3.73 \(\int \frac {x^7 (A+B x^3)}{(a+b x^3)^2} \, dx\)

Optimal. Leaf size=215 \[ -\frac {a^{2/3} (5 A b-8 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 b^{11/3}}+\frac {a^{2/3} (5 A b-8 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{11/3}}+\frac {a^{2/3} (5 A b-8 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} b^{11/3}}+\frac {x^2 (5 A b-8 a B)}{6 b^3}-\frac {x^5 (5 A b-8 a B)}{15 a b^2}+\frac {x^8 (A b-a B)}{3 a b \left (a+b x^3\right )} \]

[Out]

1/6*(5*A*b-8*B*a)*x^2/b^3-1/15*(5*A*b-8*B*a)*x^5/a/b^2+1/3*(A*b-B*a)*x^8/a/b/(b*x^3+a)+1/9*a^(2/3)*(5*A*b-8*B*
a)*ln(a^(1/3)+b^(1/3)*x)/b^(11/3)-1/18*a^(2/3)*(5*A*b-8*B*a)*ln(a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/b^(11/3
)+1/9*a^(2/3)*(5*A*b-8*B*a)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1/2))/b^(11/3)*3^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 215, 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, 292, 31, 634, 617, 204, 628} \[ -\frac {a^{2/3} (5 A b-8 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 b^{11/3}}+\frac {a^{2/3} (5 A b-8 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{11/3}}+\frac {a^{2/3} (5 A b-8 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} b^{11/3}}-\frac {x^5 (5 A b-8 a B)}{15 a b^2}+\frac {x^2 (5 A b-8 a B)}{6 b^3}+\frac {x^8 (A b-a B)}{3 a b \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x^7*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

((5*A*b - 8*a*B)*x^2)/(6*b^3) - ((5*A*b - 8*a*B)*x^5)/(15*a*b^2) + ((A*b - a*B)*x^8)/(3*a*b*(a + b*x^3)) + (a^
(2/3)*(5*A*b - 8*a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*b^(11/3)) + (a^(2/3)*(5*A*
b - 8*a*B)*Log[a^(1/3) + b^(1/3)*x])/(9*b^(11/3)) - (a^(2/3)*(5*A*b - 8*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x +
b^(2/3)*x^2])/(18*b^(11/3))

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 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^7 \left (A+B x^3\right )}{\left (a+b x^3\right )^2} \, dx &=\frac {(A b-a B) x^8}{3 a b \left (a+b x^3\right )}+\frac {(-5 A b+8 a B) \int \frac {x^7}{a+b x^3} \, dx}{3 a b}\\ &=\frac {(A b-a B) x^8}{3 a b \left (a+b x^3\right )}+\frac {(-5 A b+8 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}{3 a b}\\ &=\frac {(5 A b-8 a B) x^2}{6 b^3}-\frac {(5 A b-8 a B) x^5}{15 a b^2}+\frac {(A b-a B) x^8}{3 a b \left (a+b x^3\right )}-\frac {(a (5 A b-8 a B)) \int \frac {x}{a+b x^3} \, dx}{3 b^3}\\ &=\frac {(5 A b-8 a B) x^2}{6 b^3}-\frac {(5 A b-8 a B) x^5}{15 a b^2}+\frac {(A b-a B) x^8}{3 a b \left (a+b x^3\right )}+\frac {\left (a^{2/3} (5 A b-8 a B)\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{9 b^{10/3}}-\frac {\left (a^{2/3} (5 A b-8 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}{9 b^{10/3}}\\ &=\frac {(5 A b-8 a B) x^2}{6 b^3}-\frac {(5 A b-8 a B) x^5}{15 a b^2}+\frac {(A b-a B) x^8}{3 a b \left (a+b x^3\right )}+\frac {a^{2/3} (5 A b-8 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{11/3}}-\frac {\left (a^{2/3} (5 A b-8 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^{11/3}}-\frac {(a (5 A b-8 a B)) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{6 b^{10/3}}\\ &=\frac {(5 A b-8 a B) x^2}{6 b^3}-\frac {(5 A b-8 a B) x^5}{15 a b^2}+\frac {(A b-a B) x^8}{3 a b \left (a+b x^3\right )}+\frac {a^{2/3} (5 A b-8 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{11/3}}-\frac {a^{2/3} (5 A b-8 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 b^{11/3}}-\frac {\left (a^{2/3} (5 A b-8 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^{11/3}}\\ &=\frac {(5 A b-8 a B) x^2}{6 b^3}-\frac {(5 A b-8 a B) x^5}{15 a b^2}+\frac {(A b-a B) x^8}{3 a b \left (a+b x^3\right )}+\frac {a^{2/3} (5 A b-8 a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{3 \sqrt {3} b^{11/3}}+\frac {a^{2/3} (5 A b-8 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 b^{11/3}}-\frac {a^{2/3} (5 A b-8 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 b^{11/3}}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 185, normalized size = 0.86 \[ \frac {5 a^{2/3} (8 a B-5 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-10 a^{2/3} (8 a B-5 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-10 \sqrt {3} a^{2/3} (8 a B-5 A b) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+45 b^{2/3} x^2 (A b-2 a B)+\frac {30 a b^{2/3} x^2 (A b-a B)}{a+b x^3}+18 b^{5/3} B x^5}{90 b^{11/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

