3.920 \(\int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^6} \, dx\)

Optimal. Leaf size=235 \[ \frac {\left (b^2-4 a c\right ) \left (2 a B \left (5 b^2-4 a c\right )-A \left (7 b^3-12 a b c\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{256 a^{9/2}}-\frac {\left (a+b x+c x^2\right )^{3/2} \left (-32 a A c-50 a b B+35 A b^2\right )}{240 a^3 x^3}+\frac {(7 A b-10 a B) \left (a+b x+c x^2\right )^{3/2}}{40 a^2 x^4}+\frac {(2 a+b x) \sqrt {a+b x+c x^2} \left (8 a^2 B c-12 a A b c-10 a b^2 B+7 A b^3\right )}{128 a^4 x^2}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{5 a x^5} \]

[Out]

-1/5*A*(c*x^2+b*x+a)^(3/2)/a/x^5+1/40*(7*A*b-10*B*a)*(c*x^2+b*x+a)^(3/2)/a^2/x^4-1/240*(-32*A*a*c+35*A*b^2-50*
B*a*b)*(c*x^2+b*x+a)^(3/2)/a^3/x^3+1/256*(-4*a*c+b^2)*(2*a*B*(-4*a*c+5*b^2)-A*(-12*a*b*c+7*b^3))*arctanh(1/2*(
b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))/a^(9/2)+1/128*(-12*A*a*b*c+7*A*b^3+8*B*a^2*c-10*B*a*b^2)*(b*x+2*a)*(c*x^
2+b*x+a)^(1/2)/a^4/x^2

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Rubi [A]  time = 0.27, antiderivative size = 235, 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 = {834, 806, 720, 724, 206} \[ -\frac {\left (a+b x+c x^2\right )^{3/2} \left (-32 a A c-50 a b B+35 A b^2\right )}{240 a^3 x^3}+\frac {(2 a+b x) \sqrt {a+b x+c x^2} \left (8 a^2 B c-12 a A b c-10 a b^2 B+7 A b^3\right )}{128 a^4 x^2}+\frac {\left (b^2-4 a c\right ) \left (2 a B \left (5 b^2-4 a c\right )-A \left (7 b^3-12 a b c\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{256 a^{9/2}}+\frac {(7 A b-10 a B) \left (a+b x+c x^2\right )^{3/2}}{40 a^2 x^4}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{5 a x^5} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^6,x]

[Out]

((7*A*b^3 - 10*a*b^2*B - 12*a*A*b*c + 8*a^2*B*c)*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(128*a^4*x^2) - (A*(a + b*
x + c*x^2)^(3/2))/(5*a*x^5) + ((7*A*b - 10*a*B)*(a + b*x + c*x^2)^(3/2))/(40*a^2*x^4) - ((35*A*b^2 - 50*a*b*B
- 32*a*A*c)*(a + b*x + c*x^2)^(3/2))/(240*a^3*x^3) + ((b^2 - 4*a*c)*(2*a*B*(5*b^2 - 4*a*c) - A*(7*b^3 - 12*a*b
*c))*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(256*a^(9/2))

