3.1.31 \(\int \frac {x (A+B x)}{(a+b x^2)^{3/2}} \, dx\) [31]

Optimal. Leaf size=48 \[ -\frac {A+B x}{b \sqrt {a+b x^2}}+\frac {B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 48, 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 = {792, 223, 212} \begin {gather*} \frac {B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}}-\frac {A+B x}{b \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((A + B*x)/(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 792

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

Rubi steps

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

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Mathematica [A]
time = 0.25, size = 53, normalized size = 1.10 \begin {gather*} \frac {-A-B x}{b \sqrt {a+b x^2}}-\frac {B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 55, normalized size = 1.15

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 46, normalized size = 0.96 \begin {gather*} -\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}}} - \frac {A}{\sqrt {b x^{2} + a} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 1.02, size = 147, normalized size = 3.06 \begin {gather*} \left [\frac {{\left (B b x^{2} + B a\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (B b x + A b\right )} \sqrt {b x^{2} + a}}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}}, -\frac {{\left (B b x^{2} + B a\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (B b x + A b\right )} \sqrt {b x^{2} + a}}{b^{3} x^{2} + a b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 3.07, size = 66, normalized size = 1.38 \begin {gather*} A \left (\begin {cases} - \frac {1}{b \sqrt {a + b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x^{2}}{2 a^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + 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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A*Piecewise((-1/(b*sqrt(a + b*x**2)), Ne(b, 0)), (x**2/(2*a**(3/2)), True)) + B*(asinh(sqrt(b)*x/sqrt(a))/b**(
3/2) - x/(sqrt(a)*b*sqrt(1 + b*x**2/a)))

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Giac [A]
time = 0.68, size = 48, normalized size = 1.00 \begin {gather*} -\frac {\frac {B x}{b} + \frac {A}{b}}{\sqrt {b x^{2} + a}} - \frac {B \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.06, size = 53, normalized size = 1.10 \begin {gather*} \frac {B\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{b^{3/2}}-\frac {A}{b\,\sqrt {b\,x^2+a}}-\frac {B\,x}{b\,\sqrt {b\,x^2+a}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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