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

Optimal. Leaf size=53 \[ \frac{B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{\sqrt{c}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

[Out]

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

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Rubi [A]  time = 0.0387258, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {844, 217, 206, 266, 63, 208} \[ \frac{B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{\sqrt{c}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x*Sqrt[a + c*x^2]),x]

[Out]

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

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + c*x^2)^p, x], x] /; FreeQ[{a,
c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IGtQ[m, 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 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.0150324, size = 53, normalized size = 1. \[ \frac{B \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )}{\sqrt{c}}-\frac{A \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 52, normalized size = 1. \begin{align*}{B\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ){\frac{1}{\sqrt{c}}}}-{A\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.78089, size = 683, normalized size = 12.89 \begin{align*} \left [\frac{B a \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + A \sqrt{a} c \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right )}{2 \, a c}, -\frac{2 \, B a \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) - A \sqrt{a} c \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right )}{2 \, a c}, \frac{2 \, A \sqrt{-a} c \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) + B a \sqrt{c} \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right )}{2 \, a c}, -\frac{B a \sqrt{-c} \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) - A \sqrt{-a} c \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right )}{a c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.84692, size = 99, normalized size = 1.87 \begin{align*} - \frac{A \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{c} x} \right )}}{\sqrt{a}} + B \left (\begin{cases} \frac{\sqrt{- \frac{a}{c}} \operatorname{asin}{\left (x \sqrt{- \frac{c}{a}} \right )}}{\sqrt{a}} & \text{for}\: a > 0 \wedge c < 0 \\\frac{\sqrt{\frac{a}{c}} \operatorname{asinh}{\left (x \sqrt{\frac{c}{a}} \right )}}{\sqrt{a}} & \text{for}\: a > 0 \wedge c > 0 \\\frac{\sqrt{- \frac{a}{c}} \operatorname{acosh}{\left (x \sqrt{- \frac{c}{a}} \right )}}{\sqrt{- a}} & \text{for}\: c > 0 \wedge a < 0 \end{cases}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.14681, size = 78, normalized size = 1.47 \begin{align*} \frac{2 \, A \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - \frac{B \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right )}{\sqrt{c}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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