3.1.77 \(\int \frac {x^3 (A+B x^3)}{(a+b x^3)^2} \, dx\)

Optimal. Leaf size=190 \[ -\frac {(A b-4 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{2/3} b^{7/3}}+\frac {(A b-4 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{2/3} b^{7/3}}-\frac {(A b-4 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{2/3} b^{7/3}}-\frac {x (A b-4 a B)}{3 a b^2}+\frac {x^4 (A b-a B)}{3 a b \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.11, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {457, 321, 200, 31, 634, 617, 204, 628} \begin {gather*} -\frac {(A b-4 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{2/3} b^{7/3}}+\frac {(A b-4 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{2/3} b^{7/3}}-\frac {(A b-4 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{2/3} b^{7/3}}-\frac {x (A b-4 a B)}{3 a b^2}+\frac {x^4 (A b-a B)}{3 a b \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

-((A*b - 4*a*B)*x)/(3*a*b^2) + ((A*b - a*B)*x^4)/(3*a*b*(a + b*x^3)) - ((A*b - 4*a*B)*ArcTan[(a^(1/3) - 2*b^(1
/3)*x)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(2/3)*b^(7/3)) + ((A*b - 4*a*B)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(2/3)*b
^(7/3)) - ((A*b - 4*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(2/3)*b^(7/3))

Rule 31

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

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

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 {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx &=\frac {(A b-a B) x^4}{3 a b \left (a+b x^3\right )}+\frac {(-A b+4 a B) \int \frac {x^3}{a+b x^3} \, dx}{3 a b}\\ &=-\frac {(A b-4 a B) x}{3 a b^2}+\frac {(A b-a B) x^4}{3 a b \left (a+b x^3\right )}+\frac {(A b-4 a B) \int \frac {1}{a+b x^3} \, dx}{3 b^2}\\ &=-\frac {(A b-4 a B) x}{3 a b^2}+\frac {(A b-a B) x^4}{3 a b \left (a+b x^3\right )}+\frac {(A b-4 a B) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{9 a^{2/3} b^2}+\frac {(A b-4 a B) \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}{9 a^{2/3} b^2}\\ &=-\frac {(A b-4 a B) x}{3 a b^2}+\frac {(A b-a B) x^4}{3 a b \left (a+b x^3\right )}+\frac {(A b-4 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{2/3} b^{7/3}}-\frac {(A b-4 a B) \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}{18 a^{2/3} b^{7/3}}+\frac {(A b-4 a B) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{6 \sqrt [3]{a} b^2}\\ &=-\frac {(A b-4 a B) x}{3 a b^2}+\frac {(A b-a B) x^4}{3 a b \left (a+b x^3\right )}+\frac {(A b-4 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{2/3} b^{7/3}}-\frac {(A b-4 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{2/3} b^{7/3}}+\frac {(A b-4 a B) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{3 a^{2/3} b^{7/3}}\\ &=-\frac {(A b-4 a B) x}{3 a b^2}+\frac {(A b-a B) x^4}{3 a b \left (a+b x^3\right )}-\frac {(A b-4 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} a^{2/3} b^{7/3}}+\frac {(A b-4 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{2/3} b^{7/3}}-\frac {(A b-4 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{2/3} b^{7/3}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 160, normalized size = 0.84 \begin {gather*} \frac {\frac {(4 a B-A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{a^{2/3}}+\frac {2 (A b-4 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{a^{2/3}}+\frac {2 \sqrt {3} (4 a B-A b) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{a^{2/3}}-\frac {6 \sqrt [3]{b} x (A b-a B)}{a+b x^3}+18 \sqrt [3]{b} B x}{18 b^{7/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

(18*b^(1/3)*B*x - (6*b^(1/3)*(A*b - a*B)*x)/(a + b*x^3) + (2*Sqrt[3]*(-(A*b) + 4*a*B)*ArcTan[(1 - (2*b^(1/3)*x
)/a^(1/3))/Sqrt[3]])/a^(2/3) + (2*(A*b - 4*a*B)*Log[a^(1/3) + b^(1/3)*x])/a^(2/3) + ((-(A*b) + 4*a*B)*Log[a^(2
/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/a^(2/3))/(18*b^(7/3))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^3*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

IntegrateAlgebraic[(x^3*(A + B*x^3))/(a + b*x^3)^2, x]

