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

Optimal. Leaf size=218 \[ a^{3/2} (-A) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )+\frac {\left (\left (b^2-4 a c\right ) \left (-12 a B c-8 A b c+3 b^2 B\right )+64 a A b c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2}}-\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{64 c^2}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c} \]

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Rubi [A]  time = 0.25, antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {814, 843, 621, 206, 724} \begin {gather*} a^{3/2} (-A) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )-\frac {\sqrt {a+b x+c x^2} \left (2 c x \left (-12 a B c-8 A b c+3 b^2 B\right )-64 a A c^2-12 a b B c-8 A b^2 c+3 b^3 B\right )}{64 c^2}+\frac {\left (\left (b^2-4 a c\right ) \left (-12 a B c-8 A b c+3 b^2 B\right )+64 a A b c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 c^{5/2}}+\frac {\left (a+b x+c x^2\right )^{3/2} (8 A c+3 b B+6 B c x)}{24 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*b^3*B - 8*A*b^2*c - 12*a*b*B*c - 64*a*A*c^2 + 2*c*(3*b^2*B - 8*A*b*c - 12*a*B*c)*x)*Sqrt[a + b*x + c*x^2]
)/(64*c^2) + ((3*b*B + 8*A*c + 6*B*c*x)*(a + b*x + c*x^2)^(3/2))/(24*c) - a^(3/2)*A*ArcTanh[(2*a + b*x)/(2*Sqr
t[a]*Sqrt[a + b*x + c*x^2])] + ((64*a*A*b*c^2 + (b^2 - 4*a*c)*(3*b^2*B - 8*A*b*c - 12*a*B*c))*ArcTanh[(b + 2*c
*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(128*c^(5/2))

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 814

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((d + e*x)^(m + 1)*(c*e*f*(m + 2*p + 2) - g*(c*d + 2*c*d*p - b*e*p) + g*c*e*(m + 2*p + 1)*x)*(a + b*x + c*x^
2)^p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), x] - Dist[p/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a
+ b*x + c*x^2)^(p - 1)*Simp[c*e*f*(b*d - 2*a*e)*(m + 2*p + 2) + g*(a*e*(b*e - 2*c*d*m + b*e*m) + b*d*(b*e*p -
c*d - 2*c*d*p)) + (c*e*f*(2*c*d - b*e)*(m + 2*p + 2) + g*(b^2*e^2*(p + m + 1) - 2*c^2*d^2*(1 + 2*p) - c*e*(b*
d*(m - 2*p) + 2*a*e*(m + 2*p + 1))))*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] && GtQ[p, 0] && (IntegerQ[p] ||  !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])
) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

