3.589 \(\int \frac{x^2 (A+B x^2)}{(a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=77 \[ \frac{x^3 (A b-a B)}{3 a b \left (a+b x^2\right )^{3/2}}-\frac{B x}{b^2 \sqrt{a+b x^2}}+\frac{B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{b^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.0303405, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {452, 288, 217, 206} \[ \frac{x^3 (A b-a B)}{3 a b \left (a+b x^2\right )^{3/2}}-\frac{B x}{b^2 \sqrt{a+b x^2}}+\frac{B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{b^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

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

Rule 452

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[((b*c - a*d)
*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*(m + 1)), x] + Dist[d/b, Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /;
FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && NeQ[m, -1]

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

Mathematica [A]  time = 0.155562, size = 96, normalized size = 1.25 \[ \frac{\sqrt{b} x \left (-3 a^2 B-4 a b B x^2+A b^2 x^2\right )+3 a^{3/2} B \left (a+b x^2\right ) \sqrt{\frac{b x^2}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{3 a b^{5/2} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(A + B*x^2))/(a + b*x^2)^(5/2),x]

[Out]

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

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Maple [A]  time = 0.008, size = 92, normalized size = 1.2 \begin{align*} -{\frac{{x}^{3}B}{3\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}-{\frac{Bx}{{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{B\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}-{\frac{Ax}{3\,b} \left ( b{x}^{2}+a \right ) ^{-{\frac{3}{2}}}}+{\frac{Ax}{3\,ab}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.63859, size = 531, normalized size = 6.9 \begin{align*} \left [\frac{3 \,{\left (B a b^{2} x^{4} + 2 \, B a^{2} b x^{2} + B 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 a^{2} b x +{\left (4 \, B a b^{2} - A b^{3}\right )} x^{3}\right )} \sqrt{b x^{2} + a}}{6 \,{\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}}, -\frac{3 \,{\left (B a b^{2} x^{4} + 2 \, B a^{2} b x^{2} + B a^{3}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (3 \, B a^{2} b x +{\left (4 \, B a b^{2} - A b^{3}\right )} x^{3}\right )} \sqrt{b x^{2} + a}}{3 \,{\left (a b^{5} x^{4} + 2 \, a^{2} b^{4} x^{2} + a^{3} b^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B]  time = 12.4323, size = 352, normalized size = 4.57 \begin{align*} \frac{A x^{3}}{3 a^{\frac{5}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{3}{2}} b x^{2} \sqrt{1 + \frac{b x^{2}}{a}}} + B \left (\frac{3 a^{\frac{39}{2}} b^{11} \sqrt{1 + \frac{b x^{2}}{a}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{2} \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 a^{\frac{37}{2}} b^{12} x^{2} \sqrt{1 + \frac{b x^{2}}{a}} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{3 a^{19} b^{\frac{23}{2}} x}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{4 a^{18} b^{\frac{25}{2}} x^{3}}{3 a^{\frac{39}{2}} b^{\frac{27}{2}} \sqrt{1 + \frac{b x^{2}}{a}} + 3 a^{\frac{37}{2}} b^{\frac{29}{2}} x^{2} \sqrt{1 + \frac{b x^{2}}{a}}}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*x**3/(3*a**(5/2)*sqrt(1 + b*x**2/a) + 3*a**(3/2)*b*x**2*sqrt(1 + b*x**2/a)) + B*(3*a**(39/2)*b**11*sqrt(1 +
b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt
(1 + b*x**2/a)) + 3*a**(37/2)*b**12*x**2*sqrt(1 + b*x**2/a)*asinh(sqrt(b)*x/sqrt(a))/(3*a**(39/2)*b**(27/2)*sq
rt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 3*a**19*b**(23/2)*x/(3*a**(39/2)*b**(27/2)
*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)) - 4*a**18*b**(25/2)*x**3/(3*a**(39/2)*b**
(27/2)*sqrt(1 + b*x**2/a) + 3*a**(37/2)*b**(29/2)*x**2*sqrt(1 + b*x**2/a)))

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Giac [A]  time = 1.12293, size = 93, normalized size = 1.21 \begin{align*} -\frac{x{\left (\frac{3 \, B a}{b^{2}} + \frac{{\left (4 \, B a b^{2} - A b^{3}\right )} x^{2}}{a b^{3}}\right )}}{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}}} - \frac{B \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{b^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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