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

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 55, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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 {gather*}

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 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 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]

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 = 0.05, size = 81, normalized size = 1.47 \begin {gather*} \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 )}} \end {gather*}

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)])

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.54, size = 62, normalized size = 1.13 \begin {gather*} \frac {x \sqrt {a x+b x^4}}{3 b}-\frac {a \tanh ^{-1}\left (\frac {\sqrt {b} x \sqrt {a x+b x^4}}{a+b x^3}\right )}{3 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.57, size = 133, normalized size = 2.42 \begin {gather*} \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 {gather*}

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]

________________________________________________________________________________________

giac [A]  time = 0.30, size = 45, normalized size = 0.82 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

maple [C]  time = 0.09, size = 997, normalized size = 18.13

result too large to display

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*(-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)*x/(-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))^(1/
2)*(x-(-a*b^2)^(1/3)/b)^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)/(-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))^(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)/(-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))^(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*(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))^(
1/2)*((-a*b^2)^(1/3)/b*EllipticF(((-3/2*(-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)/(x-(-a*b^2)^(1/3)/b))^(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)*x/(-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))^(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)))

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {x^4}{\sqrt {b\,x^4+a\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a*x + b*x^4)^(1/2),x)

[Out]

int(x^4/(a*x + b*x^4)^(1/2), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{4}}{\sqrt {x \left (a + b x^{3}\right )}}\, dx \end {gather*}

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)

________________________________________________________________________________________