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

Optimal. Leaf size=227 \[ \frac {5 (4 A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{11/3} \sqrt [3]{b}}-\frac {5 (4 A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{11/3} \sqrt [3]{b}}+\frac {5 (4 A b-a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{11/3} \sqrt [3]{b}}-\frac {5 (4 A b-a B)}{18 a^3 b x^2}+\frac {4 A b-a B}{9 a^2 b x^2 \left (a+b x^3\right )}+\frac {A b-a B}{6 a b x^2 \left (a+b x^3\right )^2} \]

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Rubi [A]  time = 0.13, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.450, Rules used = {457, 290, 325, 200, 31, 634, 617, 204, 628} \begin {gather*} \frac {5 (4 A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{11/3} \sqrt [3]{b}}+\frac {4 A b-a B}{9 a^2 b x^2 \left (a+b x^3\right )}-\frac {5 (4 A b-a B)}{18 a^3 b x^2}-\frac {5 (4 A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{11/3} \sqrt [3]{b}}+\frac {5 (4 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^{11/3} \sqrt [3]{b}}+\frac {A b-a B}{6 a b x^2 \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*(4*A*b - a*B))/(18*a^3*b*x^2) + (A*b - a*B)/(6*a*b*x^2*(a + b*x^3)^2) + (4*A*b - a*B)/(9*a^2*b*x^2*(a + b*
x^3)) + (5*(4*A*b - a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(11/3)*b^(1/3)) - (5*
(4*A*b - a*B)*Log[a^(1/3) + b^(1/3)*x])/(27*a^(11/3)*b^(1/3)) + (5*(4*A*b - a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)
*x + b^(2/3)*x^2])/(54*a^(11/3)*b^(1/3))

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 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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 {A+B x^3}{x^3 \left (a+b x^3\right )^3} \, dx &=\frac {A b-a B}{6 a b x^2 \left (a+b x^3\right )^2}+\frac {(8 A b-2 a B) \int \frac {1}{x^3 \left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=\frac {A b-a B}{6 a b x^2 \left (a+b x^3\right )^2}+\frac {4 A b-a B}{9 a^2 b x^2 \left (a+b x^3\right )}+\frac {(5 (4 A b-a B)) \int \frac {1}{x^3 \left (a+b x^3\right )} \, dx}{9 a^2 b}\\ &=-\frac {5 (4 A b-a B)}{18 a^3 b x^2}+\frac {A b-a B}{6 a b x^2 \left (a+b x^3\right )^2}+\frac {4 A b-a B}{9 a^2 b x^2 \left (a+b x^3\right )}-\frac {(5 (4 A b-a B)) \int \frac {1}{a+b x^3} \, dx}{9 a^3}\\ &=-\frac {5 (4 A b-a B)}{18 a^3 b x^2}+\frac {A b-a B}{6 a b x^2 \left (a+b x^3\right )^2}+\frac {4 A b-a B}{9 a^2 b x^2 \left (a+b x^3\right )}-\frac {(5 (4 A b-a B)) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{27 a^{11/3}}-\frac {(5 (4 A b-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}{27 a^{11/3}}\\ &=-\frac {5 (4 A b-a B)}{18 a^3 b x^2}+\frac {A b-a B}{6 a b x^2 \left (a+b x^3\right )^2}+\frac {4 A b-a B}{9 a^2 b x^2 \left (a+b x^3\right )}-\frac {5 (4 A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{11/3} \sqrt [3]{b}}-\frac {(5 (4 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^{10/3}}+\frac {(5 (4 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^{11/3} \sqrt [3]{b}}\\ &=-\frac {5 (4 A b-a B)}{18 a^3 b x^2}+\frac {A b-a B}{6 a b x^2 \left (a+b x^3\right )^2}+\frac {4 A b-a B}{9 a^2 b x^2 \left (a+b x^3\right )}-\frac {5 (4 A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{11/3} \sqrt [3]{b}}+\frac {5 (4 A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{11/3} \sqrt [3]{b}}-\frac {(5 (4 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^{11/3} \sqrt [3]{b}}\\ &=-\frac {5 (4 A b-a B)}{18 a^3 b x^2}+\frac {A b-a B}{6 a b x^2 \left (a+b x^3\right )^2}+\frac {4 A b-a B}{9 a^2 b x^2 \left (a+b x^3\right )}+\frac {5 (4 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^{11/3} \sqrt [3]{b}}-\frac {5 (4 A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{11/3} \sqrt [3]{b}}+\frac {5 (4 A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{54 a^{11/3} \sqrt [3]{b}}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 189, normalized size = 0.83 \begin {gather*} \frac {\frac {5 (4 A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{b}}+\frac {9 a^{5/3} x (a B-A b)}{\left (a+b x^3\right )^2}+\frac {3 a^{2/3} x (5 a B-11 A b)}{a+b x^3}-\frac {27 a^{2/3} A}{x^2}+\frac {10 (a B-4 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}+\frac {10 \sqrt {3} (4 A b-a B) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{54 a^{11/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-27*a^(2/3)*A)/x^2 + (9*a^(5/3)*(-(A*b) + a*B)*x)/(a + b*x^3)^2 + (3*a^(2/3)*(-11*A*b + 5*a*B)*x)/(a + b*x^3
) + (10*Sqrt[3]*(4*A*b - a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]])/b^(1/3) + (10*(-4*A*b + a*B)*Log[a^
(1/3) + b^(1/3)*x])/b^(1/3) + (5*(4*A*b - a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(1/3))/(54*a^
(11/3))

