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

Optimal. Leaf size=75 \[ -\frac{\sqrt{a+c x^2} (A-B x)}{x}+A \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )-\sqrt{a} B \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0577512, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {813, 844, 217, 206, 266, 63, 208} \[ -\frac{\sqrt{a+c x^2} (A-B x)}{x}+A \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a+c x^2}}\right )-\sqrt{a} B \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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

Mathematica [A]  time = 0.159998, size = 99, normalized size = 1.32 \[ \frac{\sqrt{a+c x^2} (B x-A)}{x}+\frac{\sqrt{a} A \sqrt{c} \sqrt{\frac{c x^2}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{a}}\right )}{\sqrt{a+c x^2}}-\sqrt{a} B \tanh ^{-1}\left (\frac{\sqrt{a+c x^2}}{\sqrt{a}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-A + B*x)*Sqrt[a + c*x^2])/x + (Sqrt[a]*A*Sqrt[c]*Sqrt[1 + (c*x^2)/a]*ArcSinh[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[a +
c*x^2] - Sqrt[a]*B*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]]

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Maple [A]  time = 0.009, size = 97, normalized size = 1.3 \begin{align*} -B\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{c{x}^{2}+a} \right ) } \right ) +B\sqrt{c{x}^{2}+a}-{\frac{A}{ax} \left ( c{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{Acx}{a}\sqrt{c{x}^{2}+a}}+A\sqrt{c}\ln \left ( x\sqrt{c}+\sqrt{c{x}^{2}+a} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.74108, size = 821, normalized size = 10.95 \begin{align*} \left [\frac{A \sqrt{c} x \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + B \sqrt{a} x \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \, \sqrt{c x^{2} + a}{\left (B x - A\right )}}{2 \, x}, -\frac{2 \, A \sqrt{-c} x \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) - B \sqrt{a} x \log \left (-\frac{c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \, \sqrt{c x^{2} + a}{\left (B x - A\right )}}{2 \, x}, \frac{2 \, B \sqrt{-a} x \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) + A \sqrt{c} x \log \left (-2 \, c x^{2} - 2 \, \sqrt{c x^{2} + a} \sqrt{c} x - a\right ) + 2 \, \sqrt{c x^{2} + a}{\left (B x - A\right )}}{2 \, x}, -\frac{A \sqrt{-c} x \arctan \left (\frac{\sqrt{-c} x}{\sqrt{c x^{2} + a}}\right ) - B \sqrt{-a} x \arctan \left (\frac{\sqrt{-a}}{\sqrt{c x^{2} + a}}\right ) - \sqrt{c x^{2} + a}{\left (B x - A\right )}}{x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 4.05148, size = 124, normalized size = 1.65 \begin{align*} - \frac{A \sqrt{a}}{x \sqrt{1 + \frac{c x^{2}}{a}}} + A \sqrt{c} \operatorname{asinh}{\left (\frac{\sqrt{c} x}{\sqrt{a}} \right )} - \frac{A c x}{\sqrt{a} \sqrt{1 + \frac{c x^{2}}{a}}} - B \sqrt{a} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{c} x} \right )} + \frac{B a}{\sqrt{c} x \sqrt{\frac{a}{c x^{2}} + 1}} + \frac{B \sqrt{c} x}{\sqrt{\frac{a}{c x^{2}} + 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-A*sqrt(a)/(x*sqrt(1 + c*x**2/a)) + A*sqrt(c)*asinh(sqrt(c)*x/sqrt(a)) - A*c*x/(sqrt(a)*sqrt(1 + c*x**2/a)) -
B*sqrt(a)*asinh(sqrt(a)/(sqrt(c)*x)) + B*a/(sqrt(c)*x*sqrt(a/(c*x**2) + 1)) + B*sqrt(c)*x/sqrt(a/(c*x**2) + 1)

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Giac [A]  time = 1.1653, size = 138, normalized size = 1.84 \begin{align*} \frac{2 \, B a \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - A \sqrt{c} \log \left ({\left | -\sqrt{c} x + \sqrt{c x^{2} + a} \right |}\right ) + \sqrt{c x^{2} + a} B + \frac{2 \, A a \sqrt{c}}{{\left (\sqrt{c} x - \sqrt{c x^{2} + a}\right )}^{2} - a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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