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

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

[Out]

-3/4*a*arctanh(x^2*b^(1/2)/(b*x^4+a)^(1/2))/b^(5/2)-1/2*x^6/b/(b*x^4+a)^(1/2)+3/4*x^2*(b*x^4+a)^(1/2)/b^2

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Rubi [A]  time = 0.05, antiderivative size = 74, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.60, size = 167, normalized size = 2.26 \[ \left [\frac {3 \, {\left (a b x^{4} + a^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{4} + 2 \, \sqrt {b x^{4} + a} \sqrt {b} x^{2} - a\right ) + 2 \, {\left (b^{2} x^{6} + 3 \, a b x^{2}\right )} \sqrt {b x^{4} + a}}{8 \, {\left (b^{4} x^{4} + a b^{3}\right )}}, \frac {3 \, {\left (a b x^{4} + a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x^{2}}{\sqrt {b x^{4} + a}}\right ) + {\left (b^{2} x^{6} + 3 \, a b x^{2}\right )} \sqrt {b x^{4} + a}}{4 \, {\left (b^{4} x^{4} + a b^{3}\right )}}\right ] \]

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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giac [A]  time = 0.19, size = 55, normalized size = 0.74 \[ \frac {{\left (\frac {x^{4}}{b} + \frac {3 \, a}{b^{2}}\right )} x^{2}}{4 \, \sqrt {b x^{4} + a}} + \frac {3 \, a \log \left ({\left | -\sqrt {b} x^{2} + \sqrt {b x^{4} + a} \right |}\right )}{4 \, b^{\frac {5}{2}}} \]

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)

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maple [A]  time = 0.01, size = 61, normalized size = 0.82 \[ \frac {x^{6}}{4 \sqrt {b \,x^{4}+a}\, b}+\frac {3 a \,x^{2}}{4 \sqrt {b \,x^{4}+a}\, b^{2}}-\frac {3 a \ln \left (\sqrt {b}\, x^{2}+\sqrt {b \,x^{4}+a}\right )}{4 b^{\frac {5}{2}}} \]

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

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maxima [A]  time = 3.01, size = 103, normalized size = 1.39 \[ \frac {2 \, a b - \frac {3 \, {\left (b x^{4} + a\right )} a}{x^{4}}}{4 \, {\left (\frac {\sqrt {b x^{4} + a} b^{3}}{x^{2}} - \frac {{\left (b x^{4} + a\right )}^{\frac {3}{2}} b^{2}}{x^{6}}\right )}} + \frac {3 \, a \log \left (-\frac {\sqrt {b} - \frac {\sqrt {b x^{4} + a}}{x^{2}}}{\sqrt {b} + \frac {\sqrt {b x^{4} + a}}{x^{2}}}\right )}{8 \, b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*b - 3*(b*x^4 + a)*a/x^4)/(sqrt(b*x^4 + a)*b^3/x^2 - (b*x^4 + a)^(3/2)*b^2/x^6) + 3/8*a*log(-(sqrt(b)
- sqrt(b*x^4 + a)/x^2)/(sqrt(b) + sqrt(b*x^4 + a)/x^2))/b^(5/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^9}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^9/(a + b*x^4)^(3/2),x)

[Out]

int(x^9/(a + b*x^4)^(3/2), x)

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sympy [A]  time = 4.27, size = 75, normalized size = 1.01 \[ \frac {3 \sqrt {a} x^{2}}{4 b^{2} \sqrt {1 + \frac {b x^{4}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x^{2}}{\sqrt {a}} \right )}}{4 b^{\frac {5}{2}}} + \frac {x^{6}}{4 \sqrt {a} b \sqrt {1 + \frac {b x^{4}}{a}}} \]

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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