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) \left (a+b x+c x^2\right )^{3/2}}{x} \, dx &=\frac {(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}-\frac {\int \frac {\left (-8 a A c-\frac {1}{2} \left (8 A b c-3 B \left (b^2-4 a c\right )\right ) x\right ) \sqrt {a+b x+c x^2}}{x} \, dx}{8 c}\\ &=-\frac {\left (3 b^3 B-8 A b^2 c-12 a b B c-64 a A c^2+2 c \left (3 b^2 B-8 A b c-12 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}+\frac {\int \frac {32 a^2 A c^2+\frac {1}{4} \left (64 a A b c^2-\left (b^2-4 a c\right ) \left (8 A b c-3 B \left (b^2-4 a c\right )\right )\right ) x}{x \sqrt {a+b x+c x^2}} \, dx}{32 c^2}\\ &=-\frac {\left (3 b^3 B-8 A b^2 c-12 a b B c-64 a A c^2+2 c \left (3 b^2 B-8 A b c-12 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}+\left (a^2 A\right ) \int \frac {1}{x \sqrt {a+b x+c x^2}} \, dx+\frac {1}{128} \left (64 a A b+\frac {\left (b^2-4 a c\right ) \left (3 b^2 B-8 A b c-12 a B c\right )}{c^2}\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx\\ &=-\frac {\left (3 b^3 B-8 A b^2 c-12 a b B c-64 a A c^2+2 c \left (3 b^2 B-8 A b c-12 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}-\left (2 a^2 A\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 )+\frac {1}{64} \left (64 a A b+\frac {\left (b^2-4 a c\right ) \left (3 b^2 B-8 A b c-12 a B c\right )}{c^2}\right ) \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 (3 b^3 B-8 A b^2 c-12 a b B c-64 a A c^2+2 c \left (3 b^2 B-8 A b c-12 a B c\right ) x\right ) \sqrt {a+b x+c x^2}}{64 c^2}+\frac {(3 b B+8 A c+6 B c x) \left (a+b x+c x^2\right )^{3/2}}{24 c}-a^{3/2} A \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+b x+c x^2}}\right )+\frac {\left (64 a A b+\frac {\left (b^2-4 a c\right ) \left (3 b^2 B-8 A b c-12 a B c\right )}{c^2}\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{128 \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 206, normalized size = 0.94 \begin {gather*} a^{3/2} (-A) \tanh ^{-1}\left (\frac {2 a+b x}{2 \sqrt {a} \sqrt {a+x (b+c x)}}\right )+\frac {\left (48 a^2 B c^2+96 a A b c^2-24 a b^2 B c-8 A b^3 c+3 b^4 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{128 c^{5/2}}+\frac {\sqrt {a+x (b+c x)} \left (4 b c (15 a B+2 c x (14 A+9 B x))+8 c^2 \left (32 a A+15 a B x+8 A c x^2+6 B c x^3\right )+6 b^2 c (4 A+B x)-9 b^3 B\right )}{192 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + x*(b + c*x)]*(-9*b^3*B + 6*b^2*c*(4*A + B*x) + 8*c^2*(32*a*A + 15*a*B*x + 8*A*c*x^2 + 6*B*c*x^3) + 4
*b*c*(15*a*B + 2*c*x*(14*A + 9*B*x))))/(192*c^2) - a^(3/2)*A*ArcTanh[(2*a + b*x)/(2*Sqrt[a]*Sqrt[a + x*(b + c*
x)])] + ((3*b^4*B - 8*A*b^3*c - 24*a*b^2*B*c + 96*a*A*b*c^2 + 48*a^2*B*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr
t[a + x*(b + c*x)])])/(128*c^(5/2))

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IntegrateAlgebraic [A]  time = 0.95, size = 226, normalized size = 1.04 \begin {gather*} 2 a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}-\frac {\sqrt {a+b x+c x^2}}{\sqrt {a}}\right )+\frac {\left (-48 a^2 B c^2-96 a A b c^2+24 a b^2 B c+8 A b^3 c-3 b^4 B\right ) \log \left (-2 c^{5/2} \sqrt {a+b x+c x^2}+b c^2+2 c^3 x\right )}{128 c^{5/2}}+\frac {\sqrt {a+b x+c x^2} \left (256 a A c^2+60 a b B c+120 a B c^2 x+24 A b^2 c+112 A b c^2 x+64 A c^3 x^2-9 b^3 B+6 b^2 B c x+72 b B c^2 x^2+48 B c^3 x^3\right )}{192 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-9*b^3*B + 24*A*b^2*c + 60*a*b*B*c + 256*a*A*c^2 + 6*b^2*B*c*x + 112*A*b*c^2*x + 120*a
*B*c^2*x + 72*b*B*c^2*x^2 + 64*A*c^3*x^2 + 48*B*c^3*x^3))/(192*c^2) + 2*a^(3/2)*A*ArcTanh[(Sqrt[c]*x)/Sqrt[a]
- Sqrt[a + b*x + c*x^2]/Sqrt[a]] + ((-3*b^4*B + 8*A*b^3*c + 24*a*b^2*B*c - 96*a*A*b*c^2 - 48*a^2*B*c^2)*Log[b*
c^2 + 2*c^3*x - 2*c^(5/2)*Sqrt[a + b*x + c*x^2]])/(128*c^(5/2))

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fricas [A]  time = 5.17, size = 1023, normalized size = 4.69

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)^(3/2)/x,x, algorithm="fricas")

[Out]

