3.1.67 \(\int \frac {A+B x^3}{x^5 (a+b x^3)} \, dx\) [67]

Optimal. Leaf size=165 \[ -\frac {A}{4 a x^4}+\frac {A b-a B}{a^2 x}-\frac {\sqrt [3]{b} (A b-a B) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{7/3}}-\frac {\sqrt [3]{b} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{7/3}}+\frac {\sqrt [3]{b} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{7/3}} \]

[Out]

-1/4*A/a/x^4+(A*b-B*a)/a^2/x-1/3*b^(1/3)*(A*b-B*a)*ln(a^(1/3)+b^(1/3)*x)/a^(7/3)+1/6*b^(1/3)*(A*b-B*a)*ln(a^(2
/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/a^(7/3)-1/3*b^(1/3)*(A*b-B*a)*arctan(1/3*(a^(1/3)-2*b^(1/3)*x)/a^(1/3)*3^(1
/2))/a^(7/3)*3^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 165, 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 = {464, 331, 298, 31, 648, 631, 210, 642} \begin {gather*} -\frac {\sqrt [3]{b} (A b-a B) \text {ArcTan}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} a^{7/3}}+\frac {\sqrt [3]{b} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{7/3}}-\frac {\sqrt [3]{b} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{7/3}}+\frac {A b-a B}{a^2 x}-\frac {A}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^5*(a + b*x^3)),x]

[Out]

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

Rule 31

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

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 298

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 331

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 464

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 631

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 642

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 648

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^5 \left (a+b x^3\right )} \, dx &=-\frac {A}{4 a x^4}-\frac {(4 A b-4 a B) \int \frac {1}{x^2 \left (a+b x^3\right )} \, dx}{4 a}\\ &=-\frac {A}{4 a x^4}+\frac {A b-a B}{a^2 x}+\frac {(b (A b-a B)) \int \frac {x}{a+b x^3} \, dx}{a^2}\\ &=-\frac {A}{4 a x^4}+\frac {A b-a B}{a^2 x}-\frac {\left (b^{2/3} (A b-a B)\right ) \int \frac {1}{\sqrt [3]{a}+\sqrt [3]{b} x} \, dx}{3 a^{7/3}}+\frac {\left (b^{2/3} (A b-a B)\right ) \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^{7/3}}\\ &=-\frac {A}{4 a x^4}+\frac {A b-a B}{a^2 x}-\frac {\sqrt [3]{b} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{7/3}}+\frac {\left (\sqrt [3]{b} (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^{7/3}}+\frac {\left (b^{2/3} (A b-a B)\right ) \int \frac {1}{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2} \, dx}{2 a^2}\\ &=-\frac {A}{4 a x^4}+\frac {A b-a B}{a^2 x}-\frac {\sqrt [3]{b} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{7/3}}+\frac {\sqrt [3]{b} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{7/3}}+\frac {\left (\sqrt [3]{b} (A b-a B)\right ) \text {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{a^{7/3}}\\ &=-\frac {A}{4 a x^4}+\frac {A b-a B}{a^2 x}-\frac {\sqrt [3]{b} (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^{7/3}}-\frac {\sqrt [3]{b} (A b-a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 a^{7/3}}+\frac {\sqrt [3]{b} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 a^{7/3}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 154, normalized size = 0.93 \begin {gather*} \frac {-\frac {3 a^{4/3} A}{x^4}+\frac {12 \sqrt [3]{a} (A b-a B)}{x}-4 \sqrt {3} \sqrt [3]{b} (A b-a B) \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+4 \sqrt [3]{b} (-A b+a B) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 \sqrt [3]{b} (A b-a B) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{12 a^{7/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^5*(a + b*x^3)),x]

[Out]

((-3*a^(4/3)*A)/x^4 + (12*a^(1/3)*(A*b - a*B))/x - 4*Sqrt[3]*b^(1/3)*(A*b - a*B)*ArcTan[(1 - (2*b^(1/3)*x)/a^(
1/3))/Sqrt[3]] + 4*b^(1/3)*(-(A*b) + a*B)*Log[a^(1/3) + b^(1/3)*x] + 2*b^(1/3)*(A*b - a*B)*Log[a^(2/3) - a^(1/
3)*b^(1/3)*x + b^(2/3)*x^2])/(12*a^(7/3))

