3.9.82 \(\int \frac {A+B x}{x^3 \sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=116 \[ -\frac {\left (-4 a A c-4 a b B+3 A b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{5/2}}+\frac {(3 A b-4 a B) \sqrt {a+b x+c x^2}}{4 a^2 x}-\frac {A \sqrt {a+b x+c x^2}}{2 a x^2} \]

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Rubi [A]  time = 0.09, antiderivative size = 116, 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 = {834, 806, 724, 206} \begin {gather*} -\frac {\left (-4 a A c-4 a b B+3 A b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{5/2}}+\frac {(3 A b-4 a B) \sqrt {a+b x+c x^2}}{4 a^2 x}-\frac {A \sqrt {a+b x+c x^2}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(A*Sqrt[a + b*x + c*x^2])/(2*a*x^2) + ((3*A*b - 4*a*B)*Sqrt[a + b*x + c*x^2])/(4*a^2*x) - ((3*A*b^2 - 4*a*b*B
- 4*a*A*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(8*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 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^3 \sqrt {a+b x+c x^2}} \, dx &=-\frac {A \sqrt {a+b x+c x^2}}{2 a x^2}-\frac {\int \frac {\frac {1}{2} (3 A b-4 a B)+A c x}{x^2 \sqrt {a+b x+c x^2}} \, dx}{2 a}\\ &=-\frac {A \sqrt {a+b x+c x^2}}{2 a x^2}+\frac {(3 A b-4 a B) \sqrt {a+b x+c x^2}}{4 a^2 x}+\frac {\left (3 A b^2-4 a b B-4 a A c\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{8 a^2}\\ &=-\frac {A \sqrt {a+b x+c x^2}}{2 a x^2}+\frac {(3 A b-4 a B) \sqrt {a+b x+c x^2}}{4 a^2 x}-\frac {\left (3 A b^2-4 a b B-4 a A 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 )}{4 a^2}\\ &=-\frac {A \sqrt {a+b x+c x^2}}{2 a x^2}+\frac {(3 A b-4 a B) \sqrt {a+b x+c x^2}}{4 a^2 x}-\frac {\left (3 A b^2-4 a b B-4 a A c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 95, normalized size = 0.82 \begin {gather*} \frac {\left (4 a A c+4 a b B-3 A b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{8 a^{5/2}}+\frac {\sqrt {a+x (b+c x)} (3 A b x-2 a (A+2 B x))}{4 a^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.59, size = 131, normalized size = 1.13 \begin {gather*} \frac {3 A b^2 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {(A c+b B) \tanh ^{-1}\left (\frac {\sqrt {a+b x+c x^2}-\sqrt {c} x}{\sqrt {a}}\right )}{a^{3/2}}+\frac {\sqrt {a+b x+c x^2} (-2 a A-4 a B x+3 A b x)}{4 a^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.62, size = 235, normalized size = 2.03 \begin {gather*} \left [\frac {{\left (4 \, B a b - 3 \, A b^{2} + 4 \, A a c\right )} \sqrt {a} x^{2} \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 (2 \, A a^{2} + {\left (4 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{16 \, a^{3} x^{2}}, -\frac {{\left (4 \, B a b - 3 \, A b^{2} + 4 \, A a c\right )} \sqrt {-a} x^{2} \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 (2 \, A a^{2} + {\left (4 \, B a^{2} - 3 \, A a b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{8 \, a^{3} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/16*((4*B*a*b - 3*A*b^2 + 4*A*a*c)*sqrt(a)*x^2*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*(2*A*a^2 + (4*B*a^2 - 3*A*a*b)*x)*sqrt(c*x^2 + b*x + a))/(a^3*x^2), -1/8*
((4*B*a*b - 3*A*b^2 + 4*A*a*c)*sqrt(-a)*x^2*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*(2*A*a^2 + (4*B*a^2 - 3*A*a*b)*x)*sqrt(c*x^2 + b*x + a))/(a^3*x^2)]

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giac [B]  time = 0.29, size = 303, normalized size = 2.61 \begin {gather*} -\frac {{\left (4 \, B a b - 3 \, A b^{2} + 4 \, A a c\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{4 \, \sqrt {-a} a^{2}} + \frac {4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} B a b - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A b^{2} + 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a c + 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} B a^{2} \sqrt {c} - 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{2} b + 5 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a b^{2} + 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a^{2} c - 8 \, B a^{3} \sqrt {c} + 8 \, A a^{2} b \sqrt {c}}{4 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )}^{2} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

-1/4*(4*B*a*b - 3*A*b^2 + 4*A*a*c)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a^2) + 1/4*
(4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a*b - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*b^2 + 4*(sqrt(c)*x
- sqrt(c*x^2 + b*x + a))^3*A*a*c + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^2*sqrt(c) - 4*(sqrt(c)*x - sqrt
(c*x^2 + b*x + a))*B*a^2*b + 5*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a*b^2 + 4*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))*A*a^2*c - 8*B*a^3*sqrt(c) + 8*A*a^2*b*sqrt(c))/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^2*a^2)

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maple [A]  time = 0.06, size = 176, normalized size = 1.52 \begin {gather*} \frac {A 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 A \,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 {B b \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 \sqrt {c \,x^{2}+b x +a}\, A b}{4 a^{2} x}-\frac {\sqrt {c \,x^{2}+b x +a}\, B}{a x}-\frac {\sqrt {c \,x^{2}+b x +a}\, A}{2 a \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^3/(c*x^2+b*x+a)^(1/2),x)

[Out]

-1/2*A*(c*x^2+b*x+a)^(1/2)/a/x^2+3/4*A/a^2*b/x*(c*x^2+b*x+a)^(1/2)-3/8*A/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*A*c/a^(3/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-B/a/x*(c*x^2+b*x+a)^(1/2)+1
/2*B*b/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^3/(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^3\,\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^3*(a + b*x + c*x^2)^(1/2)),x)

[Out]

int((A + B*x)/(x^3*(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^{3} \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**3/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral((A + B*x)/(x**3*sqrt(a + b*x + c*x**2)), x)

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