3.1.74 \(\int \frac {x^6 (A+B x^3)}{(a+b x^3)^2} \, dx\)

Optimal. Leaf size=213 \[ \frac {\sqrt [3]{a} (4 A b-7 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 b^{10/3}}-\frac {\sqrt [3]{a} (4 A b-7 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{10/3}}+\frac {\sqrt [3]{a} (4 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} b^{10/3}}+\frac {x (4 A b-7 a B)}{3 b^3}-\frac {x^4 (4 A b-7 a B)}{12 a b^2}+\frac {x^7 (A b-a B)}{3 a b \left (a+b x^3\right )} \]

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Rubi [A]  time = 0.13, antiderivative size = 213, 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 = {457, 302, 200, 31, 634, 617, 204, 628} \begin {gather*} \frac {\sqrt [3]{a} (4 A b-7 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 b^{10/3}}-\frac {x^4 (4 A b-7 a B)}{12 a b^2}+\frac {x (4 A b-7 a B)}{3 b^3}-\frac {\sqrt [3]{a} (4 A b-7 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{10/3}}+\frac {\sqrt [3]{a} (4 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} b^{10/3}}+\frac {x^7 (A b-a B)}{3 a b \left (a+b x^3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^6*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

((4*A*b - 7*a*B)*x)/(3*b^3) - ((4*A*b - 7*a*B)*x^4)/(12*a*b^2) + ((A*b - a*B)*x^7)/(3*a*b*(a + b*x^3)) + (a^(1
/3)*(4*A*b - 7*a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*b^(10/3)) - (a^(1/3)*(4*A*b
- 7*a*B)*Log[a^(1/3) + b^(1/3)*x])/(9*b^(10/3)) + (a^(1/3)*(4*A*b - 7*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b
^(2/3)*x^2])/(18*b^(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 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 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^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx &=\frac {(A b-a B) x^7}{3 a b \left (a+b x^3\right )}+\frac {(-4 A b+7 a B) \int \frac {x^6}{a+b x^3} \, dx}{3 a b}\\ &=\frac {(A b-a B) x^7}{3 a b \left (a+b x^3\right )}+\frac {(-4 A b+7 a B) \int \left (-\frac {a}{b^2}+\frac {x^3}{b}+\frac {a^2}{b^2 \left (a+b x^3\right )}\right ) \, dx}{3 a b}\\ &=\frac {(4 A b-7 a B) x}{3 b^3}-\frac {(4 A b-7 a B) x^4}{12 a b^2}+\frac {(A b-a B) x^7}{3 a b \left (a+b x^3\right )}-\frac {(a (4 A b-7 a B)) \int \frac {1}{a+b x^3} \, dx}{3 b^3}\\ &=\frac {(4 A b-7 a B) x}{3 b^3}-\frac {(4 A b-7 a B) x^4}{12 a b^2}+\frac {(A b-a B) x^7}{3 a b \left (a+b x^3\right )}-\frac {\left (\sqrt [3]{a} (4 A b-7 a B)\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{9 b^3}-\frac {\left (\sqrt [3]{a} (4 A b-7 a B)\right ) \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}{9 b^3}\\ &=\frac {(4 A b-7 a B) x}{3 b^3}-\frac {(4 A b-7 a B) x^4}{12 a b^2}+\frac {(A b-a B) x^7}{3 a b \left (a+b x^3\right )}-\frac {\sqrt [3]{a} (4 A b-7 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{10/3}}+\frac {\left (\sqrt [3]{a} (4 A b-7 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}{18 b^{10/3}}-\frac {\left (a^{2/3} (4 A b-7 a B)\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{6 b^3}\\ &=\frac {(4 A b-7 a B) x}{3 b^3}-\frac {(4 A b-7 a B) x^4}{12 a b^2}+\frac {(A b-a B) x^7}{3 a b \left (a+b x^3\right )}-\frac {\sqrt [3]{a} (4 A b-7 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{10/3}}+\frac {\sqrt [3]{a} (4 A b-7 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 b^{10/3}}-\frac {\left (\sqrt [3]{a} (4 A b-7 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 )}{3 b^{10/3}}\\ &=\frac {(4 A b-7 a B) x}{3 b^3}-\frac {(4 A b-7 a B) x^4}{12 a b^2}+\frac {(A b-a B) x^7}{3 a b \left (a+b x^3\right )}+\frac {\sqrt [3]{a} (4 A b-7 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} b^{10/3}}-\frac {\sqrt [3]{a} (4 A b-7 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{10/3}}+\frac {\sqrt [3]{a} (4 A b-7 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 b^{10/3}}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 181, normalized size = 0.85 \begin {gather*} \frac {-2 \sqrt [3]{a} (7 a B-4 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+\frac {12 a \sqrt [3]{b} x (A b-a B)}{a+b x^3}+36 \sqrt [3]{b} x (A b-2 a B)+4 \sqrt [3]{a} (7 a B-4 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-4 \sqrt {3} \sqrt [3]{a} (7 a B-4 A b) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+9 b^{4/3} B x^4}{36 b^{10/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^6*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

