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

Optimal. Leaf size=201 \[ \frac {(a B+2 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{7/3} b^{5/3}}-\frac {(a B+2 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{7/3} b^{5/3}}-\frac {(a B+2 A b) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{7/3} b^{5/3}}+\frac {x^2 (a B+2 A b)}{9 a^2 b \left (a+b x^3\right )}+\frac {x^2 (A b-a B)}{6 a b \left (a+b x^3\right )^2} \]

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Rubi [A]  time = 0.12, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {457, 290, 292, 31, 634, 617, 204, 628} \begin {gather*} \frac {(a B+2 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{7/3} b^{5/3}}-\frac {(a B+2 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{7/3} b^{5/3}}-\frac {(a B+2 A b) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{7/3} b^{5/3}}+\frac {x^2 (a B+2 A b)}{9 a^2 b \left (a+b x^3\right )}+\frac {x^2 (A b-a B)}{6 a b \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x^3))/(a + b*x^3)^3,x]

[Out]

((A*b - a*B)*x^2)/(6*a*b*(a + b*x^3)^2) + ((2*A*b + a*B)*x^2)/(9*a^2*b*(a + b*x^3)) - ((2*A*b + a*B)*ArcTan[(a
^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(7/3)*b^(5/3)) - ((2*A*b + a*B)*Log[a^(1/3) + b^(1/3)*x
])/(27*a^(7/3)*b^(5/3)) + ((2*A*b + a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(54*a^(7/3)*b^(5/3))

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 290

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

Rule 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(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 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 \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx &=\frac {(A b-a B) x^2}{6 a b \left (a+b x^3\right )^2}+\frac {(4 A b+2 a B) \int \frac {x}{\left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=\frac {(A b-a B) x^2}{6 a b \left (a+b x^3\right )^2}+\frac {(2 A b+a B) x^2}{9 a^2 b \left (a+b x^3\right )}+\frac {(2 A b+a B) \int \frac {x}{a+b x^3} \, dx}{9 a^2 b}\\ &=\frac {(A b-a B) x^2}{6 a b \left (a+b x^3\right )^2}+\frac {(2 A b+a B) x^2}{9 a^2 b \left (a+b x^3\right )}-\frac {(2 A b+a B) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{27 a^{7/3} b^{4/3}}+\frac {(2 A b+a B) \int \frac {\sqrt [3]{a}+\sqrt [3]{b} x}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{27 a^{7/3} b^{4/3}}\\ &=\frac {(A b-a B) x^2}{6 a b \left (a+b x^3\right )^2}+\frac {(2 A b+a B) x^2}{9 a^2 b \left (a+b x^3\right )}-\frac {(2 A b+a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{7/3} b^{5/3}}+\frac {(2 A b+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}{54 a^{7/3} b^{5/3}}+\frac {(2 A b+a B) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{18 a^2 b^{4/3}}\\ &=\frac {(A b-a B) x^2}{6 a b \left (a+b x^3\right )^2}+\frac {(2 A b+a B) x^2}{9 a^2 b \left (a+b x^3\right )}-\frac {(2 A b+a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{7/3} b^{5/3}}+\frac {(2 A b+a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{7/3} b^{5/3}}+\frac {(2 A b+a B) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{9 a^{7/3} b^{5/3}}\\ &=\frac {(A b-a B) x^2}{6 a b \left (a+b x^3\right )^2}+\frac {(2 A b+a B) x^2}{9 a^2 b \left (a+b x^3\right )}-\frac {(2 A b+a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{7/3} b^{5/3}}-\frac {(2 A b+a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{7/3} b^{5/3}}+\frac {(2 A b+a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{7/3} b^{5/3}}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 178, normalized size = 0.89 \begin {gather*} \frac {(a B+2 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-\frac {9 a^{4/3} b^{2/3} x^2 (a B-A b)}{\left (a+b x^3\right )^2}+\frac {6 \sqrt [3]{a} b^{2/3} x^2 (a B+2 A b)}{a+b x^3}-2 (a B+2 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-2 \sqrt {3} (a B+2 A b) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{54 a^{7/3} b^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-9*a^(4/3)*b^(2/3)*(-(A*b) + a*B)*x^2)/(a + b*x^3)^2 + (6*a^(1/3)*b^(2/3)*(2*A*b + a*B)*x^2)/(a + b*x^3) - 2
*Sqrt[3]*(2*A*b + a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 2*(2*A*b + a*B)*Log[a^(1/3) + b^(1/3)*x]
+ (2*A*b + a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(54*a^(7/3)*b^(5/3))