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Maple [A]
time = 0.29, size = 130, normalized size = 0.79

method	result	size
default	\(\frac {\left (-\frac {\ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}+\frac {\sqrt {3}\, \arctan \left (\frac {\sqrt {3}\, \left (\frac {2 x}{\left (\frac {a}{b}\right )^{\frac {1}{3}}}-1\right )}{3}\right )}{3 b \left (\frac {a}{b}\right )^{\frac {1}{3}}}\right ) b \left (A b -B a \right )}{a^{2}}-\frac {A}{4 a \,x^{4}}-\frac {-A b +B a}{a^{2} x}\)	\(130\)
risch	\(\frac {\frac {\left (A b -B a \right ) x^{3}}{a^{2}}-\frac {A}{4 a}}{x^{4}}+\frac {\left (\munderset {\textit {\_R} =\RootOf \left (a^{7} \textit {\_Z}^{3}+A^{3} b^{4}-3 A^{2} B a \,b^{3}+3 A \,B^{2} a^{2} b^{2}-B^{3} a^{3} b \right )}{\sum }\textit {\_R} \ln \left (\left (-4 a^{7} \textit {\_R}^{3}-3 A^{3} b^{4}+9 A^{2} B a \,b^{3}-9 A \,B^{2} a^{2} b^{2}+3 B^{3} a^{3} b \right ) x +\left (A \,a^{5} b -B \,a^{6}\right ) \textit {\_R}^{2}\right )\right )}{3}\)	\(151\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/x^5/(b*x^3+a),x,method=_RETURNVERBOSE)

[Out]

(-1/3/b/(a/b)^(1/3)*ln(x+(a/b)^(1/3))+1/6/b/(a/b)^(1/3)*ln(x^2-(a/b)^(1/3)*x+(a/b)^(2/3))+1/3*3^(1/2)/b/(a/b)^
(1/3)*arctan(1/3*3^(1/2)*(2/(a/b)^(1/3)*x-1)))*b*(A*b-B*a)/a^2-1/4*A/a/x^4-1/a^2*(-A*b+B*a)/x

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Maxima [A]
time = 0.52, size = 147, normalized size = 0.89 \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 {1}{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 {1}{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 {1}{3}}} - \frac {4 \, {\left (B a - A b\right )} x^{3} + A a}{4 \, a^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^5/(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)^(1/3)) - 1/6*(B*a - A*
b)*log(x^2 - x*(a/b)^(1/3) + (a/b)^(2/3))/(a^2*(a/b)^(1/3)) + 1/3*(B*a - A*b)*log(x + (a/b)^(1/3))/(a^2*(a/b)^
(1/3)) - 1/4*(4*(B*a - A*b)*x^3 + A*a)/(a^2*x^4)

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Fricas [A]
time = 1.93, size = 158, normalized size = 0.96 \begin {gather*} -\frac {4 \, \sqrt {3} {\left (B a - A b\right )} x^{4} \left (-\frac {b}{a}\right )^{\frac {1}{3}} \arctan \left (\frac {2}{3} \, \sqrt {3} x \left (-\frac {b}{a}\right )^{\frac {1}{3}} + \frac {1}{3} \, \sqrt {3}\right ) - 2 \, {\left (B a - A b\right )} x^{4} \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x^{2} - a x \left (-\frac {b}{a}\right )^{\frac {2}{3}} - a \left (-\frac {b}{a}\right )^{\frac {1}{3}}\right ) + 4 \, {\left (B a - A b\right )} x^{4} \left (-\frac {b}{a}\right )^{\frac {1}{3}} \log \left (b x + a \left (-\frac {b}{a}\right )^{\frac {2}{3}}\right ) + 12 \, {\left (B a - A b\right )} x^{3} + 3 \, A a}{12 \, a^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^5/(b*x^3+a),x, algorithm="fricas")

[Out]