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.52, size = 812, normalized size = 3.58 \begin {gather*} \left [\frac {15 \, {\left (B a^{3} b^{2} - 4 \, A a^{2} b^{3}\right )} x^{6} - 27 \, A a^{4} b + 24 \, {\left (B a^{4} b - 4 \, A a^{3} b^{2}\right )} x^{3} - 15 \, \sqrt {\frac {1}{3}} {\left ({\left (B a^{2} b^{3} - 4 \, A a b^{4}\right )} x^{8} + 2 \, {\left (B a^{3} b^{2} - 4 \, A a^{2} b^{3}\right )} x^{5} + {\left (B a^{4} b - 4 \, A a^{3} b^{2}\right )} x^{2}\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 ) - 5 \, {\left ({\left (B a b^{2} - 4 \, A b^{3}\right )} x^{8} + 2 \, {\left (B a^{2} b - 4 \, A a b^{2}\right )} x^{5} + {\left (B a^{3} - 4 \, A a^{2} b\right )} x^{2}\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 ) + 10 \, {\left ({\left (B a b^{2} - 4 \, A b^{3}\right )} x^{8} + 2 \, {\left (B a^{2} b - 4 \, A a b^{2}\right )} x^{5} + {\left (B a^{3} - 4 \, A a^{2} b\right )} x^{2}\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (-a^{2} b\right )^{\frac {2}{3}}\right )}{54 \, {\left (a^{5} b^{3} x^{8} + 2 \, a^{6} b^{2} x^{5} + a^{7} b x^{2}\right )}}, \frac {15 \, {\left (B a^{3} b^{2} - 4 \, A a^{2} b^{3}\right )} x^{6} - 27 \, A a^{4} b + 24 \, {\left (B a^{4} b - 4 \, A a^{3} b^{2}\right )} x^{3} + 30 \, \sqrt {\frac {1}{3}} {\left ({\left (B a^{2} b^{3} - 4 \, A a b^{4}\right )} x^{8} + 2 \, {\left (B a^{3} b^{2} - 4 \, A a^{2} b^{3}\right )} x^{5} + {\left (B a^{4} b - 4 \, A a^{3} b^{2}\right )} x^{2}\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 ) - 5 \, {\left ({\left (B a b^{2} - 4 \, A b^{3}\right )} x^{8} + 2 \, {\left (B a^{2} b - 4 \, A a b^{2}\right )} x^{5} + {\left (B a^{3} - 4 \, A a^{2} b\right )} x^{2}\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 ) + 10 \, {\left ({\left (B a b^{2} - 4 \, A b^{3}\right )} x^{8} + 2 \, {\left (B a^{2} b - 4 \, A a b^{2}\right )} x^{5} + {\left (B a^{3} - 4 \, A a^{2} b\right )} x^{2}\right )} \left (-a^{2} b\right )^{\frac {2}{3}} \log \left (a b x + \left (-a^{2} b\right )^{\frac {2}{3}}\right )}{54 \, {\left (a^{5} b^{3} x^{8} + 2 \, a^{6} b^{2} x^{5} + a^{7} b x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/54*(15*(B*a^3*b^2 - 4*A*a^2*b^3)*x^6 - 27*A*a^4*b + 24*(B*a^4*b - 4*A*a^3*b^2)*x^3 - 15*sqrt(1/3)*((B*a^2*b
^3 - 4*A*a*b^4)*x^8 + 2*(B*a^3*b^2 - 4*A*a^2*b^3)*x^5 + (B*a^4*b - 4*A*a^3*b^2)*x^2)*sqrt((-a^2*b)^(1/3)/b)*lo
g((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)) - 5*((B*a*b^2 - 4*A*b^3)*x^8 + 2*(B*a^2*b - 4*A*a*b^2)*x^5 + (B*a^3 - 4*A*a^2
*b)*x^2)*(-a^2*b)^(2/3)*log(a*b*x^2 - (-a^2*b)^(2/3)*x - (-a^2*b)^(1/3)*a) + 10*((B*a*b^2 - 4*A*b^3)*x^8 + 2*(
B*a^2*b - 4*A*a*b^2)*x^5 + (B*a^3 - 4*A*a^2*b)*x^2)*(-a^2*b)^(2/3)*log(a*b*x + (-a^2*b)^(2/3)))/(a^5*b^3*x^8 +
2*a^6*b^2*x^5 + a^7*b*x^2), 1/54*(15*(B*a^3*b^2 - 4*A*a^2*b^3)*x^6 - 27*A*a^4*b + 24*(B*a^4*b - 4*A*a^3*b^2)*
x^3 + 30*sqrt(1/3)*((B*a^2*b^3 - 4*A*a*b^4)*x^8 + 2*(B*a^3*b^2 - 4*A*a^2*b^3)*x^5 + (B*a^4*b - 4*A*a^3*b^2)*x^
2)*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) - 5*((B*a*b^2 - 4*A*b^3)*x^8 + 2*(B*a^2*b - 4*A*a*b^2)*x^5 + (B*a^3 - 4*A*a^2*b)*x^2)*(-a^2*b)^(2/3)*log(a*
b*x^2 - (-a^2*b)^(2/3)*x - (-a^2*b)^(1/3)*a) + 10*((B*a*b^2 - 4*A*b^3)*x^8 + 2*(B*a^2*b - 4*A*a*b^2)*x^5 + (B*
a^3 - 4*A*a^2*b)*x^2)*(-a^2*b)^(2/3)*log(a*b*x + (-a^2*b)^(2/3)))/(a^5*b^3*x^8 + 2*a^6*b^2*x^5 + a^7*b*x^2)]

