/(18*a) + (4*c*x)/(9*a) + (5*b*c*x^4)/(18*a^2) + (2*b*d*x^5)/(9*a^2))/(a^2 + b^2*x^6 + 2*a*b*x^3) +
symsum(log((b*(10*c*d + 4*d^2*x + 729*root(19683*a^8*b^2*z^3 + 810*a^3*b*c*d*z - 125*b*c^3 + 8*a*d^3, z, k)^2*
a^5*b + 135*root(19683*a^8*b^2*z^3 + 810*a^3*b*c*d*z - 125*b*c^3 + 8*a*d^3, z, k)*a^2*b*c*x))/(81*a^4))*root(1
9683*a^8*b^2*z^3 + 810*a^3*b*c*d*z - 125*b*c^3 + 8*a*d^3, z, k), k, 1, 3)

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sympy [A]  time = 2.47, size = 146, normalized size = 0.68 \[ \operatorname {RootSum} {\left (19683 t^{3} a^{8} b^{2} + 810 t a^{3} b c d + 8 a d^{3} - 125 b c^{3}, \left (t \mapsto t \log {\left (x + \frac {1458 t^{2} a^{6} b d + 675 t a^{3} b c^{2} + 40 a c d^{2}}{8 a d^{3} + 125 b c^{3}} \right )} \right )\right )} + \frac {8 a c x + 7 a d x^{2} + 5 b c x^{4} + 4 b d x^{5}}{18 a^{4} + 36 a^{3} b x^{3} + 18 a^{2} b^{2} x^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

RootSum(19683*_t**3*a**8*b**2 + 810*_t*a**3*b*c*d + 8*a*d**3 - 125*b*c**3, Lambda(_t, _t*log(x + (1458*_t**2*a
**6*b*d + 675*_t*a**3*b*c**2 + 40*a*c*d**2)/(8*a*d**3 + 125*b*c**3)))) + (8*a*c*x + 7*a*d*x**2 + 5*b*c*x**4 +
4*b*d*x**5)/(18*a**4 + 36*a**3*b*x**3 + 18*a**2*b**2*x**6)

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