3.1.56 \(\int \frac {x}{\sqrt {a x+b x^4}} \, dx\)

Optimal. Leaf size=32 \[ \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x+b x^4}}\right )}{3 \sqrt {b}} \]

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Rubi [A]  time = 0.03, antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2029, 206} \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x+b x^4}}\right )}{3 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/Sqrt[a*x + b*x^4],x]

[Out]

(2*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a*x + b*x^4]])/(3*Sqrt[b])

Rule 206

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

Rule 2029

Int[(x_)^(m_.)/Sqrt[(a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.)], x_Symbol] :> Dist[-2/(n - j), Subst[Int[1/(1 - a*x^2
), x], x, x^(j/2)/Sqrt[a*x^j + b*x^n]], x] /; FreeQ[{a, b, j, n}, x] && EqQ[m, j/2 - 1] && NeQ[n, j]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {a x+b x^4}} \, dx &=\frac {2}{3} \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x^2}{\sqrt {a x+b x^4}}\right )\\ &=\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a x+b x^4}}\right )}{3 \sqrt {b}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 61, normalized size = 1.91 \begin {gather*} \frac {2 \sqrt {x} \sqrt {a+b x^3} \tanh ^{-1}\left (\frac {\sqrt {b} x^{3/2}}{\sqrt {a+b x^3}}\right )}{3 \sqrt {b} \sqrt {x \left (a+b x^3\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/Sqrt[a*x + b*x^4],x]

[Out]

(2*Sqrt[x]*Sqrt[a + b*x^3]*ArcTanh[(Sqrt[b]*x^(3/2))/Sqrt[a + b*x^3]])/(3*Sqrt[b]*Sqrt[x*(a + b*x^3)])

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IntegrateAlgebraic [A]  time = 0.41, size = 39, normalized size = 1.22 \begin {gather*} \frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} x \sqrt {a x+b x^4}}{a+b x^3}\right )}{3 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*ArcTanh[(Sqrt[b]*x*Sqrt[a*x + b*x^4])/(a + b*x^3)])/(3*Sqrt[b])

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fricas [A]  time = 0.57, size = 94, normalized size = 2.94 \begin {gather*} \left [\frac {\log \left (-8 \, b^{2} x^{6} - 8 \, a b x^{3} - a^{2} - 4 \, {\left (2 \, b x^{4} + a x\right )} \sqrt {b x^{4} + a x} \sqrt {b}\right )}{6 \, \sqrt {b}}, -\frac {\sqrt {-b} \arctan \left (\frac {2 \, \sqrt {b x^{4} + a x} \sqrt {-b} x}{2 \, b x^{3} + a}\right )}{3 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/6*log(-8*b^2*x^6 - 8*a*b*x^3 - a^2 - 4*(2*b*x^4 + a*x)*sqrt(b*x^4 + a*x)*sqrt(b))/sqrt(b), -1/3*sqrt(-b)*ar
ctan(2*sqrt(b*x^4 + a*x)*sqrt(-b)*x/(2*b*x^3 + a))/b]

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giac [A]  time = 0.21, size = 23, normalized size = 0.72 \begin {gather*} -\frac {2 \, \arctan \left (\frac {\sqrt {b + \frac {a}{x^{3}}}}{\sqrt {-b}}\right )}{3 \, \sqrt {-b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.09, size = 979, normalized size = 30.59

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*((-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b
)/(-1/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(x-(-a*b^2)^(1/3)/b)*x)^(1/2)*(x-(-a*b^2)^(1/3)/b)^2*
((-a*b^2)^(1/3)*(x+1/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(-1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(
-a*b^2)^(1/3)/b)/(x-(-a*b^2)^(1/3)/b)/b)^(1/2)*((-a*b^2)^(1/3)*(x+1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^
(1/3)/b)/(-1/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(x-(-a*b^2)^(1/3)/b)/b)^(1/2)/(-3/2*(-a*b^2)^(
1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(-a*b^2)^(1/3)*b/((x-(-a*b^2)^(1/3)/b)*(x+1/2*(-a*b^2)^(1/3)/b+1/2*I*3^
(1/2)*(-a*b^2)^(1/3)/b)*(x+1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*b*x)^(1/2)*((-a*b^2)^(1/3)/b*E
llipticF(((-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(-1/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)
^(1/3)/b)/(x-(-a*b^2)^(1/3)/b)*x)^(1/2),((3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*(1/2*(-a*b^2)^(
1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(1/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(3/2*(-a*b^2)^(1/
3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b))^(1/2))-(-a*b^2)^(1/3)/b*EllipticPi(((-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)
*(-a*b^2)^(1/3)/b)/(-1/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(x-(-a*b^2)^(1/3)/b)*x)^(1/2),(-1/2*
(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b),((3/2*
(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*(1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(1/2*(-
a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/(3/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b))^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {b x^{4} + a x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x/sqrt(b*x^4 + a*x), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {x}{\sqrt {b\,x^4+a\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x + b*x^4)^(1/2),x)

[Out]

int(x/(a*x + b*x^4)^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x/sqrt(x*(a + b*x**3)), x)

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