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

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

[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))

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Rubi [A]  time = 0.0291006, antiderivative size = 91, normalized size of antiderivative = 1., 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 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^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*}

Mathematica [A]  time = 0.129976, 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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Maple [A]  time = 0.007, size = 75, normalized size = 0.8 \begin{align*}{\frac{{x}^{5}}{2\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,a{x}^{3}}{6\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{5\,ax}{2\,{b}^{3}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{5\,a}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{7}{2}}}} \end{align*}

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/b^2*a*x^3/(b*x^2+a)^(3/2)+5/2/b^3*a*x/(b*x^2+a)^(1/2)-5/2/b^(7/2)*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^6/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.39588, size = 506, normalized size = 5.56 \begin{align*} \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 ] \end{align*}

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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Sympy [B]  time = 4.71714, size = 367, normalized size = 4.03 \begin{align*} - \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}}} \end{align*}

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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Giac [A]  time = 2.70859, size = 88, normalized size = 0.97 \begin{align*} \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}}} \end{align*}

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)