3.856 \(\int \frac{x^9}{(a+b x^4)^{3/2}} \, dx\)

Optimal. Leaf size=74 \[ \frac{3 x^2 \sqrt{a+b x^4}}{4 b^2}-\frac{3 a \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{5/2}}-\frac{x^6}{2 b \sqrt{a+b x^4}} \]

[Out]

-x^6/(2*b*Sqrt[a + b*x^4]) + (3*x^2*Sqrt[a + b*x^4])/(4*b^2) - (3*a*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4
*b^(5/2))

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Rubi [A]  time = 0.045765, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {275, 288, 321, 217, 206} \[ \frac{3 x^2 \sqrt{a+b x^4}}{4 b^2}-\frac{3 a \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{5/2}}-\frac{x^6}{2 b \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^9/(a + b*x^4)^(3/2),x]

[Out]

-x^6/(2*b*Sqrt[a + b*x^4]) + (3*x^2*Sqrt[a + b*x^4])/(4*b^2) - (3*a*ArcTanh[(Sqrt[b]*x^2)/Sqrt[a + b*x^4]])/(4
*b^(5/2))

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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^9}{\left (a+b x^4\right )^{3/2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^4}{\left (a+b x^2\right )^{3/2}} \, dx,x,x^2\right )\\ &=-\frac{x^6}{2 b \sqrt{a+b x^4}}+\frac{3 \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+b x^2}} \, dx,x,x^2\right )}{2 b}\\ &=-\frac{x^6}{2 b \sqrt{a+b x^4}}+\frac{3 x^2 \sqrt{a+b x^4}}{4 b^2}-\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,x^2\right )}{4 b^2}\\ &=-\frac{x^6}{2 b \sqrt{a+b x^4}}+\frac{3 x^2 \sqrt{a+b x^4}}{4 b^2}-\frac{(3 a) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x^2}{\sqrt{a+b x^4}}\right )}{4 b^2}\\ &=-\frac{x^6}{2 b \sqrt{a+b x^4}}+\frac{3 x^2 \sqrt{a+b x^4}}{4 b^2}-\frac{3 a \tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{4 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0349259, size = 75, normalized size = 1.01 \[ \frac{\sqrt{b} x^2 \left (3 a+b x^4\right )-3 a^{3/2} \sqrt{\frac{b x^4}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )}{4 b^{5/2} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^9/(a + b*x^4)^(3/2),x]

[Out]

(Sqrt[b]*x^2*(3*a + b*x^4) - 3*a^(3/2)*Sqrt[1 + (b*x^4)/a]*ArcSinh[(Sqrt[b]*x^2)/Sqrt[a]])/(4*b^(5/2)*Sqrt[a +
b*x^4])

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Maple [A]  time = 0.012, size = 61, normalized size = 0.8 \begin{align*}{\frac{{x}^{6}}{4\,b}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{3\,a{x}^{2}}{4\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{3\,a}{4}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ){b}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(b*x^4+a)^(3/2),x)

[Out]

1/4*x^6/b/(b*x^4+a)^(1/2)+3/4/b^2*a*x^2/(b*x^4+a)^(1/2)-3/4/b^(5/2)*a*ln(x^2*b^(1/2)+(b*x^4+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^9/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.6116, size = 373, normalized size = 5.04 \begin{align*} \left [\frac{3 \,{\left (a b x^{4} + a^{2}\right )} \sqrt{b} \log \left (-2 \, b x^{4} + 2 \, \sqrt{b x^{4} + a} \sqrt{b} x^{2} - a\right ) + 2 \,{\left (b^{2} x^{6} + 3 \, a b x^{2}\right )} \sqrt{b x^{4} + a}}{8 \,{\left (b^{4} x^{4} + a b^{3}\right )}}, \frac{3 \,{\left (a b x^{4} + a^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x^{2}}{\sqrt{b x^{4} + a}}\right ) +{\left (b^{2} x^{6} + 3 \, a b x^{2}\right )} \sqrt{b x^{4} + a}}{4 \,{\left (b^{4} x^{4} + a b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*(a*b*x^4 + a^2)*sqrt(b)*log(-2*b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(b)*x^2 - a) + 2*(b^2*x^6 + 3*a*b*x^2)*sq
rt(b*x^4 + a))/(b^4*x^4 + a*b^3), 1/4*(3*(a*b*x^4 + a^2)*sqrt(-b)*arctan(sqrt(-b)*x^2/sqrt(b*x^4 + a)) + (b^2*
x^6 + 3*a*b*x^2)*sqrt(b*x^4 + a))/(b^4*x^4 + a*b^3)]

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Sympy [A]  time = 4.32779, size = 75, normalized size = 1.01 \begin{align*} \frac{3 \sqrt{a} x^{2}}{4 b^{2} \sqrt{1 + \frac{b x^{4}}{a}}} - \frac{3 a \operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{4 b^{\frac{5}{2}}} + \frac{x^{6}}{4 \sqrt{a} b \sqrt{1 + \frac{b x^{4}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**9/(b*x**4+a)**(3/2),x)

[Out]

3*sqrt(a)*x**2/(4*b**2*sqrt(1 + b*x**4/a)) - 3*a*asinh(sqrt(b)*x**2/sqrt(a))/(4*b**(5/2)) + x**6/(4*sqrt(a)*b*
sqrt(1 + b*x**4/a))

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Giac [A]  time = 1.13446, size = 74, normalized size = 1. \begin{align*} \frac{{\left (\frac{x^{4}}{b} + \frac{3 \, a}{b^{2}}\right )} x^{2}}{4 \, \sqrt{b x^{4} + a}} + \frac{3 \, a \log \left ({\left | -\sqrt{b} x^{2} + \sqrt{b x^{4} + a} \right |}\right )}{4 \, b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^9/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

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