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

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

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 122, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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])

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 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 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]

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 = 0.22, size = 108, normalized size = 0.89 \begin {gather*} \frac {\sqrt {a+b x^2} \left (\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}}+\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 )\right )}{48 b^{5/2}} \end {gather*}

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))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.13, size = 103, normalized size = 0.84 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-3 a^2 B x+6 a A b x+2 a b B x^3+12 A b^2 x^3+8 b^2 B x^5\right )}{48 b^2}+\frac {\left (2 a^2 A b-a^3 B\right ) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{16 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(6*a*A*b*x - 3*a^2*B*x + 12*A*b^2*x^3 + 2*a*b*B*x^3 + 8*b^2*B*x^5))/(48*b^2) + ((2*a^2*A*b -
a^3*B)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*b^(5/2))

________________________________________________________________________________________

fricas [A]  time = 0.81, size = 206, normalized size = 1.69 \begin {gather*} \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 {gather*}

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]

________________________________________________________________________________________

giac [A]  time = 0.40, size = 100, normalized size = 0.82 \begin {gather*} \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 {gather*}

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)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 139, normalized size = 1.14 \begin {gather*} \frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,x^{3}}{6 b}-\frac {A \,a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {3}{2}}}+\frac {B \,a^{3} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}-\frac {\sqrt {b \,x^{2}+a}\, A a x}{8 b}+\frac {\sqrt {b \,x^{2}+a}\, B \,a^{2} x}{16 b^{2}}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} A x}{4 b}-\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} B a x}{8 b^{2}} \end {gather*}

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*a/b^2*x*(b*x^2+a)^(3/2)+1/16*B*a^2/b^2*x*(b*x^2+a)^(1/2)+1/16*B*a^3/b^(5/2)*
ln(b^(1/2)*x+(b*x^2+a)^(1/2))+1/4*A*x*(b*x^2+a)^(3/2)/b-1/8*A*a/b*x*(b*x^2+a)^(1/2)-1/8*A*a^2/b^(3/2)*ln(b^(1/
2)*x+(b*x^2+a)^(1/2))

________________________________________________________________________________________

maxima [A]  time = 1.01, size = 124, normalized size = 1.02 \begin {gather*} \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{3}}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x}{4 \, b} - \frac {\sqrt {b x^{2} + a} A a x}{8 \, b} + \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {3}{2}}} \end {gather*}

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]

1/6*(b*x^2 + a)^(3/2)*B*x^3/b - 1/8*(b*x^2 + a)^(3/2)*B*a*x/b^2 + 1/16*sqrt(b*x^2 + a)*B*a^2*x/b^2 + 1/4*(b*x^
2 + a)^(3/2)*A*x/b - 1/8*sqrt(b*x^2 + a)*A*a*x/b + 1/16*B*a^3*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 1/8*A*a^2*arcsi
nh(b*x/sqrt(a*b))/b^(3/2)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^2\,\left (B\,x^2+A\right )\,\sqrt {b\,x^2+a} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [B]  time = 12.09, size = 226, normalized size = 1.85 \begin {gather*} \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 {gather*}

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))

________________________________________________________________________________________