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

Optimal. Leaf size=205 \[ \frac {\left (a+b x+c x^2\right )^{3/2} \left (-32 a B c-6 c x (7 b B-10 A c)-50 A b c+35 b^2 B\right )}{240 c^3}+\frac {\left (b^2-4 a c\right ) \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{9/2}}-\frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right )}{128 c^4}+\frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c} \]

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Rubi [A]  time = 0.20, antiderivative size = 205, 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 = {832, 779, 612, 621, 206} \begin {gather*} \frac {\left (a+b x+c x^2\right )^{3/2} \left (-32 a B c-6 c x (7 b B-10 A c)-50 A b c+35 b^2 B\right )}{240 c^3}-\frac {(b+2 c x) \sqrt {a+b x+c x^2} \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right )}{128 c^4}+\frac {\left (b^2-4 a c\right ) \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{9/2}}+\frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((7*b^3*B - 10*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*(b + 2*c*x)*Sqrt[a + b*x + c*x^2])/(128*c^4) + (B*x^2*(a + b
*x + c*x^2)^(3/2))/(5*c) + ((35*b^2*B - 50*A*b*c - 32*a*B*c - 6*c*(7*b*B - 10*A*c)*x)*(a + b*x + c*x^2)^(3/2))
/(240*c^3) + ((b^2 - 4*a*c)*(7*b^3*B - 10*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqr
t[a + b*x + c*x^2])])/(256*c^(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 612

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p +
1)), x] - Dist[(p*(b^2 - 4*a*c))/(2*c*(2*p + 1)), Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

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 779

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] &&  !LeQ[p, -1]

Rule 832

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

Rubi steps

\begin {align*} \int x^2 (A+B x) \sqrt {a+b x+c x^2} \, dx &=\frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {\int x \left (-2 a B-\frac {1}{2} (7 b B-10 A c) x\right ) \sqrt {a+b x+c x^2} \, dx}{5 c}\\ &=\frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (35 b^2 B-50 A b c-32 a B c-6 c (7 b B-10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}-\frac {\left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) \int \sqrt {a+b x+c x^2} \, dx}{32 c^3}\\ &=-\frac {\left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^4}+\frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (35 b^2 B-50 A b c-32 a B c-6 c (7 b B-10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}+\frac {\left (\left (b^2-4 a c\right ) \left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{256 c^4}\\ &=-\frac {\left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^4}+\frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (35 b^2 B-50 A b c-32 a B c-6 c (7 b B-10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}+\frac {\left (\left (b^2-4 a c\right ) \left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right )\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 )}{128 c^4}\\ &=-\frac {\left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) (b+2 c x) \sqrt {a+b x+c x^2}}{128 c^4}+\frac {B x^2 \left (a+b x+c x^2\right )^{3/2}}{5 c}+\frac {\left (35 b^2 B-50 A b c-32 a B c-6 c (7 b B-10 A c) x\right ) \left (a+b x+c x^2\right )^{3/2}}{240 c^3}+\frac {\left (b^2-4 a c\right ) \left (7 b^3 B-10 A b^2 c-12 a b B c+8 a A c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 179, normalized size = 0.87 \begin {gather*} \frac {\frac {(a+x (b+c x))^{3/2} \left (4 c (15 A c x-8 a B)-2 b c (25 A+21 B x)+35 b^2 B\right )}{48 c^2}+\frac {5 \left (8 a A c^2-12 a b B c-10 A b^2 c+7 b^3 B\right ) \left (\left (b^2-4 a c\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )-2 \sqrt {c} (b+2 c x) \sqrt {a+x (b+c x)}\right )}{256 c^{7/2}}+B x^2 (a+x (b+c x))^{3/2}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x^2*(a + x*(b + c*x))^(3/2) + ((a + x*(b + c*x))^(3/2)*(35*b^2*B - 2*b*c*(25*A + 21*B*x) + 4*c*(-8*a*B + 15
*A*c*x)))/(48*c^2) + (5*(7*b^3*B - 10*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*(-2*Sqrt[c]*(b + 2*c*x)*Sqrt[a + x*(b
+ c*x)] + (b^2 - 4*a*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])]))/(256*c^(7/2)))/(5*c)

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IntegrateAlgebraic [A]  time = 0.94, size = 244, normalized size = 1.19 \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-256 a^2 B c^2-520 a A b c^2+240 a A c^3 x+460 a b^2 B c-232 a b B c^2 x+128 a B c^3 x^2+150 A b^3 c-100 A b^2 c^2 x+80 A b c^3 x^2+480 A c^4 x^3-105 b^4 B+70 b^3 B c x-56 b^2 B c^2 x^2+48 b B c^3 x^3+384 B c^4 x^4\right )}{1920 c^4}+\frac {\left (32 a^2 A c^3-48 a^2 b B c^2-48 a A b^2 c^2+40 a b^3 B c+10 A b^4 c-7 b^5 B\right ) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{256 c^{9/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x + c*x^2]*(-105*b^4*B + 150*A*b^3*c + 460*a*b^2*B*c - 520*a*A*b*c^2 - 256*a^2*B*c^2 + 70*b^3*B*c*
x - 100*A*b^2*c^2*x - 232*a*b*B*c^2*x + 240*a*A*c^3*x - 56*b^2*B*c^2*x^2 + 80*A*b*c^3*x^2 + 128*a*B*c^3*x^2 +
48*b*B*c^3*x^3 + 480*A*c^4*x^3 + 384*B*c^4*x^4))/(1920*c^4) + ((-7*b^5*B + 10*A*b^4*c + 40*a*b^3*B*c - 48*a*A*
b^2*c^2 - 48*a^2*b*B*c^2 + 32*a^2*A*c^3)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b*x + c*x^2]])/(256*c^(9/2))