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 720

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*
(d*b - 2*a*e + (2*c*d - b*e)*x)*(a + b*x + c*x^2)^p)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(p*(b^2 -
4*a*c))/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[
{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && EqQ[m +
2*p + 2, 0] && GtQ[p, 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) \sqrt {a+b x+c x^2}}{x^6} \, dx &=-\frac {A \left (a+b x+c x^2\right )^{3/2}}{5 a x^5}-\frac {\int \frac {\left (\frac {1}{2} (7 A b-10 a B)+2 A c x\right ) \sqrt {a+b x+c x^2}}{x^5} \, dx}{5 a}\\ &=-\frac {A \left (a+b x+c x^2\right )^{3/2}}{5 a x^5}+\frac {(7 A b-10 a B) \left (a+b x+c x^2\right )^{3/2}}{40 a^2 x^4}+\frac {\int \frac {\left (\frac {1}{4} \left (35 A b^2-50 a b B-32 a A c\right )+\frac {1}{2} (7 A b-10 a B) c x\right ) \sqrt {a+b x+c x^2}}{x^4} \, dx}{20 a^2}\\ &=-\frac {A \left (a+b x+c x^2\right )^{3/2}}{5 a x^5}+\frac {(7 A b-10 a B) \left (a+b x+c x^2\right )^{3/2}}{40 a^2 x^4}-\frac {\left (35 A b^2-50 a b B-32 a A c\right ) \left (a+b x+c x^2\right )^{3/2}}{240 a^3 x^3}-\frac {\left (7 A b^3-10 a b^2 B-12 a A b c+8 a^2 B c\right ) \int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx}{32 a^3}\\ &=\frac {\left (7 A b^3-10 a b^2 B-12 a A b c+8 a^2 B c\right ) (2 a+b x) \sqrt {a+b x+c x^2}}{128 a^4 x^2}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{5 a x^5}+\frac {(7 A b-10 a B) \left (a+b x+c x^2\right )^{3/2}}{40 a^2 x^4}-\frac {\left (35 A b^2-50 a b B-32 a A c\right ) \left (a+b x+c x^2\right )^{3/2}}{240 a^3 x^3}+\frac {\left (\left (b^2-4 a c\right ) \left (7 A b^3-10 a b^2 B-12 a A b c+8 a^2 B c\right )\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{256 a^4}\\ &=\frac {\left (7 A b^3-10 a b^2 B-12 a A b c+8 a^2 B c\right ) (2 a+b x) \sqrt {a+b x+c x^2}}{128 a^4 x^2}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{5 a x^5}+\frac {(7 A b-10 a B) \left (a+b x+c x^2\right )^{3/2}}{40 a^2 x^4}-\frac {\left (35 A b^2-50 a b B-32 a A c\right ) \left (a+b x+c x^2\right )^{3/2}}{240 a^3 x^3}-\frac {\left (\left (b^2-4 a c\right ) \left (7 A b^3-10 a b^2 B-12 a A b c+8 a^2 B 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 )}{128 a^4}\\ &=\frac {\left (7 A b^3-10 a b^2 B-12 a A b c+8 a^2 B c\right ) (2 a+b x) \sqrt {a+b x+c x^2}}{128 a^4 x^2}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{5 a x^5}+\frac {(7 A b-10 a B) \left (a+b x+c x^2\right )^{3/2}}{40 a^2 x^4}-\frac {\left (35 A b^2-50 a b B-32 a A c\right ) \left (a+b x+c x^2\right )^{3/2}}{240 a^3 x^3}+\frac {\left (b^2-4 a c\right ) \left (2 a B \left (5 b^2-4 a c\right )-A \left (7 b^3-12 a b c\right )\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{256 a^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.39, size = 206, normalized size = 0.88 \[ \frac {-\frac {5 \left (A \left (7 b^3-12 a b c\right )+2 a B \left (4 a c-5 b^2\right )\right ) \left (x^2 \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {a} (2 a+b x) \sqrt {a+x (b+c x)}\right )}{32 a^{7/2} x^2}+\frac {(a+x (b+c x))^{3/2} \left (32 a A c+50 a b B-35 A b^2\right )}{6 a^2 x^3}+\frac {(7 A b-10 a B) (a+x (b+c x))^{3/2}}{a x^4}-\frac {8 A (a+x (b+c x))^{3/2}}{x^5}}{40 a} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^6,x]

[Out]

((-8*A*(a + x*(b + c*x))^(3/2))/x^5 + ((7*A*b - 10*a*B)*(a + x*(b + c*x))^(3/2))/(a*x^4) + ((-35*A*b^2 + 50*a*
b*B + 32*a*A*c)*(a + x*(b + c*x))^(3/2))/(6*a^2*x^3) - (5*(2*a*B*(-5*b^2 + 4*a*c) + A*(7*b^3 - 12*a*b*c))*(-2*
Sqrt[a]*(2*a + b*x)*Sqrt[a + x*(b + c*x)] + (b^2 - 4*a*c)*x^2*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c
*x)])]))/(32*a^(7/2)*x^2))/(40*a)

