3.9.44 \(\int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^5} \, dx\)

Optimal. Leaf size=172 \[ \frac {\left (b^2-4 a c\right ) \left (-4 a A c-8 a b B+5 A b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{128 a^{7/2}}-\frac {(2 a+b x) \sqrt {a+b x+c x^2} \left (-4 a A c-8 a b B+5 A b^2\right )}{64 a^3 x^2}+\frac {(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4} \]

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Rubi [A]  time = 0.14, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \begin {gather*} -\frac {(2 a+b x) \sqrt {a+b x+c x^2} \left (-4 a A c-8 a b B+5 A b^2\right )}{64 a^3 x^2}+\frac {\left (b^2-4 a c\right ) \left (-4 a A c-8 a b B+5 A b^2\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{128 a^{7/2}}+\frac {(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^5,x]

[Out]

-((5*A*b^2 - 8*a*b*B - 4*a*A*c)*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(64*a^3*x^2) - (A*(a + b*x + c*x^2)^(3/2))/
(4*a*x^4) + ((5*A*b - 8*a*B)*(a + b*x + c*x^2)^(3/2))/(24*a^2*x^3) + ((b^2 - 4*a*c)*(5*A*b^2 - 8*a*b*B - 4*a*A
*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(128*a^(7/2))

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^5} \, dx &=-\frac {A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4}-\frac {\int \frac {\left (\frac {1}{2} (5 A b-8 a B)+A c x\right ) \sqrt {a+b x+c x^2}}{x^4} \, dx}{4 a}\\ &=-\frac {A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4}+\frac {(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}+\frac {\left (5 A b^2-8 a b B-4 a A c\right ) \int \frac {\sqrt {a+b x+c x^2}}{x^3} \, dx}{16 a^2}\\ &=-\frac {\left (5 A b^2-8 a b B-4 a A c\right ) (2 a+b x) \sqrt {a+b x+c x^2}}{64 a^3 x^2}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4}+\frac {(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}-\frac {\left (\left (b^2-4 a c\right ) \left (5 A b^2-8 a b B-4 a A c\right )\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx}{128 a^3}\\ &=-\frac {\left (5 A b^2-8 a b B-4 a A c\right ) (2 a+b x) \sqrt {a+b x+c x^2}}{64 a^3 x^2}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4}+\frac {(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}+\frac {\left (\left (b^2-4 a c\right ) \left (5 A b^2-8 a b B-4 a 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 )}{64 a^3}\\ &=-\frac {\left (5 A b^2-8 a b B-4 a A c\right ) (2 a+b x) \sqrt {a+b x+c x^2}}{64 a^3 x^2}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{4 a x^4}+\frac {(5 A b-8 a B) \left (a+b x+c x^2\right )^{3/2}}{24 a^2 x^3}+\frac {\left (b^2-4 a c\right ) \left (5 A b^2-8 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 )}{128 a^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 153, normalized size = 0.89 \begin {gather*} \frac {\frac {3 \left (-4 a A c-8 a b B+5 A b^2\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 )}{16 a^{3/2} x^2}+\frac {(5 A b-8 a B) (a+x (b+c x))^{3/2}}{x^3}-\frac {6 a A (a+x (b+c x))^{3/2}}{x^4}}{24 a^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^5,x]

[Out]

((-6*a*A*(a + x*(b + c*x))^(3/2))/x^4 + ((5*A*b - 8*a*B)*(a + x*(b + c*x))^(3/2))/x^3 + (3*(5*A*b^2 - 8*a*b*B
- 4*a*A*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]*Sq
rt[a + x*(b + c*x)])]))/(16*a^(3/2)*x^2))/(24*a^2)