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fricas [A]  time = 0.55, size = 573, normalized size = 3.02 \begin {gather*} \left [\frac {18 \, B a^{2} b^{2} x^{4} - 3 \, \sqrt {\frac {1}{3}} {\left (4 \, B a^{3} b - A a^{2} b^{2} + {\left (4 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3}\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}} \log \left (\frac {2 \, a b x^{3} - 3 \, \left (a^{2} b\right )^{\frac {1}{3}} a x - a^{2} + 3 \, \sqrt {\frac {1}{3}} {\left (2 \, a b x^{2} + \left (a^{2} b\right )^{\frac {2}{3}} x - \left (a^{2} b\right )^{\frac {1}{3}} a\right )} \sqrt {-\frac {\left (a^{2} b\right )^{\frac {1}{3}}}{b}}}{b x^{3} + a}\right ) + {\left ({\left (4 \, B a b - A b^{2}\right )} x^{3} + 4 \, B a^{2} - A a b\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) - 2 \, {\left ({\left (4 \, B a b - A b^{2}\right )} x^{3} + 4 \, B a^{2} - A a b\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right ) + 6 \, {\left (4 \, B a^{3} b - A a^{2} b^{2}\right )} x}{18 \, {\left (a^{2} b^{4} x^{3} + a^{3} b^{3}\right )}}, \frac {18 \, B a^{2} b^{2} x^{4} - 6 \, \sqrt {\frac {1}{3}} {\left (4 \, B a^{3} b - A a^{2} b^{2} + {\left (4 \, B a^{2} b^{2} - A a b^{3}\right )} x^{3}\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}} x - \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 ({\left (4 \, B a b - A b^{2}\right )} x^{3} + 4 \, B a^{2} - A a b\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac {2}{3}} x + \left (a^{2} b\right )^{\frac {1}{3}} a\right ) - 2 \, {\left ({\left (4 \, B a b - A b^{2}\right )} x^{3} + 4 \, B a^{2} - A a b\right )} \left (a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac {2}{3}}\right ) + 6 \, {\left (4 \, B a^{3} b - A a^{2} b^{2}\right )} x}{18 \, {\left (a^{2} b^{4} x^{3} + a^{3} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

[1/18*(18*B*a^2*b^2*x^4 - 3*sqrt(1/3)*(4*B*a^3*b - A*a^2*b^2 + (4*B*a^2*b^2 - A*a*b^3)*x^3)*sqrt(-(a^2*b)^(1/3
)/b)*log((2*a*b*x^3 - 3*(a^2*b)^(1/3)*a*x - a^2 + 3*sqrt(1/3)*(2*a*b*x^2 + (a^2*b)^(2/3)*x - (a^2*b)^(1/3)*a)*
sqrt(-(a^2*b)^(1/3)/b))/(b*x^3 + a)) + ((4*B*a*b - A*b^2)*x^3 + 4*B*a^2 - A*a*b)*(a^2*b)^(2/3)*log(a*b*x^2 - (
a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) - 2*((4*B*a*b - A*b^2)*x^3 + 4*B*a^2 - A*a*b)*(a^2*b)^(2/3)*log(a*b*x + (a^2
*b)^(2/3)) + 6*(4*B*a^3*b - A*a^2*b^2)*x)/(a^2*b^4*x^3 + a^3*b^3), 1/18*(18*B*a^2*b^2*x^4 - 6*sqrt(1/3)*(4*B*a
^3*b - A*a^2*b^2 + (4*B*a^2*b^2 - A*a*b^3)*x^3)*sqrt((a^2*b)^(1/3)/b)*arctan(sqrt(1/3)*(2*(a^2*b)^(2/3)*x - (a
^2*b)^(1/3)*a)*sqrt((a^2*b)^(1/3)/b)/a^2) + ((4*B*a*b - A*b^2)*x^3 + 4*B*a^2 - A*a*b)*(a^2*b)^(2/3)*log(a*b*x^
2 - (a^2*b)^(2/3)*x + (a^2*b)^(1/3)*a) - 2*((4*B*a*b - A*b^2)*x^3 + 4*B*a^2 - A*a*b)*(a^2*b)^(2/3)*log(a*b*x +
(a^2*b)^(2/3)) + 6*(4*B*a^3*b - A*a^2*b^2)*x)/(a^2*b^4*x^3 + a^3*b^3)]

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giac [A]  time = 0.19, size = 166, normalized size = 0.87 \begin {gather*} \frac {\sqrt {3} {\left (4 \, B a - A b\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 )}{9 \, \left (-a b^{2}\right )^{\frac {2}{3}} b} + \frac {{\left (4 \, B a - A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, \left (-a b^{2}\right )^{\frac {2}{3}} b} + \frac {B x}{b^{2}} + \frac {{\left (4 \, B a - A b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b^{2}} + \frac {B a x - A b x}{3 \, {\left (b x^{3} + a\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="giac")

