3.1.95 \(\int \frac {x^{10} (A+B x^3)}{(a+b x^3)^3} \, dx\)

Optimal. Leaf size=246 \[ -\frac {2 a^{2/3} (5 A b-11 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{27 b^{14/3}}+\frac {4 a^{2/3} (5 A b-11 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 b^{14/3}}+\frac {4 a^{2/3} (5 A b-11 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} b^{14/3}}+\frac {2 x^2 (5 A b-11 a B)}{9 b^4}-\frac {4 x^5 (5 A b-11 a B)}{45 a b^3}+\frac {x^8 (5 A b-11 a B)}{18 a b^2 \left (a+b x^3\right )}+\frac {x^{11} (A b-a B)}{6 a b \left (a+b x^3\right )^2} \]

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Rubi [A]  time = 0.16, 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, 288, 302, 292, 31, 634, 617, 204, 628} \begin {gather*} -\frac {2 a^{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 )}{27 b^{14/3}}+\frac {4 a^{2/3} (5 A b-11 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 b^{14/3}}+\frac {4 a^{2/3} (5 A b-11 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} b^{14/3}}+\frac {x^8 (5 A b-11 a B)}{18 a b^2 \left (a+b x^3\right )}-\frac {4 x^5 (5 A b-11 a B)}{45 a b^3}+\frac {2 x^2 (5 A b-11 a B)}{9 b^4}+\frac {x^{11} (A b-a B)}{6 a b \left (a+b x^3\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^10*(A + B*x^3))/(a + b*x^3)^3,x]

[Out]

(2*(5*A*b - 11*a*B)*x^2)/(9*b^4) - (4*(5*A*b - 11*a*B)*x^5)/(45*a*b^3) + ((A*b - a*B)*x^11)/(6*a*b*(a + b*x^3)
^2) + ((5*A*b - 11*a*B)*x^8)/(18*a*b^2*(a + b*x^3)) + (4*a^(2/3)*(5*A*b - 11*a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*
x)/(Sqrt[3]*a^(1/3))])/(9*Sqrt[3]*b^(14/3)) + (4*a^(2/3)*(5*A*b - 11*a*B)*Log[a^(1/3) + b^(1/3)*x])/(27*b^(14/
3)) - (2*a^(2/3)*(5*A*b - 11*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(27*b^(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 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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& 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 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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^{10} \left (A+B x^3\right )}{\left (a+b x^3\right )^3} \, dx &=\frac {(A b-a B) x^{11}}{6 a b \left (a+b x^3\right )^2}+\frac {(-5 A b+11 a B) \int \frac {x^{10}}{\left (a+b x^3\right )^2} \, dx}{6 a b}\\ &=\frac {(A b-a B) x^{11}}{6 a b \left (a+b x^3\right )^2}+\frac {(5 A b-11 a B) x^8}{18 a b^2 \left (a+b x^3\right )}-\frac {(4 (5 A b-11 a B)) \int \frac {x^7}{a+b x^3} \, dx}{9 a b^2}\\ &=\frac {(A b-a B) x^{11}}{6 a b \left (a+b x^3\right )^2}+\frac {(5 A b-11 a B) x^8}{18 a b^2 \left (a+b x^3\right )}-\frac {(4 (5 A b-11 a B)) \int \left (-\frac {a x}{b^2}+\frac {x^4}{b}+\frac {a^2 x}{b^2 \left (a+b x^3\right )}\right ) \, dx}{9 a b^2}\\ &=\frac {2 (5 A b-11 a B) x^2}{9 b^4}-\frac {4 (5 A b-11 a B) x^5}{45 a b^3}+\frac {(A b-a B) x^{11}}{6 a b \left (a+b x^3\right )^2}+\frac {(5 A b-11 a B) x^8}{18 a b^2 \left (a+b x^3\right )}-\frac {(4 a (5 A b-11 a B)) \int \frac {x}{a+b x^3} \, dx}{9 b^4}\\ &=\frac {2 (5 A b-11 a B) x^2}{9 b^4}-\frac {4 (5 A b-11 a B) x^5}{45 a b^3}+\frac {(A b-a B) x^{11}}{6 a b \left (a+b x^3\right )^2}+\frac {(5 A b-11 a B) x^8}{18 a b^2 \left (a+b x^3\right )}+\frac {\left (4 a^{2/3} (5 A b-11 a B)\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{27 b^{13/3}}-\frac {\left (4 a^{2/3} (5 A b-11 a B)\right ) \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 b^{13/3}}\\ &=\frac {2 (5 A b-11 a B) x^2}{9 b^4}-\frac {4 (5 A b-11 a B) x^5}{45 a b^3}+\frac {(A b-a B) x^{11}}{6 a b \left (a+b x^3\right )^2}+\frac {(5 A b-11 a B) x^8}{18 a b^2 \left (a+b x^3\right )}+\frac {4 a^{2/3} (5 A b-11 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 b^{14/3}}-\frac {\left (2 a^{2/3} (5 A b-11 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 b^{14/3}}-\frac {(2 a (5 A b-11 a B)) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{9 b^{13/3}}\\ &=\frac {2 (5 A b-11 a B) x^2}{9 b^4}-\frac {4 (5 A b-11 a B) x^5}{45 a b^3}+\frac {(A b-a B) x^{11}}{6 a b \left (a+b x^3\right )^2}+\frac {(5 A b-11 a B) x^8}{18 a b^2 \left (a+b x^3\right )}+\frac {4 a^{2/3} (5 A b-11 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 b^{14/3}}-\frac {2 a^{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 )}{27 b^{14/3}}-\frac {\left (4 a^{2/3} (5 A b-11 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 b^{14/3}}\\ &=\frac {2 (5 A b-11 a B) x^2}{9 b^4}-\frac {4 (5 A b-11 a B) x^5}{45 a b^3}+\frac {(A b-a B) x^{11}}{6 a b \left (a+b x^3\right )^2}+\frac {(5 A b-11 a B) x^8}{18 a b^2 \left (a+b x^3\right )}+\frac {4 a^{2/3} (5 A b-11 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{9 \sqrt {3} b^{14/3}}+\frac {4 a^{2/3} (5 A b-11 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{27 b^{14/3}}-\frac {2 a^{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 )}{27 b^{14/3}}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 216, normalized size = 0.88 \begin {gather*} \frac {20 a^{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 )-40 a^{2/3} (11 a B-5 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-40 \sqrt {3} a^{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 )+\frac {45 a^2 b^{2/3} x^2 (a B-A b)}{\left (a+b x^3\right )^2}+135 b^{2/3} x^2 (A b-3 a B)+\frac {30 a b^{2/3} x^2 (7 A b-10 a B)}{a+b x^3}+54 b^{5/3} B x^5}{270 b^{14/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^10*(A + B*x^3))/(a + b*x^3)^3,x]

