3.106 \(\int \frac {A+B x^3}{x^6 (a+b x^3)^3} \, dx\)

Optimal. Leaf size=246 \[ -\frac {2 b^{2/3} (11 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{27 a^{14/3}}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}-\frac {4 b^{2/3} (11 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{14/3}}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2} \]

[Out]

-4/45*(11*A*b-5*B*a)/a^3/b/x^5+2/9*(11*A*b-5*B*a)/a^4/x^2+1/6*(A*b-B*a)/a/b/x^5/(b*x^3+a)^2+1/18*(11*A*b-5*B*a
)/a^2/b/x^5/(b*x^3+a)+4/27*b^(2/3)*(11*A*b-5*B*a)*ln(a^(1/3)+b^(1/3)*x)/a^(14/3)-2/27*b^(2/3)*(11*A*b-5*B*a)*l
n(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(14/3)-4/27*b^(2/3)*(11*A*b-5*B*a)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)
/a^(1/3)*3^(1/2))/a^(14/3)*3^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 246, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ -\frac {2 b^{2/3} (11 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{27 a^{14/3}}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}-\frac {4 b^{2/3} (11 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{14/3}}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^6*(a + b*x^3)^3),x]

[Out]

(-4*(11*A*b - 5*a*B))/(45*a^3*b*x^5) + (2*(11*A*b - 5*a*B))/(9*a^4*x^2) + (A*b - a*B)/(6*a*b*x^5*(a + b*x^3)^2
) + (11*A*b - 5*a*B)/(18*a^2*b*x^5*(a + b*x^3)) - (4*b^(2/3)*(11*A*b - 5*a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(
Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*a^(14/3)) + (4*b^(2/3)*(11*A*b - 5*a*B)*Log[a^(1/3) + b^(1/3)*x])/(27*a^(14/3))
- (2*b^(2/3)*(11*A*b - 5*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(27*a^(14/3))

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^6 \left (a+b x^3\right )^3} \, dx &=\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {(11 A b-5 a B) \int \frac {1}{x^6 \left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {(4 (11 A b-5 a B)) \int \frac {1}{x^6 \left (a+b x^3\right )} \, dx}{9 a^2 b}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}-\frac {(4 (11 A b-5 a B)) \int \frac {1}{x^3 \left (a+b x^3\right )} \, dx}{9 a^3}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {(4 b (11 A b-5 a B)) \int \frac {1}{a+b x^3} \, dx}{9 a^4}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {(4 b (11 A b-5 a B)) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{27 a^{14/3}}+\frac {(4 b (11 A b-5 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^{14/3}}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}-\frac {\left (2 b^{2/3} (11 A b-5 a B)\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}{27 a^{14/3}}+\frac {(2 b (11 A b-5 a B)) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{9 a^{13/3}}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}-\frac {2 b^{2/3} (11 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{27 a^{14/3}}+\frac {\left (4 b^{2/3} (11 A b-5 a B)\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^{14/3}}\\ &=-\frac {4 (11 A b-5 a B)}{45 a^3 b x^5}+\frac {2 (11 A b-5 a B)}{9 a^4 x^2}+\frac {A b-a B}{6 a b x^5 \left (a+b x^3\right )^2}+\frac {11 A b-5 a B}{18 a^2 b x^5 \left (a+b x^3\right )}-\frac {4 b^{2/3} (11 A b-5 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} a^{14/3}}+\frac {4 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 a^{14/3}}-\frac {2 b^{2/3} (11 A b-5 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{27 a^{14/3}}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 210, normalized size = 0.85 \[ \frac {20 b^{2/3} (5 a B-11 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-\frac {45 a^{5/3} b x (a B-A b)}{\left (a+b x^3\right )^2}-\frac {15 a^{2/3} b x (11 a B-17 A b)}{a+b x^3}-\frac {135 a^{2/3} (a B-3 A b)}{x^2}-\frac {54 a^{5/3} A}{x^5}+40 b^{2/3} (11 A b-5 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-40 \sqrt {3} b^{2/3} (11 A b-5 a B) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{270 a^{14/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^6*(a + b*x^3)^3),x]

[Out]

((-54*a^(5/3)*A)/x^5 - (135*a^(2/3)*(-3*A*b + a*B))/x^2 - (45*a^(5/3)*b*(-(A*b) + a*B)*x)/(a + b*x^3)^2 - (15*
a^(2/3)*b*(-17*A*b + 11*a*B)*x)/(a + b*x^3) - 40*Sqrt[3]*b^(2/3)*(11*A*b - 5*a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^
(1/3))/Sqrt[3]] + 40*b^(2/3)*(11*A*b - 5*a*B)*Log[a^(1/3) + b^(1/3)*x] + 20*b^(2/3)*(-11*A*b + 5*a*B)*Log[a^(2
/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(270*a^(14/3))

