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

Optimal. Leaf size=56 \[ \frac {\sqrt {a+b x^2} (2 A+B x)}{2 b}-\frac {a B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {780, 217, 206} \[ \frac {\sqrt {a+b x^2} (2 A+B x)}{2 b}-\frac {a B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(A + B*x))/Sqrt[a + b*x^2],x]

[Out]

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

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 780

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(((e*f + d*g)*(2*p
+ 3) + 2*e*g*(p + 1)*x)*(a + c*x^2)^(p + 1))/(2*c*(p + 1)*(2*p + 3)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(c*
(2*p + 3)), Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, p}, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {x (A+B x)}{\sqrt {a+b x^2}} \, dx &=\frac {(2 A+B x) \sqrt {a+b x^2}}{2 b}-\frac {(a B) \int \frac {1}{\sqrt {a+b x^2}} \, dx}{2 b}\\ &=\frac {(2 A+B x) \sqrt {a+b x^2}}{2 b}-\frac {(a B) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )}{2 b}\\ &=\frac {(2 A+B x) \sqrt {a+b x^2}}{2 b}-\frac {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 = 0.04, size = 57, normalized size = 1.02 \[ \frac {\sqrt {b} \sqrt {a+b x^2} (2 A+B x)-a B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.19, size = 109, normalized size = 1.95 \[ \left [\frac {B a \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (B b x + 2 \, A b\right )} \sqrt {b x^{2} + a}}{4 \, b^{2}}, \frac {B a \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (B b x + 2 \, A b\right )} \sqrt {b x^{2} + a}}{2 \, b^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.48, size = 50, normalized size = 0.89 \[ \frac {1}{2} \, \sqrt {b x^{2} + a} {\left (\frac {B x}{b} + \frac {2 \, A}{b}\right )} + \frac {B a \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 55, normalized size = 0.98 \[ -\frac {B a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {\sqrt {b \,x^{2}+a}\, B x}{2 b}+\frac {\sqrt {b \,x^{2}+a}\, A}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.33, size = 47, normalized size = 0.84 \[ \frac {\sqrt {b x^{2} + a} B x}{2 \, b} - \frac {B a \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {\sqrt {b x^{2} + a} A}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+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) + sqrt(b*x^2 + a)*A/b

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mupad [B]  time = 1.24, size = 82, normalized size = 1.46 \[ \left \{\begin {array}{cl} \frac {2\,B\,x^3+3\,A\,x^2}{6\,\sqrt {a}} & \text {\ if\ \ }b=0\\ \frac {A\,\sqrt {b\,x^2+a}}{b}-\frac {B\,a\,\ln \left (2\,\sqrt {b}\,x+2\,\sqrt {b\,x^2+a}\right )}{2\,b^{3/2}}+\frac {B\,x\,\sqrt {b\,x^2+a}}{2\,b} & \text {\ if\ \ }b\neq 0 \end {array}\right . \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

piecewise(b == 0, (3*A*x^2 + 2*B*x^3)/(6*a^(1/2)), b ~= 0, (A*(a + b*x^2)^(1/2))/b - (B*a*log(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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sympy [A]  time = 6.26, size = 70, normalized size = 1.25 \[ A \left (\begin {cases} \frac {x^{2}}{2 \sqrt {a}} & \text {for}\: b = 0 \\\frac {\sqrt {a + b x^{2}}}{b} & \text {otherwise} \end {cases}\right ) + \frac {B \sqrt {a} x \sqrt {1 + \frac {b x^{2}}{a}}}{2 b} - \frac {B a \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{2 b^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*Piecewise((x**2/(2*sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**2)/b, True)) + 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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