3.379 \(\int x^2 (a+b x^2)^{3/2} \, dx\)

Optimal. Leaf size=91 \[ -\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{3/2}}+\frac{a^2 x \sqrt{a+b x^2}}{16 b}+\frac{1}{8} a x^3 \sqrt{a+b x^2}+\frac{1}{6} x^3 \left (a+b x^2\right )^{3/2} \]

[Out]

(a^2*x*Sqrt[a + b*x^2])/(16*b) + (a*x^3*Sqrt[a + b*x^2])/8 + (x^3*(a + b*x^2)^(3/2))/6 - (a^3*ArcTanh[(Sqrt[b]
*x)/Sqrt[a + b*x^2]])/(16*b^(3/2))

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Rubi [A]  time = 0.0306598, 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 = {279, 321, 217, 206} \[ -\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{3/2}}+\frac{a^2 x \sqrt{a+b x^2}}{16 b}+\frac{1}{8} a x^3 \sqrt{a+b x^2}+\frac{1}{6} x^3 \left (a+b x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*x^2)^(3/2),x]

[Out]

(a^2*x*Sqrt[a + b*x^2])/(16*b) + (a*x^3*Sqrt[a + b*x^2])/8 + (x^3*(a + b*x^2)^(3/2))/6 - (a^3*ArcTanh[(Sqrt[b]
*x)/Sqrt[a + b*x^2]])/(16*b^(3/2))

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 x^2 \left (a+b x^2\right )^{3/2} \, dx &=\frac{1}{6} x^3 \left (a+b x^2\right )^{3/2}+\frac{1}{2} a \int x^2 \sqrt{a+b x^2} \, dx\\ &=\frac{1}{8} a x^3 \sqrt{a+b x^2}+\frac{1}{6} x^3 \left (a+b x^2\right )^{3/2}+\frac{1}{8} a^2 \int \frac{x^2}{\sqrt{a+b x^2}} \, dx\\ &=\frac{a^2 x \sqrt{a+b x^2}}{16 b}+\frac{1}{8} a x^3 \sqrt{a+b x^2}+\frac{1}{6} x^3 \left (a+b x^2\right )^{3/2}-\frac{a^3 \int \frac{1}{\sqrt{a+b x^2}} \, dx}{16 b}\\ &=\frac{a^2 x \sqrt{a+b x^2}}{16 b}+\frac{1}{8} a x^3 \sqrt{a+b x^2}+\frac{1}{6} x^3 \left (a+b x^2\right )^{3/2}-\frac{a^3 \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{16 b}\\ &=\frac{a^2 x \sqrt{a+b x^2}}{16 b}+\frac{1}{8} a x^3 \sqrt{a+b x^2}+\frac{1}{6} x^3 \left (a+b x^2\right )^{3/2}-\frac{a^3 \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.115548, size = 83, normalized size = 0.91 \[ \frac{\sqrt{a+b x^2} \left (\sqrt{b} x \left (3 a^2+14 a b x^2+8 b^2 x^4\right )-\frac{3 a^{5/2} \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{\frac{b x^2}{a}+1}}\right )}{48 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*x^2)^(3/2),x]

[Out]

(Sqrt[a + b*x^2]*(Sqrt[b]*x*(3*a^2 + 14*a*b*x^2 + 8*b^2*x^4) - (3*a^(5/2)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/Sqrt[1
+ (b*x^2)/a]))/(48*b^(3/2))

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Maple [A]  time = 0.006, size = 75, normalized size = 0.8 \begin{align*}{\frac{x}{6\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{5}{2}}}}-{\frac{ax}{24\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{{a}^{2}x}{16\,b}\sqrt{b{x}^{2}+a}}-{\frac{{a}^{3}}{16}\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)^(3/2),x)

[Out]

1/6*x*(b*x^2+a)^(5/2)/b-1/24/b*a*x*(b*x^2+a)^(3/2)-1/16*a^2*x*(b*x^2+a)^(1/2)/b-1/16/b^(3/2)*a^3*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^2+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61282, size = 346, normalized size = 3.8 \begin{align*} \left [\frac{3 \, a^{3} \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + 2 \,{\left (8 \, b^{3} x^{5} + 14 \, a b^{2} x^{3} + 3 \, a^{2} b x\right )} \sqrt{b x^{2} + a}}{96 \, b^{2}}, \frac{3 \, a^{3} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left (8 \, b^{3} x^{5} + 14 \, a b^{2} x^{3} + 3 \, a^{2} b x\right )} \sqrt{b x^{2} + a}}{48 \, b^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

[1/96*(3*a^3*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8*b^3*x^5 + 14*a*b^2*x^3 + 3*a^2*b*x
)*sqrt(b*x^2 + a))/b^2, 1/48*(3*a^3*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (8*b^3*x^5 + 14*a*b^2*x^3 +
3*a^2*b*x)*sqrt(b*x^2 + a))/b^2]

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Sympy [A]  time = 4.97693, size = 119, normalized size = 1.31 \begin{align*} \frac{a^{\frac{5}{2}} x}{16 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{17 a^{\frac{3}{2}} x^{3}}{48 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{11 \sqrt{a} b x^{5}}{24 \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{16 b^{\frac{3}{2}}} + \frac{b^{2} x^{7}}{6 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(b*x**2+a)**(3/2),x)

[Out]

a**(5/2)*x/(16*b*sqrt(1 + b*x**2/a)) + 17*a**(3/2)*x**3/(48*sqrt(1 + b*x**2/a)) + 11*sqrt(a)*b*x**5/(24*sqrt(1
+ b*x**2/a)) - a**3*asinh(sqrt(b)*x/sqrt(a))/(16*b**(3/2)) + b**2*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 2.66679, size = 85, normalized size = 0.93 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (4 \, b x^{2} + 7 \, a\right )} x^{2} + \frac{3 \, a^{2}}{b}\right )} \sqrt{b x^{2} + a} x + \frac{a^{3} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{16 \, b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

1/48*(2*(4*b*x^2 + 7*a)*x^2 + 3*a^2/b)*sqrt(b*x^2 + a)*x + 1/16*a^3*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(
3/2)