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

Optimal. Leaf size=171 \[ \frac{(2 a B+A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{4/3} b^{5/3}}-\frac{(2 a B+A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{4/3} b^{5/3}}-\frac{(2 a B+A b) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{4/3} b^{5/3}}+\frac{x^2 (A b-a B)}{3 a b \left (a+b x^3\right )} \]

[Out]

((A*b - a*B)*x^2)/(3*a*b*(a + b*x^3)) - ((A*b + 2*a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*S
qrt[3]*a^(4/3)*b^(5/3)) - ((A*b + 2*a*B)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(4/3)*b^(5/3)) + ((A*b + 2*a*B)*Log[a^
(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(4/3)*b^(5/3))

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Rubi [A]  time = 0.0879305, antiderivative size = 171, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.389, Rules used = {457, 292, 31, 634, 617, 204, 628} \[ \frac{(2 a B+A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{4/3} b^{5/3}}-\frac{(2 a B+A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{4/3} b^{5/3}}-\frac{(2 a B+A b) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{4/3} b^{5/3}}+\frac{x^2 (A b-a B)}{3 a b \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x^3))/(a + b*x^3)^2,x]

[Out]

((A*b - a*B)*x^2)/(3*a*b*(a + b*x^3)) - ((A*b + 2*a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/(Sqrt[3]*a^(1/3))])/(3*S
qrt[3]*a^(4/3)*b^(5/3)) - ((A*b + 2*a*B)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(4/3)*b^(5/3)) + ((A*b + 2*a*B)*Log[a^
(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(4/3)*b^(5/3))

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 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 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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]

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.0903798, size = 146, normalized size = 0.85 \[ \frac{(2 a B+A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-\frac{6 \sqrt [3]{a} b^{2/3} x^2 (a B-A b)}{a+b x^3}-2 (2 a B+A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-2 \sqrt{3} (2 a B+A b) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{18 a^{4/3} b^{5/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 223, normalized size = 1.3 \begin{align*}{\frac{ \left ( Ab-Ba \right ){x}^{2}}{3\,ab \left ( b{x}^{3}+a \right ) }}-{\frac{A}{9\,ab}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{2\,B}{9\,{b}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{A}{18\,ab}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{B}{9\,{b}^{2}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{\sqrt{3}A}{9\,ab}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{2\,\sqrt{3}B}{9\,{b}^{2}}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(A*b-B*a)*x^2/a/b/(b*x^3+a)-1/9/a/b/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*A-2/9/b^2/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*
B+1/18/a/b/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*A+1/9/b^2/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3
))*B+1/9/a/b*3^(1/2)/(a/b)^(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*A+2/9/b^2*3^(1/2)/(a/b)^(1/3)*arctan(
1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*B

