3.1.89 \(\int \frac {x^4}{\sqrt {a^2+2 a b x^3+b^2 x^6}} \, dx\)

Optimal. Leaf size=240 \[ \frac {x^2 \left (a+b x^3\right )}{2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{2/3} \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{2/3} \left (a+b x^3\right ) \tan ^{-1}\left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a^{2/3} \left (a+b x^3\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}} \]

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Rubi [A]  time = 0.12, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {1355, 321, 292, 31, 634, 617, 204, 628} \begin {gather*} \frac {x^2 \left (a+b x^3\right )}{2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{2/3} \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a^{2/3} \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 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{2/3} \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} b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

(x^2*(a + b*x^3))/(2*b*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) + (a^(2/3)*(a + b*x^3)*ArcTan[(a^(1/3) - 2*b^(1/3)*x)/
(Sqrt[3]*a^(1/3))])/(Sqrt[3]*b^(5/3)*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) + (a^(2/3)*(a + b*x^3)*Log[a^(1/3) + b^(
1/3)*x])/(3*b^(5/3)*Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6]) - (a^(2/3)*(a + b*x^3)*Log[a^(2/3) - a^(1/3)*b^(1/3)*x +
b^(2/3)*x^2])/(6*b^(5/3)*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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, 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^4}{\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^4}{a b+b^2 x^3} \, dx}{\sqrt {a^2+2 a b x^3+b^2 x^6}}\\ &=\frac {x^2 \left (a+b x^3\right )}{2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a \left (a b+b^2 x^3\right )\right ) \int \frac {x}{a b+b^2 x^3} \, dx}{b \sqrt {a^2+2 a b x^3+b^2 x^6}}\\ &=\frac {x^2 \left (a+b x^3\right )}{2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {\left (a^{2/3} \left (a b+b^2 x^3\right )\right ) \int \frac {1}{\sqrt [3]{a} \sqrt [3]{b}+b^{2/3} x} \, dx}{3 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a^{2/3} \left (a b+b^2 x^3\right )\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 b^2 \sqrt {a^2+2 a b x^3+b^2 x^6}}\\ &=\frac {x^2 \left (a+b x^3\right )}{2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{2/3} \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a^{2/3} \left (a b+b^2 x^3\right )\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 b^{8/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a \left (a b+b^2 x^3\right )\right ) \int \frac {1}{a^{2/3} b^{2/3}-\sqrt [3]{a} b x+b^{4/3} x^2} \, dx}{2 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}\\ &=\frac {x^2 \left (a+b x^3\right )}{2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{2/3} \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a^{2/3} \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 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {\left (a^{2/3} \left (a b+b^2 x^3\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-3-x^2} \, dx,x,1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}\right )}{b^{8/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}\\ &=\frac {x^2 \left (a+b x^3\right )}{2 b \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{2/3} \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} b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}+\frac {a^{2/3} \left (a+b x^3\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}-\frac {a^{2/3} \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 b^{5/3} \sqrt {a^2+2 a b x^3+b^2 x^6}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 131, normalized size = 0.55 \begin {gather*} \frac {\left (a+b x^3\right ) \left (-a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )+2 a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )+2 \sqrt {3} a^{2/3} \tan ^{-1}\left (\frac {1-\frac {2 \sqrt [3]{b} x}{\sqrt [3]{a}}}{\sqrt {3}}\right )+3 b^{2/3} x^2\right )}{6 b^{5/3} \sqrt {\left (a+b x^3\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

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

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IntegrateAlgebraic [A]  time = 7.36, size = 144, normalized size = 0.60 \begin {gather*} \frac {\left (a+b x^3\right ) \left (-\frac {a^{2/3} \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2\right )}{6 b^{5/3}}+\frac {a^{2/3} \log \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{3 b^{5/3}}+\frac {a^{2/3} \tan ^{-1}\left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} x}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} b^{5/3}}+\frac {x^2}{2 b}\right )}{\sqrt {\left (a+b x^3\right )^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^4/Sqrt[a^2 + 2*a*b*x^3 + b^2*x^6],x]

[Out]

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

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fricas [A]  time = 0.85, size = 123, normalized size = 0.51 \begin {gather*} \frac {3 \, x^{2} - 2 \, \sqrt {3} \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} - \sqrt {3} a}{3 \, a}\right ) - \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x^{2} - b x \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}} + a \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}}\right ) + 2 \, \left (\frac {a^{2}}{b^{2}}\right )^{\frac {1}{3}} \log \left (a x + b \left (\frac {a^{2}}{b^{2}}\right )^{\frac {2}{3}}\right )}{6 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/((b*x^3+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.37, size = 146, normalized size = 0.61 \begin {gather*} \frac {x^{2} \mathrm {sgn}\left (b x^{3} + a\right )}{2 \, b} + \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 \, b} + \frac {\sqrt {3} \left (-a b^{2}\right )^{\frac {2}{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 \, b^{3}} - \frac {\left (-a b^{2}\right )^{\frac {2}{3}} \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 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/((b*x^3+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 113, normalized size = 0.47 \begin {gather*} \frac {\left (b \,x^{3}+a \right ) \left (3 \left (\frac {a}{b}\right )^{\frac {1}{3}} b \,x^{2}+2 \sqrt {3}\, a \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 a \ln \left (x +\left (\frac {a}{b}\right )^{\frac {1}{3}}\right )-a \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^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/((b*x^3+a)^2)^(1/2),x)

[Out]

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

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maxima [A]  time = 2.18, size = 109, normalized size = 0.45 \begin {gather*} \frac {x^{2}}{2 \, b} - \frac {\sqrt {3} a \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^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} - \frac {a \log \left (x^{2} - x \left (\frac {a}{b}\right )^{\frac {1}{3}} + \left (\frac {a}{b}\right )^{\frac {2}{3}}\right )}{6 \, b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} + \frac {a \log \left (x + \left (\frac {a}{b}\right )^{\frac {1}{3}}\right )}{3 \, b^{2} \left (\frac {a}{b}\right )^{\frac {1}{3}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4/((b*x^3+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4}{\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^4/((a + b*x^3)^2)^(1/2),x)

[Out]

int(x^4/((a + b*x^3)^2)^(1/2), x)

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sympy [A]  time = 0.21, size = 32, normalized size = 0.13 \begin {gather*} \operatorname {RootSum} {\left (27 t^{3} b^{5} - a^{2}, \left (t \mapsto t \log {\left (\frac {9 t^{2} b^{3}}{a} + x \right )} \right )\right )} + \frac {x^{2}}{2 b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**4/((b*x**3+a)**2)**(1/2),x)

[Out]

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

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