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

Optimal. Leaf size=43 \[ \frac {A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}+\frac {B \sqrt {a+b x^2}}{b} \]

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Rubi [A]  time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {641, 217, 206} \begin {gather*} \frac {A \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{\sqrt {b}}+\frac {B \sqrt {a+b x^2}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/Sqrt[a + b*x^2],x]

[Out]

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

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 641

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

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 46, normalized size = 1.07 \begin {gather*} \frac {A \log \left (\sqrt {b} \sqrt {a+b x^2}+b x\right )}{\sqrt {b}}+\frac {B \sqrt {a+b x^2}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.24, size = 46, normalized size = 1.07 \begin {gather*} \frac {B \sqrt {a+b x^2}}{b}-\frac {A \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{\sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.16, size = 92, normalized size = 2.14 \begin {gather*} \left [\frac {A \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, \sqrt {b x^{2} + a} B}{2 \, b}, -\frac {A \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - \sqrt {b x^{2} + a} B}{b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.55, size = 39, normalized size = 0.91 \begin {gather*} -\frac {A \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{\sqrt {b}} + \frac {\sqrt {b x^{2} + a} B}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 37, normalized size = 0.86 \begin {gather*} \frac {A \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+\frac {\sqrt {b \,x^{2}+a}\, B}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.36, size = 29, normalized size = 0.67 \begin {gather*} \frac {A \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} + \frac {\sqrt {b x^{2} + a} B}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.14, size = 36, normalized size = 0.84 \begin {gather*} \frac {B\,\sqrt {b\,x^2+a}}{b}+\frac {A\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(a + b*x^2)^(1/2),x)

[Out]

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

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sympy [B]  time = 2.64, size = 102, normalized size = 2.37 \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 ) + B \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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