3.1.68 \(\int \frac {A+B x^3}{x^6 (a+b x^3)} \, dx\)

Optimal. Leaf size=168 \[ -\frac {b^{2/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{8/3}}+\frac {b^{2/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{8/3}}-\frac {b^{2/3} (A b-a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{8/3}}+\frac {A b-a B}{2 a^2 x^2}-\frac {A}{5 a x^5} \]

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Rubi [A]  time = 0.12, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {453, 325, 200, 31, 634, 617, 204, 628} \begin {gather*} -\frac {b^{2/3} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{8/3}}+\frac {b^{2/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{8/3}}-\frac {b^{2/3} (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^{8/3}}+\frac {A b-a B}{2 a^2 x^2}-\frac {A}{5 a x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^6*(a + b*x^3)),x]

[Out]

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

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

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Mathematica [A]  time = 0.13, size = 154, normalized size = 0.92 \begin {gather*} \frac {5 b^{2/3} (a B-A b) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+\frac {15 a^{2/3} (A b-a B)}{x^2}-\frac {6 a^{5/3} A}{x^5}+10 b^{2/3} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-10 \sqrt {3} b^{2/3} (A b-a B) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )}{30 a^{8/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^6*(a + b*x^3)),x]

[Out]

((-6*a^(5/3)*A)/x^5 + (15*a^(2/3)*(A*b - a*B))/x^2 - 10*Sqrt[3]*b^(2/3)*(A*b - a*B)*ArcTan[(1 - (2*b^(1/3)*x)/
a^(1/3))/Sqrt[3]] + 10*b^(2/3)*(A*b - a*B)*Log[a^(1/3) + b^(1/3)*x] + 5*b^(2/3)*(-(A*b) + a*B)*Log[a^(2/3) - a
^(1/3)*b^(1/3)*x + b^(2/3)*x^2])/(30*a^(8/3))

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A+B x^3}{x^6 \left (a+b x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x^3)/(x^6*(a + b*x^3)),x]

[Out]

IntegrateAlgebraic[(A + B*x^3)/(x^6*(a + b*x^3)), x]

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fricas [A]  time = 0.60, size = 176, normalized size = 1.05 \begin {gather*} -\frac {10 \, \sqrt {3} {\left (B a - A b\right )} x^{5} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} a x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}} - \sqrt {3} b}{3 \, b}\right ) - 5 \, {\left (B a - A b\right )} x^{5} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b^{2} x^{2} - a b x \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} + a^{2} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {2}{3}}\right ) + 10 \, {\left (B a - A b\right )} x^{5} \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}} \log \left (b x + a \left (\frac {b^{2}}{a^{2}}\right )^{\frac {1}{3}}\right ) + 15 \, {\left (B a - A b\right )} x^{3} + 6 \, A a}{30 \, a^{2} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(b*x^3+a),x, algorithm="fricas")

[Out]

-1/30*(10*sqrt(3)*(B*a - A*b)*x^5*(b^2/a^2)^(1/3)*arctan(1/3*(2*sqrt(3)*a*x*(b^2/a^2)^(2/3) - sqrt(3)*b)/b) -
5*(B*a - A*b)*x^5*(b^2/a^2)^(1/3)*log(b^2*x^2 - a*b*x*(b^2/a^2)^(1/3) + a^2*(b^2/a^2)^(2/3)) + 10*(B*a - A*b)*
x^5*(b^2/a^2)^(1/3)*log(b*x + a*(b^2/a^2)^(1/3)) + 15*(B*a - A*b)*x^3 + 6*A*a)/(a^2*x^5)

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giac [A]  time = 0.18, size = 176, normalized size = 1.05 \begin {gather*} -\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^{3}} + \frac {{\left (B a b - A b^{2}\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^{3}} - \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^{3}} - \frac {5 \, B a x^{3} - 5 \, A b x^{3} + 2 \, A a}{10 \, a^{2} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(b*x^3+a),x, algorithm="giac")

[Out]