[Out]

1/9*sqrt(3)*(4*B*a - A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/((-a*b^2)^(2/3)*b) + 1/18*(4*B
*a - A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/((-a*b^2)^(2/3)*b) + B*x/b^2 + 1/9*(4*B*a - A*b)*(-a/b)^(1/
3)*log(abs(x - (-a/b)^(1/3)))/(a*b^2) + 1/3*(B*a*x - A*b*x)/((b*x^3 + a)*b^2)

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maple [A]  time = 0.04, size = 228, normalized size = 1.20 \begin {gather*} -\frac {A x}{3 \left (b \,x^{3}+a \right ) b}+\frac {B a x}{3 \left (b \,x^{3}+a \right ) b^{2}}+\frac {\sqrt {3}\, A \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}+\frac {A \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}-\frac {A \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{2}}-\frac {4 \sqrt {3}\, B a \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{3}}-\frac {4 B a \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{3}}+\frac {2 B a \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{3}}+\frac {B x}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x^3+A)/(b*x^3+a)^2,x)

[Out]

B/b^2*x-1/3/b*x/(b*x^3+a)*A+1/3/b^2*x/(b*x^3+a)*B*a-4/9/b^3*B*a/(a/b)^(2/3)*ln(x+(a/b)^(1/3))+2/9/b^3*B*a/(a/b
)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))-4/9/b^3*B*a/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-
1))+1/9/b^2*A/(a/b)^(2/3)*ln(x+(a/b)^(1/3))-1/18/b^2*A/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+1/9/b^2*A
/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))

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maxima [A]  time = 1.22, size = 157, normalized size = 0.83 \begin {gather*} \frac {{\left (B a - A b\right )} x}{3 \, {\left (b^{3} x^{3} + a b^{2}\right )}} + \frac {B x}{b^{2}} - \frac {\sqrt {3} {\left (4 \, B a - A b\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 )}{9 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (4 \, B a - A b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (4 \, B a - A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{3} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

1/3*(B*a - A*b)*x/(b^3*x^3 + a*b^2) + B*x/b^2 - 1/9*sqrt(3)*(4*B*a - A*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3
))/(a/b)^(1/3))/(b^3*(a/b)^(2/3)) + 1/18*(4*B*a - A*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^3*(a/b)^(2/3)
) - 1/9*(4*B*a - A*b)*log(x + (a/b)^(1/3))/(b^3*(a/b)^(2/3))

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mupad [B]  time = 2.58, size = 150, normalized size = 0.79 \begin {gather*} \frac {B\,x}{b^2}-\frac {x\,\left (\frac {A\,b}{3}-\frac {B\,a}{3}\right )}{b^3\,x^3+a\,b^2}+\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (A\,b-4\,B\,a\right )}{9\,a^{2/3}\,b^{7/3}}-\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-4\,B\,a\right )}{9\,a^{2/3}\,b^{7/3}}+\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-4\,B\,a\right )}{9\,a^{2/3}\,b^{7/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(A + B*x^3))/(a + b*x^3)^2,x)

[Out]

(B*x)/b^2 - (x*((A*b)/3 - (B*a)/3))/(a*b^2 + b^3*x^3) + (log(b^(1/3)*x + a^(1/3))*(A*b - 4*B*a))/(9*a^(2/3)*b^
(7/3)) - (log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(A*b - 4*B*a))/(9*a^(2/3)*b^(
7/3)) + (log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(A*b - 4*B*a))/(9*a^(2/3)*b^(7
/3))

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sympy [A]  time = 1.67, size = 102, normalized size = 0.54 \begin {gather*} \frac {B x}{b^{2}} + \frac {x \left (- A b + B a\right )}{3 a b^{2} + 3 b^{3} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} a^{2} b^{7} - A^{3} b^{3} + 12 A^{2} B a b^{2} - 48 A B^{2} a^{2} b + 64 B^{3} a^{3}, \left (t \mapsto t \log {\left (- \frac {9 t a b^{2}}{- A b + 4 B a} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x**3+A)/(b*x**3+a)**2,x)

[Out]

B*x/b**2 + x*(-A*b + B*a)/(3*a*b**2 + 3*b**3*x**3) + RootSum(729*_t**3*a**2*b**7 - A**3*b**3 + 12*A**2*B*a*b**
2 - 48*A*B**2*a**2*b + 64*B**3*a**3, Lambda(_t, _t*log(-9*_t*a*b**2/(-A*b + 4*B*a) + x)))

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