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giac [A]  time = 0.19, size = 209, normalized size = 0.92 \begin {gather*} -\frac {5 \, {\left (B a - 4 \, 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 )}{27 \, a^{4}} + \frac {5 \, \sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {1}{3}} B a - 4 \, \left (-a b^{2}\right )^{\frac {1}{3}} 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^{4} b} + \frac {5 \, {\left (\left (-a b^{2}\right )^{\frac {1}{3}} B a - 4 \, \left (-a b^{2}\right )^{\frac {1}{3}} 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^{4} b} + \frac {5 \, B a b x^{6} - 20 \, A b^{2} x^{6} + 8 \, B a^{2} x^{3} - 32 \, A a b x^{3} - 9 \, A a^{2}}{18 \, {\left (b x^{4} + a x\right )}^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/27*(B*a - 4*A*b)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^4 + 5/27*sqrt(3)*((-a*b^2)^(1/3)*B*a - 4*(-a*b^2
)^(1/3)*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^4*b) + 5/54*((-a*b^2)^(1/3)*B*a - 4*(-a*
b^2)^(1/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^4*b) + 1/18*(5*B*a*b*x^6 - 20*A*b^2*x^6 + 8*B*a^2*
x^3 - 32*A*a*b*x^3 - 9*A*a^2)/((b*x^4 + a*x)^2*a^3)

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maple [A]  time = 0.05, size = 277, normalized size = 1.22 \begin {gather*} -\frac {11 A \,b^{2} x^{4}}{18 \left (b \,x^{3}+a \right )^{2} a^{3}}+\frac {5 B b \,x^{4}}{18 \left (b \,x^{3}+a \right )^{2} a^{2}}-\frac {7 A b x}{9 \left (b \,x^{3}+a \right )^{2} a^{2}}+\frac {4 B x}{9 \left (b \,x^{3}+a \right )^{2} a}-\frac {20 \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 {2}{3}} a^{3}}-\frac {20 A \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{3}}+\frac {10 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 {2}{3}} a^{3}}+\frac {5 \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 {2}{3}} a^{2} b}+\frac {5 B \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 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 {2}{3}} a^{2} b}-\frac {A}{2 a^{3} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/18/a^3/(b*x^3+a)^2*A*x^4*b^2+5/18/a^2/(b*x^3+a)^2*B*x^4*b-7/9/a^2/(b*x^3+a)^2*b*A*x+4/9/a/(b*x^3+a)^2*B*x-
20/27/a^3*A/(a/b)^(2/3)*ln(x+(a/b)^(1/3))+10/27/a^3*A/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))-20/27/a^3*
A/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))+5/27/a^2*B/b/(a/b)^(2/3)*ln(x+(a/b)^(1/3))-5/54/
a^2*B/b/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+5/27/a^2*B/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(
a/b)^(1/3)*x-1))-1/2*A/a^3/x^2