-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
^3 + 1/3*(B*a*b - A*b^2)*(-a/b)^(1/3)*log(abs(x - (-a/b)^(1/3)))/a^3 - 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^3 - 1/10*(5*B*a*x^3 - 5*A*b*x^3 + 2*A*a)/(a^2*x^5)

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maple [A]  time = 0.05, size = 217, normalized size = 1.29 \begin {gather*} \frac {\sqrt {3}\, A 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}} a^{2}}+\frac {A b \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} a^{2}}-\frac {A 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}} a^{2}}-\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}} a}-\frac {B \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \left (\frac {a}{b}\right )^{\frac {2}{3}} a}+\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}} a}+\frac {A b}{2 a^{2} x^{2}}-\frac {B}{2 a \,x^{2}}-\frac {A}{5 a \,x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^6/(b*x^3+a),x)

[Out]

1/3/a^2*b/(a/b)^(2/3)*ln(x+(a/b)^(1/3))*A-1/3/a/(a/b)^(2/3)*ln(x+(a/b)^(1/3))*B-1/6/a^2*b/(a/b)^(2/3)*ln(x^2-(
a/b)^(1/3)*x+(a/b)^(2/3))*A+1/6/a/(a/b)^(2/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))*B+1/3/a^2*b/(a/b)^(2/3)*3^(1/2
)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*A-1/3/a/(a/b)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1))*
B-1/5*A/a/x^5+1/2/a^2/x^2*A*b-1/2/a/x^2*B

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maxima [A]  time = 1.28, size = 148, normalized size = 0.88 \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^{2} \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^{2} \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^{2} \left (\frac {a}{b}\right )^{\frac {2}{3}}} - \frac {5 \, {\left (B a - A b\right )} x^{3} + 2 \, A a}{10 \, a^{2} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(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^2*(a/b)^(2/3)) + 1/6*(B*a - A*
b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*(a/b)^(2/3)) - 1/3*(B*a - A*b)*log(x + (a/b)^(1/3))/(a^2*(a/b)^
(2/3)) - 1/10*(5*(B*a - A*b)*x^3 + 2*A*a)/(a^2*x^5)

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mupad [B]  time = 2.56, size = 145, normalized size = 0.86 \begin {gather*} \frac {b^{2/3}\,\ln \left (b^{1/3}\,x+a^{1/3}\right )\,\left (A\,b-B\,a\right )}{3\,a^{8/3}}-\frac {\frac {A}{5\,a}-\frac {x^3\,\left (A\,b-B\,a\right )}{2\,a^2}}{x^5}-\frac {b^{2/3}\,\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^{8/3}}+\frac {b^{2/3}\,\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^{8/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^3)/(x^6*(a + b*x^3)),x)

[Out]

(b^(2/3)*log(b^(1/3)*x + a^(1/3))*(A*b - B*a))/(3*a^(8/3)) - (A/(5*a) - (x^3*(A*b - B*a))/(2*a^2))/x^5 - (b^(2
/3)*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^(8/3)) + (b^(2/3)
*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^(8/3))

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sympy [A]  time = 0.93, size = 99, normalized size = 0.59 \begin {gather*} \operatorname {RootSum} {\left (27 t^{3} a^{8} - A^{3} b^{5} + 3 A^{2} B a b^{4} - 3 A B^{2} a^{2} b^{3} + B^{3} a^{3} b^{2}, \left (t \mapsto t \log {\left (- \frac {3 t a^{3}}{- A b^{2} + B a b} + x \right )} \right )\right )} + \frac {- 2 A a + x^{3} \left (5 A b - 5 B a\right )}{10 a^{2} x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/x**6/(b*x**3+a),x)

[Out]

RootSum(27*_t**3*a**8 - A**3*b**5 + 3*A**2*B*a*b**4 - 3*A*B**2*a**2*b**3 + B**3*a**3*b**2, Lambda(_t, _t*log(-
3*_t*a**3/(-A*b**2 + B*a*b) + x))) + (-2*A*a + x**3*(5*A*b - 5*B*a))/(10*a**2*x**5)

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