3.1.70 \(\int \frac {A+B x^3}{x^8 (a+b x^3)} \, dx\)

Optimal. Leaf size=184 \[ -\frac {b^{4/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{10/3}}+\frac {b^{4/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{10/3}}+\frac {b^{4/3} (A b-a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{10/3}}-\frac {b (A b-a B)}{a^3 x}+\frac {A b-a B}{4 a^2 x^4}-\frac {A}{7 a x^7} \]

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Rubi [A]  time = 0.13, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {453, 325, 292, 31, 634, 617, 204, 628} \begin {gather*} -\frac {b^{4/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{10/3}}+\frac {b^{4/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{10/3}}+\frac {b^{4/3} (A b-a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{10/3}}+\frac {A b-a B}{4 a^2 x^4}-\frac {b (A b-a B)}{a^3 x}-\frac {A}{7 a x^7} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^8*(a + b*x^3)),x]

[Out]

-A/(7*a*x^7) + (A*b - a*B)/(4*a^2*x^4) - (b*(A*b - a*B))/(a^3*x) + (b^(4/3)*(A*b - a*B)*ArcTan[(a^(1/3) - 2*b^
(1/3)*x)/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*a^(10/3)) + (b^(4/3)*(A*b - a*B)*Log[a^(1/3) + b^(1/3)*x])/(3*a^(10/3))
- (b^(4/3)*(A*b - a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(6*a^(10/3))

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 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 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 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[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^8 \left (a+b x^3\right )} \, dx &=-\frac {A}{7 a x^7}-\frac {(7 A b-7 a B) \int \frac {1}{x^5 \left (a+b x^3\right )} \, dx}{7 a}\\ &=-\frac {A}{7 a x^7}+\frac {A b-a B}{4 a^2 x^4}+\frac {(b (A b-a B)) \int \frac {1}{x^2 \left (a+b x^3\right )} \, dx}{a^2}\\ &=-\frac {A}{7 a x^7}+\frac {A b-a B}{4 a^2 x^4}-\frac {b (A b-a B)}{a^3 x}-\frac {\left (b^2 (A b-a B)\right ) \int \frac {x}{a+b x^3} \, dx}{a^3}\\ &=-\frac {A}{7 a x^7}+\frac {A b-a B}{4 a^2 x^4}-\frac {b (A b-a B)}{a^3 x}+\frac {\left (b^{5/3} (A b-a B)\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a^{10/3}}-\frac {\left (b^{5/3} (A b-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}{3 a^{10/3}}\\ &=-\frac {A}{7 a x^7}+\frac {A b-a B}{4 a^2 x^4}-\frac {b (A b-a B)}{a^3 x}+\frac {b^{4/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{10/3}}-\frac {\left (b^{4/3} (A b-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}{6 a^{10/3}}-\frac {\left (b^{5/3} (A b-a B)\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 a^3}\\ &=-\frac {A}{7 a x^7}+\frac {A b-a B}{4 a^2 x^4}-\frac {b (A b-a B)}{a^3 x}+\frac {b^{4/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{10/3}}-\frac {b^{4/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{10/3}}-\frac {\left (b^{4/3} (A b-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 )}{a^{10/3}}\\ &=-\frac {A}{7 a x^7}+\frac {A b-a B}{4 a^2 x^4}-\frac {b (A b-a B)}{a^3 x}+\frac {b^{4/3} (A b-a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{10/3}}+\frac {b^{4/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{10/3}}-\frac {b^{4/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{10/3}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 173, normalized size = 0.94 \begin {gather*} \frac {14 b^{4/3} (a B-A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+\frac {21 a^{4/3} (A b-a B)}{x^4}-\frac {12 a^{7/3} A}{x^7}+28 b^{4/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+28 \sqrt {3} b^{4/3} (A b-a B) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+\frac {84 \sqrt [3]{a} b (a B-A b)}{x}}{84 a^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^8*(a + b*x^3)),x]

[Out]