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.79, size = 752, normalized size = 3.74 \begin {gather*} \left [\frac {6 \, {\left (B a^{2} b^{3} + 2 \, A a b^{4}\right )} x^{5} - 3 \, {\left (B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{2} + 3 \, \sqrt {\frac {1}{3}} {\left ({\left (B a^{2} b^{3} + 2 \, A a b^{4}\right )} x^{6} + B a^{4} b + 2 \, A a^{3} b^{2} + 2 \, {\left (B a^{3} b^{2} + 2 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \log \left (\frac {2 \, b^{2} x^{3} - a b + 3 \, \sqrt {\frac {1}{3}} {\left (a b x + 2 \, \left (-a b^{2}\right )^{\frac {2}{3}} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} a\right )} \sqrt {\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} - 3 \, \left (-a b^{2}\right )^{\frac {2}{3}} x}{b x^{3} + a}\right ) + {\left ({\left (B a b^{2} + 2 \, A b^{3}\right )} x^{6} + B a^{3} + 2 \, A a^{2} b + 2 \, {\left (B a^{2} b + 2 \, A a b^{2}\right )} x^{3}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} b x + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, {\left ({\left (B a b^{2} + 2 \, A b^{3}\right )} x^{6} + B a^{3} + 2 \, A a^{2} b + 2 \, {\left (B a^{2} b + 2 \, A a b^{2}\right )} x^{3}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{54 \, {\left (a^{3} b^{5} x^{6} + 2 \, a^{4} b^{4} x^{3} + a^{5} b^{3}\right )}}, \frac {6 \, {\left (B a^{2} b^{3} + 2 \, A a b^{4}\right )} x^{5} - 3 \, {\left (B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{2} + 6 \, \sqrt {\frac {1}{3}} {\left ({\left (B a^{2} b^{3} + 2 \, A a b^{4}\right )} x^{6} + B a^{4} b + 2 \, A a^{3} b^{2} + 2 \, {\left (B a^{3} b^{2} + 2 \, A a^{2} b^{3}\right )} x^{3}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, b x + \left (-a b^{2}\right )^{\frac {1}{3}}\right )} \sqrt {-\frac {\left (-a b^{2}\right )^{\frac {1}{3}}}{a}}}{b}\right ) + {\left ({\left (B a b^{2} + 2 \, A b^{3}\right )} x^{6} + B a^{3} + 2 \, A a^{2} b + 2 \, {\left (B a^{2} b + 2 \, A a b^{2}\right )} x^{3}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac {1}{3}} b x + \left (-a b^{2}\right )^{\frac {2}{3}}\right ) - 2 \, {\left ({\left (B a b^{2} + 2 \, A b^{3}\right )} x^{6} + B a^{3} + 2 \, A a^{2} b + 2 \, {\left (B a^{2} b + 2 \, A a b^{2}\right )} x^{3}\right )} \left (-a b^{2}\right )^{\frac {2}{3}} \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{54 \, {\left (a^{3} b^{5} x^{6} + 2 \, a^{4} b^{4} x^{3} + a^{5} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/54*(6*(B*a^2*b^3 + 2*A*a*b^4)*x^5 - 3*(B*a^3*b^2 - 7*A*a^2*b^3)*x^2 + 3*sqrt(1/3)*((B*a^2*b^3 + 2*A*a*b^4)*
x^6 + B*a^4*b + 2*A*a^3*b^2 + 2*(B*a^3*b^2 + 2*A*a^2*b^3)*x^3)*sqrt((-a*b^2)^(1/3)/a)*log((2*b^2*x^3 - a*b + 3
*sqrt(1/3)*(a*b*x + 2*(-a*b^2)^(2/3)*x^2 + (-a*b^2)^(1/3)*a)*sqrt((-a*b^2)^(1/3)/a) - 3*(-a*b^2)^(2/3)*x)/(b*x
^3 + a)) + ((B*a*b^2 + 2*A*b^3)*x^6 + B*a^3 + 2*A*a^2*b + 2*(B*a^2*b + 2*A*a*b^2)*x^3)*(-a*b^2)^(2/3)*log(b^2*
x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3)) - 2*((B*a*b^2 + 2*A*b^3)*x^6 + B*a^3 + 2*A*a^2*b + 2*(B*a^2*b + 2*A
*a*b^2)*x^3)*(-a*b^2)^(2/3)*log(b*x - (-a*b^2)^(1/3)))/(a^3*b^5*x^6 + 2*a^4*b^4*x^3 + a^5*b^3), 1/54*(6*(B*a^2
*b^3 + 2*A*a*b^4)*x^5 - 3*(B*a^3*b^2 - 7*A*a^2*b^3)*x^2 + 6*sqrt(1/3)*((B*a^2*b^3 + 2*A*a*b^4)*x^6 + B*a^4*b +
 2*A*a^3*b^2 + 2*(B*a^3*b^2 + 2*A*a^2*b^3)*x^3)*sqrt(-(-a*b^2)^(1/3)/a)*arctan(sqrt(1/3)*(2*b*x + (-a*b^2)^(1/
3))*sqrt(-(-a*b^2)^(1/3)/a)/b) + ((B*a*b^2 + 2*A*b^3)*x^6 + B*a^3 + 2*A*a^2*b + 2*(B*a^2*b + 2*A*a*b^2)*x^3)*(
-a*b^2)^(2/3)*log(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3)) - 2*((B*a*b^2 + 2*A*b^3)*x^6 + B*a^3 + 2*A*a^
2*b + 2*(B*a^2*b + 2*A*a*b^2)*x^3)*(-a*b^2)^(2/3)*log(b*x - (-a*b^2)^(1/3)))/(a^3*b^5*x^6 + 2*a^4*b^4*x^3 + a^
5*b^3)]