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IntegrateAlgebraic [A]  time = 1.49, size = 229, normalized size = 1.33 \begin {gather*} -\frac {5 A b^4 \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )}{64 a^{7/2}}+\frac {\left (2 a A c^2+4 a b B c-3 A b^2 c+b^3 (-B)\right ) \tanh ^{-1}\left (\frac {\sqrt {a+b x+c x^2}-\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2}}+\frac {\sqrt {a+b x+c x^2} \left (-48 a^3 A-64 a^3 B x-8 a^2 A b x-24 a^2 A c x^2-16 a^2 b B x^2-64 a^2 B c x^3+10 a A b^2 x^2+52 a A b c x^3+24 a b^2 B x^3-15 A b^3 x^3\right )}{192 a^3 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^5,x]

[Out]

(Sqrt[a + b*x + c*x^2]*(-48*a^3*A - 8*a^2*A*b*x - 64*a^3*B*x + 10*a*A*b^2*x^2 - 16*a^2*b*B*x^2 - 24*a^2*A*c*x^
2 - 15*A*b^3*x^3 + 24*a*b^2*B*x^3 + 52*a*A*b*c*x^3 - 64*a^2*B*c*x^3))/(192*a^3*x^4) + ((-(b^3*B) - 3*A*b^2*c +
4*a*b*B*c + 2*a*A*c^2)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + b*x + c*x^2])/Sqrt[a]])/(8*a^(5/2)) - (5*A*b^4*ArcTan
h[(Sqrt[c]*x)/Sqrt[a] - Sqrt[a + b*x + c*x^2]/Sqrt[a]])/(64*a^(7/2))

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fricas [A]  time = 1.02, size = 425, normalized size = 2.47 \begin {gather*} \left [-\frac {3 \, {\left (8 \, B a b^{3} - 5 \, A b^{4} - 16 \, A a^{2} c^{2} - 8 \, {\left (4 \, B a^{2} b - 3 \, A a b^{2}\right )} c\right )} \sqrt {a} x^{4} \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 (48 \, A a^{4} - {\left (24 \, B a^{2} b^{2} - 15 \, A a b^{3} - 4 \, {\left (16 \, B a^{3} - 13 \, A a^{2} b\right )} c\right )} x^{3} + 2 \, {\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2} + 12 \, A a^{3} c\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + A a^{3} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, a^{4} x^{4}}, \frac {3 \, {\left (8 \, B a b^{3} - 5 \, A b^{4} - 16 \, A a^{2} c^{2} - 8 \, {\left (4 \, B a^{2} b - 3 \, A a b^{2}\right )} c\right )} \sqrt {-a} x^{4} \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 (48 \, A a^{4} - {\left (24 \, B a^{2} b^{2} - 15 \, A a b^{3} - 4 \, {\left (16 \, B a^{3} - 13 \, A a^{2} b\right )} c\right )} x^{3} + 2 \, {\left (8 \, B a^{3} b - 5 \, A a^{2} b^{2} + 12 \, A a^{3} c\right )} x^{2} + 8 \, {\left (8 \, B a^{4} + A a^{3} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, a^{4} x^{4}}\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^5,x, algorithm="fricas")

[Out]

[-1/768*(3*(8*B*a*b^3 - 5*A*b^4 - 16*A*a^2*c^2 - 8*(4*B*a^2*b - 3*A*a*b^2)*c)*sqrt(a)*x^4*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*(48*A*a^4 - (24*B*a^2*b^2 - 15*A
*a*b^3 - 4*(16*B*a^3 - 13*A*a^2*b)*c)*x^3 + 2*(8*B*a^3*b - 5*A*a^2*b^2 + 12*A*a^3*c)*x^2 + 8*(8*B*a^4 + A*a^3*
b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^4), 1/384*(3*(8*B*a*b^3 - 5*A*b^4 - 16*A*a^2*c^2 - 8*(4*B*a^2*b - 3*A*a*b^
2)*c)*sqrt(-a)*x^4*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*(48*A*a^
4 - (24*B*a^2*b^2 - 15*A*a*b^3 - 4*(16*B*a^3 - 13*A*a^2*b)*c)*x^3 + 2*(8*B*a^3*b - 5*A*a^2*b^2 + 12*A*a^3*c)*x
^2 + 8*(8*B*a^4 + A*a^3*b)*x)*sqrt(c*x^2 + b*x + a))/(a^4*x^4)]

