3.1.92 \(\int \frac {x}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\)

Optimal. Leaf size=202 \[ -\frac {\left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} b^{2/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {\left (a+b x^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} b^{2/3} \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {1355, 292, 31, 634, 617, 204, 628} \begin {gather*} -\frac {\left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {\left (a+b x^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} b^{2/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a+b x^3\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} b^{2/3} \sqrt {a^2+2 a b x^3+b^2 x^6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

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

Rule 31

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

Rule 1355

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.))^(p_), x_Symbol] :> Dist[(a + b*x^n + c*x^
(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^n)^(2*p), x], x] /; Fr
eeQ[{a, b, c, d, m, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.02, size = 109, normalized size = 0.54 \begin {gather*} \frac {\left (a+b x^3\right ) \left (\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )-2 \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )-2 \sqrt {3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )\right )}{6 \sqrt [3]{a} b^{2/3} \sqrt {\left (a+b x^3\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 8.36, size = 135, normalized size = 0.67 \begin {gather*} \frac {\left (a+b x^3\right ) \left (\frac {\log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 \sqrt [3]{a} b^{2/3}}-\frac {\log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 \sqrt [3]{a} b^{2/3}}-\frac {\tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \sqrt [3]{a} b^{2/3}}\right )}{\sqrt {\left (a+b x^3\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.76, size = 304, normalized size = 1.50 \begin {gather*} \left [\frac {3 \, \sqrt {\frac {1}{3}} a b \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}} \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}} \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{6 \, a b^{2}}, \frac {6 \, \sqrt {\frac {1}{3}} a b \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}} \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}} \log \left (b x - \left (-a b^{2}\right )^{\frac {1}{3}}\right )}{6 \, a b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/((b*x^3+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.42, size = 124, normalized size = 0.61 \begin {gather*} \frac {\sqrt {3} \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 ) \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, \left (-a b^{2}\right )^{\frac {1}{3}}} - \frac {\log \left (x^{2} + x \left (-\frac {a}{b}\right )^{\frac {1}{3}} + \left (-\frac {a}{b}\right )^{\frac {2}{3}}\right ) \mathrm {sgn}\left (b x^{3} + a\right )}{6 \, \left (-a b^{2}\right )^{\frac {1}{3}}} - \frac {\left (-\frac {a}{b}\right )^{\frac {2}{3}} \log \left ({\left | x - \left (-\frac {a}{b}\right )^{\frac {1}{3}} \right |}\right ) \mathrm {sgn}\left (b x^{3} + a\right )}{3 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/((b*x^3+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/3*sqrt(3)*arctan(1/3*sqrt(3)*(2*x + (-a/b)^(1/3))/(-a/b)^(1/3))*sgn(b*x^3 + a)/(-a*b^2)^(1/3) - 1/6*log(x^2
+ x*(-a/b)^(1/3) + (-a/b)^(2/3))*sgn(b*x^3 + a)/(-a*b^2)^(1/3) - 1/3*(-a/b)^(2/3)*log(abs(x - (-a/b)^(1/3)))*s
gn(b*x^3 + a)/a

________________________________________________________________________________________

maple [A]  time = 0.00, size = 97, normalized size = 0.48 \begin {gather*} -\frac {\left (b \,x^{3}+a \right ) \left (2 \sqrt {3}\, \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 )+2 \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )-\ln \left (x^{2}-\left (\frac {a}{b}\right )^{\frac {1}{3}} x +\left (\frac {a}{b}\right )^{\frac {2}{3}}\right )\right )}{6 \sqrt {\left (b \,x^{3}+a \right )^{2}}\, \left (\frac {a}{b}\right )^{\frac {1}{3}} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((b*x^3+a)^2)^(1/2),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.28, size = 98, normalized size = 0.49 \begin {gather*} \frac {\sqrt {3} \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 \, b \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {\log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {\log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b \left (\frac {a}{b}\right )^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/((b*x^3+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x}{\sqrt {{\left (b\,x^3+a\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((a + b*x^3)^2)^(1/2),x)

[Out]

int(x/((a + b*x^3)^2)^(1/2), x)

________________________________________________________________________________________

sympy [A]  time = 0.18, size = 24, normalized size = 0.12 \begin {gather*} \operatorname {RootSum} {\left (27 t^{3} a b^{2} + 1, \left (t \mapsto t \log {\left (9 t^{2} a b + x \right )} \right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/((b*x**3+a)**2)**(1/2),x)

[Out]

RootSum(27*_t**3*a*b**2 + 1, Lambda(_t, _t*log(9*_t**2*a*b + x)))

________________________________________________________________________________________