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giac [A]  time = 0.21, size = 207, normalized size = 1.03 \begin {gather*} \frac {\sqrt {3} {\left (B a + 2 \, 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 )}{27 \, \left (-a b^{2}\right )^{\frac {1}{3}} a^{2} b} - \frac {{\left (B a + 2 \, 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 )}{54 \, \left (-a b^{2}\right )^{\frac {1}{3}} a^{2} b} - \frac {{\left (B a \left (-\frac {a}{b}\right )^{\frac {1}{3}} + 2 \, A b \left (-\frac {a}{b}\right )^{\frac {1}{3}}\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} b} + \frac {2 \, B a b x^{5} + 4 \, A b^{2} x^{5} - B a^{2} x^{2} + 7 \, A a b x^{2}}{18 \, {\left (b x^{3} + a\right )}^{2} a^{2} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 251, normalized size = 1.25 \begin {gather*} \frac {2 \sqrt {3}\, A \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 A \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 {A \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}+\frac {\sqrt {3}\, B \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 \,b^{2}}-\frac {B \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {1}{3}} a \,b^{2}}+\frac {B \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 {1}{3}} a \,b^{2}}+\frac {\frac {\left (2 A b +B a \right ) x^{5}}{9 a^{2}}+\frac {\left (7 A b -B a \right ) x^{2}}{18 a b}}{\left (b \,x^{3}+a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1/9*(2*A*b+B*a)/a^2*x^5+1/18*(7*A*b-B*a)/a/b*x^2)/(b*x^3+a)^2-2/27/a^2/b/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*A-1/27
/a/b^2/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*B+1/27/a^2/b/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*A+1/54/a/b^2/(
a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*B+2/27/a^2/b*3^(1/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3
)*x-1))*A+1/27/a/b^2*3^(1/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*B

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maxima [A]  time = 1.40, size = 195, normalized size = 0.97 \begin {gather*} \frac {2 \, {\left (B a b + 2 \, A b^{2}\right )} x^{5} - {\left (B a^{2} - 7 \, A a b\right )} x^{2}}{18 \, {\left (a^{2} b^{3} x^{6} + 2 \, a^{3} b^{2} x^{3} + a^{4} b\right )}} + \frac {\sqrt {3} {\left (B a + 2 \, 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 )}{27 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (B a + 2 \, 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 )}{54 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (B a + 2 \, A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, a^{2} b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.27, size = 175, normalized size = 0.87 \begin {gather*} \frac {\frac {x^5\,\left (2\,A\,b+B\,a\right )}{9\,a^2}+\frac {x^2\,\left (7\,A\,b-B\,a\right )}{18\,a\,b}}{a^2+2\,a\,b\,x^3+b^2\,x^6}-\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (2\,A\,b+B\,a\right )}{27\,a^{7/3}\,b^{5/3}}-\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 (2\,A\,b+B\,a\right )}{27\,a^{7/3}\,b^{5/3}}+\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 (2\,A\,b+B\,a\right )}{27\,a^{7/3}\,b^{5/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(A + B*x^3))/(a + b*x^3)^3,x)

[Out]

((x^5*(2*A*b + B*a))/(9*a^2) + (x^2*(7*A*b - B*a))/(18*a*b))/(a^2 + b^2*x^6 + 2*a*b*x^3) - (log(b^(1/3)*x + a^
(1/3))*(2*A*b + B*a))/(27*a^(7/3)*b^(5/3)) - (log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2
- 1/2)*(2*A*b + B*a))/(27*a^(7/3)*b^(5/3)) + (log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2
+ 1/2)*(2*A*b + B*a))/(27*a^(7/3)*b^(5/3))

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sympy [A]  time = 2.03, size = 153, normalized size = 0.76 \begin {gather*} \frac {x^{5} \left (4 A b^{2} + 2 B a b\right ) + x^{2} \left (7 A a b - B a^{2}\right )}{18 a^{4} b + 36 a^{3} b^{2} x^{3} + 18 a^{2} b^{3} x^{6}} + \operatorname {RootSum} {\left (19683 t^{3} a^{7} b^{5} + 8 A^{3} b^{3} + 12 A^{2} B a b^{2} + 6 A B^{2} a^{2} b + B^{3} a^{3}, \left (t \mapsto t \log {\left (\frac {729 t^{2} a^{5} b^{3}}{4 A^{2} b^{2} + 4 A B a b + B^{2} a^{2}} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x**5*(4*A*b**2 + 2*B*a*b) + x**2*(7*A*a*b - B*a**2))/(18*a**4*b + 36*a**3*b**2*x**3 + 18*a**2*b**3*x**6) + Ro
otSum(19683*_t**3*a**7*b**5 + 8*A**3*b**3 + 12*A**2*B*a*b**2 + 6*A*B**2*a**2*b + B**3*a**3, Lambda(_t, _t*log(
729*_t**2*a**5*b**3/(4*A**2*b**2 + 4*A*B*a*b + B**2*a**2) + x)))

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