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giac [B]  time = 0.27, size = 991, normalized size = 5.76 \begin {gather*} \frac {{\left (8 \, B a b^{3} - 5 \, A b^{4} - 32 \, B a^{2} b c + 24 \, A a b^{2} c - 16 \, A a^{2} c^{2}\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{64 \, \sqrt {-a} a^{3}} - \frac {24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{7} B a b^{3} - 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{7} A b^{4} - 96 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{7} B a^{2} b c + 72 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{7} A a b^{2} c - 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{7} A a^{2} c^{2} - 384 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{6} B a^{3} c^{\frac {3}{2}} - 88 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} B a^{2} b^{3} + 55 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} A a b^{4} - 288 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} B a^{3} b c - 264 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} A a^{2} b^{2} c - 336 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} A a^{3} c^{2} - 384 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} B a^{3} b^{2} \sqrt {c} + 384 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} B a^{4} c^{\frac {3}{2}} - 1152 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} A a^{3} b c^{\frac {3}{2}} + 40 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} B a^{3} b^{3} - 73 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a^{2} b^{4} + 96 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} B a^{4} b c - 648 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a^{3} b^{2} c - 336 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a^{4} c^{2} + 384 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} B a^{4} b^{2} \sqrt {c} - 384 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} A a^{3} b^{3} \sqrt {c} - 128 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} B a^{5} c^{\frac {3}{2}} - 256 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} A a^{4} b c^{\frac {3}{2}} + 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{4} b^{3} - 15 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a^{3} b^{4} + 288 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{5} b c - 312 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a^{4} b^{2} c - 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a^{5} c^{2} + 128 \, B a^{6} c^{\frac {3}{2}} - 128 \, A a^{5} b c^{\frac {3}{2}}}{192 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )}^{4} a^{3}} \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^5,x, algorithm="giac")

[Out]

1/64*(8*B*a*b^3 - 5*A*b^4 - 32*B*a^2*b*c + 24*A*a*b^2*c - 16*A*a^2*c^2)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x
+ a))/sqrt(-a))/(sqrt(-a)*a^3) - 1/192*(24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a*b^3 - 15*(sqrt(c)*x - sqr
t(c*x^2 + b*x + a))^7*A*b^4 - 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*B*a^2*b*c + 72*(sqrt(c)*x - sqrt(c*x^2
+ b*x + a))^7*A*a*b^2*c - 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^7*A*a^2*c^2 - 384*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))^6*B*a^3*c^(3/2) - 88*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^2*b^3 + 55*(sqrt(c)*x - sqrt(c*x^2 + b
*x + a))^5*A*a*b^4 - 288*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^3*b*c - 264*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^5*A*a^2*b^2*c - 336*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a^3*c^2 - 384*(sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^4*B*a^3*b^2*sqrt(c) + 384*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^4*c^(3/2) - 1152*(sqrt(c)*x - sqrt(c*
x^2 + b*x + a))^4*A*a^3*b*c^(3/2) + 40*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^3*b^3 - 73*(sqrt(c)*x - sqrt(
c*x^2 + b*x + a))^3*A*a^2*b^4 + 96*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*B*a^4*b*c - 648*(sqrt(c)*x - sqrt(c*x
^2 + b*x + a))^3*A*a^3*b^2*c - 336*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^4*c^2 + 384*(sqrt(c)*x - sqrt(c*x
^2 + b*x + a))^2*B*a^4*b^2*sqrt(c) - 384*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^3*b^3*sqrt(c) - 128*(sqrt(c
)*x - sqrt(c*x^2 + b*x + a))^2*B*a^5*c^(3/2) - 256*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^4*b*c^(3/2) + 24*
(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^4*b^3 - 15*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^3*b^4 + 288*(sqrt(c
)*x - sqrt(c*x^2 + b*x + a))*B*a^5*b*c - 312*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^4*b^2*c - 48*(sqrt(c)*x -
sqrt(c*x^2 + b*x + a))*A*a^5*c^2 + 128*B*a^6*c^(3/2) - 128*A*a^5*b*c^(3/2))/(((sqrt(c)*x - sqrt(c*x^2 + b*x +
a))^2 - a)^4*a^3)

