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

Optimal. Leaf size=147 \[ -\frac{(A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{4/3} b^{2/3}}+\frac{(A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} b^{2/3}}+\frac{(A b-a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} b^{2/3}}-\frac{A}{a x} \]

[Out]

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

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Rubi [A]  time = 0.0856454, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {453, 292, 31, 634, 617, 204, 628} \[ -\frac{(A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{4/3} b^{2/3}}+\frac{(A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} b^{2/3}}+\frac{(A b-a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} b^{2/3}}-\frac{A}{a x} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^2*(a + b*x^3)),x]

[Out]

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

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[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{A+B x^3}{x^2 \left (a+b x^3\right )} \, dx &=-\frac{A}{a x}-\frac{(A b-a B) \int \frac{x}{a+b x^3} \, dx}{a}\\ &=-\frac{A}{a x}+\frac{(A b-a B) \int \frac{1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a^{4/3} \sqrt [3]{b}}-\frac{(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}{3 a^{4/3} \sqrt [3]{b}}\\ &=-\frac{A}{a x}+\frac{(A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} b^{2/3}}-\frac{(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}{6 a^{4/3} b^{2/3}}-\frac{(A b-a B) \int \frac{1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 a \sqrt [3]{b}}\\ &=-\frac{A}{a x}+\frac{(A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} b^{2/3}}-\frac{(A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{4/3} b^{2/3}}-\frac{(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 )}{a^{4/3} b^{2/3}}\\ &=-\frac{A}{a x}+\frac{(A b-a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{\sqrt{3} a^{4/3} b^{2/3}}+\frac{(A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{4/3} b^{2/3}}-\frac{(A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{4/3} b^{2/3}}\\ \end{align*}

Mathematica [A]  time = 0.0807601, size = 134, normalized size = 0.91 \[ \frac{-x (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-6 \sqrt [3]{a} A b^{2/3}+2 x (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 \sqrt{3} x (A b-a B) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{6 a^{4/3} b^{2/3} x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.004, size = 195, normalized size = 1.3 \begin{align*} -{\frac{A}{ax}}+{\frac{A}{3\,a}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{B}{3\,b}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ){\frac{1}{\sqrt [3]{{\frac{a}{b}}}}}}-{\frac{A}{6\,a}\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}{6\,b}\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}{3\,a}\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{\sqrt{3}B}{3\,b}\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((B*x^3+A)/x^2/(b*x^3+a),x)

[Out]

-A/a/x+1/3/a/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*A-1/3/b/(a/b)^(1/3)*ln(x+(a/b)^(1/3))*B-1/6/a/(a/b)^(1/3)*ln(x^2-(a
/b)^(1/3)*x+(a/b)^(2/3))*A+1/6/b/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*B-1/3/a*3^(1/2)/(a/b)^(1/3)*arc
tan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*A+1/3*3^(1/2)/b/(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((B*x^3+A)/x^2/(b*x^3+a),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.50286, size = 891, normalized size = 6.06 \begin{align*} \left [-\frac{6 \, A a b^{2} + 3 \, \sqrt{\frac{1}{3}}{\left (B a^{2} b - A a b^{2}\right )} x \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 (a b^{2}\right )^{\frac{2}{3}}{\left (B a - A b\right )} x \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 (a b^{2}\right )^{\frac{2}{3}}{\left (B a - A b\right )} x \log \left (b x + \left (a b^{2}\right )^{\frac{1}{3}}\right )}{6 \, a^{2} b^{2} x}, -\frac{6 \, A a b^{2} + 6 \, \sqrt{\frac{1}{3}}{\left (B a^{2} b - A a b^{2}\right )} x \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 (a b^{2}\right )^{\frac{2}{3}}{\left (B a - A b\right )} x \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 (a b^{2}\right )^{\frac{2}{3}}{\left (B a - A b\right )} x \log \left (b x + \left (a b^{2}\right )^{\frac{1}{3}}\right )}{6 \, a^{2} b^{2} x}\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),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.742758, size = 90, normalized size = 0.61 \begin{align*} - \frac{A}{a x} + \operatorname{RootSum}{\left (27 t^{3} a^{4} b^{2} - A^{3} b^{3} + 3 A^{2} B a b^{2} - 3 A B^{2} a^{2} b + B^{3} a^{3}, \left ( t \mapsto t \log{\left (\frac{9 t^{2} a^{3} b}{A^{2} b^{2} - 2 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),x)

[Out]

-A/(a*x) + RootSum(27*_t**3*a**4*b**2 - A**3*b**3 + 3*A**2*B*a*b**2 - 3*A*B**2*a**2*b + B**3*a**3, Lambda(_t,
_t*log(9*_t**2*a**3*b/(A**2*b**2 - 2*A*B*a*b + B**2*a**2) + x)))

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Giac [A]  time = 1.18995, size = 239, normalized size = 1.63 \begin{align*} -\frac{{\left (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 )}{3 \, a^{2}} - \frac{A}{a x} - \frac{\sqrt{3}{\left (\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 )}{3 \, a^{2} b^{2}} + \frac{{\left (\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 )}{6 \, a^{2} 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),x, algorithm="giac")

[Out]

-1/3*(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 - A/(a*x) - 1/3*sqrt(3)
*((-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^2) + 1
/6*((-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^2)