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

Optimal. Leaf size=122 \[ -\frac{a^2 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{5/2}}+\frac{a x \sqrt{a+b x^2} (2 A b-a B)}{16 b^2}+\frac{x^3 \sqrt{a+b x^2} (2 A b-a B)}{8 b}+\frac{B x^3 \left (a+b x^2\right )^{3/2}}{6 b} \]

[Out]

(a*(2*A*b - a*B)*x*Sqrt[a + b*x^2])/(16*b^2) + ((2*A*b - a*B)*x^3*Sqrt[a + b*x^2])/(8*b) + (B*x^3*(a + b*x^2)^
(3/2))/(6*b) - (a^2*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(5/2))

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Rubi [A]  time = 0.0591792, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {459, 279, 321, 217, 206} \[ -\frac{a^2 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{5/2}}+\frac{a x \sqrt{a+b x^2} (2 A b-a B)}{16 b^2}+\frac{x^3 \sqrt{a+b x^2} (2 A b-a B)}{8 b}+\frac{B x^3 \left (a+b x^2\right )^{3/2}}{6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(2*A*b - a*B)*x*Sqrt[a + b*x^2])/(16*b^2) + ((2*A*b - a*B)*x^3*Sqrt[a + b*x^2])/(8*b) + (B*x^3*(a + b*x^2)^
(3/2))/(6*b) - (a^2*(2*A*b - a*B)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(16*b^(5/2))

Rule 459

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

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 \sqrt{a+b x^2} \left (A+B x^2\right ) \, dx &=\frac{B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac{(-6 A b+3 a B) \int x^2 \sqrt{a+b x^2} \, dx}{6 b}\\ &=\frac{(2 A b-a B) x^3 \sqrt{a+b x^2}}{8 b}+\frac{B x^3 \left (a+b x^2\right )^{3/2}}{6 b}+\frac{(a (2 A b-a B)) \int \frac{x^2}{\sqrt{a+b x^2}} \, dx}{8 b}\\ &=\frac{a (2 A b-a B) x \sqrt{a+b x^2}}{16 b^2}+\frac{(2 A b-a B) x^3 \sqrt{a+b x^2}}{8 b}+\frac{B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac{\left (a^2 (2 A b-a B)\right ) \int \frac{1}{\sqrt{a+b x^2}} \, dx}{16 b^2}\\ &=\frac{a (2 A b-a B) x \sqrt{a+b x^2}}{16 b^2}+\frac{(2 A b-a B) x^3 \sqrt{a+b x^2}}{8 b}+\frac{B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac{\left (a^2 (2 A b-a B)\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )}{16 b^2}\\ &=\frac{a (2 A b-a B) x \sqrt{a+b x^2}}{16 b^2}+\frac{(2 A b-a B) x^3 \sqrt{a+b x^2}}{8 b}+\frac{B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac{a^2 (2 A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{16 b^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.210085, size = 108, normalized size = 0.89 \[ \frac{\sqrt{a+b x^2} \left (\sqrt{b} x \left (-3 a^2 B+2 a b \left (3 A+B x^2\right )+4 b^2 x^2 \left (3 A+2 B x^2\right )\right )+\frac{3 a^{3/2} (a B-2 A b) \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )}{\sqrt{\frac{b x^2}{a}+1}}\right )}{48 b^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 139, normalized size = 1.1 \begin{align*}{\frac{B{x}^{3}}{6\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{Bax}{8\,{b}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{{a}^{2}Bx}{16\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{B{a}^{3}}{16}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{5}{2}}}}+{\frac{Ax}{4\,b} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{aAx}{8\,b}\sqrt{b{x}^{2}+a}}-{\frac{A{a}^{2}}{8}\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)^(1/2),x)

[Out]

