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

Optimal. Leaf size=96 \[ \frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {A \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{a^{3/2}} \]

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Rubi [A]  time = 0.06, antiderivative size = 96, 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, 12, 724, 206} \begin {gather*} \frac {2 \left (c x (A b-2 a B)-2 a A c-a b B+A b^2\right )}{a \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {A \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

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Mathematica [A]  time = 0.12, size = 104, normalized size = 1.08 \begin {gather*} \frac {\frac {2 \sqrt {a} \left (a B (b+2 c x)-A \left (-2 a c+b^2+b c x\right )\right )}{\sqrt {a+x (b+c x)}}+A \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 )}{a^{3/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.53, size = 103, normalized size = 1.07 \begin {gather*} \frac {2 A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {2 \left (2 a A c+a b B+2 a B c x-A b^2-A b c x\right )}{a \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*(a + b*x + c*x^2)^(3/2)),x]

[Out]

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

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fricas [B]  time = 0.66, size = 412, normalized size = 4.29 \begin {gather*} \left [\frac {{\left (A a b^{2} - 4 \, A a^{2} c + {\left (A b^{2} c - 4 \, A a c^{2}\right )} x^{2} + {\left (A b^{3} - 4 \, A a b 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 (B a^{2} b - A a b^{2} + 2 \, A a^{2} c + {\left (2 \, B a^{2} - A a b\right )} c x\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (a^{3} b^{2} - 4 \, a^{4} c + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{2} + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x\right )}}, \frac {{\left (A a b^{2} - 4 \, A a^{2} c + {\left (A b^{2} c - 4 \, A a c^{2}\right )} x^{2} + {\left (A b^{3} - 4 \, A a b 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 (B a^{2} b - A a b^{2} + 2 \, A a^{2} c + {\left (2 \, B a^{2} - A a b\right )} c x\right )} \sqrt {c x^{2} + b x + a}}{a^{3} b^{2} - 4 \, a^{4} c + {\left (a^{2} b^{2} c - 4 \, a^{3} c^{2}\right )} x^{2} + {\left (a^{2} b^{3} - 4 \, a^{3} b c\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.27, size = 125, normalized size = 1.30 \begin {gather*} -\frac {2 \, {\left (\frac {{\left (2 \, B a^{2} c - A a b c\right )} x}{a^{2} b^{2} - 4 \, a^{3} c} + \frac {B a^{2} b - A a b^{2} + 2 \, A a^{2} c}{a^{2} b^{2} - 4 \, a^{3} c}\right )}}{\sqrt {c x^{2} + b x + a}} + \frac {2 \, A \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 153, normalized size = 1.59 \begin {gather*} -\frac {2 A b c x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a}-\frac {A \,b^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, a}+\frac {2 \left (2 c x +b \right ) B}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {A \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{a^{\frac {3}{2}}}+\frac {A}{\sqrt {c \,x^{2}+b x +a}\, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*B*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+A/a/(c*x^2+b*x+a)^(1/2)-2*A*b/a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*
c*x-A*b^2/a/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-A/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/(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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+B\,x}{x\,{\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*(a + b*x + c*x^2)^(3/2)),x)

[Out]

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

[Out]

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

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