3.10.18 \(\int \frac {(A+B x) \sqrt {a+b x+c x^2}}{x^4} \, dx\) [918]

Optimal. Leaf size=121 \[ \frac {(A b-2 a B) (2 a+b x) \sqrt {a+b x+c x^2}}{8 a^2 x^2}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{3 a x^3}-\frac {(A b-2 a B) \left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 a^{5/2}} \]

[Out]

-1/3*A*(c*x^2+b*x+a)^(3/2)/a/x^3-1/16*(A*b-2*B*a)*(-4*a*c+b^2)*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/
2))/a^(5/2)+1/8*(A*b-2*B*a)*(b*x+2*a)*(c*x^2+b*x+a)^(1/2)/a^2/x^2

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Rubi [A]
time = 0.04, antiderivative size = 121, 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 = {820, 734, 738, 212} \begin {gather*} -\frac {\left (b^2-4 a c\right ) (A b-2 a B) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )}{16 a^{5/2}}+\frac {(2 a+b x) (A b-2 a B) \sqrt {a+b x+c x^2}}{8 a^2 x^2}-\frac {A \left (a+b x+c x^2\right )^{3/2}}{3 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^4,x]

[Out]

((A*b - 2*a*B)*(2*a + b*x)*Sqrt[a + b*x + c*x^2])/(8*a^2*x^2) - (A*(a + b*x + c*x^2)^(3/2))/(3*a*x^3) - ((A*b
- 2*a*B)*(b^2 - 4*a*c)*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + b*x + c*x^2])])/(16*a^(5/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 734

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] /; Free
Q[{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 738

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 820

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)/(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[S
implify[m + 2*p + 3], 0]

Rubi steps

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

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Mathematica [A]
time = 0.84, size = 165, normalized size = 1.36 \begin {gather*} \frac {\sqrt {a} \sqrt {a+x (b+c x)} \left (3 A b^2 x^2-4 a^2 (2 A+3 B x)-2 a x (3 b B x+A (b+4 c x))\right )+3 \left (A b^3+8 a^2 B c\right ) x^3 \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )+6 a b (b B+2 A c) x^3 \tanh ^{-1}\left (\frac {-\sqrt {c} x+\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )}{24 a^{5/2} x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a + b*x + c*x^2])/x^4,x]

[Out]

(Sqrt[a]*Sqrt[a + x*(b + c*x)]*(3*A*b^2*x^2 - 4*a^2*(2*A + 3*B*x) - 2*a*x*(3*b*B*x + A*(b + 4*c*x))) + 3*(A*b^
3 + 8*a^2*B*c)*x^3*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqrt[a]] + 6*a*b*(b*B + 2*A*c)*x^3*ArcTanh[(-(S
qrt[c]*x) + Sqrt[a + x*(b + c*x)])/Sqrt[a]])/(24*a^(5/2)*x^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(604\) vs. \(2(103)=206\).
time = 0.79, size = 605, normalized size = 5.00

method	result	size
risch	\(-\frac {\sqrt {c \,x^{2}+b x +a}\, \left (8 a A c \,x^{2}-3 b^{2} A \,x^{2}+6 B a b \,x^{2}+2 a b A x +12 a^{2} B x +8 a^{2} A \right )}{24 x^{3} a^{2}}+\frac {\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) A b c}{4 a^{\frac {3}{2}}}-\frac {\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) A \,b^{3}}{16 a^{\frac {5}{2}}}-\frac {\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) B c}{2 \sqrt {a}}+\frac {\ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right ) B \,b^{2}}{8 a^{\frac {3}{2}}}\)	\(216\)
default	\(B \left (-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{2 a \,x^{2}}-\frac {b \left (-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{a x}+\frac {b \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )}{2 a}+\frac {2 c \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{a}\right )}{4 a}+\frac {c \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )}{2 a}\right )+A \left (-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3 a \,x^{3}}-\frac {b \left (-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{2 a \,x^{2}}-\frac {b \left (-\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{a x}+\frac {b \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )}{2 a}+\frac {2 c \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{a}\right )}{4 a}+\frac {c \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )}{2 a}\right )}{2 a}\right )\)	\(605\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x+a)^(1/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