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maple [B]  time = 0.06, size = 569, normalized size = 3.31 \begin {gather*} \frac {A \,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 A \,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 A \,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 {B b c \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )}{4 a^{\frac {3}{2}}}-\frac {B \,b^{3} \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 {\sqrt {c \,x^{2}+b x +a}\, A b \,c^{2} x}{16 a^{3}}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,b^{3} c x}{64 a^{4}}+\frac {\sqrt {c \,x^{2}+b x +a}\, B \,b^{2} c x}{8 a^{3}}-\frac {\sqrt {c \,x^{2}+b x +a}\, A \,c^{2}}{8 a^{2}}+\frac {7 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2} c}{32 a^{3}}-\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,b^{4}}{64 a^{4}}-\frac {\sqrt {c \,x^{2}+b x +a}\, B b c}{4 a^{2}}+\frac {\sqrt {c \,x^{2}+b x +a}\, B \,b^{3}}{8 a^{3}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b c}{16 a^{3} x}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{3}}{64 a^{4} x}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{2}}{8 a^{3} x}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A c}{8 a^{2} x^{2}}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A \,b^{2}}{32 a^{3} x^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b}{4 a^{2} x^{2}}+\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b}{24 a^{2} x^{3}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B}{3 a \,x^{3}}-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A}{4 a \,x^{4}} \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^5,x)

[Out]

-1/4*A*(c*x^2+b*x+a)^(3/2)/a/x^4+5/24*A/a^2*b/x^3*(c*x^2+b*x+a)^(3/2)-5/32*A/a^3*b^2/x^2*(c*x^2+b*x+a)^(3/2)+5
/64*A/a^4*b^3/x*(c*x^2+b*x+a)^(3/2)-5/64*A/a^4*b^4*(c*x^2+b*x+a)^(1/2)+5/128*A/a^(7/2)*b^4*ln((b*x+2*a+2*(c*x^
2+b*x+a)^(1/2)*a^(1/2))/x)-5/64*A/a^4*b^3*c*(c*x^2+b*x+a)^(1/2)*x+7/32*A/a^3*b^2*c*(c*x^2+b*x+a)^(1/2)-3/16*A/
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*A*c/a^2/x^2*(c*x^2+b*x+a)^(3/2)-1/16*A*c/a^3*b
/x*(c*x^2+b*x+a)^(3/2)+1/16*A*c^2/a^3*b*(c*x^2+b*x+a)^(1/2)*x-1/8*A*c^2/a^2*(c*x^2+b*x+a)^(1/2)+1/8*A*c^2/a^(3
/2)*ln((b*x+2*a+2*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)-1/3*B/a/x^3*(c*x^2+b*x+a)^(3/2)+1/4*B/a^2*b/x^2*(c*x^2+b*x+a
)^(3/2)-1/8*B/a^3*b^2/x*(c*x^2+b*x+a)^(3/2)+1/8*B/a^3*b^3*(c*x^2+b*x+a)^(1/2)-1/16*B/a^(5/2)*b^3*ln((b*x+2*a+2
*(c*x^2+b*x+a)^(1/2)*a^(1/2))/x)+1/8*B/a^3*b^2*c*(c*x^2+b*x+a)^(1/2)*x-1/4*B/a^2*b*c*(c*x^2+b*x+a)^(1/2)+1/4*B
/a^(3/2)*b*c*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)*(c*x^2+b*x+a)^(1/2)/x^5,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^5} \,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^5,x)

[Out]

int(((A + B*x)*(a + b*x + c*x^2)^(1/2))/x^5, 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^{5}}\, 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**5,x)

[Out]

Integral((A + B*x)*sqrt(a + b*x + c*x**2)/x**5, x)

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