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

Optimal. Leaf size=167 \[ -\frac {\sqrt {a+b x+c x^2} \left (-16 a A c-18 a b B+15 A b^2\right )}{24 a^3 x}+\frac {(5 A b-6 a B) \sqrt {a+b x+c x^2}}{12 a^2 x^2}+\frac {\left (8 a^2 B c-12 a A b c-6 a b^2 B+5 A b^3\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 a^{7/2}}-\frac {A \sqrt {a+b x+c x^2}}{3 a x^3} \]

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Rubi [A]  time = 0.16, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {834, 806, 724, 206} \begin {gather*} -\frac {\sqrt {a+b x+c x^2} \left (-16 a A c-18 a b B+15 A b^2\right )}{24 a^3 x}+\frac {\left (8 a^2 B c-12 a A b c-6 a b^2 B+5 A b^3\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 a^{7/2}}+\frac {(5 A b-6 a B) \sqrt {a+b x+c x^2}}{12 a^2 x^2}-\frac {A \sqrt {a+b x+c x^2}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[a + b*x + c*x^2])/(3*a*x^3) + ((5*A*b - 6*a*B)*Sqrt[a + b*x + c*x^2])/(12*a^2*x^2) - ((15*A*b^2 - 18*
a*b*B - 16*a*A*c)*Sqrt[a + b*x + c*x^2])/(24*a^3*x) + ((5*A*b^3 - 6*a*b^2*B - 12*a*A*b*c + 8*a^2*B*c)*ArcTanh[
(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(7/2))

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 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]

Rule 834

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

Rubi steps

\begin {align*} \int \frac {A+B x}{x^4 \sqrt {a+b x+c x^2}} \, dx &=-\frac {A \sqrt {a+b x+c x^2}}{3 a x^3}-\frac {\int \frac {\frac {1}{2} (5 A b-6 a B)+2 A c x}{x^3 \sqrt {a+b x+c x^2}} \, dx}{3 a}\\ &=-\frac {A \sqrt {a+b x+c x^2}}{3 a x^3}+\frac {(5 A b-6 a B) \sqrt {a+b x+c x^2}}{12 a^2 x^2}+\frac {\int \frac {\frac {1}{4} \left (15 A b^2-18 a b B-16 a A c\right )+\frac {1}{2} (5 A b-6 a B) c x}{x^2 \sqrt {a+b x+c x^2}} \, dx}{6 a^2}\\ &=-\frac {A \sqrt {a+b x+c x^2}}{3 a x^3}+\frac {(5 A b-6 a B) \sqrt {a+b x+c x^2}}{12 a^2 x^2}-\frac {\left (15 A b^2-18 a b B-16 a A c\right ) \sqrt {a+b x+c x^2}}{24 a^3 x}-\frac {\left (5 A b^3-6 a b^2 B-12 a A b c+8 a^2 B c\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{16 a^3}\\ &=-\frac {A \sqrt {a+b x+c x^2}}{3 a x^3}+\frac {(5 A b-6 a B) \sqrt {a+b x+c x^2}}{12 a^2 x^2}-\frac {\left (15 A b^2-18 a b B-16 a A c\right ) \sqrt {a+b x+c x^2}}{24 a^3 x}+\frac {\left (5 A b^3-6 a b^2 B-12 a A b c+8 a^2 B c\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^3}\\ &=-\frac {A \sqrt {a+b x+c x^2}}{3 a x^3}+\frac {(5 A b-6 a B) \sqrt {a+b x+c x^2}}{12 a^2 x^2}-\frac {\left (15 A b^2-18 a b B-16 a A c\right ) \sqrt {a+b x+c x^2}}{24 a^3 x}+\frac {\left (5 A b^3-6 a b^2 B-12 a A b c+8 a^2 B c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 a^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.14, size = 132, normalized size = 0.79 \begin {gather*} \frac {\left (8 a^2 B c-12 a A b c-6 a b^2 B+5 A b^3\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{16 a^{7/2}}-\frac {\sqrt {a+x (b+c x)} \left (4 a^2 (2 A+3 B x)-2 a x (5 A b+8 A c x+9 b B x)+15 A b^2 x^2\right )}{24 a^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24*(Sqrt[a + x*(b + c*x)]*(15*A*b^2*x^2 + 4*a^2*(2*A + 3*B*x) - 2*a*x*(5*A*b + 9*b*B*x + 8*A*c*x)))/(a^3*x^
3) + ((5*A*b^3 - 6*a*b^2*B - 12*a*A*b*c + 8*a^2*B*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*x)])])/(
16*a^(7/2))

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IntegrateAlgebraic [A]  time = 0.85, size = 174, normalized size = 1.04 \begin {gather*} -\frac {3 \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^{5/2}}+\frac {\left (-8 a^2 B c-5 A b^3\right ) \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )}{8 a^{7/2}}+\frac {\sqrt {a+b x+c x^2} \left (-8 a^2 A-12 a^2 B x+10 a A b x+16 a A c x^2+18 a b B x^2-15 A b^2 x^2\right )}{24 a^3 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-8*a^2*A + 10*a*A*b*x - 12*a^2*B*x - 15*A*b^2*x^2 + 18*a*b*B*x^2 + 16*a*A*c*x^2))/(24*
a^3*x^3) + ((-5*A*b^3 - 8*a^2*B*c)*ArcTanh[(Sqrt[c]*x - Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(8*a^(7/2)) - (3*(b^2
*B + 2*A*b*c)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(4*a^(5/2))

