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

Optimal. Leaf size=158 \[ \frac {(3 A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{2 a^{5/2}}-\frac {\sqrt {a+b x+c x^2} \left (-8 a A c-2 a b B+3 A b^2\right )}{a^2 x \left (b^2-4 a c\right )}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{a x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 158, 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 = {822, 806, 724, 206} \begin {gather*} -\frac {\sqrt {a+b x+c x^2} \left (-8 a A c-2 a b B+3 A b^2\right )}{a^2 x \left (b^2-4 a c\right )}+\frac {(3 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 a^{5/2}}+\frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{a x \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b^2 - a*b*B - 2*a*A*c + (A*b - 2*a*B)*c*x))/(a*(b^2 - 4*a*c)*x*Sqrt[a + b*x + c*x^2]) - ((3*A*b^2 - 2*a*
b*B - 8*a*A*c)*Sqrt[a + b*x + c*x^2])/(a^2*(b^2 - 4*a*c)*x) + ((3*A*b - 2*a*B)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*
Sqrt[a + b*x + c*x^2])])/(2*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 822

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

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 150, normalized size = 0.95 \begin {gather*} \frac {\frac {2 \sqrt {a} \left (-4 a^2 c (A-B x)+a A \left (b^2-10 b c x-8 c^2 x^2\right )-2 a b B x (b+c x)+3 A b^2 x (b+c x)\right )}{x \sqrt {a+x (b+c x)}}-\left (b^2-4 a c\right ) (3 A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{2 a^{5/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[a]*(-4*a^2*c*(A - B*x) + 3*A*b^2*x*(b + c*x) - 2*a*b*B*x*(b + c*x) + a*A*(b^2 - 10*b*c*x - 8*c^2*x^2)
))/(x*Sqrt[a + x*(b + c*x)]) - (3*A*b - 2*a*B)*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*
x)])])/(2*a^(5/2)*(-b^2 + 4*a*c))

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IntegrateAlgebraic [A]  time = 0.77, size = 153, normalized size = 0.97 \begin {gather*} \frac {(2 a B-3 A b) \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )}{a^{5/2}}+\frac {-4 a^2 A c+4 a^2 B c x+a A b^2-10 a A b c x-8 a A c^2 x^2-2 a b^2 B x-2 a b B c x^2+3 A b^3 x+3 A b^2 c x^2}{a^2 x \left (4 a c-b^2\right ) \sqrt {a+b x+c x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*A*b^2 - 4*a^2*A*c + 3*A*b^3*x - 2*a*b^2*B*x - 10*a*A*b*c*x + 4*a^2*B*c*x + 3*A*b^2*c*x^2 - 2*a*b*B*c*x^2 -
8*a*A*c^2*x^2)/(a^2*(-b^2 + 4*a*c)*x*Sqrt[a + b*x + c*x^2]) + ((-3*A*b + 2*a*B)*ArcTanh[(Sqrt[c]*x - Sqrt[a +
b*x + c*x^2])/Sqrt[a]])/a^(5/2)

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fricas [B]  time = 0.86, size = 657, normalized size = 4.16 \begin {gather*} \left [\frac {{\left ({\left (4 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} c^{2} - {\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} c\right )} x^{3} - {\left (2 \, B a b^{3} - 3 \, A b^{4} - 4 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} c\right )} x^{2} - {\left (2 \, B a^{2} b^{2} - 3 \, A a b^{3} - 4 \, {\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} c\right )} x\right )} \sqrt {a} \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 (A a^{2} b^{2} - 4 \, A a^{3} c - {\left (8 \, A a^{2} c^{2} + {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} c\right )} x^{2} - {\left (2 \, B a^{2} b^{2} - 3 \, A a b^{3} - 2 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} c\right )} x\right )} \sqrt {c x^{2} + b x + a}}{4 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{3} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{2} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x\right )}}, -\frac {{\left ({\left (4 \, {\left (2 \, B a^{2} - 3 \, A a b\right )} c^{2} - {\left (2 \, B a b^{2} - 3 \, A b^{3}\right )} c\right )} x^{3} - {\left (2 \, B a b^{3} - 3 \, A b^{4} - 4 \, {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} c\right )} x^{2} - {\left (2 \, B a^{2} b^{2} - 3 \, A a b^{3} - 4 \, {\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} c\right )} x\right )} \sqrt {-a} \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 (A a^{2} b^{2} - 4 \, A a^{3} c - {\left (8 \, A a^{2} c^{2} + {\left (2 \, B a^{2} b - 3 \, A a b^{2}\right )} c\right )} x^{2} - {\left (2 \, B a^{2} b^{2} - 3 \, A a b^{3} - 2 \, {\left (2 \, B a^{3} - 5 \, A a^{2} b\right )} c\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left ({\left (a^{3} b^{2} c - 4 \, a^{4} c^{2}\right )} x^{3} + {\left (a^{3} b^{3} - 4 \, a^{4} b c\right )} x^{2} + {\left (a^{4} b^{2} - 4 \, a^{5} c\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 220, normalized size = 1.39 \begin {gather*} \frac {2 \, {\left (\frac {{\left (B a^{3} b c - A a^{2} b^{2} c + 2 \, A a^{3} c^{2}\right )} x}{a^{4} b^{2} - 4 \, a^{5} c} + \frac {B a^{3} b^{2} - A a^{2} b^{3} - 2 \, B a^{4} c + 3 \, A a^{3} b c}{a^{4} b^{2} - 4 \, a^{5} c}\right )}}{\sqrt {c x^{2} + b x + a}} + \frac {{\left (2 \, B a - 3 \, A b\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A b + 2 \, A a \sqrt {c}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.06, size = 330, normalized size = 2.09 \begin {gather*} -\frac {8 A \,c^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a}+\frac {3 A \,b^{2} c x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2}}-\frac {2 B b c x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a}-\frac {4 A b c}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a}+\frac {3 A \,b^{3}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a^{2}}-\frac {B \,b^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a}+\frac {3 A b \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {5}{2}}}-\frac {B \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{a^{\frac {3}{2}}}-\frac {3 A b}{2 \sqrt {c \,x^{2}+b x +a}\, a^{2}}+\frac {B}{\sqrt {c \,x^{2}+b x +a}\, a}-\frac {A}{\sqrt {c \,x^{2}+b x +a}\, a x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{2} \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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