(45*b^(2/3)*(A*b - 2*a*B)*x^2 + 18*b^(5/3)*B*x^5 + (30*a*b^(2/3)*(A*b - a*B)*x^2)/(a + b*x^3) - 10*Sqrt[3]*a^(
2/3)*(-5*A*b + 8*a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(1/3))/Sqrt[3]] - 10*a^(2/3)*(-5*A*b + 8*a*B)*Log[a^(1/3) +
b^(1/3)*x] + 5*a^(2/3)*(-5*A*b + 8*a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(90*b^(11/3))

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fricas [A]  time = 0.88, size = 257, normalized size = 1.20 \[ \frac {18 \, B b^{2} x^{8} - 9 \, {\left (8 \, B a b - 5 \, A b^{2}\right )} x^{5} - 15 \, {\left (8 \, B a^{2} - 5 \, A a b\right )} x^{2} + 10 \, \sqrt {3} {\left ({\left (8 \, B a b - 5 \, A b^{2}\right )} x^{3} + 8 \, B a^{2} - 5 \, A a b\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) + 5 \, {\left ({\left (8 \, B a b - 5 \, A b^{2}\right )} x^{3} + 8 \, B a^{2} - 5 \, A a b\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) - 10 \, {\left ({\left (8 \, B a b - 5 \, A b^{2}\right )} x^{3} + 8 \, B a^{2} - 5 \, A a b\right )} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{90 \, {\left (b^{4} x^{3} + a b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

1/90*(18*B*b^2*x^8 - 9*(8*B*a*b - 5*A*b^2)*x^5 - 15*(8*B*a^2 - 5*A*a*b)*x^2 + 10*sqrt(3)*((8*B*a*b - 5*A*b^2)*
x^3 + 8*B*a^2 - 5*A*a*b)*(a^2/b^2)^(1/3)*arctan(1/3*(2*sqrt(3)*b*x*(a^2/b^2)^(1/3) - sqrt(3)*a)/a) + 5*((8*B*a
*b - 5*A*b^2)*x^3 + 8*B*a^2 - 5*A*a*b)*(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)) -
10*((8*B*a*b - 5*A*b^2)*x^3 + 8*B*a^2 - 5*A*a*b)*(a^2/b^2)^(1/3)*log(a*x + b*(a^2/b^2)^(2/3)))/(b^4*x^3 + a*b^
3)

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giac [A]  time = 0.18, size = 236, normalized size = 1.10 \[ -\frac {{\left (8 \, B a^{2} \left (-\frac {a}{b}\right )^{\frac {1}{3}} - 5 \, A a b \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )} \left (-\frac {a}{b}\right )^{\frac {1}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right )}{9 \, a b^{3}} - \frac {\sqrt {3} {\left (8 \, \left (-a b^{2}\right )^{\frac {2}{3}} B a - 5 \, \left (-a b^{2}\right )^{\frac {2}{3}} A b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x + \left (-\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (-\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, b^{5}} - \frac {B a^{2} x^{2} - A a b x^{2}}{3 \, {\left (b x^{3} + a\right )} b^{3}} + \frac {{\left (8 \, \left (-a b^{2}\right )^{\frac {2}{3}} B a - 5 \, \left (-a b^{2}\right )^{\frac {2}{3}} A b\right )} \log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, b^{5}} + \frac {2 \, B b^{8} x^{5} - 10 \, B a b^{7} x^{2} + 5 \, A b^{8} x^{2}}{10 \, b^{10}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="giac")

[Out]

-1/9*(8*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^3) - 1/9*sqrt(
3)*(8*(-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^5 - 1
/3*(B*a^2*x^2 - A*a*b*x^2)/((b*x^3 + a)*b^3) + 1/18*(8*(-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^5 + 1/10*(2*B*b^8*x^5 - 10*B*a*b^7*x^2 + 5*A*b^8*x^2)/b^10

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maple [A]  time = 0.05, size = 266, normalized size = 1.24 \[ \frac {B \,x^{5}}{5 b^{2}}+\frac {A a \,x^{2}}{3 \left (b \,x^{3}+a \right ) b^{2}}-\frac {B \,a^{2} x^{2}}{3 \left (b \,x^{3}+a \right ) b^{3}}+\frac {A \,x^{2}}{2 b^{2}}-\frac {B a \,x^{2}}{b^{3}}-\frac {5 \sqrt {3}\, A a \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{3}}+\frac {5 A a \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{3}}-\frac {5 A a \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{3}}+\frac {8 \sqrt {3}\, B \,a^{2} \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{9 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{4}}-\frac {8 B \,a^{2} \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{4}}+\frac {4 B \,a^{2} \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{9 \left (\frac {a}{b}\right )^{\frac {1}{3}} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(B*x^3+A)/(b*x^3+a)^2,x)