((-12*a^(7/3)*A)/x^7 + (21*a^(4/3)*(A*b - a*B))/x^4 + (84*a^(1/3)*b*(-(A*b) + a*B))/x + 28*Sqrt[3]*b^(4/3)*(A*
b - a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + 28*b^(4/3)*(A*b - a*B)*Log[a^(1/3) + b^(1/3)*x] + 14*b^
(4/3)*(-(A*b) + a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(84*a^(10/3))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^3}{x^8 \left (a+b x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x^3)/(x^8*(a + b*x^3)),x]

[Out]

IntegrateAlgebraic[(A + B*x^3)/(x^8*(a + b*x^3)), x]

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fricas [A]  time = 0.85, size = 180, normalized size = 0.98 \begin {gather*} \frac {28 \, \sqrt {3} {\left (B a b - A b^{2}\right )} x^{7} \left (\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x \left (\frac {b}{a}\right )^{\frac {1}{3}} - \frac {1}{3} \, \sqrt {3}\right ) + 14 \, {\left (B a b - A b^{2}\right )} x^{7} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x^{2} - a x \left (\frac {b}{a}\right )^{\frac {2}{3}} + a \left (\frac {b}{a}\right )^{\frac {1}{3}}\right ) - 28 \, {\left (B a b - A b^{2}\right )} x^{7} \left (\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 84 \, {\left (B a b - A b^{2}\right )} x^{6} - 21 \, {\left (B a^{2} - A a b\right )} x^{3} - 12 \, A a^{2}}{84 \, a^{3} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^8/(b*x^3+a),x, algorithm="fricas")

[Out]

1/84*(28*sqrt(3)*(B*a*b - A*b^2)*x^7*(b/a)^(1/3)*arctan(2/3*sqrt(3)*x*(b/a)^(1/3) - 1/3*sqrt(3)) + 14*(B*a*b -
A*b^2)*x^7*(b/a)^(1/3)*log(b*x^2 - a*x*(b/a)^(2/3) + a*(b/a)^(1/3)) - 28*(B*a*b - A*b^2)*x^7*(b/a)^(1/3)*log(
b*x + a*(b/a)^(2/3)) + 84*(B*a*b - A*b^2)*x^6 - 21*(B*a^2 - A*a*b)*x^3 - 12*A*a^2)/(a^3*x^7)

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giac [A]  time = 0.18, size = 216, normalized size = 1.17 \begin {gather*} -\frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {2}{3}} B a - \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 )}{3 \, a^{4}} - \frac {{\left (B a b^{2} \left (-\frac {a}{b}\right )^{\frac {1}{3}} - A b^{3} \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 )}{3 \, a^{4}} + \frac {{\left (\left (-a b^{2}\right )^{\frac {2}{3}} B a - \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 )}{6 \, a^{4}} + \frac {28 \, B a b x^{6} - 28 \, A b^{2} x^{6} - 7 \, B a^{2} x^{3} + 7 \, A a b x^{3} - 4 \, A a^{2}}{28 \, a^{3} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^8/(b*x^3+a),x, algorithm="giac")

[Out]

-1/3*sqrt(3)*((-a*b^2)^(2/3)*B*a - (-a*b^2)^(2/3)*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/a
^4 - 1/3*(B*a*b^2*(-a/b)^(1/3) - A*b^3*(-a/b)^(1/3))*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^4 + 1/6*((-a*b^
2)^(2/3)*B*a - (-a*b^2)^(2/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/a^4 + 1/28*(28*B*a*b*x^6 - 28*A*b^
2*x^6 - 7*B*a^2*x^3 + 7*A*a*b*x^3 - 4*A*a^2)/(a^3*x^7)

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maple [A]  time = 0.06, size = 247, normalized size = 1.34 \begin {gather*} -\frac {\sqrt {3}\, A \,b^{2} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{3}}+\frac {A \,b^{2} \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{3}}-\frac {A \,b^{2} \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{3}}+\frac {\sqrt {3}\, B b \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{2}}-\frac {B b \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{2}}+\frac {B b \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {1}{3}} a^{2}}-\frac {A \,b^{2}}{a^{3} x}+\frac {B b}{a^{2} x}+\frac {A b}{4 a^{2} x^{4}}-\frac {B}{4 a \,x^{4}}-\frac {A}{7 a \,x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^8/(b*x^3+a),x)

