3.6.60 \(\int \frac {A+B x^2}{\sqrt {a+b x^2}} \, dx\) [560]

Optimal. Leaf size=58 \[ \frac {B x \sqrt {a+b x^2}}{2 b}+\frac {(2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {396, 223, 212} \begin {gather*} \frac {(2 A b-a B) \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}+\frac {B x \sqrt {a+b x^2}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/Sqrt[a + b*x^2],x]

[Out]

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

Rubi steps

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

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Mathematica [A]
time = 0.05, size = 59, normalized size = 1.02 \begin {gather*} \frac {B x \sqrt {a+b x^2}}{2 b}+\frac {(-2 A b+a B) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 63, normalized size = 1.09

method	result	size
risch	\(\frac {B x \sqrt {b \,x^{2}+a}}{2 b}+\frac {A \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}-\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right ) B a}{2 b^{\frac {3}{2}}}\)	\(62\)
default	\(B \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+\frac {A \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.31, size = 47, normalized size = 0.81 \begin {gather*} \frac {\sqrt {b x^{2} + a} B x}{2 \, b} - \frac {B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {A \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.27, size = 110, normalized size = 1.90 \begin {gather*} \left [\frac {2 \, \sqrt {b x^{2} + a} B b x - {\left (B a - 2 \, A b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right )}{4 \, b^{2}}, \frac {\sqrt {b x^{2} + a} B b x + {\left (B a - 2 \, A b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right )}{2 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 1.52, size = 126, normalized size = 2.17 \begin {gather*} A \left (\begin {cases} \frac {\sqrt {- \frac {a}{b}} \operatorname {asin}{\left (x \sqrt {- \frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b < 0 \\\frac {\sqrt {\frac {a}{b}} \operatorname {asinh}{\left (x \sqrt {\frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b > 0 \\\frac {\sqrt {- \frac {a}{b}} \operatorname {acosh}{\left (x \sqrt {- \frac {b}{a}} \right )}}{\sqrt {- a}} & \text {for}\: b > 0 \wedge a < 0 \end {cases}\right ) + \frac {B \sqrt {a} x \sqrt {1 + \frac {b x^{2}}{a}}}{2 b} - \frac {B a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*Piecewise((sqrt(-a/b)*asin(x*sqrt(-b/a))/sqrt(a), (a > 0) & (b < 0)), (sqrt(a/b)*asinh(x*sqrt(b/a))/sqrt(a),
(a > 0) & (b > 0)), (sqrt(-a/b)*acosh(x*sqrt(-b/a))/sqrt(-a), (b > 0) & (a < 0))) + B*sqrt(a)*x*sqrt(1 + b*x*
*2/a)/(2*b) - B*a*asinh(sqrt(b)*x/sqrt(a))/(2*b**(3/2))

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Giac [A]
time = 1.73, size = 48, normalized size = 0.83 \begin {gather*} \frac {\sqrt {b x^{2} + a} B x}{2 \, b} + \frac {{\left (B a - 2 \, A b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.50, size = 86, normalized size = 1.48 \begin {gather*} \left \{\begin {array}{cl} \frac {B\,x^3+3\,A\,x}{3\,\sqrt {a}} & \text {\ if\ \ }b=0\\ \frac {A\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{\sqrt {b}}-\frac {B\,a\,\ln \left (2\,\sqrt {b}\,x+2\,\sqrt {b\,x^2+a}\right )}{2\,b^{3/2}}+\frac {B\,x\,\sqrt {b\,x^2+a}}{2\,b} & \text {\ if\ \ }b\neq 0 \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(a + b*x^2)^(1/2),x)

[Out]

piecewise(b == 0, (3*A*x + B*x^3)/(3*a^(1/2)), b ~= 0, (A*log(b^(1/2)*x + (a + b*x^2)^(1/2)))/b^(1/2) - (B*a*l
og(2*b^(1/2)*x + 2*(a + b*x^2)^(1/2)))/(2*b^(3/2)) + (B*x*(a + b*x^2)^(1/2))/(2*b))

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