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fricas [A]  time = 4.69, size = 553, normalized size = 2.35 \[ \left [-\frac {15 \, {\left (10 \, B a b^{4} - 7 \, A b^{5} + 16 \, {\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} c^{2} - 8 \, {\left (6 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} c\right )} \sqrt {a} x^{5} \log \left (-\frac {8 \, a b x + {\left (b^{2} + 4 \, a c\right )} x^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {a} + 8 \, a^{2}}{x^{2}}\right ) + 4 \, {\left (384 \, A a^{5} + {\left (150 \, B a^{2} b^{3} - 105 \, A a b^{4} - 256 \, A a^{3} c^{2} - 20 \, {\left (26 \, B a^{3} b - 23 \, A a^{2} b^{2}\right )} c\right )} x^{4} - 2 \, {\left (50 \, B a^{3} b^{2} - 35 \, A a^{2} b^{3} - 4 \, {\left (30 \, B a^{4} - 29 \, A a^{3} b\right )} c\right )} x^{3} + 8 \, {\left (10 \, B a^{4} b - 7 \, A a^{3} b^{2} + 16 \, A a^{4} c\right )} x^{2} + 48 \, {\left (10 \, B a^{5} + A a^{4} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{7680 \, a^{5} x^{5}}, -\frac {15 \, {\left (10 \, B a b^{4} - 7 \, A b^{5} + 16 \, {\left (2 \, B a^{3} - 3 \, A a^{2} b\right )} c^{2} - 8 \, {\left (6 \, B a^{2} b^{2} - 5 \, A a b^{3}\right )} c\right )} \sqrt {-a} x^{5} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (b x + 2 \, a\right )} \sqrt {-a}}{2 \, {\left (a c x^{2} + a b x + a^{2}\right )}}\right ) + 2 \, {\left (384 \, A a^{5} + {\left (150 \, B a^{2} b^{3} - 105 \, A a b^{4} - 256 \, A a^{3} c^{2} - 20 \, {\left (26 \, B a^{3} b - 23 \, A a^{2} b^{2}\right )} c\right )} x^{4} - 2 \, {\left (50 \, B a^{3} b^{2} - 35 \, A a^{2} b^{3} - 4 \, {\left (30 \, B a^{4} - 29 \, A a^{3} b\right )} c\right )} x^{3} + 8 \, {\left (10 \, B a^{4} b - 7 \, A a^{3} b^{2} + 16 \, A a^{4} c\right )} x^{2} + 48 \, {\left (10 \, B a^{5} + A a^{4} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3840 \, a^{5} x^{5}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(1/2)/x^6,x, algorithm="fricas")

[Out]

[-1/7680*(15*(10*B*a*b^4 - 7*A*b^5 + 16*(2*B*a^3 - 3*A*a^2*b)*c^2 - 8*(6*B*a^2*b^2 - 5*A*a*b^3)*c)*sqrt(a)*x^5
*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*(384*A*a^5
+ (150*B*a^2*b^3 - 105*A*a*b^4 - 256*A*a^3*c^2 - 20*(26*B*a^3*b - 23*A*a^2*b^2)*c)*x^4 - 2*(50*B*a^3*b^2 - 35*
A*a^2*b^3 - 4*(30*B*a^4 - 29*A*a^3*b)*c)*x^3 + 8*(10*B*a^4*b - 7*A*a^3*b^2 + 16*A*a^4*c)*x^2 + 48*(10*B*a^5 +
A*a^4*b)*x)*sqrt(c*x^2 + b*x + a))/(a^5*x^5), -1/3840*(15*(10*B*a*b^4 - 7*A*b^5 + 16*(2*B*a^3 - 3*A*a^2*b)*c^2
- 8*(6*B*a^2*b^2 - 5*A*a*b^3)*c)*sqrt(-a)*x^5*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*(384*A*a^5 + (150*B*a^2*b^3 - 105*A*a*b^4 - 256*A*a^3*c^2 - 20*(26*B*a^3*b - 23*A*a^2*b^2)
*c)*x^4 - 2*(50*B*a^3*b^2 - 35*A*a^2*b^3 - 4*(30*B*a^4 - 29*A*a^3*b)*c)*x^3 + 8*(10*B*a^4*b - 7*A*a^3*b^2 + 16
*A*a^4*c)*x^2 + 48*(10*B*a^5 + A*a^4*b)*x)*sqrt(c*x^2 + b*x + a))/(a^5*x^5)]

