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

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

[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))

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Rubi [A]  time = 0.0225984, antiderivative size = 68, normalized size of antiderivative = 1., 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 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]

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 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}{\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*}

Mathematica [A]  time = 0.0325632, 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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Maple [A]  time = 0.004, size = 57, normalized size = 0.8 \begin{align*}{\frac{{x}^{3}}{2\,b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{3\,ax}{2\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{3\,a}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.34425, size = 362, normalized size = 5.32 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 3.08214, size = 71, normalized size = 1.04 \begin{align*} \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}}} \end{align*}

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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Giac [A]  time = 1.68139, size = 69, normalized size = 1.01 \begin{align*} \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}}} \end{align*}

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)