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fricas [A]  time = 1.01, size = 384, normalized size = 1.56 \[ -\frac {60 \, {\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{9} + 96 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{6} + 54 \, A a^{3} + 27 \, {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{3} + 40 \, \sqrt {3} {\left ({\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{11} + 2 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{8} + {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{5}\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - 20 \, {\left ({\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{11} + 2 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{8} + {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{5}\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} x^{2} - a b x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + a^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right ) + 40 \, {\left ({\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{11} + 2 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{8} + {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{5}\right )} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right )}{270 \, {\left (a^{4} b^{2} x^{11} + 2 \, a^{5} b x^{8} + a^{6} x^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(b*x^3+a)^3,x, algorithm="fricas")

[Out]

-1/270*(60*(5*B*a*b^2 - 11*A*b^3)*x^9 + 96*(5*B*a^2*b - 11*A*a*b^2)*x^6 + 54*A*a^3 + 27*(5*B*a^3 - 11*A*a^2*b)
*x^3 + 40*sqrt(3)*((5*B*a*b^2 - 11*A*b^3)*x^11 + 2*(5*B*a^2*b - 11*A*a*b^2)*x^8 + (5*B*a^3 - 11*A*a^2*b)*x^5)*
(b^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x*(b^2/a^2)^(2/3) - sqrt(3)*b)/b) - 20*((5*B*a*b^2 - 11*A*b^3)*x^11 +
2*(5*B*a^2*b - 11*A*a*b^2)*x^8 + (5*B*a^3 - 11*A*a^2*b)*x^5)*(b^2/a^2)^(1/3)*log(b^2*x^2 - a*b*x*(b^2/a^2)^(1/
3) + a^2*(b^2/a^2)^(2/3)) + 40*((5*B*a*b^2 - 11*A*b^3)*x^11 + 2*(5*B*a^2*b - 11*A*a*b^2)*x^8 + (5*B*a^3 - 11*A
*a^2*b)*x^5)*(b^2/a^2)^(1/3)*log(b*x + a*(b^2/a^2)^(1/3)))/(a^4*b^2*x^11 + 2*a^5*b*x^8 + a^6*x^5)

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giac [A]  time = 0.19, size = 229, normalized size = 0.93 \[ -\frac {4 \, \sqrt {3} {\left (5 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a - 11 \, \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, a^{5}} + \frac {4 \, {\left (5 \, B a b - 11 \, A b^{2}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{27 \, a^{5}} - \frac {2 \, {\left (5 \, \left (-a b^{2}\right )^{\frac {1}{3}} B a - 11 \, \left (-a b^{2}\right )^{\frac {1}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{27 \, a^{5}} - \frac {11 \, B a b^{2} x^{4} - 17 \, A b^{3} x^{4} + 14 \, B a^{2} b x - 20 \, A a b^{2} x}{18 \, {\left (b x^{3} + a\right )}^{2} a^{4}} - \frac {5 \, B a x^{3} - 15 \, A b x^{3} + 2 \, A a}{10 \, a^{4} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(b*x^3+a)^3,x, algorithm="giac")

[Out]

-4/27*sqrt(3)*(5*(-a*b^2)^(1/3)*B*a - 11*(-a*b^2)^(1/3)*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1
/3))/a^5 + 4/27*(5*B*a*b - 11*A*b^2)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^5 - 2/27*(5*(-a*b^2)^(1/3)*B*a
- 11*(-a*b^2)^(1/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/a^5 - 1/18*(11*B*a*b^2*x^4 - 17*A*b^3*x^4 +
14*B*a^2*b*x - 20*A*a*b^2*x)/((b*x^3 + a)^2*a^4) - 1/10*(5*B*a*x^3 - 15*A*b*x^3 + 2*A*a)/(a^4*x^5)

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maple [A]  time = 0.06, size = 295, normalized size = 1.20 \[ \frac {17 A \,b^{3} x^{4}}{18 \left (b \,x^{3}+a \right )^{2} a^{4}}-\frac {11 B \,b^{2} x^{4}}{18 \left (b \,x^{3}+a \right )^{2} a^{3}}+\frac {10 A \,b^{2} x}{9 \left (b \,x^{3}+a \right )^{2} a^{3}}-\frac {7 B b x}{9 \left (b \,x^{3}+a \right )^{2} a^{2}}+\frac {44 \sqrt {3}\, A b \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{4}}+\frac {44 A b \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{4}}-\frac {22 A b \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{4}}-\frac {20 \sqrt {3}\, B \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{3}}-\frac {20 B \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{3}}+\frac {10 B \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{3}}+\frac {3 A b}{2 a^{4} x^{2}}-\frac {B}{2 a^{3} x^{2}}-\frac {A}{5 a^{3} x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^6/(b*x^3+a)^3,x)

[Out]

17/18/a^4*b^3/(b*x^3+a)^2*A*x^4-11/18/a^3*b^2/(b*x^3+a)^2*B*x^4+10/9/a^3*b^2/(b*x^3+a)^2*A*x-7/9/a^2*b/(b*x^3+
a)^2*B*x+44/27/a^4*b*A/(a/b)^(2/3)*ln(x+(a/b)^(1/3))-22/27/a^4*b*A/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3
))+44/27/a^4*b*A/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))-20/27/a^3*B/(a/b)^(2/3)*ln(x+(a/b
)^(1/3))+10/27/a^3*B/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))-20/27/a^3*B/(a/b)^(2/3)*3^(1/2)*arctan(1/3*
3^(1/2)*(2/(a/b)^(1/3)*x-1))-1/5/a^3*A/x^5+3/2/a^4/x^2*A*b-1/2/a^3/x^2*B

