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

Optimal. Leaf size=196 \[ -\frac{(4 A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{7/3} b^{2/3}}+\frac{(4 A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{7/3} b^{2/3}}+\frac{(4 A b-a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{7/3} b^{2/3}}-\frac{4 A b-a B}{3 a^2 b x}+\frac{A b-a B}{3 a b x \left (a+b x^3\right )} \]

[Out]

-(4*A*b - a*B)/(3*a^2*b*x) + (A*b - a*B)/(3*a*b*x*(a + b*x^3)) + ((4*A*b - a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)
/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(7/3)*b^(2/3)) + ((4*A*b - a*B)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(7/3)*b^(2/3)
) - ((4*A*b - a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(7/3)*b^(2/3))

________________________________________________________________________________________

Rubi [A]  time = 0.105552, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {457, 325, 292, 31, 634, 617, 204, 628} \[ -\frac{(4 A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{7/3} b^{2/3}}+\frac{(4 A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{7/3} b^{2/3}}+\frac{(4 A b-a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{7/3} b^{2/3}}-\frac{4 A b-a B}{3 a^2 b x}+\frac{A b-a B}{3 a b x \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(4*A*b - a*B)/(3*a^2*b*x) + (A*b - a*B)/(3*a*b*x*(a + b*x^3)) + ((4*A*b - a*B)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)
/(Sqrt[3]*a^(1/3))])/(3*Sqrt[3]*a^(7/3)*b^(2/3)) + ((4*A*b - a*B)*Log[a^(1/3) + b^(1/3)*x])/(9*a^(7/3)*b^(2/3)
) - ((4*A*b - a*B)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(18*a^(7/3)*b^(2/3))

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

Mathematica [A]  time = 0.123271, size = 164, normalized size = 0.84 \[ \frac{\frac{(a B-4 A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{b^{2/3}}+\frac{2 (4 A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{b^{2/3}}+\frac{2 \sqrt{3} (4 A b-a B) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{b^{2/3}}+\frac{6 \sqrt [3]{a} x^2 (a B-A b)}{a+b x^3}-\frac{18 \sqrt [3]{a} A}{x}}{18 a^{7/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-18*a^(1/3)*A)/x + (6*a^(1/3)*(-(A*b) + a*B)*x^2)/(a + b*x^3) + (2*Sqrt[3]*(4*A*b - a*B)*ArcTan[(1 - (2*b^(1
/3)*x)/a^(1/3))/Sqrt[3]])/b^(2/3) + (2*(4*A*b - a*B)*Log[a^(1/3) + b^(1/3)*x])/b^(2/3) + ((-4*A*b + a*B)*Log[a
^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/b^(2/3))/(18*a^(7/3))

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 241, normalized size = 1.2 \begin{align*} -{\frac{A{x}^{2}b}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}+{\frac{{x}^{2}B}{3\,a \left ( b{x}^{3}+a \right ) }}+{\frac{4\,A}{9\,{a}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{2\,A}{9\,{a}^{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{4\,A\sqrt{3}}{9\,{a}^{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}}}}}}-{\frac{B}{9\,ab}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}+{\frac{B}{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\sqrt{3}}{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{A}{{a}^{2}x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.84136, size = 1296, normalized size = 6.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 1.36558, size = 122, normalized size = 0.62 \begin{align*} \frac{- 3 A a + x^{3} \left (- 4 A b + B a\right )}{3 a^{3} x + 3 a^{2} b x^{4}} + \operatorname{RootSum}{\left (729 t^{3} a^{7} b^{2} - 64 A^{3} b^{3} + 48 A^{2} B a b^{2} - 12 A B^{2} a^{2} b + B^{3} a^{3}, \left ( t \mapsto t \log{\left (\frac{81 t^{2} a^{5} b}{16 A^{2} b^{2} - 8 A B a b + B^{2} a^{2}} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*A*a + x**3*(-4*A*b + B*a))/(3*a**3*x + 3*a**2*b*x**4) + RootSum(729*_t**3*a**7*b**2 - 64*A**3*b**3 + 48*A*
*2*B*a*b**2 - 12*A*B**2*a**2*b + B**3*a**3, Lambda(_t, _t*log(81*_t**2*a**5*b/(16*A**2*b**2 - 8*A*B*a*b + B**2
*a**2) + x)))

________________________________________________________________________________________

Giac [A]  time = 1.14221, size = 273, normalized size = 1.39 \begin{align*} -\frac{{\left (B a \left (-\frac{a}{b}\right )^{\frac{1}{3}} - 4 \, 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^{3}} + \frac{B a x^{3} - 4 \, A b x^{3} - 3 \, A a}{3 \,{\left (b x^{4} + a x\right )} a^{2}} - \frac{\sqrt{3}{\left (\left (-a b^{2}\right )^{\frac{2}{3}} B a - 4 \, \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^{3} b^{2}} + \frac{{\left (\left (-a b^{2}\right )^{\frac{2}{3}} B a - 4 \, \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^{3} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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