[1/768*(384*A*a^(3/2)*c^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) + 3*(3*B*b^4 + 48*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*sqrt(c)*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*(48*B*c^4*x^3 - 9*B*b^3*c + 256*A*a*c^3 + 12*
(5*B*a*b + 2*A*b^2)*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 4*(15*B*a + 14*A*b)*c^3)*x)*sqrt(c*x^
2 + b*x + a))/c^3, 1/384*(192*A*a^(3/2)*c^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) - 3*(3*B*b^4 + 48*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*sqrt(-c)*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)) + 2*(48*B*c^4*x^3 - 9*B*b^3*c + 256*A
*a*c^3 + 12*(5*B*a*b + 2*A*b^2)*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 4*(15*B*a + 14*A*b)*c^3)*
x)*sqrt(c*x^2 + b*x + a))/c^3, 1/768*(768*A*sqrt(-a)*a*c^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)) + 3*(3*B*b^4 + 48*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*sqrt(c)*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*(48*B*c^4*x^3 - 9*B*b^3*c +
256*A*a*c^3 + 12*(5*B*a*b + 2*A*b^2)*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 4*(15*B*a + 14*A*b)
*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^3, 1/384*(384*A*sqrt(-a)*a*c^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)) - 3*(3*B*b^4 + 48*(B*a^2 + 2*A*a*b)*c^2 - 8*(3*B*a*b^2 + A*b^3)*c)*sqrt(-c)*
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)) + 2*(48*B*c^4*x^3 - 9*B*b^3*c +
256*A*a*c^3 + 12*(5*B*a*b + 2*A*b^2)*c^2 + 8*(9*B*b*c^3 + 8*A*c^4)*x^2 + 2*(3*B*b^2*c^2 + 4*(15*B*a + 14*A*b)
*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^3]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a sub
stitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perha
ps be purged.index.cc index_m operator + Error: Bad Argument Value

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maple [B]  time = 0.05, size = 390, normalized size = 1.79 \begin {gather*} -A \,a^{\frac {3}{2}} \ln \left (\frac {b x +2 a +2 \sqrt {c \,x^{2}+b x +a}\, \sqrt {a}}{x}\right )+\frac {3 A a b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{4 \sqrt {c}}-\frac {A \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {3}{2}}}+\frac {3 B \,a^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 \sqrt {c}}-\frac {3 B a \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {3}{2}}}+\frac {3 B \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {5}{2}}}+\frac {\sqrt {c \,x^{2}+b x +a}\, A b x}{4}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B a x}{8}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, B \,b^{2} x}{32 c}+\sqrt {c \,x^{2}+b x +a}\, A a +\frac {\sqrt {c \,x^{2}+b x +a}\, A \,b^{2}}{8 c}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B a b}{16 c}-\frac {3 \sqrt {c \,x^{2}+b x +a}\, B \,b^{3}}{64 c^{2}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B x}{4}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A}{3}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b}{8 c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*B*x*(c*x^2+b*x+a)^(3/2)+1/8*B/c*(c*x^2+b*x+a)^(3/2)*b+3/8*B*(c*x^2+b*x+a)^(1/2)*x*a-3/32*B/c*(c*x^2+b*x+a)
^(1/2)*x*b^2+3/16*B/c*(c*x^2+b*x+a)^(1/2)*b*a-3/64*B/c^2*(c*x^2+b*x+a)^(1/2)*b^3+3/8*B/c^(1/2)*ln((c*x+1/2*b)/
c^(1/2)+(c*x^2+b*x+a)^(1/2))*a^2-3/16*B/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^2*a+3/128*B/c^(5
/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^4+1/3*A*(c*x^2+b*x+a)^(3/2)+1/4*A*b*(c*x^2+b*x+a)^(1/2)*x+1/
8*A/c*(c*x^2+b*x+a)^(1/2)*b^2+3/4*A*b/c^(1/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*a-1/16*A/c^(3/2)*ln(
(c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*b^3+A*a*(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)*(c*x^2+b*x+a)^(3/2)/x,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 \begin {gather*} \int \frac {\left (A+B\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{x} \,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)^(3/2))/x,x)

[Out]

int(((A + B*x)*(a + b*x + c*x^2)^(3/2))/x, 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 ) \left (a + b x + c x^{2}\right )^{\frac {3}{2}}}{x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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