3.23 \(\int \frac{x^2 (A+B x)}{\sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=81 \[ -\frac{\sqrt{a+b x^2} (4 a B-3 A b x)}{6 b^2}-\frac{a A \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}}+\frac{B x^2 \sqrt{a+b x^2}}{3 b} \]

[Out]

(B*x^2*Sqrt[a + b*x^2])/(3*b) - ((4*a*B - 3*A*b*x)*Sqrt[a + b*x^2])/(6*b^2) - (a*A*ArcTanh[(Sqrt[b]*x)/Sqrt[a
+ b*x^2]])/(2*b^(3/2))

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Rubi [A]  time = 0.0421102, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {833, 780, 217, 206} \[ -\frac{\sqrt{a+b x^2} (4 a B-3 A b x)}{6 b^2}-\frac{a A \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}}+\frac{B x^2 \sqrt{a+b x^2}}{3 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x^2*Sqrt[a + b*x^2])/(3*b) - ((4*a*B - 3*A*b*x)*Sqrt[a + b*x^2])/(6*b^2) - (a*A*ArcTanh[(Sqrt[b]*x)/Sqrt[a
+ b*x^2]])/(2*b^(3/2))

Rule 833

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(g*(d + e*x)
^m*(a + c*x^2)^(p + 1))/(c*(m + 2*p + 2)), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m - 1)*(a + c*x^2)^p*
Simp[c*d*f*(m + 2*p + 2) - a*e*g*m + c*(e*f*(m + 2*p + 2) + d*g*m)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, p
}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p]) &&  !(IGtQ[m, 0] && EqQ[f, 0])

Rule 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

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

Mathematica [A]  time = 0.035785, size = 64, normalized size = 0.79 \[ \frac{\sqrt{a+b x^2} (b x (3 A+2 B x)-4 a B)-3 a A \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{6 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-4*a*B + b*x*(3*A + 2*B*x)) - 3*a*A*Sqrt[b]*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(6*b^2)

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Maple [A]  time = 0.005, size = 75, normalized size = 0.9 \begin{align*}{\frac{B{x}^{2}}{3\,b}\sqrt{b{x}^{2}+a}}-{\frac{2\,Ba}{3\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{Ax}{2\,b}\sqrt{b{x}^{2}+a}}-{\frac{Aa}{2}\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+A)/(b*x^2+a)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57618, size = 320, normalized size = 3.95 \begin{align*} \left [\frac{3 \, A a \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (2 \, B b x^{2} + 3 \, A b x - 4 \, B a\right )} \sqrt{b x^{2} + a}}{12 \, b^{2}}, \frac{3 \, A a \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (2 \, B b x^{2} + 3 \, A b x - 4 \, B a\right )} \sqrt{b x^{2} + a}}{6 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.52792, size = 94, normalized size = 1.16 \begin{align*} \frac{A \sqrt{a} x \sqrt{1 + \frac{b x^{2}}{a}}}{2 b} - \frac{A a \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2 b^{\frac{3}{2}}} + B \left (\begin{cases} - \frac{2 a \sqrt{a + b x^{2}}}{3 b^{2}} + \frac{x^{2} \sqrt{a + b x^{2}}}{3 b} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 \sqrt{a}} & \text{otherwise} \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(B*x+A)/(b*x**2+a)**(1/2),x)

[Out]

A*sqrt(a)*x*sqrt(1 + b*x**2/a)/(2*b) - A*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(3/2)) + B*Piecewise((-2*a*sqrt(a +
b*x**2)/(3*b**2) + x**2*sqrt(a + b*x**2)/(3*b), Ne(b, 0)), (x**4/(4*sqrt(a)), True))

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Giac [A]  time = 1.21206, size = 82, normalized size = 1.01 \begin{align*} \frac{1}{6} \, \sqrt{b x^{2} + a}{\left ({\left (\frac{2 \, B x}{b} + \frac{3 \, A}{b}\right )} x - \frac{4 \, B a}{b^{2}}\right )} + \frac{A a \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{2 \, b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(B*x+A)/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

1/6*sqrt(b*x^2 + a)*((2*B*x/b + 3*A/b)*x - 4*B*a/b^2) + 1/2*A*a*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2)