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maxima [A]  time = 1.46, size = 221, normalized size = 0.90 \[ -\frac {20 \, {\left (5 \, B a b^{2} - 11 \, A b^{3}\right )} x^{9} + 32 \, {\left (5 \, B a^{2} b - 11 \, A a b^{2}\right )} x^{6} + 18 \, A a^{3} + 9 \, {\left (5 \, B a^{3} - 11 \, A a^{2} b\right )} x^{3}}{90 \, {\left (a^{4} b^{2} x^{11} + 2 \, a^{5} b x^{8} + a^{6} x^{5}\right )}} - \frac {4 \, \sqrt {3} {\left (5 \, B a - 11 \, A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{27 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {2 \, {\left (5 \, B a - 11 \, A b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{27 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {4 \, {\left (5 \, B a - 11 \, A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, a^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(b*x^3+a)^3,x, algorithm="maxima")

[Out]

-1/90*(20*(5*B*a*b^2 - 11*A*b^3)*x^9 + 32*(5*B*a^2*b - 11*A*a*b^2)*x^6 + 18*A*a^3 + 9*(5*B*a^3 - 11*A*a^2*b)*x
^3)/(a^4*b^2*x^11 + 2*a^5*b*x^8 + a^6*x^5) - 4/27*sqrt(3)*(5*B*a - 11*A*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/
3))/(a/b)^(1/3))/(a^4*(a/b)^(2/3)) + 2/27*(5*B*a - 11*A*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^4*(a/b)^(
2/3)) - 4/27*(5*B*a - 11*A*b)*log(x + (a/b)^(1/3))/(a^4*(a/b)^(2/3))

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mupad [B]  time = 2.58, size = 207, normalized size = 0.84 \[ \frac {\frac {x^3\,\left (11\,A\,b-5\,B\,a\right )}{10\,a^2}-\frac {A}{5\,a}+\frac {2\,b^2\,x^9\,\left (11\,A\,b-5\,B\,a\right )}{9\,a^4}+\frac {16\,b\,x^6\,\left (11\,A\,b-5\,B\,a\right )}{45\,a^3}}{a^2\,x^5+2\,a\,b\,x^8+b^2\,x^{11}}+\frac {4\,b^{2/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (11\,A\,b-5\,B\,a\right )}{27\,a^{14/3}}-\frac {4\,b^{2/3}\,\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (11\,A\,b-5\,B\,a\right )}{27\,a^{14/3}}+\frac {4\,b^{2/3}\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (11\,A\,b-5\,B\,a\right )}{27\,a^{14/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^3)/(x^6*(a + b*x^3)^3),x)

[Out]

((x^3*(11*A*b - 5*B*a))/(10*a^2) - A/(5*a) + (2*b^2*x^9*(11*A*b - 5*B*a))/(9*a^4) + (16*b*x^6*(11*A*b - 5*B*a)
)/(45*a^3))/(a^2*x^5 + b^2*x^11 + 2*a*b*x^8) + (4*b^(2/3)*log(b^(1/3)*x + a^(1/3))*(11*A*b - 5*B*a))/(27*a^(14
/3)) - (4*b^(2/3)*log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(11*A*b - 5*B*a))/(27
*a^(14/3)) + (4*b^(2/3)*log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x - a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(11*A*b - 5*B*a
))/(27*a^(14/3))

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sympy [A]  time = 2.18, size = 173, normalized size = 0.70 \[ \operatorname {RootSum} {\left (19683 t^{3} a^{14} - 85184 A^{3} b^{5} + 116160 A^{2} B a b^{4} - 52800 A B^{2} a^{2} b^{3} + 8000 B^{3} a^{3} b^{2}, \left (t \mapsto t \log {\left (- \frac {27 t a^{5}}{- 44 A b^{2} + 20 B a b} + x \right )} \right )\right )} + \frac {- 18 A a^{3} + x^{9} \left (220 A b^{3} - 100 B a b^{2}\right ) + x^{6} \left (352 A a b^{2} - 160 B a^{2} b\right ) + x^{3} \left (99 A a^{2} b - 45 B a^{3}\right )}{90 a^{6} x^{5} + 180 a^{5} b x^{8} + 90 a^{4} b^{2} x^{11}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/x**6/(b*x**3+a)**3,x)

[Out]

RootSum(19683*_t**3*a**14 - 85184*A**3*b**5 + 116160*A**2*B*a*b**4 - 52800*A*B**2*a**2*b**3 + 8000*B**3*a**3*b
**2, Lambda(_t, _t*log(-27*_t*a**5/(-44*A*b**2 + 20*B*a*b) + x))) + (-18*A*a**3 + x**9*(220*A*b**3 - 100*B*a*b
**2) + x**6*(352*A*a*b**2 - 160*B*a**2*b) + x**3*(99*A*a**2*b - 45*B*a**3))/(90*a**6*x**5 + 180*a**5*b*x**8 +
90*a**4*b**2*x**11)

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