[Out]

(135*b^(2/3)*(A*b - 3*a*B)*x^2 + 54*b^(5/3)*B*x^5 + (45*a^2*b^(2/3)*(-(A*b) + a*B)*x^2)/(a + b*x^3)^2 + (30*a*
b^(2/3)*(7*A*b - 10*a*B)*x^2)/(a + b*x^3) - 40*Sqrt[3]*a^(2/3)*(-5*A*b + 11*a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(
1/3))/Sqrt[3]] - 40*a^(2/3)*(-5*A*b + 11*a*B)*Log[a^(1/3) + b^(1/3)*x] + 20*a^(2/3)*(-5*A*b + 11*a*B)*Log[a^(2
/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(270*b^(14/3))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{10} \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^10*(A + B*x^3))/(a + b*x^3)^3,x]

[Out]

IntegrateAlgebraic[(x^10*(A + B*x^3))/(a + b*x^3)^3, x]

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fricas [A]  time = 0.65, size = 364, normalized size = 1.48 \begin {gather*} \frac {54 \, B b^{3} x^{11} - 27 \, {\left (11 \, B a b^{2} - 5 \, A b^{3}\right )} x^{8} - 96 \, {\left (11 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{5} - 60 \, {\left (11 \, B a^{3} - 5 \, A a^{2} b\right )} x^{2} + 40 \, \sqrt {3} {\left ({\left (11 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 11 \, B a^{3} - 5 \, A a^{2} b + 2 \, {\left (11 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{3}\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) + 20 \, {\left ({\left (11 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 11 \, B a^{3} - 5 \, A a^{2} b + 2 \, {\left (11 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{3}\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) - 40 \, {\left ({\left (11 \, B a b^{2} - 5 \, A b^{3}\right )} x^{6} + 11 \, B a^{3} - 5 \, A a^{2} b + 2 \, {\left (11 \, B a^{2} b - 5 \, A a b^{2}\right )} x^{3}\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{270 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 259, normalized size = 1.05 \begin {gather*} -\frac {4 \, {\left (11 \, B a^{2} \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, A 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 b^{4}} - \frac {4 \, \sqrt {3} {\left (11 \, \left (-a b^{2}\right )^{\frac {2}{3}} B a - 5 \, \left (-a b^{2}\right )^{\frac {2}{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 \, b^{6}} + \frac {2 \, {\left (11 \, \left (-a b^{2}\right )^{\frac {2}{3}} B a - 5 \, \left (-a b^{2}\right )^{\frac {2}{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 \, b^{6}} - \frac {20 \, B a^{2} b x^{5} - 14 \, A a b^{2} x^{5} + 17 \, B a^{3} x^{2} - 11 \, A a^{2} b x^{2}}{18 \, {\left (b x^{3} + a\right )}^{2} b^{4}} + \frac {2 \, B b^{12} x^{5} - 15 \, B a b^{11} x^{2} + 5 \, A b^{12} x^{2}}{10 \, b^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="giac")

[Out]

-4/27*(11*B*a^2*(-a/b)^(1/3) - 5*A*a*b*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/(a*b^4) - 4/27*sq
rt(3)*(11*(-a*b^2)^(2/3)*B*a - 5*(-a*b^2)^(2/3)*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/b^6
+ 2/27*(11*(-a*b^2)^(2/3)*B*a - 5*(-a*b^2)^(2/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/b^6 - 1/18*(20
*B*a^2*b*x^5 - 14*A*a*b^2*x^5 + 17*B*a^3*x^2 - 11*A*a^2*b*x^2)/((b*x^3 + a)^2*b^4) + 1/10*(2*B*b^12*x^5 - 15*B
*a*b^11*x^2 + 5*A*b^12*x^2)/b^15

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maple [A]  time = 0.06, size = 308, normalized size = 1.25 \begin {gather*} \frac {7 A a \,x^{5}}{9 \left (b \,x^{3}+a \right )^{2} b^{2}}-\frac {10 B \,a^{2} x^{5}}{9 \left (b \,x^{3}+a \right )^{2} b^{3}}+\frac {B \,x^{5}}{5 b^{3}}+\frac {11 A \,a^{2} x^{2}}{18 \left (b \,x^{3}+a \right )^{2} b^{3}}-\frac {17 B \,a^{3} x^{2}}{18 \left (b \,x^{3}+a \right )^{2} b^{4}}+\frac {A \,x^{2}}{2 b^{3}}-\frac {3 B a \,x^{2}}{2 b^{4}}-\frac {20 \sqrt {3}\, A 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}} b^{4}}+\frac {20 A a \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{4}}-\frac {10 A 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}} b^{4}}+\frac {44 \sqrt {3}\, B \,a^{2} \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}} b^{5}}-\frac {44 B \,a^{2} \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{5}}+\frac {22 B \,a^{2} \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}} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^10*(B*x^3+A)/(b*x^3+a)^3,x)

[Out]

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

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maxima [A]  time = 1.37, size = 228, normalized size = 0.93 \begin {gather*} -\frac {2 \, {\left (10 \, B a^{2} b - 7 \, A a b^{2}\right )} x^{5} + {\left (17 \, B a^{3} - 11 \, A a^{2} b\right )} x^{2}}{18 \, {\left (b^{6} x^{6} + 2 \, a b^{5} x^{3} + a^{2} b^{4}\right )}} + \frac {4 \, \sqrt {3} {\left (11 \, B a^{2} - 5 \, 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 )}{27 \, b^{5} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {2 \, B b x^{5} - 5 \, {\left (3 \, B a - A b\right )} x^{2}}{10 \, b^{4}} + \frac {2 \, {\left (11 \, B a^{2} - 5 \, 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 )}{27 \, b^{5} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {4 \, {\left (11 \, B a^{2} - 5 \, A a b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{27 \, b^{5} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^10*(B*x^3+A)/(b*x^3+a)^3,x, algorithm="maxima")

[Out]