(36*b^(1/3)*(A*b - 2*a*B)*x + 9*b^(4/3)*B*x^4 + (12*a*b^(1/3)*(A*b - a*B)*x)/(a + b*x^3) - 4*Sqrt[3]*a^(1/3)*(
-4*A*b + 7*a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] + 4*a^(1/3)*(-4*A*b + 7*a*B)*Log[a^(1/3) + b^(1/3)
*x] - 2*a^(1/3)*(-4*A*b + 7*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(36*b^(10/3))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^6 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(x^6*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

IntegrateAlgebraic[(x^6*(A + B*x^3))/(a + b*x^3)^2, x]

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fricas [A]  time = 0.85, size = 240, normalized size = 1.13 \begin {gather*} \frac {9 \, B b^{2} x^{7} - 9 \, {\left (7 \, B a b - 4 \, A b^{2}\right )} x^{4} - 4 \, \sqrt {3} {\left ({\left (7 \, B a b - 4 \, A b^{2}\right )} x^{3} + 7 \, B a^{2} - 4 \, A a b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (-\frac {a}{b}\right )^{\frac {2}{3}} - \sqrt {3} a}{3 \, a}\right ) + 2 \, {\left ({\left (7 \, B a b - 4 \, A b^{2}\right )} x^{3} + 7 \, B a^{2} - 4 \, A a b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right ) - 4 \, {\left ({\left (7 \, B a b - 4 \, A b^{2}\right )} x^{3} + 7 \, B a^{2} - 4 \, A a b\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left (x - \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right ) - 12 \, {\left (7 \, B a^{2} - 4 \, A a b\right )} x}{36 \, {\left (b^{4} x^{3} + a b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

1/36*(9*B*b^2*x^7 - 9*(7*B*a*b - 4*A*b^2)*x^4 - 4*sqrt(3)*((7*B*a*b - 4*A*b^2)*x^3 + 7*B*a^2 - 4*A*a*b)*(-a/b)
^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*(-a/b)^(2/3) - sqrt(3)*a)/a) + 2*((7*B*a*b - 4*A*b^2)*x^3 + 7*B*a^2 - 4*A*a*b
)*(-a/b)^(1/3)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3)) - 4*((7*B*a*b - 4*A*b^2)*x^3 + 7*B*a^2 - 4*A*a*b)*(-a/
b)^(1/3)*log(x - (-a/b)^(1/3)) - 12*(7*B*a^2 - 4*A*a*b)*x)/(b^4*x^3 + a*b^3)

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giac [A]  time = 0.18, size = 211, normalized size = 0.99 \begin {gather*} \frac {\sqrt {3} {\left (7 \, \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 )}{9 \, b^{4}} - \frac {{\left (7 \, B a^{2} - 4 \, A 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 )}{9 \, a b^{3}} + \frac {{\left (7 \, \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 )}{18 \, b^{4}} - \frac {B a^{2} x - A a b x}{3 \, {\left (b x^{3} + a\right )} b^{3}} + \frac {B b^{6} x^{4} - 8 \, B a b^{5} x + 4 \, A b^{6} x}{4 \, b^{8}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 257, normalized size = 1.21 \begin {gather*} \frac {B \,x^{4}}{4 b^{2}}+\frac {A a x}{3 \left (b \,x^{3}+a \right ) b^{2}}-\frac {B \,a^{2} x}{3 \left (b \,x^{3}+a \right ) b^{3}}-\frac {4 \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 )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{3}}-\frac {4 A a \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{3}}+\frac {2 A a \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{3}}+\frac {A x}{b^{2}}+\frac {7 \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 )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{4}}+\frac {7 B \,a^{2} \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{4}}-\frac {7 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 )}{18 \left (\frac {a}{b}\right )^{\frac {2}{3}} b^{4}}-\frac {2 B a x}{b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(B*x^3+A)/(b*x^3+a)^2,x)

[Out]