1/6*B*x^3*(b*x^2+a)^(3/2)/b-1/8*B/b^2*a*x*(b*x^2+a)^(3/2)+1/16*B/b^2*a^2*x*(b*x^2+a)^(1/2)+1/16*B/b^(5/2)*a^3*
ln(x*b^(1/2)+(b*x^2+a)^(1/2))+1/4*A*x*(b*x^2+a)^(3/2)/b-1/8*A/b*a*x*(b*x^2+a)^(1/2)-1/8*A/b^(3/2)*a^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^2+A)*(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.88832, size = 475, normalized size = 3.89 \begin{align*} \left [-\frac{3 \,{\left (B a^{3} - 2 \, A a^{2} b\right )} \sqrt{b} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (8 \, B b^{3} x^{5} + 2 \,{\left (B a b^{2} + 6 \, A b^{3}\right )} x^{3} - 3 \,{\left (B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{96 \, b^{3}}, -\frac{3 \,{\left (B a^{3} - 2 \, A a^{2} b\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (8 \, B b^{3} x^{5} + 2 \,{\left (B a b^{2} + 6 \, A b^{3}\right )} x^{3} - 3 \,{\left (B a^{2} b - 2 \, A a b^{2}\right )} x\right )} \sqrt{b x^{2} + a}}{48 \, b^{3}}\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)^(1/2),x, algorithm="fricas")

[Out]

[-1/96*(3*(B*a^3 - 2*A*a^2*b)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(8*B*b^3*x^5 + 2*(B*
a*b^2 + 6*A*b^3)*x^3 - 3*(B*a^2*b - 2*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^3, -1/48*(3*(B*a^3 - 2*A*a^2*b)*sqrt(-b)*
arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*B*b^3*x^5 + 2*(B*a*b^2 + 6*A*b^3)*x^3 - 3*(B*a^2*b - 2*A*a*b^2)*x)*sqr
t(b*x^2 + a))/b^3]

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Sympy [B]  time = 8.62363, size = 226, normalized size = 1.85 \begin{align*} \frac{A a^{\frac{3}{2}} x}{8 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{3 A \sqrt{a} x^{3}}{8 \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{A a^{2} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{8 b^{\frac{3}{2}}} + \frac{A b x^{5}}{4 \sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{B a^{\frac{5}{2}} x}{16 b^{2} \sqrt{1 + \frac{b x^{2}}{a}}} - \frac{B a^{\frac{3}{2}} x^{3}}{48 b \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{5 B \sqrt{a} x^{5}}{24 \sqrt{1 + \frac{b x^{2}}{a}}} + \frac{B a^{3} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{16 b^{\frac{5}{2}}} + \frac{B b 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)*(b*x**2+a)**(1/2),x)

[Out]

A*a**(3/2)*x/(8*b*sqrt(1 + b*x**2/a)) + 3*A*sqrt(a)*x**3/(8*sqrt(1 + b*x**2/a)) - A*a**2*asinh(sqrt(b)*x/sqrt(
a))/(8*b**(3/2)) + A*b*x**5/(4*sqrt(a)*sqrt(1 + b*x**2/a)) - B*a**(5/2)*x/(16*b**2*sqrt(1 + b*x**2/a)) - B*a**
(3/2)*x**3/(48*b*sqrt(1 + b*x**2/a)) + 5*B*sqrt(a)*x**5/(24*sqrt(1 + b*x**2/a)) + B*a**3*asinh(sqrt(b)*x/sqrt(
a))/(16*b**(5/2)) + B*b*x**7/(6*sqrt(a)*sqrt(1 + b*x**2/a))

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Giac [A]  time = 1.13321, size = 135, normalized size = 1.11 \begin{align*} \frac{1}{48} \,{\left (2 \,{\left (4 \, B x^{2} + \frac{B a b^{3} + 6 \, A b^{4}}{b^{4}}\right )} x^{2} - \frac{3 \,{\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )}}{b^{4}}\right )} \sqrt{b x^{2} + a} x - \frac{{\left (B a^{3} - 2 \, A a^{2} b\right )} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right )}{16 \, 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)^(1/2),x, algorithm="giac")

[Out]

1/48*(2*(4*B*x^2 + (B*a*b^3 + 6*A*b^4)/b^4)*x^2 - 3*(B*a^2*b^2 - 2*A*a*b^3)/b^4)*sqrt(b*x^2 + a)*x - 1/16*(B*a
^3 - 2*A*a^2*b)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)