3.918 \(\int \frac{(A+B x) \sqrt{a+b x+c x^2}}{x^4} \, dx\)

Optimal. Leaf size=121 \[ -\frac{\left (b^2-4 a c\right ) (A b-2 a B) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 a^{5/2}}+\frac{(2 a+b x) (A b-2 a B) \sqrt{a+b x+c x^2}}{8 a^2 x^2}-\frac{A \left (a+b x+c x^2\right )^{3/2}}{3 a x^3} \]

[Out]

((A*b - 2*a*B)*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(8*a^2*x^2) - (A*(a + b*x + c*x^2)^(3/2))/(3*a*x^3) - ((A*b
- 2*a*B)*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(5/2))

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Rubi [A]  time = 0.0663229, antiderivative size = 121, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {806, 720, 724, 206} \[ -\frac{\left (b^2-4 a c\right ) (A b-2 a B) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 a^{5/2}}+\frac{(2 a+b x) (A b-2 a B) \sqrt{a+b x+c x^2}}{8 a^2 x^2}-\frac{A \left (a+b x+c x^2\right )^{3/2}}{3 a x^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^4,x]

[Out]

((A*b - 2*a*B)*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(8*a^2*x^2) - (A*(a + b*x + c*x^2)^(3/2))/(3*a*x^3) - ((A*b
- 2*a*B)*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(5/2))

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 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 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) \sqrt{a+b x+c x^2}}{x^4} \, dx &=-\frac{A \left (a+b x+c x^2\right )^{3/2}}{3 a x^3}-\frac{(A b-2 a B) \int \frac{\sqrt{a+b x+c x^2}}{x^3} \, dx}{2 a}\\ &=\frac{(A b-2 a B) (2 a+b x) \sqrt{a+b x+c x^2}}{8 a^2 x^2}-\frac{A \left (a+b x+c x^2\right )^{3/2}}{3 a x^3}+\frac{\left ((A b-2 a B) \left (b^2-4 a c\right )\right ) \int \frac{1}{x \sqrt{a+b x+c x^2}} \, dx}{16 a^2}\\ &=\frac{(A b-2 a B) (2 a+b x) \sqrt{a+b x+c x^2}}{8 a^2 x^2}-\frac{A \left (a+b x+c x^2\right )^{3/2}}{3 a x^3}-\frac{\left ((A b-2 a B) \left (b^2-4 a c\right )\right ) \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 )}{8 a^2}\\ &=\frac{(A b-2 a B) (2 a+b x) \sqrt{a+b x+c x^2}}{8 a^2 x^2}-\frac{A \left (a+b x+c x^2\right )^{3/2}}{3 a x^3}-\frac{(A b-2 a B) \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+b x+c x^2}}\right )}{16 a^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.161608, size = 116, normalized size = 0.96 \[ \frac{3 x (A b-2 a B) \left (2 \sqrt{a} (2 a+b x) \sqrt{a+x (b+c x)}-x^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac{2 a+b x}{2 \sqrt{a} \sqrt{a+x (b+c x)}}\right )\right )-16 a^{3/2} A (a+x (b+c x))^{3/2}}{48 a^{5/2} x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^4,x]

[Out]

(-16*a^(3/2)*A*(a + x*(b + c*x))^(3/2) + 3*(A*b - 2*a*B)*x*(2*Sqrt[a]*(2*a + b*x)*Sqrt[a + x*(b + c*x)] - (b^2
- 4*a*c)*x^2*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])]))/(48*a^(5/2)*x^3)

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Maple [B]  time = 0.011, size = 386, normalized size = 3.2 \begin{align*} -{\frac{A}{3\,a{x}^{3}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{Ab}{4\,{a}^{2}{x}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{A{b}^{2}}{8\,{a}^{3}x} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{A{b}^{3}}{8\,{a}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{A{b}^{3}}{16}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}+{\frac{A{b}^{2}cx}{8\,{a}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{Abc}{4\,{a}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{Abc}{4}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{B}{2\,a{x}^{2}} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}+{\frac{bB}{4\,{a}^{2}x} \left ( c{x}^{2}+bx+a \right ) ^{{\frac{3}{2}}}}-{\frac{{b}^{2}B}{4\,{a}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{{b}^{2}B}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+bx+2\,\sqrt{a}\sqrt{c{x}^{2}+bx+a} \right ) } \right ){a}^{-{\frac{3}{2}}}}-{\frac{Bcbx}{4\,{a}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{Bc}{2\,a}\sqrt{c{x}^{2}+bx+a}}-{\frac{Bc}{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}}}} \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^4,x)

[Out]

