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

Optimal. Leaf size=169 \[ -\frac{(a B+2 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{5/3} b^{4/3}}+\frac{(a B+2 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{5/3} b^{4/3}}-\frac{(a B+2 A b) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{5/3} b^{4/3}}+\frac{x (A b-a B)}{3 a b \left (a+b x^3\right )} \]

[Out]

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

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Rubi [A]  time = 0.0862685, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {385, 200, 31, 634, 617, 204, 628} \[ -\frac{(a B+2 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{5/3} b^{4/3}}+\frac{(a B+2 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{5/3} b^{4/3}}-\frac{(a B+2 A b) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{5/3} b^{4/3}}+\frac{x (A b-a B)}{3 a b \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(a + b*x^3)^2,x]

[Out]

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

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d)*x*(a + b*x^n)^(p +
1))/(a*b*n*(p + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/n + p, 0])

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

Mathematica [A]  time = 0.0895817, size = 145, normalized size = 0.86 \[ \frac{-(a B+2 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-\frac{6 a^{2/3} \sqrt [3]{b} x (a B-A b)}{a+b x^3}+2 (a B+2 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-2 \sqrt{3} (a B+2 A b) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{18 a^{5/3} b^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 221, normalized size = 1.3 \begin{align*}{\frac{ \left ( Ab-Ba \right ) x}{3\,ab \left ( b{x}^{3}+a \right ) }}+{\frac{2\,A}{9\,ab}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{B}{9\,{b}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{A}{9\,ab}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{B}{18\,{b}^{2}}\ln \left ({x}^{2}-\sqrt [3]{{\frac{a}{b}}}x+ \left ({\frac{a}{b}} \right ) ^{{\frac{2}{3}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{2\,\sqrt{3}A}{9\,ab}\arctan \left ({\frac{\sqrt{3}}{3} \left ( 2\,{x{\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-1 \right ) } \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{\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 ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(A*b-B*a)*x/a/b/(b*x^3+a)+2/9/a/b/(a/b)^(2/3)*ln(x+(a/b)^(1/3))*A+1/9/b^2/(a/b)^(2/3)*ln(x+(a/b)^(1/3))*B-
1/9/a/b/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*A-1/18/b^2/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))
*B+2/9/a/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*A+1/9/b^2/(a/b)^(2/3)*3^(1/2)*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((B*x^3+A)/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.89835, size = 1233, normalized size = 7.3 \begin{align*} \left [\frac{3 \, \sqrt{\frac{1}{3}}{\left (B a^{3} b + 2 \, A a^{2} b^{2} +{\left (B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3}\right )} \sqrt{-\frac{\left (a^{2} b\right )^{\frac{1}{3}}}{b}} \log \left (\frac{2 \, a b x^{3} - 3 \, \left (a^{2} b\right )^{\frac{1}{3}} a x - a^{2} + 3 \, \sqrt{\frac{1}{3}}{\left (2 \, a b x^{2} + \left (a^{2} b\right )^{\frac{2}{3}} x - \left (a^{2} b\right )^{\frac{1}{3}} a\right )} \sqrt{-\frac{\left (a^{2} b\right )^{\frac{1}{3}}}{b}}}{b x^{3} + a}\right ) -{\left ({\left (B a b + 2 \, A b^{2}\right )} x^{3} + B a^{2} + 2 \, A a b\right )} \left (a^{2} b\right )^{\frac{2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac{2}{3}} x + \left (a^{2} b\right )^{\frac{1}{3}} a\right ) + 2 \,{\left ({\left (B a b + 2 \, A b^{2}\right )} x^{3} + B a^{2} + 2 \, A a b\right )} \left (a^{2} b\right )^{\frac{2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac{2}{3}}\right ) - 6 \,{\left (B a^{3} b - A a^{2} b^{2}\right )} x}{18 \,{\left (a^{3} b^{3} x^{3} + a^{4} b^{2}\right )}}, \frac{6 \, \sqrt{\frac{1}{3}}{\left (B a^{3} b + 2 \, A a^{2} b^{2} +{\left (B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3}\right )} \sqrt{\frac{\left (a^{2} b\right )^{\frac{1}{3}}}{b}} \arctan \left (\frac{\sqrt{\frac{1}{3}}{\left (2 \, \left (a^{2} b\right )^{\frac{2}{3}} x - \left (a^{2} b\right )^{\frac{1}{3}} a\right )} \sqrt{\frac{\left (a^{2} b\right )^{\frac{1}{3}}}{b}}}{a^{2}}\right ) -{\left ({\left (B a b + 2 \, A b^{2}\right )} x^{3} + B a^{2} + 2 \, A a b\right )} \left (a^{2} b\right )^{\frac{2}{3}} \log \left (a b x^{2} - \left (a^{2} b\right )^{\frac{2}{3}} x + \left (a^{2} b\right )^{\frac{1}{3}} a\right ) + 2 \,{\left ({\left (B a b + 2 \, A b^{2}\right )} x^{3} + B a^{2} + 2 \, A a b\right )} \left (a^{2} b\right )^{\frac{2}{3}} \log \left (a b x + \left (a^{2} b\right )^{\frac{2}{3}}\right ) - 6 \,{\left (B a^{3} b - A a^{2} b^{2}\right )} x}{18 \,{\left (a^{3} b^{3} x^{3} + a^{4} b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.25105, size = 97, normalized size = 0.57 \begin{align*} - \frac{x \left (- A b + B a\right )}{3 a^{2} b + 3 a b^{2} x^{3}} + \operatorname{RootSum}{\left (729 t^{3} a^{5} b^{4} - 8 A^{3} b^{3} - 12 A^{2} B a b^{2} - 6 A B^{2} a^{2} b - B^{3} a^{3}, \left ( t \mapsto t \log{\left (\frac{9 t a^{2} b}{2 A b + B a} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-x*(-A*b + B*a)/(3*a**2*b + 3*a*b**2*x**3) + RootSum(729*_t**3*a**5*b**4 - 8*A**3*b**3 - 12*A**2*B*a*b**2 - 6*
A*B**2*a**2*b - B**3*a**3, Lambda(_t, _t*log(9*_t*a**2*b/(2*A*b + B*a) + x)))

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Giac [A]  time = 1.14177, size = 246, normalized size = 1.46 \begin{align*} -\frac{{\left (B a + 2 \, 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^{2} b} + \frac{\sqrt{3}{\left (\left (-a b^{2}\right )^{\frac{1}{3}} B a + 2 \, \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 \, a^{2} b^{2}} - \frac{B a x - A b x}{3 \,{\left (b x^{3} + a\right )} a b} + \frac{{\left (\left (-a b^{2}\right )^{\frac{1}{3}} B a + 2 \, \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 \, a^{2} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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