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

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Fricas [A]
time = 2.89, size = 317, normalized size = 2.62 \begin {gather*} \left [\frac {3 \, {\left (2 \, B a b^{2} - A b^{3} - 4 \, {\left (2 \, B a^{2} - A a b\right )} c\right )} \sqrt {a} x^{3} \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 (8 \, A a^{3} + {\left (6 \, B a^{2} b - 3 \, A a b^{2} + 8 \, A a^{2} c\right )} x^{2} + 2 \, {\left (6 \, B a^{3} + A a^{2} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{96 \, a^{3} x^{3}}, -\frac {3 \, {\left (2 \, B a b^{2} - A b^{3} - 4 \, {\left (2 \, B a^{2} - A a b\right )} c\right )} \sqrt {-a} x^{3} \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 (8 \, A a^{3} + {\left (6 \, B a^{2} b - 3 \, A a b^{2} + 8 \, A a^{2} c\right )} x^{2} + 2 \, {\left (6 \, B a^{3} + A a^{2} b\right )} x\right )} \sqrt {c x^{2} + b x + a}}{48 \, a^{3} x^{3}}\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^4,x, algorithm="fricas")

[Out]

[1/96*(3*(2*B*a*b^2 - A*b^3 - 4*(2*B*a^2 - A*a*b)*c)*sqrt(a)*x^3*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*(8*A*a^3 + (6*B*a^2*b - 3*A*a*b^2 + 8*A*a^2*c)*x^2 + 2*(6
*B*a^3 + A*a^2*b)*x)*sqrt(c*x^2 + b*x + a))/(a^3*x^3), -1/48*(3*(2*B*a*b^2 - A*b^3 - 4*(2*B*a^2 - A*a*b)*c)*sq
rt(-a)*x^3*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*(8*A*a^3 + (6*B*
a^2*b - 3*A*a*b^2 + 8*A*a^2*c)*x^2 + 2*(6*B*a^3 + A*a^2*b)*x)*sqrt(c*x^2 + b*x + a))/(a^3*x^3)]

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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^{4}}\, 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**4,x)

[Out]

Integral((A + B*x)*sqrt(a + b*x + c*x**2)/x**4, x)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 524 vs. \(2 (105) = 210\).
time = 1.40, size = 524, normalized size = 4.33 \begin {gather*} -\frac {{\left (2 \, B a b^{2} - A b^{3} - 8 \, B a^{2} c + 4 \, A a b c\right )} \arctan \left (-\frac {\sqrt {c} x - \sqrt {c x^{2} + b x + a}}{\sqrt {-a}}\right )}{8 \, \sqrt {-a} a^{2}} + \frac {6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} B a b^{2} - 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} A b^{3} + 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} B a^{2} c + 12 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{5} A a b c + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} B a^{2} b \sqrt {c} + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{4} A a^{2} c^{\frac {3}{2}} + 8 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a b^{3} + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{3} A a^{2} b c - 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} B a^{3} b \sqrt {c} + 48 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} A a^{2} b^{2} \sqrt {c} - 6 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{3} b^{2} + 3 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a^{2} b^{3} - 24 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} B a^{4} c + 36 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} A a^{3} b c + 16 \, A a^{4} c^{\frac {3}{2}}}{24 \, {\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )}^{2} - a\right )}^{3} a^{2}} \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^4,x, algorithm="giac")

[Out]

-1/8*(2*B*a*b^2 - A*b^3 - 8*B*a^2*c + 4*A*a*b*c)*arctan(-(sqrt(c)*x - sqrt(c*x^2 + b*x + a))/sqrt(-a))/(sqrt(-
a)*a^2) + 1/24*(6*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a*b^2 - 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*b^
3 + 24*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*B*a^2*c + 12*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^5*A*a*b*c + 48*(
sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*B*a^2*b*sqrt(c) + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^4*A*a^2*c^(3/2)
+ 8*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a*b^3 + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^3*A*a^2*b*c - 48*(s
qrt(c)*x - sqrt(c*x^2 + b*x + a))^2*B*a^3*b*sqrt(c) + 48*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2*A*a^2*b^2*sqrt(
c) - 6*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^3*b^2 + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^2*b^3 - 24*(s
qrt(c)*x - sqrt(c*x^2 + b*x + a))*B*a^4*c + 36*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*A*a^3*b*c + 16*A*a^4*c^(3/2
))/(((sqrt(c)*x - sqrt(c*x^2 + b*x + a))^2 - a)^3*a^2)

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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^4} \,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^4,x)

[Out]

int(((A + B*x)*(a + b*x + c*x^2)^(1/2))/x^4, x)

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