-1/12*(4*sqrt(3)*(B*a - A*b)*x^4*(-b/a)^(1/3)*arctan(2/3*sqrt(3)*x*(-b/a)^(1/3) + 1/3*sqrt(3)) - 2*(B*a - A*b)
*x^4*(-b/a)^(1/3)*log(b*x^2 - a*x*(-b/a)^(2/3) - a*(-b/a)^(1/3)) + 4*(B*a - A*b)*x^4*(-b/a)^(1/3)*log(b*x + a*
(-b/a)^(2/3)) + 12*(B*a - A*b)*x^3 + 3*A*a)/(a^2*x^4)

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Sympy [A]
time = 0.69, size = 112, normalized size = 0.68 \begin {gather*} \operatorname {RootSum} {\left (27 t^{3} a^{7} + A^{3} b^{4} - 3 A^{2} B a b^{3} + 3 A B^{2} a^{2} b^{2} - B^{3} a^{3} b, \left ( t \mapsto t \log {\left (\frac {9 t^{2} a^{5}}{A^{2} b^{3} - 2 A B a b^{2} + B^{2} a^{2} b} + x \right )} \right )\right )} + \frac {- A a + x^{3} \cdot \left (4 A b - 4 B a\right )}{4 a^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/x**5/(b*x**3+a),x)

[Out]

RootSum(27*_t**3*a**7 + A**3*b**4 - 3*A**2*B*a*b**3 + 3*A*B**2*a**2*b**2 - B**3*a**3*b, Lambda(_t, _t*log(9*_t
**2*a**5/(A**2*b**3 - 2*A*B*a*b**2 + B**2*a**2*b) + x))) + (-A*a + x**3*(4*A*b - 4*B*a))/(4*a**2*x**4)

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Giac [A]
time = 0.86, size = 197, normalized size = 1.19 \begin {gather*} \frac {{\left (B a b \left (-\frac {a}{b}\right )^{\frac {1}{3}} - A b^{2} \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^{3}} + \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^{3} b} - \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^{3} b} - \frac {4 \, B a x^{3} - 4 \, A b x^{3} + A a}{4 \, a^{2} x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^5/(b*x^3+a),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 2.59, size = 178, normalized size = 1.08 \begin {gather*} \frac {{\left (-b\right )}^{1/3}\,\ln \left (a^{1/3}\,{\left (-b\right )}^{8/3}+b^3\,x\right )\,\left (A\,b-B\,a\right )}{3\,a^{7/3}}-\frac {\frac {A}{4\,a}-\frac {x^3\,\left (A\,b-B\,a\right )}{a^2}}{x^4}+\frac {{\left (-b\right )}^{1/3}\,\ln \left (a^{1/3}\,{\left (-b\right )}^{8/3}-2\,b^3\,x+\sqrt {3}\,a^{1/3}\,{\left (-b\right )}^{8/3}\,1{}\mathrm {i}\right )\,\left (-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,a^{7/3}}-\frac {{\left (-b\right )}^{1/3}\,\ln \left (2\,b^3\,x-a^{1/3}\,{\left (-b\right )}^{8/3}+\sqrt {3}\,a^{1/3}\,{\left (-b\right )}^{8/3}\,1{}\mathrm {i}\right )\,\left (\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}\right )\,\left (A\,b-B\,a\right )}{3\,a^{7/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^3)/(x^5*(a + b*x^3)),x)

[Out]

((-b)^(1/3)*log(a^(1/3)*(-b)^(8/3) + b^3*x)*(A*b - B*a))/(3*a^(7/3)) - (A/(4*a) - (x^3*(A*b - B*a))/a^2)/x^4 +
((-b)^(1/3)*log(a^(1/3)*(-b)^(8/3) - 2*b^3*x + 3^(1/2)*a^(1/3)*(-b)^(8/3)*1i)*((3^(1/2)*1i)/2 - 1/2)*(A*b - B
*a))/(3*a^(7/3)) - ((-b)^(1/3)*log(2*b^3*x - a^(1/3)*(-b)^(8/3) + 3^(1/2)*a^(1/3)*(-b)^(8/3)*1i)*((3^(1/2)*1i)
/2 + 1/2)*(A*b - B*a))/(3*a^(7/3))

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