\(\int \frac {x^2 (A+B x^2)}{(a+b x^2)^{3/2}} \, dx\) [573]

Optimal result

Integrand size = 22, antiderivative size = 83 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {(A b-a B) x}{b^2 \sqrt {a+b x^2}}+\frac {B x \sqrt {a+b x^2}}{2 b^2}+\frac {(2 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 83, 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 = {466, 396, 223, 212} \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {(2 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}}-\frac {x (A b-a B)}{b^2 \sqrt {a+b x^2}}+\frac {B x \sqrt {a+b x^2}}{2 b^2} \]

[In]

Int[(x^2*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

-(((A*b - a*B)*x)/(b^2*Sqrt[a + b*x^2])) + (B*x*Sqrt[a + b*x^2])/(2*b^2) + ((2*A*b - 3*a*B)*ArcTanh[(Sqrt[b]*x
)/Sqrt[a + b*x^2]])/(2*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 396

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

Rule 466

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-a)^(m/2 - 1)*(b*c - a*d)*x
*((a + b*x^2)^(p + 1)/(2*b^(m/2 + 1)*(p + 1))), x] + Dist[1/(2*b^(m/2 + 1)*(p + 1)), Int[(a + b*x^2)^(p + 1)*E
xpandToSum[2*b*(p + 1)*x^2*Together[(b^(m/2)*x^(m - 2)*(c + d*x^2) - (-a)^(m/2 - 1)*(b*c - a*d))/(a + b*x^2)]
- (-a)^(m/2 - 1)*(b*c - a*d), x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && IGtQ[
m/2, 0] && (IntegerQ[p] || EqQ[m + 2*p + 1, 0])

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

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-2 A b x+3 a B x+b B x^3}{2 b^2 \sqrt {a+b x^2}}+\frac {(2 A b-3 a B) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{b^{5/2}} \]

[In]

Integrate[(x^2*(A + B*x^2))/(a + b*x^2)^(3/2),x]

[Out]

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

Maple [A] (verified)

Time = 2.96 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.88

method	result	size
pseudoelliptic	\(\frac {\sqrt {b \,x^{2}+a}\, \left (A b -\frac {3 B a}{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-x \left (\left (-\frac {x^{2} B}{2}+A \right ) b^{\frac {3}{2}}-\frac {3 B \sqrt {b}\, a}{2}\right )}{\sqrt {b \,x^{2}+a}\, b^{\frac {5}{2}}}\)	\(73\)
risch	\(\frac {B x \sqrt {b \,x^{2}+a}}{2 b^{2}}+\frac {-\frac {a B x}{\sqrt {b \,x^{2}+a}}+\left (2 b^{2} A -3 a b B \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b^{2}}\)	\(87\)
default	\(B \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+A \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )\)	\(102\)

[In]

int(x^2*(B*x^2+A)/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 213, normalized size of antiderivative = 2.57 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\left [-\frac {{\left (3 \, B a^{2} - 2 \, A a b + {\left (3 \, B a b - 2 \, A b^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (B b^{2} x^{3} + {\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{4 \, {\left (b^{4} x^{2} + a b^{3}\right )}}, \frac {{\left (3 \, B a^{2} - 2 \, A a b + {\left (3 \, B a b - 2 \, A b^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (B b^{2} x^{3} + {\left (3 \, B a b - 2 \, A b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{2 \, {\left (b^{4} x^{2} + a b^{3}\right )}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 3.68 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.37 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=A \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {x}{\sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) + B \left (\frac {3 \sqrt {a} x}{2 b^{2} \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {3 a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {5}{2}}} + \frac {x^{3}}{2 \sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \]

[In]

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

[Out]

A*(asinh(sqrt(b)*x/sqrt(a))/b**(3/2) - x/(sqrt(a)*b*sqrt(1 + b*x**2/a))) + B*(3*sqrt(a)*x/(2*b**2*sqrt(1 + b*x
**2/a)) - 3*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(5/2)) + x**3/(2*sqrt(a)*b*sqrt(1 + b*x**2/a)))

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.99 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B x^{3}}{2 \, \sqrt {b x^{2} + a} b} + \frac {3 \, B a x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {A x}{\sqrt {b x^{2} + a} b} - \frac {3 \, B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {A \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.84 \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left (\frac {B x^{2}}{b} + \frac {3 \, B a b - 2 \, A b^{2}}{b^{3}}\right )} x}{2 \, \sqrt {b x^{2} + a}} + \frac {{\left (3 \, B a - 2 \, A b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 \left (A+B x^2\right )}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {x^2\,\left (B\,x^2+A\right )}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

[In]

int((x^2*(A + B*x^2))/(a + b*x^2)^(3/2),x)

[Out]

int((x^2*(A + B*x^2))/(a + b*x^2)^(3/2), x)