1/4/b^2*B*x^4+1/b^2*A*x-2/b^3*B*a*x+1/3*a/b^2*x/(b*x^3+a)*A-1/3*a^2/b^3*x/(b*x^3+a)*B-4/9*a/b^3*A/(a/b)^(2/3)*
ln(x+(a/b)^(1/3))+2/9*a/b^3*A/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))-4/9*a/b^3*A/(a/b)^(2/3)*3^(1/2)*ar
ctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))+7/9*a^2/b^4*B/(a/b)^(2/3)*ln(x+(a/b)^(1/3))-7/18*a^2/b^4*B/(a/b)^(2/3)*l
n(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+7/9*a^2/b^4*B/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))

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maxima [A]  time = 1.01, size = 187, normalized size = 0.88 \begin {gather*} -\frac {{\left (B a^{2} - A a b\right )} x}{3 \, {\left (b^{4} x^{3} + a b^{3}\right )}} + \frac {B b x^{4} - 4 \, {\left (2 \, B a - A b\right )} x}{4 \, b^{3}} + \frac {\sqrt {3} {\left (7 \, B a^{2} - 4 \, 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 )}{9 \, b^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (7 \, B a^{2} - 4 \, 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 )}{18 \, b^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (7 \, B a^{2} - 4 \, A a b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{4} \left (\frac {a}{b}\right )^{\frac {2}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.62, size = 193, normalized size = 0.91 \begin {gather*} x\,\left (\frac {A}{b^2}-\frac {2\,B\,a}{b^3}\right )-\frac {x\,\left (\frac {B\,a^2}{3}-\frac {A\,a\,b}{3}\right )}{b^4\,x^3+a\,b^3}+\frac {B\,x^4}{4\,b^2}+\frac {{\left (-a\right )}^{1/3}\,\ln \left ({\left (-a\right )}^{4/3}+a\,b^{1/3}\,x\right )\,\left (4\,A\,b-7\,B\,a\right )}{9\,b^{10/3}}-\frac {{\left (-a\right )}^{1/3}\,\ln \left ({\left (-a\right )}^{4/3}-2\,a\,b^{1/3}\,x+\sqrt {3}\,{\left (-a\right )}^{4/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (4\,A\,b-7\,B\,a\right )}{9\,b^{10/3}}+\frac {{\left (-a\right )}^{1/3}\,\ln \left (2\,a\,b^{1/3}\,x-{\left (-a\right )}^{4/3}+\sqrt {3}\,{\left (-a\right )}^{4/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (4\,A\,b-7\,B\,a\right )}{9\,b^{10/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^6*(A + B*x^3))/(a + b*x^3)^2,x)

[Out]

x*(A/b^2 - (2*B*a)/b^3) - (x*((B*a^2)/3 - (A*a*b)/3))/(a*b^3 + b^4*x^3) + (B*x^4)/(4*b^2) + ((-a)^(1/3)*log((-
a)^(4/3) + a*b^(1/3)*x)*(4*A*b - 7*B*a))/(9*b^(10/3)) - ((-a)^(1/3)*log((-a)^(4/3) + 3^(1/2)*(-a)^(4/3)*1i - 2
*a*b^(1/3)*x)*((3^(1/2)*1i)/2 + 1/2)*(4*A*b - 7*B*a))/(9*b^(10/3)) + ((-a)^(1/3)*log(3^(1/2)*(-a)^(4/3)*1i - (
-a)^(4/3) + 2*a*b^(1/3)*x)*((3^(1/2)*1i)/2 - 1/2)*(4*A*b - 7*B*a))/(9*b^(10/3))

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sympy [A]  time = 2.54, size = 126, normalized size = 0.59 \begin {gather*} \frac {B x^{4}}{4 b^{2}} + x \left (\frac {A}{b^{2}} - \frac {2 B a}{b^{3}}\right ) + \frac {x \left (A a b - B a^{2}\right )}{3 a b^{3} + 3 b^{4} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} b^{10} + 64 A^{3} a b^{3} - 336 A^{2} B a^{2} b^{2} + 588 A B^{2} a^{3} b - 343 B^{3} a^{4}, \left (t \mapsto t \log {\left (\frac {9 t b^{3}}{- 4 A b + 7 B a} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6*(B*x**3+A)/(b*x**3+a)**2,x)

[Out]

B*x**4/(4*b**2) + x*(A/b**2 - 2*B*a/b**3) + x*(A*a*b - B*a**2)/(3*a*b**3 + 3*b**4*x**3) + RootSum(729*_t**3*b*
*10 + 64*A**3*a*b**3 - 336*A**2*B*a**2*b**2 + 588*A*B**2*a**3*b - 343*B**3*a**4, Lambda(_t, _t*log(9*_t*b**3/(
-4*A*b + 7*B*a) + x)))

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