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

Optimal. Leaf size=149 \[ \frac {(A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{5/3} \sqrt [3]{b}}-\frac {(A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{5/3} \sqrt [3]{b}}+\frac {(A b-a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{5/3} \sqrt [3]{b}}-\frac {A}{2 a x^2} \]

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Rubi [A]  time = 0.09, antiderivative size = 149, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {453, 200, 31, 634, 617, 204, 628} \begin {gather*} \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^{5/3} \sqrt [3]{b}}-\frac {(A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{5/3} \sqrt [3]{b}}+\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^{5/3} \sqrt [3]{b}}-\frac {A}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^3*(a + b*x^3)),x]

[Out]

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

Rule 31

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.10, size = 135, normalized size = 0.91 \begin {gather*} \frac {\frac {(A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{\sqrt [3]{b}}-\frac {3 a^{2/3} A}{x^2}+\frac {2 (a B-A b) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\sqrt [3]{b}}+\frac {2 \sqrt {3} (A b-a B) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{\sqrt [3]{b}}}{6 a^{5/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^3}{x^3 \left (a+b x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.50, size = 411, normalized size = 2.76 \begin {gather*} \left [-\frac {3 \, \sqrt {\frac {1}{3}} {\left (B a^{2} b - A a b^{2}\right )} x^{2} \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 (-a^{2} b\right )^{\frac {2}{3}} {\left (B a - A b\right )} x^{2} \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 (-a^{2} b\right )^{\frac {2}{3}} {\left (B a - A b\right )} x^{2} \log \left (a b x + \left (-a^{2} b\right )^{\frac {2}{3}}\right ) + 3 \, A a^{2} b}{6 \, a^{3} b x^{2}}, \frac {6 \, \sqrt {\frac {1}{3}} {\left (B a^{2} b - A a b^{2}\right )} x^{2} \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 (-a^{2} b\right )^{\frac {2}{3}} {\left (B a - A b\right )} x^{2} \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 (-a^{2} b\right )^{\frac {2}{3}} {\left (B a - A b\right )} x^{2} \log \left (a b x + \left (-a^{2} b\right )^{\frac {2}{3}}\right ) - 3 \, A a^{2} b}{6 \, a^{3} b x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 161, normalized size = 1.08 \begin {gather*} -\frac {{\left (B a - 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 )}{3 \, a^{2}} + \frac {\sqrt {3} {\left (\left (-a b^{2}\right )^{\frac {1}{3}} B a - \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 )}{3 \, a^{2} b} + \frac {{\left (\left (-a b^{2}\right )^{\frac {1}{3}} B a - \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 )}{6 \, a^{2} b} - \frac {A}{2 \, a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 195, normalized size = 1.31 \begin {gather*} -\frac {\sqrt {3}\, A \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} a}-\frac {A \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} a}+\frac {A \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} a}+\frac {\sqrt {3}\, B \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}+\frac {B \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}-\frac {B \ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \left (\frac {a}{b}\right )^{\frac {2}{3}} b}-\frac {A}{2 a \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/a/(a/b)^(2/3)*ln(x+(a/b)^(1/3))*A+1/3/b/(a/b)^(2/3)*ln(x+(a/b)^(1/3))*B+1/6/a/(a/b)^(2/3)*ln(x^2-(a/b)^(1
/3)*x+(a/b)^(2/3))*A-1/6/b/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*B-1/3/a/(a/b)^(2/3)*3^(1/2)*arctan(1/
3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*A+1/3/b/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*B-1/2*A/a/x
^2

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maxima [A]  time = 1.22, size = 140, normalized size = 0.94 \begin {gather*} \frac {\sqrt {3} {\left (B a - 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 b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {{\left (B a - 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 b \left (\frac {a}{b}\right )^{\frac {2}{3}}} + \frac {{\left (B a - A b\right )} \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, a b \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {A}{2 \, a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.24, size = 126, normalized size = 0.85 \begin {gather*} -\frac {A}{2\,a\,x^2}-\frac {\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (A\,b-B\,a\right )}{3\,a^{5/3}\,b^{1/3}}+\frac {\ln \left (a^{1/3}-2\,b^{1/3}\,x+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,a^{5/3}\,b^{1/3}}-\frac {\ln \left (2\,b^{1/3}\,x-a^{1/3}+\sqrt {3}\,a^{1/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,a^{5/3}\,b^{1/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^3)/(x^3*(a + b*x^3)),x)

[Out]

(log(3^(1/2)*a^(1/3)*1i - 2*b^(1/3)*x + a^(1/3))*((3^(1/2)*1i)/2 + 1/2)*(A*b - B*a))/(3*a^(5/3)*b^(1/3)) - (lo
g(b^(1/3)*x + a^(1/3))*(A*b - B*a))/(3*a^(5/3)*b^(1/3)) - A/(2*a*x^2) - (log(3^(1/2)*a^(1/3)*1i + 2*b^(1/3)*x
- a^(1/3))*((3^(1/2)*1i)/2 - 1/2)*(A*b - B*a))/(3*a^(5/3)*b^(1/3))

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sympy [A]  time = 1.23, size = 73, normalized size = 0.49 \begin {gather*} - \frac {A}{2 a x^{2}} + \operatorname {RootSum} {\left (27 t^{3} a^{5} b + 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 {3 t a^{2}}{- A b + B a} + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A/(2*a*x**2) + RootSum(27*_t**3*a**5*b + 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(3*_t*a**2/(-A*b + B*a) + x)))

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