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

Optimal. Leaf size=73 \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}}-\frac {3 a x \sqrt {a+b x^2}}{8 b^2}+\frac {x^3 \sqrt {a+b x^2}}{4 b} \]

[Out]

3/8*a^2*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5/2)-3/8*a*x*(b*x^2+a)^(1/2)/b^2+1/4*x^3*(b*x^2+a)^(1/2)/b

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Rubi [A]  time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {321, 217, 206} \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}}-\frac {3 a x \sqrt {a+b x^2}}{8 b^2}+\frac {x^3 \sqrt {a+b x^2}}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^4/Sqrt[a + b*x^2],x]

[Out]

(-3*a*x*Sqrt[a + b*x^2])/(8*b^2) + (x^3*Sqrt[a + b*x^2])/(4*b) + (3*a^2*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/
(8*b^(5/2))

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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]

Rubi steps

\begin {align*} \int \frac {x^4}{\sqrt {a+b x^2}} \, dx &=\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {(3 a) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{4 b}\\ &=-\frac {3 a x \sqrt {a+b x^2}}{8 b^2}+\frac {x^3 \sqrt {a+b x^2}}{4 b}+\frac {\left (3 a^2\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{8 b^2}\\ &=-\frac {3 a x \sqrt {a+b x^2}}{8 b^2}+\frac {x^3 \sqrt {a+b x^2}}{4 b}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{8 b^2}\\ &=-\frac {3 a x \sqrt {a+b x^2}}{8 b^2}+\frac {x^3 \sqrt {a+b x^2}}{4 b}+\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 62, normalized size = 0.85 \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )+\sqrt {b} x \sqrt {a+b x^2} \left (2 b x^2-3 a\right )}{8 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/Sqrt[a + b*x^2],x]

[Out]

(Sqrt[b]*x*Sqrt[a + b*x^2]*(-3*a + 2*b*x^2) + 3*a^2*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*b^(5/2))

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fricas [A]  time = 0.73, size = 124, normalized size = 1.70 \[ \left [\frac {3 \, a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, b^{2} x^{3} - 3 \, a b x\right )} \sqrt {b x^{2} + a}}{16 \, b^{3}}, -\frac {3 \, a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, b^{2} x^{3} - 3 \, a b x\right )} \sqrt {b x^{2} + a}}{8 \, b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*a^2*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*b^2*x^3 - 3*a*b*x)*sqrt(b*x^2 + a)
)/b^3, -1/8*(3*a^2*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*b^2*x^3 - 3*a*b*x)*sqrt(b*x^2 + a))/b^3]

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giac [A]  time = 1.04, size = 54, normalized size = 0.74 \[ \frac {1}{8} \, \sqrt {b x^{2} + a} x {\left (\frac {2 \, x^{2}}{b} - \frac {3 \, a}{b^{2}}\right )} - \frac {3 \, a^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(b*x^2 + a)*x*(2*x^2/b - 3*a/b^2) - 3/8*a^2*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)

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maple [A]  time = 0.01, size = 59, normalized size = 0.81 \[ \frac {\sqrt {b \,x^{2}+a}\, x^{3}}{4 b}+\frac {3 a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}-\frac {3 \sqrt {b \,x^{2}+a}\, a x}{8 b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^3*(b*x^2+a)^(1/2)/b-3/8*a*x*(b*x^2+a)^(1/2)/b^2+3/8*a^2/b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.35, size = 51, normalized size = 0.70 \[ \frac {\sqrt {b x^{2} + a} x^{3}}{4 \, b} - \frac {3 \, \sqrt {b x^{2} + a} a x}{8 \, b^{2}} + \frac {3 \, a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sqrt(b*x^2 + a)*x^3/b - 3/8*sqrt(b*x^2 + a)*a*x/b^2 + 3/8*a^2*arcsinh(b*x/sqrt(a*b))/b^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{\sqrt {b\,x^2+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b*x^2)^(1/2),x)

[Out]

int(x^4/(a + b*x^2)^(1/2), x)

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sympy [A]  time = 4.03, size = 95, normalized size = 1.30 \[ - \frac {3 a^{\frac {3}{2}} x}{8 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {\sqrt {a} x^{3}}{8 b \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 b^{\frac {5}{2}}} + \frac {x^{5}}{4 \sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*a**(3/2)*x/(8*b**2*sqrt(1 + b*x**2/a)) - sqrt(a)*x**3/(8*b*sqrt(1 + b*x**2/a)) + 3*a**2*asinh(sqrt(b)*x/sqr
t(a))/(8*b**(5/2)) + x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a))

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