[Out]

1/5/b^2*B*x^5+1/2/b^2*A*x^2-1/b^3*B*x^2*a+1/3*a/b^2*x^2/(b*x^3+a)*A-1/3*a^2/b^3*x^2/(b*x^3+a)*B+5/9*a/b^3*A/(a
/b)^(1/3)*ln(x+(a/b)^(1/3))-5/18*a/b^3*A/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))-5/9*a/b^3*A*3^(1/2)/(a/
b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))-8/9*a^2/b^4*B/(a/b)^(1/3)*ln(x+(a/b)^(1/3))+4/9*a^2/b^4*B/(a/
b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+8/9*a^2/b^4*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 = 1.10, size = 192, normalized size = 0.89 \[ -\frac {{\left (B a^{2} - A a b\right )} x^{2}}{3 \, {\left (b^{4} x^{3} + a b^{3}\right )}} + \frac {\sqrt {3} {\left (8 \, B a^{2} - 5 \, A a b\right )} \arctan \left (\frac {\sqrt {3} {\left (2 \, x - \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}}{3 \, \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right )}{9 \, b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {2 \, B b x^{5} - 5 \, {\left (2 \, B a - A b\right )} x^{2}}{10 \, b^{3}} + \frac {{\left (8 \, B a^{2} - 5 \, A a b\right )} \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{18 \, b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {{\left (8 \, B a^{2} - 5 \, A a b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{9 \, b^{4} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(B*x^3+A)/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

-1/3*(B*a^2 - A*a*b)*x^2/(b^4*x^3 + a*b^3) + 1/9*sqrt(3)*(8*B*a^2 - 5*A*a*b)*arctan(1/3*sqrt(3)*(2*x - (a/b)^(
1/3))/(a/b)^(1/3))/(b^4*(a/b)^(1/3)) + 1/10*(2*B*b*x^5 - 5*(2*B*a - A*b)*x^2)/b^3 + 1/18*(8*B*a^2 - 5*A*a*b)*l
og(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(b^4*(a/b)^(1/3)) - 1/9*(8*B*a^2 - 5*A*a*b)*log(x + (a/b)^(1/3))/(b^4*(a
/b)^(1/3))

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mupad [B]  time = 0.27, size = 179, normalized size = 0.83 \[ x^2\,\left (\frac {A}{2\,b^2}-\frac {B\,a}{b^3}\right )+\frac {B\,x^5}{5\,b^2}-\frac {x^2\,\left (\frac {B\,a^2}{3}-\frac {A\,a\,b}{3}\right )}{b^4\,x^3+a\,b^3}+\frac {a^{2/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (5\,A\,b-8\,B\,a\right )}{9\,b^{11/3}}+\frac {a^{2/3}\,\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (5\,A\,b-8\,B\,a\right )}{9\,b^{11/3}}-\frac {a^{2/3}\,\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (5\,A\,b-8\,B\,a\right )}{9\,b^{11/3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^7*(A + B*x^3))/(a + b*x^3)^2,x)

[Out]

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

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sympy [A]  time = 2.12, size = 151, normalized size = 0.70 \[ \frac {B x^{5}}{5 b^{2}} + x^{2} \left (\frac {A}{2 b^{2}} - \frac {B a}{b^{3}}\right ) + \frac {x^{2} \left (A a b - B a^{2}\right )}{3 a b^{3} + 3 b^{4} x^{3}} + \operatorname {RootSum} {\left (729 t^{3} b^{11} - 125 A^{3} a^{2} b^{3} + 600 A^{2} B a^{3} b^{2} - 960 A B^{2} a^{4} b + 512 B^{3} a^{5}, \left (t \mapsto t \log {\left (\frac {81 t^{2} b^{7}}{25 A^{2} a b^{2} - 80 A B a^{2} b + 64 B^{2} a^{3}} + x \right )} \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7*(B*x**3+A)/(b*x**3+a)**2,x)

[Out]

B*x**5/(5*b**2) + x**2*(A/(2*b**2) - B*a/b**3) + x**2*(A*a*b - B*a**2)/(3*a*b**3 + 3*b**4*x**3) + RootSum(729*
_t**3*b**11 - 125*A**3*a**2*b**3 + 600*A**2*B*a**3*b**2 - 960*A*B**2*a**4*b + 512*B**3*a**5, Lambda(_t, _t*log
(81*_t**2*b**7/(25*A**2*a*b**2 - 80*A*B*a**2*b + 64*B**2*a**3) + x)))

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