-1/18*(2*(10*B*a^2*b - 7*A*a*b^2)*x^5 + (17*B*a^3 - 11*A*a^2*b)*x^2)/(b^6*x^6 + 2*a*b^5*x^3 + a^2*b^4) + 4/27*
sqrt(3)*(11*B*a^2 - 5*A*a*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(b^5*(a/b)^(1/3)) + 1/10*(2*B
*b*x^5 - 5*(3*B*a - A*b)*x^2)/b^4 + 2/27*(11*B*a^2 - 5*A*a*b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^5*(a/b
)^(1/3)) - 4/27*(11*B*a^2 - 5*A*a*b)*log(x + (a/b)^(1/3))/(b^5*(a/b)^(1/3))

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mupad [B]  time = 2.58, size = 213, normalized size = 0.87 \begin {gather*} \frac {x^5\,\left (\frac {7\,A\,a\,b^2}{9}-\frac {10\,B\,a^2\,b}{9}\right )-x^2\,\left (\frac {17\,B\,a^3}{18}-\frac {11\,A\,a^2\,b}{18}\right )}{a^2\,b^4+2\,a\,b^5\,x^3+b^6\,x^6}+x^2\,\left (\frac {A}{2\,b^3}-\frac {3\,B\,a}{2\,b^4}\right )+\frac {B\,x^5}{5\,b^3}+\frac {4\,a^{2/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (5\,A\,b-11\,B\,a\right )}{27\,b^{14/3}}+\frac {4\,a^{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 (5\,A\,b-11\,B\,a\right )}{27\,b^{14/3}}-\frac {4\,a^{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 (5\,A\,b-11\,B\,a\right )}{27\,b^{14/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^10*(A + B*x^3))/(a + b*x^3)^3,x)

[Out]

(x^5*((7*A*a*b^2)/9 - (10*B*a^2*b)/9) - x^2*((17*B*a^3)/18 - (11*A*a^2*b)/18))/(a^2*b^4 + b^6*x^6 + 2*a*b^5*x^
3) + x^2*(A/(2*b^3) - (3*B*a)/(2*b^4)) + (B*x^5)/(5*b^3) + (4*a^(2/3)*log(b^(1/3)*x + a^(1/3))*(5*A*b - 11*B*a
))/(27*b^(14/3)) + (4*a^(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)*(5*A*b -
11*B*a))/(27*b^(14/3)) - (4*a^(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)*(5*
A*b - 11*B*a))/(27*b^(14/3))

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sympy [A]  time = 6.24, size = 192, normalized size = 0.78 \begin {gather*} \frac {B x^{5}}{5 b^{3}} + x^{2} \left (\frac {A}{2 b^{3}} - \frac {3 B a}{2 b^{4}}\right ) + \frac {x^{5} \left (14 A a b^{2} - 20 B a^{2} b\right ) + x^{2} \left (11 A a^{2} b - 17 B a^{3}\right )}{18 a^{2} b^{4} + 36 a b^{5} x^{3} + 18 b^{6} x^{6}} + \operatorname {RootSum} {\left (19683 t^{3} b^{14} - 8000 A^{3} a^{2} b^{3} + 52800 A^{2} B a^{3} b^{2} - 116160 A B^{2} a^{4} b + 85184 B^{3} a^{5}, \left (t \mapsto t \log {\left (\frac {729 t^{2} b^{9}}{400 A^{2} a b^{2} - 1760 A B a^{2} b + 1936 B^{2} a^{3}} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**10*(B*x**3+A)/(b*x**3+a)**3,x)

[Out]

B*x**5/(5*b**3) + x**2*(A/(2*b**3) - 3*B*a/(2*b**4)) + (x**5*(14*A*a*b**2 - 20*B*a**2*b) + x**2*(11*A*a**2*b -
17*B*a**3))/(18*a**2*b**4 + 36*a*b**5*x**3 + 18*b**6*x**6) + RootSum(19683*_t**3*b**14 - 8000*A**3*a**2*b**3
+ 52800*A**2*B*a**3*b**2 - 116160*A*B**2*a**4*b + 85184*B**3*a**5, Lambda(_t, _t*log(729*_t**2*b**9/(400*A**2*
a*b**2 - 1760*A*B*a**2*b + 1936*B**2*a**3) + x)))

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