[Out]

1/3/a^3*b^2/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*A-1/3/a^2*b/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*B-1/6/a^3*b^2/(a/b)^(1/3)*
ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*A+1/6/a^2*b/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*B-1/3/a^3*b^2*3^(1
/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*A+1/3/a^2*b*3^(1/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/
(a/b)^(1/3)*x-1))*B-1/7*A/a/x^7+1/4/a^2/x^4*A*b-1/4/a/x^4*B-b^2/a^3/x*A+b/a^2/x*B

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maxima [A]  time = 1.29, size = 178, normalized size = 0.97 \begin {gather*} \frac {\sqrt {3} {\left (B a b - A b^{2}\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 )}{3 \, a^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {{\left (B a b - A b^{2}\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, a^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (B a b - A b^{2}\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a^{3} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {28 \, {\left (B a b - A b^{2}\right )} x^{6} - 7 \, {\left (B a^{2} - A a b\right )} x^{3} - 4 \, A a^{2}}{28 \, a^{3} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^8/(b*x^3+a),x, algorithm="maxima")

[Out]

1/3*sqrt(3)*(B*a*b - A*b^2)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(1/3))/(a/b)^(1/3))/(a^3*(a/b)^(1/3)) + 1/6*(B*a*b
- A*b^2)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^3*(a/b)^(1/3)) - 1/3*(B*a*b - A*b^2)*log(x + (a/b)^(1/3))/
(a^3*(a/b)^(1/3)) + 1/28*(28*(B*a*b - A*b^2)*x^6 - 7*(B*a^2 - A*a*b)*x^3 - 4*A*a^2)/(a^3*x^7)

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mupad [B]  time = 2.58, size = 161, normalized size = 0.88 \begin {gather*} \frac {b^{4/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (A\,b-B\,a\right )}{3\,a^{10/3}}-\frac {\frac {A}{7\,a}-\frac {x^3\,\left (A\,b-B\,a\right )}{4\,a^2}+\frac {b\,x^6\,\left (A\,b-B\,a\right )}{a^3}}{x^7}+\frac {b^{4/3}\,\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 (A\,b-B\,a\right )}{3\,a^{10/3}}-\frac {b^{4/3}\,\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 (A\,b-B\,a\right )}{3\,a^{10/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^3)/(x^8*(a + b*x^3)),x)

[Out]

(b^(4/3)*log(b^(1/3)*x + a^(1/3))*(A*b - B*a))/(3*a^(10/3)) - (A/(7*a) - (x^3*(A*b - B*a))/(4*a^2) + (b*x^6*(A
*b - B*a))/a^3)/x^7 + (b^(4/3)*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 - B
*a))/(3*a^(10/3)) - (b^(4/3)*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 - B*a
))/(3*a^(10/3))

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sympy [A]  time = 1.13, size = 139, normalized size = 0.76 \begin {gather*} \operatorname {RootSum} {\left (27 t^{3} a^{10} - A^{3} b^{7} + 3 A^{2} B a b^{6} - 3 A B^{2} a^{2} b^{5} + B^{3} a^{3} b^{4}, \left (t \mapsto t \log {\left (\frac {9 t^{2} a^{7}}{A^{2} b^{5} - 2 A B a b^{4} + B^{2} a^{2} b^{3}} + x \right )} \right )\right )} + \frac {- 4 A a^{2} + x^{6} \left (- 28 A b^{2} + 28 B a b\right ) + x^{3} \left (7 A a b - 7 B a^{2}\right )}{28 a^{3} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/x**8/(b*x**3+a),x)

[Out]

RootSum(27*_t**3*a**10 - A**3*b**7 + 3*A**2*B*a*b**6 - 3*A*B**2*a**2*b**5 + B**3*a**3*b**4, Lambda(_t, _t*log(
9*_t**2*a**7/(A**2*b**5 - 2*A*B*a*b**4 + B**2*a**2*b**3) + x))) + (-4*A*a**2 + x**6*(-28*A*b**2 + 28*B*a*b) +
x**3*(7*A*a*b - 7*B*a**2))/(28*a**3*x**7)

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