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

Optimal. Leaf size=196 \[ \frac{(5 A b-2 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{8/3} \sqrt [3]{b}}-\frac{5 A b-2 a B}{6 a^2 b x^2}-\frac{(5 A b-2 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{8/3} \sqrt [3]{b}}+\frac{(5 A b-2 a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{8/3} \sqrt [3]{b}}+\frac{A b-a B}{3 a b x^2 \left (a+b x^3\right )} \]

[Out]

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

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Rubi [A]  time = 0.102756, 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, 200, 31, 634, 617, 204, 628} \[ \frac{(5 A b-2 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{18 a^{8/3} \sqrt [3]{b}}-\frac{5 A b-2 a B}{6 a^2 b x^2}-\frac{(5 A b-2 a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{9 a^{8/3} \sqrt [3]{b}}+\frac{(5 A b-2 a B) \tan ^{-1}\left (\frac{\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt{3} \sqrt [3]{a}}\right )}{3 \sqrt{3} a^{8/3} \sqrt [3]{b}}+\frac{A b-a B}{3 a b x^2 \left (a+b x^3\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.12531, size = 163, normalized size = 0.83 \[ \frac{\frac{(5 A b-2 a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{b}}+\frac{6 a^{2/3} x (a B-A b)}{a+b x^3}-\frac{9 a^{2/3} A}{x^2}+\frac{2 (2 a B-5 A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}+\frac{2 \sqrt{3} (5 A b-2 a B) \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt{3}}\right )}{\sqrt [3]{b}}}{18 a^{8/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.012, size = 237, normalized size = 1.2 \begin{align*} -{\frac{Abx}{3\,{a}^{2} \left ( b{x}^{3}+a \right ) }}+{\frac{xB}{3\,a \left ( b{x}^{3}+a \right ) }}-{\frac{5\,A}{9\,{a}^{2}}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{5\,A}{18\,{a}^{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{5\,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 ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}+{\frac{2\,B}{9\,ab}\ln \left ( x+\sqrt [3]{{\frac{a}{b}}} \right ) \left ({\frac{a}{b}} \right ) ^{-{\frac{2}{3}}}}-{\frac{B}{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{2\,B\sqrt{3}}{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{A}{2\,{a}^{2}{x}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^3/(b*x^3+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.77428, size = 1396, normalized size = 7.12 \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^3/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

[-1/18*(9*A*a^3*b - 3*(2*B*a^3*b - 5*A*a^2*b^2)*x^3 + 3*sqrt(1/3)*((2*B*a^2*b^2 - 5*A*a*b^3)*x^5 + (2*B*a^3*b
- 5*A*a^2*b^2)*x^2)*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)) + ((2*B*a*b - 5*A*b^2)*x^5 + (2*
B*a^2 - 5*A*a*b)*x^2)*(-a^2*b)^(2/3)*log(a*b*x^2 - (-a^2*b)^(2/3)*x - (-a^2*b)^(1/3)*a) - 2*((2*B*a*b - 5*A*b^
2)*x^5 + (2*B*a^2 - 5*A*a*b)*x^2)*(-a^2*b)^(2/3)*log(a*b*x + (-a^2*b)^(2/3)))/(a^4*b^2*x^5 + a^5*b*x^2), -1/18
*(9*A*a^3*b - 3*(2*B*a^3*b - 5*A*a^2*b^2)*x^3 - 6*sqrt(1/3)*((2*B*a^2*b^2 - 5*A*a*b^3)*x^5 + (2*B*a^3*b - 5*A*
a^2*b^2)*x^2)*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) + ((2*B*a*b - 5*A*b^2)*x^5 + (2*B*a^2 - 5*A*a*b)*x^2)*(-a^2*b)^(2/3)*log(a*b*x^2 - (-a^2*b)^(2/3
)*x - (-a^2*b)^(1/3)*a) - 2*((2*B*a*b - 5*A*b^2)*x^5 + (2*B*a^2 - 5*A*a*b)*x^2)*(-a^2*b)^(2/3)*log(a*b*x + (-a
^2*b)^(2/3)))/(a^4*b^2*x^5 + a^5*b*x^2)]

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Sympy [A]  time = 0.944601, size = 109, normalized size = 0.56 \begin{align*} \frac{- 3 A a + x^{3} \left (- 5 A b + 2 B a\right )}{6 a^{3} x^{2} + 6 a^{2} b x^{5}} + \operatorname{RootSum}{\left (729 t^{3} a^{8} b + 125 A^{3} b^{3} - 150 A^{2} B a b^{2} + 60 A B^{2} a^{2} b - 8 B^{3} a^{3}, \left ( t \mapsto t \log{\left (\frac{9 t a^{3}}{- 5 A b + 2 B a} + x \right )} \right )\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-3*A*a + x**3*(-5*A*b + 2*B*a))/(6*a**3*x**2 + 6*a**2*b*x**5) + RootSum(729*_t**3*a**8*b + 125*A**3*b**3 - 15
0*A**2*B*a*b**2 + 60*A*B**2*a**2*b - 8*B**3*a**3, Lambda(_t, _t*log(9*_t*a**3/(-5*A*b + 2*B*a) + x)))

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Giac [A]  time = 1.14669, size = 254, normalized size = 1.3 \begin{align*} -\frac{{\left (2 \, B a - 5 \, 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^{3}} + \frac{\sqrt{3}{\left (2 \, \left (-a b^{2}\right )^{\frac{1}{3}} B a - 5 \, \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^{3} b} + \frac{B a x - A b x}{3 \,{\left (b x^{3} + a\right )} a^{2}} + \frac{{\left (2 \, \left (-a b^{2}\right )^{\frac{1}{3}} B a - 5 \, \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^{3} b} - \frac{A}{2 \, a^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/9*(2*B*a - 5*A*b)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^3 + 1/9*sqrt(3)*(2*(-a*b^2)^(1/3)*B*a - 5*(-a*b
^2)^(1/3)*A*b)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))/(a^3*b) + 1/3*(B*a*x - A*b*x)/((b*x^3 + a
)*a^2) + 1/18*(2*(-a*b^2)^(1/3)*B*a - 5*(-a*b^2)^(1/3)*A*b)*log(x^2 + x*(-a/b)^(1/3) + (-a/b)^(2/3))/(a^3*b) -
1/2*A/(a^2*x^2)