3.7 \(\int \frac{(A+B x) \sqrt{a+b x^2}}{x^3} \, dx\)

Optimal. Leaf size=80 \[ -\frac{\sqrt{a+b x^2} (A+2 B x)}{2 x^2}-\frac{A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 \sqrt{a}}+\sqrt{b} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0604033, antiderivative size = 80, 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 = {811, 844, 217, 206, 266, 63, 208} \[ -\frac{\sqrt{a+b x^2} (A+2 B x)}{2 x^2}-\frac{A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 \sqrt{a}}+\sqrt{b} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 811

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((d + e*x)^
(m + 1)*(a + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 + a*e^2) - 2*c*d^2*p*(e*f - d*g) - e*(g*(m + 1)*(c*d^2 + a*e
^2) + 2*c*d*p*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2 + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2
+ a*e^2)), Int[(d + e*x)^(m + 2)*(a + c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) - c*(2*c*d*(d*g*(2*p + 1
) - e*f*(m + 2*p + 2)) - 2*a*e^2*g*(m + 1))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2
, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

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+b x^2}}{x^3} \, dx &=-\frac{(A+2 B x) \sqrt{a+b x^2}}{2 x^2}-\frac{\int \frac{-2 a A b-4 a b B x}{x \sqrt{a+b x^2}} \, dx}{4 a}\\ &=-\frac{(A+2 B x) \sqrt{a+b x^2}}{2 x^2}+\frac{1}{2} (A b) \int \frac{1}{x \sqrt{a+b x^2}} \, dx+(b B) \int \frac{1}{\sqrt{a+b x^2}} \, dx\\ &=-\frac{(A+2 B x) \sqrt{a+b x^2}}{2 x^2}+\frac{1}{4} (A b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^2\right )+(b B) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )\\ &=-\frac{(A+2 B x) \sqrt{a+b x^2}}{2 x^2}+\sqrt{b} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )+\frac{1}{2} A \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )\\ &=-\frac{(A+2 B x) \sqrt{a+b x^2}}{2 x^2}+\sqrt{b} B \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )-\frac{A b \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 \sqrt{a}}\\ \end{align*}

Mathematica [A]  time = 0.0911153, size = 108, normalized size = 1.35 \[ -\frac{\sqrt{a+b x^2} \left (a \sqrt{\frac{b x^2}{a}+1} (A+2 B x)+A b x^2 \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )-2 \sqrt{a} \sqrt{b} B x^2 \sinh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a}}\right )\right )}{2 a x^2 \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 121, normalized size = 1.5 \begin{align*} -{\frac{A}{2\,a{x}^{2}} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}-{\frac{Ab}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}+{\frac{Ab}{2\,a}\sqrt{b{x}^{2}+a}}-{\frac{B}{ax} \left ( b{x}^{2}+a \right ) ^{{\frac{3}{2}}}}+{\frac{bBx}{a}\sqrt{b{x}^{2}+a}}+B\sqrt{b}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b*x^2+a)^(1/2)/x^3,x)

[Out]

-1/2*A/a/x^2*(b*x^2+a)^(3/2)-1/2*A*b/a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)+1/2*A*b/a*(b*x^2+a)^(1/2)-B
/a/x*(b*x^2+a)^(3/2)+B*b/a*x*(b*x^2+a)^(1/2)+B*b^(1/2)*ln(x*b^(1/2)+(b*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)*(b*x^2+a)^(1/2)/x^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62037, size = 934, normalized size = 11.68 \begin{align*} \left [\frac{2 \, B a \sqrt{b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) + A \sqrt{a} b x^{2} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (2 \, B a x + A a\right )} \sqrt{b x^{2} + a}}{4 \, a x^{2}}, -\frac{4 \, B a \sqrt{-b} x^{2} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) - A \sqrt{a} b x^{2} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (2 \, B a x + A a\right )} \sqrt{b x^{2} + a}}{4 \, a x^{2}}, \frac{A \sqrt{-a} b x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) + B a \sqrt{b} x^{2} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) -{\left (2 \, B a x + A a\right )} \sqrt{b x^{2} + a}}{2 \, a x^{2}}, -\frac{2 \, B a \sqrt{-b} x^{2} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) - A \sqrt{-a} b x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (2 \, B a x + A a\right )} \sqrt{b x^{2} + a}}{2 \, a x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(1/2)/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.75492, size = 107, normalized size = 1.34 \begin{align*} - \frac{A \sqrt{b} \sqrt{\frac{a}{b x^{2}} + 1}}{2 x} - \frac{A b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2 \sqrt{a}} - \frac{B \sqrt{a}}{x \sqrt{1 + \frac{b x^{2}}{a}}} + B \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )} - \frac{B b x}{\sqrt{a} \sqrt{1 + \frac{b x^{2}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x**2+a)**(1/2)/x**3,x)

[Out]

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

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Giac [B]  time = 1.22417, size = 220, normalized size = 2.75 \begin{align*} \frac{A b \arctan \left (-\frac{\sqrt{b} x - \sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a}} - B \sqrt{b} \log \left ({\left | -\sqrt{b} x + \sqrt{b x^{2} + a} \right |}\right ) + \frac{{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{3} A b + 2 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} B a \sqrt{b} +{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )} A a b - 2 \, B a^{2} \sqrt{b}}{{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(b*x^2+a)^(1/2)/x^3,x, algorithm="giac")

[Out]

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