3.10.27 \(\int \frac {(A+B x) (a+b x+c x^2)^{3/2}}{x} \, dx\) [927]

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

[Out]

1/24*(6*B*c*x+8*A*c+3*B*b)*(c*x^2+b*x+a)^(3/2)/c-a^(3/2)*A*arctanh(1/2*(b*x+2*a)/a^(1/2)/(c*x^2+b*x+a)^(1/2))+
1/128*(64*a*A*b*c^2+(-4*a*c+b^2)*(-8*A*b*c-12*B*a*c+3*B*b^2))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2
))/c^(5/2)-1/64*(3*b^3*B-8*A*b^2*c-12*a*b*B*c-64*a*A*c^2+2*c*(-8*A*b*c-12*B*a*c+3*B*b^2)*x)*(c*x^2+b*x+a)^(1/2
)/c^2

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Rubi [A]
time = 0.17, 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 = {828, 857, 635, 212, 738} \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 {\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} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*((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])/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 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 635

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 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 828

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 857

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 ) \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 )+\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 ) \text {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 = 0.91, size = 208, normalized size = 0.95 \begin {gather*} \frac {\sqrt {a+x (b+c x)} \left (-9 b^3 B+6 b^2 c (4 A+B x)+8 c^2 \left (32 a A+15 a B x+8 A c x^2+6 B c x^3\right )+4 b c (15 a B+2 c x (14 A+9 B x))\right )}{192 c^2}+2 a^{3/2} A \tanh ^{-1}\left (\frac {\sqrt {c} x-\sqrt {a+x (b+c x)}}{\sqrt {a}}\right )-\frac {\left (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\right ) \log \left (c^2 \left (b+2 c x-2 \sqrt {c} \sqrt {a+x (b+c x)}\right )\right )}{128 c^{5/2}} \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) + 2*a^(3/2)*A*ArcTanh[(Sqrt[c]*x - Sqrt[a + x*(b + c*x)])/Sqr
t[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[c^2*(b + 2*c*x - 2*Sqrt[c]*Sqr
t[a + x*(b + c*x)])])/(128*c^(5/2))

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Maple [A]
time = 0.85, size = 271, normalized size = 1.24

method	result	size
default	\(B \left (\frac {\left (2 c x +b \right ) \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{8 c}+\frac {3 \left (4 a c -b^{2}\right ) \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{16 c}\right )+A \left (\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}}}{3}+\frac {b \left (\frac {\left (2 c x +b \right ) \sqrt {c \,x^{2}+b x +a}}{4 c}+\frac {\left (4 a c -b^{2}\right ) \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}\right )}{2}+a \left (\sqrt {c \,x^{2}+b x +a}+\frac {b \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 \sqrt {c}}-\sqrt {a}\, \ln \left (\frac {2 a +b x +2 \sqrt {a}\, \sqrt {c \,x^{2}+b x +a}}{x}\right )\right )\right )\)	\(271\)

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,method=_RETURNVERBOSE)

[Out]

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

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Fricas [A]
time = 16.47, size = 1023, normalized size = 4.69 \begin {gather*} \left [\frac {384 \, A a^{\frac {3}{2}} c^{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 ) + 3 \, {\left (3 \, B b^{4} + 48 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} c\right )} \sqrt {c} \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 ) + 4 \, {\left (48 \, B c^{4} x^{3} - 9 \, B b^{3} c + 256 \, A a c^{3} + 12 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} c^{2} + 4 \, {\left (15 \, B a + 14 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{3}}, \frac {192 \, A a^{\frac {3}{2}} c^{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 ) - 3 \, {\left (3 \, B b^{4} + 48 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} c\right )} \sqrt {-c} \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 ) + 2 \, {\left (48 \, B c^{4} x^{3} - 9 \, B b^{3} c + 256 \, A a c^{3} + 12 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} c^{2} + 4 \, {\left (15 \, B a + 14 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{3}}, \frac {768 \, A \sqrt {-a} a c^{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 ) + 3 \, {\left (3 \, B b^{4} + 48 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} c\right )} \sqrt {c} \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 ) + 4 \, {\left (48 \, B c^{4} x^{3} - 9 \, B b^{3} c + 256 \, A a c^{3} + 12 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} c^{2} + 4 \, {\left (15 \, B a + 14 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{768 \, c^{3}}, \frac {384 \, A \sqrt {-a} a c^{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 ) - 3 \, {\left (3 \, B b^{4} + 48 \, {\left (B a^{2} + 2 \, A a b\right )} c^{2} - 8 \, {\left (3 \, B a b^{2} + A b^{3}\right )} c\right )} \sqrt {-c} \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 ) + 2 \, {\left (48 \, B c^{4} x^{3} - 9 \, B b^{3} c + 256 \, A a c^{3} + 12 \, {\left (5 \, B a b + 2 \, A b^{2}\right )} c^{2} + 8 \, {\left (9 \, B b c^{3} + 8 \, A c^{4}\right )} x^{2} + 2 \, {\left (3 \, B b^{2} c^{2} + 4 \, {\left (15 \, B a + 14 \, A b\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{384 \, c^{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)^(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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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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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,sageVARx):;OUTP
UT:Error: Bad Argument Type

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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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