\(\int \frac {A+B x^2}{(a+b x^2)^{3/2}} \, dx\) [575]

Optimal result

Integrand size = 19, antiderivative size = 54 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {(A b-a B) x}{a b \sqrt {a+b x^2}}+\frac {B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {393, 223, 212} \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {x (A b-a B)}{a b \sqrt {a+b x^2}}+\frac {B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}} \]

[In]

Int[(A + B*x^2)/(a + b*x^2)^(3/2),x]

[Out]

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

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 393

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

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

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {A b x-a B x}{a b \sqrt {a+b x^2}}-\frac {B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 2.84 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.02

method	result	size
default	\(\frac {A x}{a \sqrt {b \,x^{2}+a}}+B \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )\)	\(55\)
pseudoelliptic	\(\frac {A \,b^{\frac {3}{2}} x +B \sqrt {b \,x^{2}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) a -B \sqrt {b}\, a x}{b^{\frac {3}{2}} \sqrt {b \,x^{2}+a}\, a}\)	\(61\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 168, normalized size of antiderivative = 3.11 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2}} \, dx=\left [-\frac {2 \, {\left (B a b - A b^{2}\right )} \sqrt {b x^{2} + a} x - {\left (B a b x^{2} + B a^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right )}{2 \, {\left (a b^{3} x^{2} + a^{2} b^{2}\right )}}, -\frac {{\left (B a b - A b^{2}\right )} \sqrt {b x^{2} + a} x + {\left (B a b x^{2} + B a^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right )}{a b^{3} x^{2} + a^{2} b^{2}}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 2.19 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.11 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {A x}{a^{\frac {3}{2}} \sqrt {1 + \frac {b x^{2}}{a}}} + B \left (\frac {\operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{b^{\frac {3}{2}}} - \frac {x}{\sqrt {a} b \sqrt {1 + \frac {b x^{2}}{a}}}\right ) \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.85 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {A x}{\sqrt {b x^{2} + a} a} - \frac {B x}{\sqrt {b x^{2} + a} b} + \frac {B \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.94 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {B \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {3}{2}}} - \frac {{\left (B a - A b\right )} x}{\sqrt {b x^{2} + a} a b} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 5.56 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.98 \[ \int \frac {A+B x^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{b^{3/2}}+\frac {A\,x}{a\,\sqrt {b\,x^2+a}}-\frac {B\,x}{b\,\sqrt {b\,x^2+a}} \]

[In]

int((A + B*x^2)/(a + b*x^2)^(3/2),x)

[Out]

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