3.506 \(\int \frac {x^6}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=91 \[ -\frac {5 a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}}+\frac {5 x \sqrt {a+b x^2}}{2 b^3}-\frac {5 x^3}{3 b^2 \sqrt {a+b x^2}}-\frac {x^5}{3 b \left (a+b x^2\right )^{3/2}} \]

[Out]

-1/3*x^5/b/(b*x^2+a)^(3/2)-5/2*a*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(7/2)-5/3*x^3/b^2/(b*x^2+a)^(1/2)+5/2*x*
(b*x^2+a)^(1/2)/b^3

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Rubi [A]  time = 0.03, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {288, 321, 217, 206} \[ -\frac {5 x^3}{3 b^2 \sqrt {a+b x^2}}+\frac {5 x \sqrt {a+b x^2}}{2 b^3}-\frac {5 a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}}-\frac {x^5}{3 b \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^6/(a + b*x^2)^(5/2),x]

[Out]

-x^5/(3*b*(a + b*x^2)^(3/2)) - (5*x^3)/(3*b^2*Sqrt[a + b*x^2]) + (5*x*Sqrt[a + b*x^2])/(2*b^3) - (5*a*ArcTanh[
(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(7/2))

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^6}{\left (a+b x^2\right )^{5/2}} \, dx &=-\frac {x^5}{3 b \left (a+b x^2\right )^{3/2}}+\frac {5 \int \frac {x^4}{\left (a+b x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac {x^5}{3 b \left (a+b x^2\right )^{3/2}}-\frac {5 x^3}{3 b^2 \sqrt {a+b x^2}}+\frac {5 \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{b^2}\\ &=-\frac {x^5}{3 b \left (a+b x^2\right )^{3/2}}-\frac {5 x^3}{3 b^2 \sqrt {a+b x^2}}+\frac {5 x \sqrt {a+b x^2}}{2 b^3}-\frac {(5 a) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{2 b^3}\\ &=-\frac {x^5}{3 b \left (a+b x^2\right )^{3/2}}-\frac {5 x^3}{3 b^2 \sqrt {a+b x^2}}+\frac {5 x \sqrt {a+b x^2}}{2 b^3}-\frac {(5 a) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{2 b^3}\\ &=-\frac {x^5}{3 b \left (a+b x^2\right )^{3/2}}-\frac {5 x^3}{3 b^2 \sqrt {a+b x^2}}+\frac {5 x \sqrt {a+b x^2}}{2 b^3}-\frac {5 a \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 90, normalized size = 0.99 \[ \frac {\sqrt {b} x \left (15 a^2+20 a b x^2+3 b^2 x^4\right )-15 a^{3/2} \left (a+b x^2\right ) \sqrt {\frac {b x^2}{a}+1} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{6 b^{7/2} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^6/(a + b*x^2)^(5/2),x]

[Out]

(Sqrt[b]*x*(15*a^2 + 20*a*b*x^2 + 3*b^2*x^4) - 15*a^(3/2)*(a + b*x^2)*Sqrt[1 + (b*x^2)/a]*ArcSinh[(Sqrt[b]*x)/
Sqrt[a]])/(6*b^(7/2)*(a + b*x^2)^(3/2))

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fricas [A]  time = 0.65, size = 227, normalized size = 2.49 \[ \left [\frac {15 \, {\left (a b^{2} x^{4} + 2 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (3 \, b^{3} x^{5} + 20 \, a b^{2} x^{3} + 15 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{12 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}, \frac {15 \, {\left (a b^{2} x^{4} + 2 \, a^{2} b x^{2} + a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, b^{3} x^{5} + 20 \, a b^{2} x^{3} + 15 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (b^{6} x^{4} + 2 \, a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(15*(a*b^2*x^4 + 2*a^2*b*x^2 + a^3)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(3*b^3*x
^5 + 20*a*b^2*x^3 + 15*a^2*b*x)*sqrt(b*x^2 + a))/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4), 1/6*(15*(a*b^2*x^4 + 2*a^2
*b*x^2 + a^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (3*b^3*x^5 + 20*a*b^2*x^3 + 15*a^2*b*x)*sqrt(b*x^2
+ a))/(b^6*x^4 + 2*a*b^5*x^2 + a^2*b^4)]

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giac [A]  time = 1.21, size = 65, normalized size = 0.71 \[ \frac {{\left (x^{2} {\left (\frac {3 \, x^{2}}{b} + \frac {20 \, a}{b^{2}}\right )} + \frac {15 \, a^{2}}{b^{3}}\right )} x}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {5 \, a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/6*(x^2*(3*x^2/b + 20*a/b^2) + 15*a^2/b^3)*x/(b*x^2 + a)^(3/2) + 5/2*a*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))
/b^(7/2)

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maple [A]  time = 0.01, size = 75, normalized size = 0.82 \[ \frac {x^{5}}{2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b}+\frac {5 a \,x^{3}}{6 \left (b \,x^{2}+a \right )^{\frac {3}{2}} b^{2}}+\frac {5 a x}{2 \sqrt {b \,x^{2}+a}\, b^{3}}-\frac {5 a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(b*x^2+a)^(5/2),x)

[Out]

1/2*x^5/b/(b*x^2+a)^(3/2)+5/6*a/b^2*x^3/(b*x^2+a)^(3/2)+5/2*a/b^3*x/(b*x^2+a)^(1/2)-5/2*a/b^(7/2)*ln(b^(1/2)*x
+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.41, size = 89, normalized size = 0.98 \[ \frac {x^{5}}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {5 \, a x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )}}{6 \, b} + \frac {5 \, a x}{6 \, \sqrt {b x^{2} + a} b^{3}} - \frac {5 \, a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^6/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

1/2*x^5/((b*x^2 + a)^(3/2)*b) + 5/6*a*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2))/b + 5/6*a*
x/(sqrt(b*x^2 + a)*b^3) - 5/2*a*arcsinh(b*x/sqrt(a*b))/b^(7/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^6}{{\left (b\,x^2+a\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6/(a + b*x^2)^(5/2),x)

[Out]

int(x^6/(a + b*x^2)^(5/2), x)

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sympy [B]  time = 5.12, size = 367, normalized size = 4.03 \[ - \frac {15 a^{\frac {81}{2}} b^{22} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{6 a^{\frac {79}{2}} b^{\frac {51}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 6 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {15 a^{\frac {79}{2}} b^{23} x^{2} \sqrt {1 + \frac {b x^{2}}{a}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{6 a^{\frac {79}{2}} b^{\frac {51}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 6 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {15 a^{40} b^{\frac {45}{2}} x}{6 a^{\frac {79}{2}} b^{\frac {51}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 6 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {20 a^{39} b^{\frac {47}{2}} x^{3}}{6 a^{\frac {79}{2}} b^{\frac {51}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 6 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} + \frac {3 a^{38} b^{\frac {49}{2}} x^{5}}{6 a^{\frac {79}{2}} b^{\frac {51}{2}} \sqrt {1 + \frac {b x^{2}}{a}} + 6 a^{\frac {77}{2}} b^{\frac {53}{2}} x^{2} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**6/(b*x**2+a)**(5/2),x)

[Out]

-15*a**(81/2)*b**22*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*
a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) - 15*a**(79/2)*b**23*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt
(a))/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) + 15*a**40*b**
(45/2)*x/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) + 20*a**39
*b**(47/2)*x**3/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a)) + 3
*a**38*b**(49/2)*x**5/(6*a**(79/2)*b**(51/2)*sqrt(1 + b*x**2/a) + 6*a**(77/2)*b**(53/2)*x**2*sqrt(1 + b*x**2/a
))

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