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.79243, size = 1235, normalized size = 7.22 \begin{align*} \left [-\frac{6 \,{\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 3 \, \sqrt{\frac{1}{3}}{\left (2 \, B a^{3} b + A a^{2} b^{2} +{\left (2 \, B a^{2} b^{2} + A a b^{3}\right )} x^{3}\right )} \sqrt{\frac{\left (-a b^{2}\right )^{\frac{1}{3}}}{a}} \log \left (\frac{2 \, b^{2} x^{3} - a b + 3 \, \sqrt{\frac{1}{3}}{\left (a b x + 2 \, \left (-a b^{2}\right )^{\frac{2}{3}} x^{2} + \left (-a b^{2}\right )^{\frac{1}{3}} a\right )} \sqrt{\frac{\left (-a b^{2}\right )^{\frac{1}{3}}}{a}} - 3 \, \left (-a b^{2}\right )^{\frac{2}{3}} x}{b x^{3} + a}\right ) -{\left ({\left (2 \, B a b + A b^{2}\right )} x^{3} + 2 \, B a^{2} + A a b\right )} \left (-a b^{2}\right )^{\frac{2}{3}} \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac{1}{3}} b x + \left (-a b^{2}\right )^{\frac{2}{3}}\right ) + 2 \,{\left ({\left (2 \, B a b + A b^{2}\right )} x^{3} + 2 \, B a^{2} + A a b\right )} \left (-a b^{2}\right )^{\frac{2}{3}} \log \left (b x - \left (-a b^{2}\right )^{\frac{1}{3}}\right )}{18 \,{\left (a^{2} b^{4} x^{3} + a^{3} b^{3}\right )}}, -\frac{6 \,{\left (B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 6 \, \sqrt{\frac{1}{3}}{\left (2 \, B a^{3} b + A a^{2} b^{2} +{\left (2 \, B a^{2} b^{2} + A a b^{3}\right )} x^{3}\right )} \sqrt{-\frac{\left (-a b^{2}\right )^{\frac{1}{3}}}{a}} \arctan \left (\frac{\sqrt{\frac{1}{3}}{\left (2 \, b x + \left (-a b^{2}\right )^{\frac{1}{3}}\right )} \sqrt{-\frac{\left (-a b^{2}\right )^{\frac{1}{3}}}{a}}}{b}\right ) -{\left ({\left (2 \, B a b + A b^{2}\right )} x^{3} + 2 \, B a^{2} + A a b\right )} \left (-a b^{2}\right )^{\frac{2}{3}} \log \left (b^{2} x^{2} + \left (-a b^{2}\right )^{\frac{1}{3}} b x + \left (-a b^{2}\right )^{\frac{2}{3}}\right ) + 2 \,{\left ({\left (2 \, B a b + A b^{2}\right )} x^{3} + 2 \, B a^{2} + A a b\right )} \left (-a b^{2}\right )^{\frac{2}{3}} \log \left (b x - \left (-a b^{2}\right )^{\frac{1}{3}}\right )}{18 \,{\left (a^{2} b^{4} x^{3} + a^{3} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/18*(6*(B*a^2*b^2 - A*a*b^3)*x^2 - 3*sqrt(1/3)*(2*B*a^3*b + A*a^2*b^2 + (2*B*a^2*b^2 + A*a*b^3)*x^3)*sqrt((
-a*b^2)^(1/3)/a)*log((2*b^2*x^3 - a*b + 3*sqrt(1/3)*(a*b*x + 2*(-a*b^2)^(2/3)*x^2 + (-a*b^2)^(1/3)*a)*sqrt((-a
*b^2)^(1/3)/a) - 3*(-a*b^2)^(2/3)*x)/(b*x^3 + a)) - ((2*B*a*b + A*b^2)*x^3 + 2*B*a^2 + A*a*b)*(-a*b^2)^(2/3)*l
og(b^2*x^2 + (-a*b^2)^(1/3)*b*x + (-a*b^2)^(2/3)) + 2*((2*B*a*b + A*b^2)*x^3 + 2*B*a^2 + A*a*b)*(-a*b^2)^(2/3)
*log(b*x - (-a*b^2)^(1/3)))/(a^2*b^4*x^3 + a^3*b^3), -1/18*(6*(B*a^2*b^2 - A*a*b^3)*x^2 - 6*sqrt(1/3)*(2*B*a^3
*b + A*a^2*b^2 + (2*B*a^2*b^2 + A*a*b^3)*x^3)*sqrt(-(-a*b^2)^(1/3)/a)*arctan(sqrt(1/3)*(2*b*x + (-a*b^2)^(1/3)
)*sqrt(-(-a*b^2)^(1/3)/a)/b) - ((2*B*a*b + A*b^2)*x^3 + 2*B*a^2 + A*a*b)*(-a*b^2)^(2/3)*log(b^2*x^2 + (-a*b^2)
^(1/3)*b*x + (-a*b^2)^(2/3)) + 2*((2*B*a*b + A*b^2)*x^3 + 2*B*a^2 + A*a*b)*(-a*b^2)^(2/3)*log(b*x - (-a*b^2)^(
1/3)))/(a^2*b^4*x^3 + a^3*b^3)]

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Sympy [A]  time = 1.04782, size = 117, normalized size = 0.68 \begin{align*} - \frac{x^{2} \left (- A b + B a\right )}{3 a^{2} b + 3 a b^{2} x^{3}} + \operatorname{RootSum}{\left (729 t^{3} a^{4} b^{5} + A^{3} b^{3} + 6 A^{2} B a b^{2} + 12 A B^{2} a^{2} b + 8 B^{3} a^{3}, \left ( t \mapsto t \log{\left (\frac{81 t^{2} a^{3} b^{3}}{A^{2} b^{2} + 4 A B a b + 4 B^{2} a^{2}} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x**2*(-A*b + B*a)/(3*a**2*b + 3*a*b**2*x**3) + RootSum(729*_t**3*a**4*b**5 + A**3*b**3 + 6*A**2*B*a*b**2 + 12
*A*B**2*a**2*b + 8*B**3*a**3, Lambda(_t, _t*log(81*_t**2*a**3*b**3/(A**2*b**2 + 4*A*B*a*b + 4*B**2*a**2) + x))
)

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Giac [A]  time = 1.12135, size = 273, normalized size = 1.6 \begin{align*} -\frac{{\left (2 \, B a \left (-\frac{a}{b}\right )^{\frac{1}{3}} + 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^{2} b} - \frac{B a x^{2} - A b x^{2}}{3 \,{\left (b x^{3} + a\right )} a b} - \frac{\sqrt{3}{\left (2 \, \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 )}{9 \, a^{2} b^{3}} + \frac{{\left (2 \, \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 )}{18 \, a^{2} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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