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

Optimal. Leaf size=55 \[ \frac{x \sqrt{a x+b x^4}}{3 b}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a x+b x^4}}\right )}{3 b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0698222, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {2024, 2029, 206} \[ \frac{x \sqrt{a x+b x^4}}{3 b}-\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a x+b x^4}}\right )}{3 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a*x + b*x^4],x]

[Out]

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

Rule 2024

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^(n - j)*(m + j*p - n + j + 1))/(b*(m + n*p + 1)
), Int[(c*x)^(m - (n - j))*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, p}, x] &&  !IntegerQ[p] && LtQ[0, j
, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[m + j*p + 1 - n + j, 0] && NeQ[m + n*p + 1, 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]

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])

Rubi steps

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

Mathematica [A]  time = 0.0327577, size = 81, normalized size = 1.47 \[ \frac{\sqrt{b} x^2 \left (a+b x^3\right )-a \sqrt{x} \sqrt{a+b x^3} \tanh ^{-1}\left (\frac{\sqrt{b} x^{3/2}}{\sqrt{a+b x^3}}\right )}{3 b^{3/2} \sqrt{x \left (a+b x^3\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a*x + b*x^4],x]

[Out]

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

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Maple [C]  time = 0.032, size = 997, normalized size = 18.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x*(b*x^4+a*x)^(1/2)/b-a*(1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*((-3/2/b*(-b^2*a)^(1/3)+1/2*
I*3^(1/2)/b*(-b^2*a)^(1/3))*x/(-1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(x-1/b*(-b^2*a)^(1/3)))^(
1/2)*(x-1/b*(-b^2*a)^(1/3))^2*(1/b*(-b^2*a)^(1/3)*(x+1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(-1/
2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(x-1/b*(-b^2*a)^(1/3)))^(1/2)*(1/b*(-b^2*a)^(1/3)*(x+1/2/b*
(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(-1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(x-1/b*(
-b^2*a)^(1/3)))^(1/2)/(-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(-b^2*a)^(1/3)/(b*x*(x-1/b*(-b^2*
a)^(1/3))*(x+1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*(x+1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^
2*a)^(1/3)))^(1/2)*(1/b*(-b^2*a)^(1/3)*EllipticF(((-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*x/(-1
/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(x-1/b*(-b^2*a)^(1/3)))^(1/2),((3/2/b*(-b^2*a)^(1/3)+1/2*I
*3^(1/2)/b*(-b^2*a)^(1/3))*(1/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(1/2/b*(-b^2*a)^(1/3)+1/2*I*3
^(1/2)/b*(-b^2*a)^(1/3))/(3/2/b*(-b^2*a)^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3)))^(1/2))-1/b*(-b^2*a)^(1/3)*Elli
pticPi(((-3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*x/(-1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*
a)^(1/3))/(x-1/b*(-b^2*a)^(1/3)))^(1/2),(-1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(-3/2/b*(-b^2*a
)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3)),((3/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))*(1/2/b*(-b^2*a)
^(1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(1/2/b*(-b^2*a)^(1/3)+1/2*I*3^(1/2)/b*(-b^2*a)^(1/3))/(3/2/b*(-b^2*a)^(
1/3)-1/2*I*3^(1/2)/b*(-b^2*a)^(1/3)))^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^4/sqrt(b*x^4 + a*x), x)

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Fricas [A]  time = 1.97972, size = 315, normalized size = 5.73 \begin{align*} \left [\frac{4 \, \sqrt{b x^{4} + a x} b x + a \sqrt{b} \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 )}{12 \, b^{2}}, \frac{2 \, \sqrt{b x^{4} + a x} b x + a \sqrt{-b} \arctan \left (\frac{2 \, \sqrt{b x^{4} + a x} \sqrt{-b} x}{2 \, b x^{3} + a}\right )}{6 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(4*sqrt(b*x^4 + a*x)*b*x + a*sqrt(b)*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)))/b^2, 1/6*(2*sqrt(b*x^4 + a*x)*b*x + a*sqrt(-b)*arctan(2*sqrt(b*x^4 + a*x)*sqrt(-b)*x/(2*b*x^3 +
a)))/b^2]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{4}}{\sqrt{x \left (a + b x^{3}\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4/sqrt(x*(a + b*x**3)), x)

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Giac [A]  time = 1.32322, size = 61, normalized size = 1.11 \begin{align*} \frac{\sqrt{b x^{4} + a x} x}{3 \, b} + \frac{a \arctan \left (\frac{\sqrt{b + \frac{a}{x^{3}}}}{\sqrt{-b}}\right )}{3 \, \sqrt{-b} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*sqrt(b*x^4 + a*x)*x/b + 1/3*a*arctan(sqrt(b + a/x^3)/sqrt(-b))/(sqrt(-b)*b)