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

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

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Rubi [A]  time = 0.07, antiderivative size = 121, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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} \end {gather*}

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 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 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 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]

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*}

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Mathematica [A]  time = 0.16, size = 116, normalized size = 0.96 \begin {gather*} \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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.98, size = 173, normalized size = 1.43 \begin {gather*} \frac {\left (2 A b c+b^2 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x+c x^2}-\sqrt {c} x}{\sqrt {a}}\right )}{4 a^{3/2}}+\frac {\sqrt {a+b x+c x^2} \left (-8 a^2 A-12 a^2 B x-2 a A b x-8 a A c x^2-6 a b B x^2+3 A b^2 x^2\right )}{24 a^2 x^3}+\frac {\left (8 a^2 B c+A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )}{8 a^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-8*a^2*A - 2*a*A*b*x - 12*a^2*B*x + 3*A*b^2*x^2 - 6*a*b*B*x^2 - 8*a*A*c*x^2))/(24*a^2*
x^3) + ((A*b^3 + 8*a^2*B*c)*ArcTanh[(Sqrt[c]*x - Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(8*a^(5/2)) + ((b^2*B + 2*A*
b*c)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(4*a^(3/2))

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fricas [A]  time = 0.70, size = 317, normalized size = 2.62 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.22, size = 524, normalized size = 4.33 \begin {gather*} -\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 {gather*}

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)

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maple [B]  time = 0.06, size = 386, normalized size = 3.19 \begin {gather*} \frac {A b c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{4 a^{\frac {3}{2}}}-\frac {A \,b^{3} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {5}{2}}}-\frac {B c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2 \sqrt {a}}+\frac {B \,b^{2} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{8 a^{\frac {3}{2}}}+\frac {\sqrt {c \,x^{2}+b x +a}\, A \,b^{2} c x}{8 a^{3}}-\frac {\sqrt {c \,x^{2}+b x +a}\, B b c x}{4 a^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, A b c}{4 a^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, A \,b^{3}}{8 a^{3}}+\frac {\sqrt {c \,x^{2}+b x +a}\, B c}{2 a}-\frac {\sqrt {c \,x^{2}+b x +a}\, B \,b^{2}}{4 a^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{2}}{8 a^{3} x}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b}{4 a^{2} x}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b}{4 a^{2} x^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B}{2 a \,x^{2}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A}{3 a \,x^{3}} \end {gather*}

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((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*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((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*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((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-1/4*B/a^2*b*c*(c*x^2+b*x+a)^(1/2)*x+1/2*B*c/a*
(c*x^2+b*x+a)^(1/2)-1/2*B*c/a^(1/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 positive, negative or zero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+B\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x + c*x^2)^(1/2))/x^4,x)

[Out]

int(((A + B*x)*(a + b*x + c*x^2)^(1/2))/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B x\right ) \sqrt {a + b x + c x^{2}}}{x^{4}}\, dx \end {gather*}

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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