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

Optimal. Leaf size=72 \[ \frac{(A b-2 a B) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 a^{3/2}}-\frac{A \sqrt{a+b x+c x^2}}{a x} \]

[Out]

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

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Rubi [A]  time = 0.0388087, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {806, 724, 206} \[ \frac{(A b-2 a B) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 a^{3/2}}-\frac{A \sqrt{a+b x+c x^2}}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x],
x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[m + 2*p + 3], 0]

Rule 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 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])

Rubi steps

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

Mathematica [A]  time = 0.0873857, size = 70, normalized size = 0.97 \[ \frac{(A b-2 a B) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{2 a^{3/2}}-\frac{A \sqrt{a+x (b+c x)}}{a x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.008, size = 94, normalized size = 1.3 \begin{align*} -{B\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{A}{ax}\sqrt{c{x}^{2}+bx+a}}+{\frac{Ab}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.16946, size = 428, normalized size = 5.94 \begin{align*} \left [-\frac{{\left (2 \, B a - A b\right )} \sqrt{a} x \log \left (-\frac{8 \, a b x +{\left (b^{2} + 4 \, a c\right )} x^{2} + 4 \, \sqrt{c x^{2} + b x + a}{\left (b x + 2 \, a\right )} \sqrt{a} + 8 \, a^{2}}{x^{2}}\right ) + 4 \, \sqrt{c x^{2} + b x + a} A a}{4 \, a^{2} x}, \frac{{\left (2 \, B a - A b\right )} \sqrt{-a} x \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (b x + 2 \, a\right )} \sqrt{-a}}{2 \,{\left (a c x^{2} + a b x + a^{2}\right )}}\right ) - 2 \, \sqrt{c x^{2} + b x + a} A a}{2 \, a^{2} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((2*B*a - A*b)*sqrt(a)*x*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 + 4*sqrt(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a)
+ 8*a^2)/x^2) + 4*sqrt(c*x^2 + b*x + a)*A*a)/(a^2*x), 1/2*((2*B*a - A*b)*sqrt(-a)*x*arctan(1/2*sqrt(c*x^2 + b
*x + a)*(b*x + 2*a)*sqrt(-a)/(a*c*x^2 + a*b*x + a^2)) - 2*sqrt(c*x^2 + b*x + a)*A*a)/(a^2*x)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x}{x^{2} \sqrt{a + b x + c x^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + B*x)/(x**2*sqrt(a + b*x + c*x**2)), x)

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Giac [A]  time = 1.16644, size = 149, normalized size = 2.07 \begin{align*} \frac{{\left (2 \, B a - A b\right )} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + b x + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} A b + 2 \, A a \sqrt{c}}{{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} - a\right )} a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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