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maxima [A]  time = 1.20, size = 201, normalized size = 0.89 \begin {gather*} \frac {5 \, {\left (B a b - 4 \, A b^{2}\right )} x^{6} + 8 \, {\left (B a^{2} - 4 \, A a b\right )} x^{3} - 9 \, A a^{2}}{18 \, {\left (a^{3} b^{2} x^{8} + 2 \, a^{4} b x^{5} + a^{5} x^{2}\right )}} + \frac {5 \, \sqrt {3} {\left (B a - 4 \, 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^{3} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {5 \, {\left (B a - 4 \, 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^{3} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {5 \, {\left (B a - 4 \, A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, a^{3} b \left (\frac {a}{b}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18*(5*(B*a*b - 4*A*b^2)*x^6 + 8*(B*a^2 - 4*A*a*b)*x^3 - 9*A*a^2)/(a^3*b^2*x^8 + 2*a^4*b*x^5 + a^5*x^2) + 5/2
7*sqrt(3)*(B*a - 4*A*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a^3*b*(a/b)^(2/3)) - 5/54*(B*a -
4*A*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^3*b*(a/b)^(2/3)) + 5/27*(B*a - 4*A*b)*log(x + (a/b)^(1/3))/(a
^3*b*(a/b)^(2/3))

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mupad [B]  time = 2.58, size = 188, normalized size = 0.83 \begin {gather*} -\frac {\frac {A}{2\,a}+\frac {4\,x^3\,\left (4\,A\,b-B\,a\right )}{9\,a^2}+\frac {5\,b\,x^6\,\left (4\,A\,b-B\,a\right )}{18\,a^3}}{a^2\,x^2+2\,a\,b\,x^5+b^2\,x^8}-\frac {5\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (4\,A\,b-B\,a\right )}{27\,a^{11/3}\,b^{1/3}}+\frac {5\,\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 (4\,A\,b-B\,a\right )}{27\,a^{11/3}\,b^{1/3}}-\frac {5\,\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 (4\,A\,b-B\,a\right )}{27\,a^{11/3}\,b^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(4*A*b - B*a))/(27*a^(11/3)*b^(1/3))
- (5*log(b^(1/3)*x + a^(1/3))*(4*A*b - B*a))/(27*a^(11/3)*b^(1/3)) - (A/(2*a) + (4*x^3*(4*A*b - B*a))/(9*a^2)
+ (5*b*x^6*(4*A*b - B*a))/(18*a^3))/(a^2*x^2 + b^2*x^8 + 2*a*b*x^5) - (5*log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x
- a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(4*A*b - B*a))/(27*a^(11/3)*b^(1/3))

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sympy [A]  time = 1.73, size = 143, normalized size = 0.63 \begin {gather*} \frac {- 9 A a^{2} + x^{6} \left (- 20 A b^{2} + 5 B a b\right ) + x^{3} \left (- 32 A a b + 8 B a^{2}\right )}{18 a^{5} x^{2} + 36 a^{4} b x^{5} + 18 a^{3} b^{2} x^{8}} + \operatorname {RootSum} {\left (19683 t^{3} a^{11} b + 8000 A^{3} b^{3} - 6000 A^{2} B a b^{2} + 1500 A B^{2} a^{2} b - 125 B^{3} a^{3}, \left (t \mapsto t \log {\left (\frac {27 t a^{4}}{- 20 A b + 5 B a} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-9*A*a**2 + x**6*(-20*A*b**2 + 5*B*a*b) + x**3*(-32*A*a*b + 8*B*a**2))/(18*a**5*x**2 + 36*a**4*b*x**5 + 18*a*
*3*b**2*x**8) + RootSum(19683*_t**3*a**11*b + 8000*A**3*b**3 - 6000*A**2*B*a*b**2 + 1500*A*B**2*a**2*b - 125*B
**3*a**3, Lambda(_t, _t*log(27*_t*a**4/(-20*A*b + 5*B*a) + x)))

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