3.573 \(\int \frac{x^2 (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=83 \[ -\frac{x (A b-a B)}{b^2 \sqrt{a+b x^2}}+\frac{(2 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{5/2}}+\frac{B x \sqrt{a+b x^2}}{2 b^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0568336, antiderivative size = 83, 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 = {455, 388, 217, 206} \[ -\frac{x (A b-a B)}{b^2 \sqrt{a+b x^2}}+\frac{(2 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{5/2}}+\frac{B x \sqrt{a+b x^2}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^2*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

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

Rule 455

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[((-a)^(m/2 - 1)*(b*c - a*d)*
x*(a + b*x^2)^(p + 1))/(2*b^(m/2 + 1)*(p + 1)), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 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 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 )^{3/2}} \, dx &=-\frac{(A b-a B) x}{b^2 \sqrt{a+b x^2}}-\frac{\int \frac{-A b+a B-b B x^2}{\sqrt{a+b x^2}} \, dx}{b^2}\\ &=-\frac{(A b-a B) x}{b^2 \sqrt{a+b x^2}}+\frac{B x \sqrt{a+b x^2}}{2 b^2}+\frac{(2 A b-3 a B) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{2 b^2}\\ &=-\frac{(A b-a B) x}{b^2 \sqrt{a+b x^2}}+\frac{B x \sqrt{a+b x^2}}{2 b^2}+\frac{(2 A b-3 a B) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{2 b^2}\\ &=-\frac{(A b-a B) x}{b^2 \sqrt{a+b x^2}}+\frac{B x \sqrt{a+b x^2}}{2 b^2}+\frac{(2 A b-3 a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0984349, size = 86, normalized size = 1.04 \[ \frac{\sqrt{b} x \left (3 a B-2 A b+b B x^2\right )-\sqrt{a} \sqrt{\frac{b x^2}{a}+1} (3 a B-2 A b) \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{2 b^{5/2} \sqrt{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^2*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 97, normalized size = 1.2 \begin{align*}{\frac{{x}^{3}B}{2\,b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{\frac{3\,Bax}{2\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{2}+a}}}}-{\frac{3\,Ba}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}-{\frac{Ax}{b}{\frac{1}{\sqrt{b{x}^{2}+a}}}}+{A\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(B*x^2+A)/(b*x^2+a)^(3/2),x)

[Out]

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

________________________________________________________________________________________

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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.62837, size = 482, normalized size = 5.81 \begin{align*} \left [-\frac{{\left (3 \, B a^{2} - 2 \, A a b +{\left (3 \, B a b - 2 \, A b^{2}\right )} x^{2}\right )} \sqrt{b} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (B b^{2} x^{3} +{\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{4 \,{\left (b^{4} x^{2} + a b^{3}\right )}}, \frac{{\left (3 \, B a^{2} - 2 \, A a b +{\left (3 \, B a b - 2 \, A b^{2}\right )} x^{2}\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (B b^{2} x^{3} +{\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{2 \,{\left (b^{4} x^{2} + a 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)^(3/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 6.29665, size = 114, normalized size = 1.37 \begin{align*} A \left (\frac{\operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{b^{\frac{3}{2}}} - \frac{x}{\sqrt{a} b \sqrt{1 + \frac{b x^{2}}{a}}}\right ) + B \left (\frac{3 \sqrt{a} x}{2 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{3 a \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2 b^{\frac{5}{2}}} + \frac{x^{3}}{2 \sqrt{a} b \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)**(3/2),x)

[Out]

A*(asinh(sqrt(b)*x/sqrt(a))/b**(3/2) - x/(sqrt(a)*b*sqrt(1 + b*x**2/a))) + B*(3*sqrt(a)*x/(2*b**2*sqrt(1 + b*x
**2/a)) - 3*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(5/2)) + x**3/(2*sqrt(a)*b*sqrt(1 + b*x**2/a)))

________________________________________________________________________________________

Giac [A]  time = 1.12964, size = 95, normalized size = 1.14 \begin{align*} \frac{{\left (\frac{B x^{2}}{b} + \frac{3 \, B a b - 2 \, A b^{2}}{b^{3}}\right )} x}{2 \, \sqrt{b x^{2} + a}} + \frac{{\left (3 \, B a - 2 \, A b\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{2 \, 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)^(3/2),x, algorithm="giac")

[Out]

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