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giac [B]  time = 0.26, size = 1407, normalized size = 5.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(1/2)/x^6,x, algorithm="giac")

[Out]

-1/128*(10*B*a*b^4 - 7*A*b^5 - 48*B*a^2*b^2*c + 40*A*a*b^3*c + 32*B*a^3*c^2 - 48*A*a^2*b*c^2)*arctan(-(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-a)*a^4) + 1/1920*(150*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a*b
^4 - 105*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*b^5 - 720*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a^2*b^2*c +
600*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a*b^3*c + 480*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*B*a^3*c^2 - 7
20*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^9*A*a^2*b*c^2 - 700*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^2*b^4 + 4
90*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a*b^5 + 3360*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^3*b^2*c - 28
00*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^2*b^3*c + 2880*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^4*c^2 +
3360*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^3*b*c^2 + 11520*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*B*a^4*b*c
^(3/2) + 7680*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^6*A*a^4*c^(5/2) + 1280*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5
*B*a^3*b^4 - 896*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^2*b^5 + 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*
B*a^4*b^2*c + 5120*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^3*b^3*c + 15360*(sqrt(c)*x - sqrt(c*x^2 + b*x + a
))^5*A*a^4*b*c^2 + 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^4*b^3*sqrt(c) - 8960*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^4*B*a^5*b*c^(3/2) + 24320*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^4*b^2*c^(3/2) + 2560*(sqrt(c)
*x - sqrt(c*x^2 + b*x + a))^4*A*a^5*c^(5/2) - 580*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^4*b^4 + 790*(sqrt(
c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^3*b^5 - 3360*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^5*b^2*c + 9200*(sqr
t(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^4*b^3*c - 2880*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^6*c^2 + 12000*(
sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^5*b*c^2 - 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^5*b^3*sqrt(c
) + 3840*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^4*b^4*sqrt(c) - 1280*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*
B*a^6*b*c^(3/2) + 5120*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^5*b^2*c^(3/2) + 2560*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^2*A*a^6*c^(5/2) - 150*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^5*b^4 + 105*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))*A*a^4*b^5 - 3120*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^6*b^2*c + 3240*(sqrt(c)*x - sqrt(c*x^2 +
b*x + a))*A*a^5*b^3*c - 480*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^7*c^2 + 720*(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))*A*a^6*b*c^2 - 1280*B*a^7*b*c^(3/2) + 1280*A*a^6*b^2*c^(3/2) - 512*A*a^7*c^(5/2))/(((sqrt(c)*x - sqrt(c*x
^2 + b*x + a))^2 - a)^5*a^4)

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maple [B]  time = 0.07, size = 777, normalized size = 3.31 \[ -\frac {3 A b \,c^{2} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {5}{2}}}+\frac {5 A \,b^{3} c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{32 a^{\frac {7}{2}}}-\frac {7 A \,b^{5} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{256 a^{\frac {9}{2}}}+\frac {B \,c^{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 {3 B \,b^{2} c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{16 a^{\frac {5}{2}}}+\frac {5 B \,b^{4} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{128 a^{\frac {7}{2}}}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2} c^{2} x}{32 a^{4}}+\frac {7 \sqrt {c \,x^{2}+b x +a}\, A \,b^{4} c x}{128 a^{5}}+\frac {\sqrt {c \,x^{2}+b x +a}\, B b \,c^{2} x}{16 a^{3}}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, B \,b^{3} c x}{64 a^{4}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, A b \,c^{2}}{16 a^{3}}-\frac {13 \sqrt {c \,x^{2}+b x +a}\, A \,b^{3} c}{64 a^{4}}+\frac {7 \sqrt {c \,x^{2}+b x +a}\, A \,b^{5}}{128 a^{5}}-\frac {\sqrt {c \,x^{2}+b x +a}\, B \,c^{2}}{8 a^{2}}+\frac {7 \sqrt {c \,x^{2}+b x +a}\, B \,b^{2} c}{32 a^{3}}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, B \,b^{4}}{64 a^{4}}+\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{2} c}{32 a^{4} x}-\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{4}}{128 a^{5} x}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b c}{16 a^{3} x}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{3}}{64 a^{4} x}-\frac {3 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b c}{16 a^{3} x^{2}}+\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{3}}{64 a^{4} x^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B c}{8 a^{2} x^{2}}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{2}}{32 a^{3} x^{2}}+\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A c}{15 a^{2} x^{3}}-\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{2}}{48 a^{3} x^{3}}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b}{24 a^{2} x^{3}}+\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b}{40 a^{2} x^{4}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B}{4 a \,x^{4}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A}{5 a \,x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^(1/2)/x^6,x)