-1/3*A*(c*x^2+b*x+a)^(3/2)/a/x^3+1/4*A/a^2*b/x^2*(c*x^2+b*x+a)^(3/2)-1/8*A/a^3*b^2/x*(c*x^2+b*x+a)^(3/2)+1/8*A
/a^3*b^3*(c*x^2+b*x+a)^(1/2)-1/16*A/a^(5/2)*b^3*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)+1/8*A/a^3*b^2*c*
(c*x^2+b*x+a)^(1/2)*x-1/4*A/a^2*b*c*(c*x^2+b*x+a)^(1/2)+1/4*A/a^(3/2)*b*c*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^
(1/2))/x)-1/2*B/a/x^2*(c*x^2+b*x+a)^(3/2)+1/4*B/a^2*b/x*(c*x^2+b*x+a)^(3/2)-1/4*B/a^2*b^2*(c*x^2+b*x+a)^(1/2)+
1/8*B/a^(3/2)*b^2*ln((2*a+b*x+2*a^(1/2)*(c*x^2+b*x+a)^(1/2))/x)-1/4*B/a^2*b*c*(c*x^2+b*x+a)^(1/2)*x+1/2*B/a*c*
(c*x^2+b*x+a)^(1/2)-1/2*B/a^(1/2)*c*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)*(c*x^2+b*x+a)^(1/2)/x^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 3.36488, size = 728, normalized size = 6.02 \begin{align*} \left [\frac{3 \,{\left (2 \, B a b^{2} - A b^{3} - 4 \,{\left (2 \, B a^{2} - A a b\right )} c\right )} \sqrt{a} x^{3} \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 \,{\left (8 \, A a^{3} +{\left (6 \, B a^{2} b - 3 \, A a b^{2} + 8 \, A a^{2} c\right )} x^{2} + 2 \,{\left (6 \, B a^{3} + A a^{2} b\right )} x\right )} \sqrt{c x^{2} + b x + a}}{96 \, a^{3} x^{3}}, -\frac{3 \,{\left (2 \, B a b^{2} - A b^{3} - 4 \,{\left (2 \, B a^{2} - A a b\right )} c\right )} \sqrt{-a} x^{3} \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 \,{\left (8 \, A a^{3} +{\left (6 \, B a^{2} b - 3 \, A a b^{2} + 8 \, A a^{2} c\right )} x^{2} + 2 \,{\left (6 \, B a^{3} + A a^{2} b\right )} x\right )} \sqrt{c x^{2} + b x + a}}{48 \, a^{3} x^{3}}\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^4,x, algorithm="fricas")

[Out]

[1/96*(3*(2*B*a*b^2 - A*b^3 - 4*(2*B*a^2 - A*a*b)*c)*sqrt(a)*x^3*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*(8*A*a^3 + (6*B*a^2*b - 3*A*a*b^2 + 8*A*a^2*c)*x^2 + 2*(6
*B*a^3 + A*a^2*b)*x)*sqrt(c*x^2 + b*x + a))/(a^3*x^3), -1/48*(3*(2*B*a*b^2 - A*b^3 - 4*(2*B*a^2 - A*a*b)*c)*sq
rt(-a)*x^3*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*(8*A*a^3 + (6*B*
a^2*b - 3*A*a*b^2 + 8*A*a^2*c)*x^2 + 2*(6*B*a^3 + A*a^2*b)*x)*sqrt(c*x^2 + b*x + a))/(a^3*x^3)]

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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^{4}}\, 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**4,x)

[Out]

Integral((A + B*x)*sqrt(a + b*x + c*x**2)/x**4, x)

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Giac [B]  time = 1.44593, size = 707, normalized size = 5.84 \begin{align*} -\frac{{\left (2 \, B a b^{2} - A b^{3} - 8 \, B a^{2} c + 4 \, A a b c\right )} \arctan \left (-\frac{\sqrt{c} x - \sqrt{c x^{2} + b x + a}}{\sqrt{-a}}\right )}{8 \, \sqrt{-a} a^{2}} + \frac{6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} B a b^{2} - 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} A b^{3} + 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} B a^{2} c + 12 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{5} A a b c + 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{4} B a^{2} b \sqrt{c} + 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{4} A a^{2} c^{\frac{3}{2}} + 8 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} A a b^{3} + 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{3} A a^{2} b c - 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} B a^{3} b \sqrt{c} + 48 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} A a^{2} b^{2} \sqrt{c} - 6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} B a^{3} b^{2} + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} A a^{2} b^{3} - 24 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} B a^{4} c + 36 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} A a^{3} b c + 16 \, A a^{4} c^{\frac{3}{2}}}{24 \,{\left ({\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )}^{2} - a\right )}^{3} a^{2}} \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^4,x, algorithm="giac")

[Out]

-1/8*(2*B*a*b^2 - A*b^3 - 8*B*a^2*c + 4*A*a*b*c)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-
a)*a^2) + 1/24*(6*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a*b^2 - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*b^
3 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^2*c + 12*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a*b*c + 48*(
sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^2*b*sqrt(c) + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^2*c^(3/2)
+ 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a*b^3 + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^2*b*c - 48*(s
qrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^3*b*sqrt(c) + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^2*b^2*sqrt(
c) - 6*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^3*b^2 + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^2*b^3 - 24*(s
qrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^4*c + 36*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^3*b*c + 16*A*a^4*c^(3/2
))/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^3*a^2)