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fricas [A]  time = 0.73, size = 321, normalized size = 1.92 \begin {gather*} \left [\frac {3 \, {\left (6 \, B a b^{2} - 5 \, A b^{3} - 4 \, {\left (2 \, B a^{2} - 3 \, 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 (18 \, B a^{2} b - 15 \, A a b^{2} + 16 \, A a^{2} c\right )} x^{2} + 2 \, {\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{96 \, a^{4} x^{3}}, \frac {3 \, {\left (6 \, B a b^{2} - 5 \, A b^{3} - 4 \, {\left (2 \, B a^{2} - 3 \, 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 (18 \, B a^{2} b - 15 \, A a b^{2} + 16 \, A a^{2} c\right )} x^{2} + 2 \, {\left (6 \, B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{48 \, a^{4} x^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/96*(3*(6*B*a*b^2 - 5*A*b^3 - 4*(2*B*a^2 - 3*A*a*b)*c)*sqrt(a)*x^3*log(-(8*a*b*x + (b^2 + 4*a*c)*x^2 - 4*sqr
t(c*x^2 + b*x + a)*(b*x + 2*a)*sqrt(a) + 8*a^2)/x^2) - 4*(8*A*a^3 - (18*B*a^2*b - 15*A*a*b^2 + 16*A*a^2*c)*x^2
+ 2*(6*B*a^3 - 5*A*a^2*b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^3), 1/48*(3*(6*B*a*b^2 - 5*A*b^3 - 4*(2*B*a^2 - 3*
A*a*b)*c)*sqrt(-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 - (18*B*a^2*b - 15*A*a*b^2 + 16*A*a^2*c)*x^2 + 2*(6*B*a^3 - 5*A*a^2*b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^3
)]

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giac [B]  time = 0.24, size = 511, normalized size = 3.06 \begin {gather*} \frac {{\left (6 \, B a b^{2} - 5 \, A b^{3} - 8 \, B a^{2} c + 12 \, 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^{3}} - \frac {18 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} B a b^{2} - 15 \, {\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 + 36 \, {\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 )}^{3} B a^{2} b^{2} + 40 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a b^{3} - 96 \, {\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} - 96 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} A a^{3} c^{\frac {3}{2}} + 30 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{3} b^{2} - 33 \, {\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 + 48 \, B a^{4} b \sqrt {c} - 48 \, A a^{3} b^{2} \sqrt {c} + 32 \, 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^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/8*(6*B*a*b^2 - 5*A*b^3 - 8*B*a^2*c + 12*A*a*b*c)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt
(-a)*a^3) - 1/24*(18*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a*b^2 - 15*(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 + 36*(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))^3*B*a^2*b^2 + 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a*b^3 - 96*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^2*b*c - 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^3*b*sqrt(c) - 96*(
sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^3*c^(3/2) + 30*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^3*b^2 - 33*(sq
rt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^2*b^3 + 24*(sqrt(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 + 48*B*a^4*b*sqrt(c) - 48*A*a^3*b^2*sqrt(c) + 32*A*a^4*c^(3/2))/(((sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^2 - a)^3*a^3)

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maple [A]  time = 0.07, size = 283, normalized size = 1.69 \begin {gather*} -\frac {3 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 {5}{2}}}+\frac {5 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 {7}{2}}}+\frac {B c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {3}{2}}}-\frac {3 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 {5}{2}}}+\frac {2 \sqrt {c \,x^{2}+b x +a}\, A c}{3 a^{2} x}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2}}{8 a^{3} x}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B b}{4 a^{2} x}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, A b}{12 a^{2} x^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, B}{2 a \,x^{2}}-\frac {\sqrt {c \,x^{2}+b x +a}\, A}{3 a \,x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*A*(c*x^2+b*x+a)^(1/2)/a/x^3+5/12*A/a^2*b/x^2*(c*x^2+b*x+a)^(1/2)-5/8*A/a^3*b^2/x*(c*x^2+b*x+a)^(1/2)+5/16
*A/a^(7/2)*b^3*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-3/4*A/a^(5/2)*b*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/
2)*a^(1/2))/x)+2/3*A*c/a^2/x*(c*x^2+b*x+a)^(1/2)-1/2*B/a/x^2*(c*x^2+b*x+a)^(1/2)+3/4*B/a^2*b/x*(c*x^2+b*x+a)^(
1/2)-3/8*B/a^(5/2)*b^2*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+1/2*B*c/a^(3/2)*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)/x^4/(c*x^2+b*x+a)^(1/2),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 {A+B\,x}{x^4\,\sqrt {c\,x^2+b\,x+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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