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

Optimal. Leaf size=133 \[ \frac {\left (-4 a A c-4 a b B+A b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{8 a^{3/2}}-\frac {\sqrt {a+b x+c x^2} (x (4 a B+A b)+2 a A)}{4 a x^2}+B \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \]

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Rubi [A]  time = 0.10, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {810, 843, 621, 206, 724} \begin {gather*} \frac {\left (-4 a A c-4 a b B+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^{3/2}}-\frac {\sqrt {a+b x+c x^2} (x (4 a B+A b)+2 a A)}{4 a x^2}+B \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((2*a*A + (A*b + 4*a*B)*x)*Sqrt[a + b*x + c*x^2])/(4*a*x^2) + ((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^(3/2)) + B*Sqrt[c]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*
x^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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 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 810

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

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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]
&&  !IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.35, size = 129, normalized size = 0.97 \begin {gather*} \frac {\left (A \left (b^2-4 a c\right )-4 a b B\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )}{8 a^{3/2}}-\frac {\sqrt {a+x (b+c x)} (2 a (A+2 B x)+A b x)}{4 a x^2}+B \sqrt {c} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.70, size = 131, normalized size = 0.98 \begin {gather*} \frac {\left (-4 a A c-4 a b B+A b^2\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} (-2 a A-4 a B x-A b x)}{4 a x^2}-B \sqrt {c} \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.96, size = 699, normalized size = 5.26 \begin {gather*} \left [\frac {8 \, B a^{2} \sqrt {c} x^{2} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + {\left (4 \, B a b - 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} + A a b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{16 \, a^{2} x^{2}}, -\frac {16 \, B a^{2} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - {\left (4 \, B a b - 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} + A a b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{16 \, a^{2} x^{2}}, \frac {4 \, B a^{2} \sqrt {c} x^{2} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + {\left (4 \, B a b - 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} + A a b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{8 \, a^{2} x^{2}}, -\frac {8 \, B a^{2} \sqrt {-c} x^{2} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) - {\left (4 \, B a b - 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} + A a b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{8 \, a^{2} x^{2}}\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^3,x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.30, size = 359, normalized size = 2.70 \begin {gather*} -B \sqrt {c} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} c - b \sqrt {c} \right |}\right ) + \frac {{\left (4 \, B a b - 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} + \frac {4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} B a b + {\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} + 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} A a b \sqrt {c} - 4 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{2} b + {\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}}{4 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )}^{2} a} \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^3,x, algorithm="giac")

[Out]

-B*sqrt(c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*c - b*sqrt(c))) + 1/4*(4*B*a*b - A*b^2 + 4*A*a*c)*ar
ctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a) + 1/4*(4*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^
3*B*a*b + (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*(sqr
t(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^2*sqrt(c) + 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a*b*sqrt(c) - 4*(s
qrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^2*b + (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))/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^2*a)

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maple [B]  time = 0.06, size = 304, normalized size = 2.29 \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 \sqrt {a}}+\frac {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 {3}{2}}}-\frac {B b \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{2 \sqrt {a}}+B \sqrt {c}\, \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )-\frac {\sqrt {c \,x^{2}+b x +a}\, A b c x}{4 a^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, B c x}{a}+\frac {\sqrt {c \,x^{2}+b x +a}\, A c}{2 a}-\frac {\sqrt {c \,x^{2}+b x +a}\, A \,b^{2}}{4 a^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, B b}{a}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b}{4 a^{2} x}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B}{a x}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A}{2 a \,x^{2}} \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^3,x)

[Out]

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

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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^3,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^3} \,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^3,x)

[Out]

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

[Out]

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

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