\(\int \frac {x^2 (A+B x^2)}{\sqrt {a+b x^2}} \, dx\) [558]

Optimal result

Integrand size = 22, antiderivative size = 89 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {(4 A b-3 a B) x \sqrt {a+b x^2}}{8 b^2}+\frac {B x^3 \sqrt {a+b x^2}}{4 b}-\frac {a (4 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {470, 327, 223, 212} \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=-\frac {a (4 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}}+\frac {x \sqrt {a+b x^2} (4 A b-3 a B)}{8 b^2}+\frac {B x^3 \sqrt {a+b x^2}}{4 b} \]

[In]

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

[Out]

((4*A*b - 3*a*B)*x*Sqrt[a + b*x^2])/(8*b^2) + (B*x^3*Sqrt[a + b*x^2])/(4*b) - (a*(4*A*b - 3*a*B)*ArcTanh[(Sqrt
[b]*x)/Sqrt[a + b*x^2]])/(8*b^(5/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 327

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 470

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*} \text {integral}& = \frac {B x^3 \sqrt {a+b x^2}}{4 b}-\frac {(-4 A b+3 a B) \int \frac {x^2}{\sqrt {a+b x^2}} \, dx}{4 b} \\ & = \frac {(4 A b-3 a B) x \sqrt {a+b x^2}}{8 b^2}+\frac {B x^3 \sqrt {a+b x^2}}{4 b}-\frac {(a (4 A b-3 a B)) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{8 b^2} \\ & = \frac {(4 A b-3 a B) x \sqrt {a+b x^2}}{8 b^2}+\frac {B x^3 \sqrt {a+b x^2}}{4 b}-\frac {(a (4 A b-3 a B)) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{8 b^2} \\ & = \frac {(4 A b-3 a B) x \sqrt {a+b x^2}}{8 b^2}+\frac {B x^3 \sqrt {a+b x^2}}{4 b}-\frac {a (4 A b-3 a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{5/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.94 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {x \sqrt {a+b x^2} \left (4 A b-3 a B+2 b B x^2\right )}{8 b^2}+\frac {a (-4 A b+3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{4 b^{5/2}} \]

[In]

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

[Out]

(x*Sqrt[a + b*x^2]*(4*A*b - 3*a*B + 2*b*B*x^2))/(8*b^2) + (a*(-4*A*b + 3*a*B)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] +
Sqrt[a + b*x^2])])/(4*b^(5/2))

Maple [A] (verified)

Time = 2.83 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.72

method	result	size
risch	\(\frac {x \left (2 b B \,x^{2}+4 A b -3 B a \right ) \sqrt {b \,x^{2}+a}}{8 b^{2}}-\frac {a \left (4 A b -3 B a \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{8 b^{\frac {5}{2}}}\)	\(64\)
pseudoelliptic	\(\frac {\left (-a b A +\frac {3}{4} a^{2} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\left (\left (\frac {x^{2} B}{2}+A \right ) b^{\frac {3}{2}}-\frac {3 B \sqrt {b}\, a}{4}\right ) x \sqrt {b \,x^{2}+a}}{2 b^{\frac {5}{2}}}\)	\(68\)
default	\(B \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+A \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\)	\(106\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.82 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\left [-\frac {{\left (3 \, B a^{2} - 4 \, A a b\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (2 \, B b^{2} x^{3} - {\left (3 \, B a b - 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, b^{3}}, -\frac {{\left (3 \, B a^{2} - 4 \, A a b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, B b^{2} x^{3} - {\left (3 \, B a b - 4 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, b^{3}}\right ] \]

[In]

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

[Out]

[-1/16*((3*B*a^2 - 4*A*a*b)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(2*B*b^2*x^3 - (3*B*a*
b - 4*A*b^2)*x)*sqrt(b*x^2 + a))/b^3, -1/8*((3*B*a^2 - 4*A*a*b)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) -
(2*B*b^2*x^3 - (3*B*a*b - 4*A*b^2)*x)*sqrt(b*x^2 + a))/b^3]

Sympy [A] (verification not implemented)

Time = 0.33 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.22 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} - \frac {a \left (A - \frac {3 B a}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{2}} \left (\frac {B x^{3}}{4 b} + \frac {x \left (A - \frac {3 B a}{4 b}\right )}{2 b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {A x^{3}}{3} + \frac {B x^{5}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-a*(A - 3*B*a/(4*b))*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(
x)/sqrt(b*x**2), True))/(2*b) + sqrt(a + b*x**2)*(B*x**3/(4*b) + x*(A - 3*B*a/(4*b))/(2*b)), Ne(b, 0)), ((A*x*
*3/3 + B*x**5/5)/sqrt(a), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.97 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} B x^{3}}{4 \, b} - \frac {3 \, \sqrt {b x^{2} + a} B a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A x}{2 \, b} + \frac {3 \, B a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {A a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.84 \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{8} \, \sqrt {b x^{2} + a} {\left (\frac {2 \, B x^{2}}{b} - \frac {3 \, B a b - 4 \, A b^{2}}{b^{3}}\right )} x - \frac {{\left (3 \, B a^{2} - 4 \, A a b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {5}{2}}} \]

[In]

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

[Out]

1/8*sqrt(b*x^2 + a)*(2*B*x^2/b - (3*B*a*b - 4*A*b^2)/b^3)*x - 1/8*(3*B*a^2 - 4*A*a*b)*log(abs(-sqrt(b)*x + sqr
t(b*x^2 + a)))/b^(5/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (A+B x^2\right )}{\sqrt {a+b x^2}} \, dx=\int \frac {x^2\,\left (B\,x^2+A\right )}{\sqrt {b\,x^2+a}} \,d x \]

[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)