3.497 \(\int \frac {x^4}{(a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=68 \[ -\frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}}+\frac {3 x \sqrt {a+b x^2}}{2 b^2}-\frac {x^3}{b \sqrt {a+b x^2}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {288, 321, 217, 206} \[ \frac {3 x \sqrt {a+b x^2}}{2 b^2}-\frac {3 a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}}-\frac {x^3}{b \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[x^4/(a + b*x^2)^(3/2),x]

[Out]

-(x^3/(b*Sqrt[a + b*x^2])) + (3*x*Sqrt[a + b*x^2])/(2*b^2) - (3*a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, 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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 71, normalized size = 1.04 \[ \frac {\sqrt {b} x \left (3 a+b x^2\right )-3 a^{3/2} \sqrt {\frac {b x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{2 b^{5/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^4/(a + b*x^2)^(3/2),x]

[Out]

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

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fricas [A]  time = 1.00, size = 159, normalized size = 2.34 \[ \left [\frac {3 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (b^{2} x^{3} + 3 \, a b x\right )} \sqrt {b x^{2} + a}}{4 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, \frac {3 \, {\left (a b x^{2} + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (b^{2} x^{3} + 3 \, a b x\right )} \sqrt {b x^{2} + a}}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.15, size = 51, normalized size = 0.75 \[ \frac {x {\left (\frac {x^{2}}{b} + \frac {3 \, a}{b^{2}}\right )}}{2 \, \sqrt {b x^{2} + a}} + \frac {3 \, a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 57, normalized size = 0.84 \[ \frac {x^{3}}{2 \sqrt {b \,x^{2}+a}\, b}+\frac {3 a x}{2 \sqrt {b \,x^{2}+a}\, b^{2}}-\frac {3 a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.32, size = 49, normalized size = 0.72 \[ \frac {x^{3}}{2 \, \sqrt {b x^{2} + a} b} + \frac {3 \, a x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {3 \, a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^4}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4/(a + b*x^2)^(3/2),x)

[Out]

int(x^4/(a + b*x^2)^(3/2), x)

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sympy [A]  time = 3.26, size = 71, normalized size = 1.04 \[ \frac {3 \sqrt {a} x}{2 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3*sqrt(a)*x/(2*b**2*sqrt(1 + b*x**2/a)) - 3*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(5/2)) + x**3/(2*sqrt(a)*b*sqrt(1
+ b*x**2/a))

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