[Out]

-13/64*A/a^4*b^3*c*(c*x^2+b*x+a)^(1/2)+5/32*A/a^(7/2)*b^3*c*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+3/16
*A/a^3*b*c^2*(c*x^2+b*x+a)^(1/2)-3/16*A/a^(5/2)*b*c^2*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+2/15*A*c/a
^2/x^3*(c*x^2+b*x+a)^(3/2)+7/32*B/a^3*b^2*c*(c*x^2+b*x+a)^(1/2)-3/16*B/a^(5/2)*b^2*c*ln((b*x+2*a+2*(c*x^2+b*x+
a)^(1/2)*a^(1/2))/x)+1/8*B*c/a^2/x^2*(c*x^2+b*x+a)^(3/2)+7/128*A/a^5*b^4*c*(c*x^2+b*x+a)^(1/2)*x-3/16*A/a^3*b*
c/x^2*(c*x^2+b*x+a)^(3/2)+3/32*A/a^4*b^2*c/x*(c*x^2+b*x+a)^(3/2)-3/32*A/a^4*b^2*c^2*(c*x^2+b*x+a)^(1/2)*x-1/16
*B*c/a^3*b/x*(c*x^2+b*x+a)^(3/2)+1/16*B*c^2/a^3*b*(c*x^2+b*x+a)^(1/2)*x-5/64*B/a^4*b^3*c*(c*x^2+b*x+a)^(1/2)*x
-5/32*B/a^3*b^2/x^2*(c*x^2+b*x+a)^(3/2)+5/24*B/a^2*b/x^3*(c*x^2+b*x+a)^(3/2)+7/128*A/a^5*b^5*(c*x^2+b*x+a)^(1/
2)-7/256*A/a^(9/2)*b^5*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-1/4*B/a/x^4*(c*x^2+b*x+a)^(3/2)-5/64*B/a^
4*b^4*(c*x^2+b*x+a)^(1/2)+5/128*B/a^(7/2)*b^4*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-1/8*B*c^2/a^2*(c*x
^2+b*x+a)^(1/2)+1/8*B*c^2/a^(3/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+7/40*A/a^2*b/x^4*(c*x^2+b*x+a)
^(3/2)-7/48*A/a^3*b^2/x^3*(c*x^2+b*x+a)^(3/2)+7/64*A/a^4*b^3/x^2*(c*x^2+b*x+a)^(3/2)-7/128*A/a^5*b^4/x*(c*x^2+
b*x+a)^(3/2)+5/64*B/a^4*b^3/x*(c*x^2+b*x+a)^(3/2)-1/5*A*(c*x^2+b*x+a)^(3/2)/a/x^5

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)^(1/2)/x^6,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 = -0.00 \[ \int \frac {\left (A+B\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{x^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a + b*x + c*x^2)^(1/2))/x^6,x)

[Out]

int(((A + B*x)*(a + b*x + c*x^2)^(1/2))/x^6, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B x\right ) \sqrt {a + b x + c x^{2}}}{x^{6}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x+a)**(1/2)/x**6,x)

[Out]

Integral((A + B*x)*sqrt(a + b*x + c*x**2)/x**6, x)

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