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

Optimal. Leaf size=121 \[ -\frac{(A-B x) \sqrt{a+b x+c x^2}}{x}-\frac{(2 a B+A b) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{a}}+\frac{(2 A c+b B) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c}} \]

[Out]

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

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Rubi [A]  time = 0.0885888, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {812, 843, 621, 206, 724} \[ -\frac{(A-B x) \sqrt{a+b x+c x^2}}{x}-\frac{(2 a B+A b) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{a}}+\frac{(2 A c+b B) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{2 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 812

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + b*x + c*x^2)^p)/(e^2*(m + 1)*(m
+ 2*p + 2)), x] + Dist[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1)*Simp[g*(
b*d + 2*a*e + 2*a*e*m + 2*b*d*p) - f*b*e*(m + 2*p + 2) + (g*(2*c*d + b*e + b*e*m + 4*c*d*p) - 2*c*e*f*(m + 2*p
+ 2))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + 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 843

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

Rule 621

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

Rubi steps

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

Mathematica [A]  time = 0.189048, size = 118, normalized size = 0.98 \[ \frac{(B x-A) \sqrt{a+x (b+c x)}}{x}-\frac{(2 a B+A b) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )}{2 \sqrt{a}}+\frac{(2 A c+b B) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{2 \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.008, size = 207, normalized size = 1.7 \begin{align*} B\sqrt{c{x}^{2}+bx+a}+{\frac{bB}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){\frac{1}{\sqrt{c}}}}-B\sqrt{a}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ) -{\frac{A}{ax} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{Ab}{a}\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 ){\frac{1}{\sqrt{a}}}}+{\frac{Acx}{a}\sqrt{c{x}^{2}+bx+a}}+A\sqrt{c}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.5624, size = 1555, normalized size = 12.85 \begin{align*} \left [\frac{{\left (2 \, B a + A b\right )} \sqrt{a} c 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 ) +{\left (B a b + 2 \, A a c\right )} \sqrt{c} x \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) + 4 \,{\left (B a c x - A a c\right )} \sqrt{c x^{2} + b x + a}}{4 \, a c x}, \frac{{\left (2 \, B a + A b\right )} \sqrt{a} c 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 ) - 2 \,{\left (B a b + 2 \, A a c\right )} \sqrt{-c} x \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 4 \,{\left (B a c x - A a c\right )} \sqrt{c x^{2} + b x + a}}{4 \, a c x}, \frac{2 \,{\left (2 \, B a + A b\right )} \sqrt{-a} c 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 ) +{\left (B a b + 2 \, A a c\right )} \sqrt{c} x \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{c} - 4 \, a c\right ) + 4 \,{\left (B a c x - A a c\right )} \sqrt{c x^{2} + b x + a}}{4 \, a c x}, \frac{{\left (2 \, B a + A b\right )} \sqrt{-a} c 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 ) -{\left (B a b + 2 \, A a c\right )} \sqrt{-c} x \arctan \left (\frac{\sqrt{c x^{2} + b x + a}{\left (2 \, c x + b\right )} \sqrt{-c}}{2 \,{\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \,{\left (B a c x - A a c\right )} \sqrt{c x^{2} + b x + a}}{2 \, a c x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.42124, size = 215, normalized size = 1.78 \begin{align*} \sqrt{c x^{2} + b x + a} B + \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}} - \frac{{\left (B b + 2 \, A c\right )} \log \left ({\left | 2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} + b \right |}\right )}{2 \, \sqrt{c}} + \frac{{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} A b + 2 \, A a \sqrt{c}}{{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} - a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sqrt(c*x^2 + b*x + a)*B + (2*B*a + A*b)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/sqrt(-a) - 1/2*(
B*b + 2*A*c)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/sqrt(c) + ((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)