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fricas [A]  time = 0.50, size = 517, normalized size = 2.52 \begin {gather*} \left [-\frac {15 \, {\left (7 \, B b^{5} - 32 \, A a^{2} c^{3} + 48 \, {\left (B a^{2} b + A a b^{2}\right )} c^{2} - 10 \, {\left (4 \, B a b^{3} + A b^{4}\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 (384 \, B c^{5} x^{4} - 105 \, B b^{4} c - 8 \, {\left (32 \, B a^{2} + 65 \, A a b\right )} c^{3} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 10 \, {\left (46 \, B a b^{2} + 15 \, A b^{3}\right )} c^{2} - 8 \, {\left (7 \, B b^{2} c^{3} - 2 \, {\left (8 \, B a + 5 \, A b\right )} c^{4}\right )} x^{2} + 2 \, {\left (35 \, B b^{3} c^{2} + 120 \, A a c^{4} - 2 \, {\left (58 \, B a b + 25 \, A b^{2}\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{7680 \, c^{5}}, -\frac {15 \, {\left (7 \, B b^{5} - 32 \, A a^{2} c^{3} + 48 \, {\left (B a^{2} b + A a b^{2}\right )} c^{2} - 10 \, {\left (4 \, B a b^{3} + A b^{4}\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 (384 \, B c^{5} x^{4} - 105 \, B b^{4} c - 8 \, {\left (32 \, B a^{2} + 65 \, A a b\right )} c^{3} + 48 \, {\left (B b c^{4} + 10 \, A c^{5}\right )} x^{3} + 10 \, {\left (46 \, B a b^{2} + 15 \, A b^{3}\right )} c^{2} - 8 \, {\left (7 \, B b^{2} c^{3} - 2 \, {\left (8 \, B a + 5 \, A b\right )} c^{4}\right )} x^{2} + 2 \, {\left (35 \, B b^{3} c^{2} + 120 \, A a c^{4} - 2 \, {\left (58 \, B a b + 25 \, A b^{2}\right )} c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3840 \, c^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/7680*(15*(7*B*b^5 - 32*A*a^2*c^3 + 48*(B*a^2*b + A*a*b^2)*c^2 - 10*(4*B*a*b^3 + A*b^4)*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*(384*B*c^5*x^4 - 105*B*b^4*c
- 8*(32*B*a^2 + 65*A*a*b)*c^3 + 48*(B*b*c^4 + 10*A*c^5)*x^3 + 10*(46*B*a*b^2 + 15*A*b^3)*c^2 - 8*(7*B*b^2*c^3
- 2*(8*B*a + 5*A*b)*c^4)*x^2 + 2*(35*B*b^3*c^2 + 120*A*a*c^4 - 2*(58*B*a*b + 25*A*b^2)*c^3)*x)*sqrt(c*x^2 + b
*x + a))/c^5, -1/3840*(15*(7*B*b^5 - 32*A*a^2*c^3 + 48*(B*a^2*b + A*a*b^2)*c^2 - 10*(4*B*a*b^3 + A*b^4)*c)*sqr
t(-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*(384*B*c^5*x^4 - 105*
B*b^4*c - 8*(32*B*a^2 + 65*A*a*b)*c^3 + 48*(B*b*c^4 + 10*A*c^5)*x^3 + 10*(46*B*a*b^2 + 15*A*b^3)*c^2 - 8*(7*B*
b^2*c^3 - 2*(8*B*a + 5*A*b)*c^4)*x^2 + 2*(35*B*b^3*c^2 + 120*A*a*c^4 - 2*(58*B*a*b + 25*A*b^2)*c^3)*x)*sqrt(c*
x^2 + b*x + a))/c^5]

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giac [A]  time = 0.23, size = 245, normalized size = 1.20 \begin {gather*} \frac {1}{1920} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, B x + \frac {B b c^{3} + 10 \, A c^{4}}{c^{4}}\right )} x - \frac {7 \, B b^{2} c^{2} - 16 \, B a c^{3} - 10 \, A b c^{3}}{c^{4}}\right )} x + \frac {35 \, B b^{3} c - 116 \, B a b c^{2} - 50 \, A b^{2} c^{2} + 120 \, A a c^{3}}{c^{4}}\right )} x - \frac {105 \, B b^{4} - 460 \, B a b^{2} c - 150 \, A b^{3} c + 256 \, B a^{2} c^{2} + 520 \, A a b c^{2}}{c^{4}}\right )} - \frac {{\left (7 \, B b^{5} - 40 \, B a b^{3} c - 10 \, A b^{4} c + 48 \, B a^{2} b c^{2} + 48 \, A a b^{2} c^{2} - 32 \, A a^{2} c^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {9}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1920*sqrt(c*x^2 + b*x + a)*(2*(4*(6*(8*B*x + (B*b*c^3 + 10*A*c^4)/c^4)*x - (7*B*b^2*c^2 - 16*B*a*c^3 - 10*A*
b*c^3)/c^4)*x + (35*B*b^3*c - 116*B*a*b*c^2 - 50*A*b^2*c^2 + 120*A*a*c^3)/c^4)*x - (105*B*b^4 - 460*B*a*b^2*c
- 150*A*b^3*c + 256*B*a^2*c^2 + 520*A*a*b*c^2)/c^4) - 1/256*(7*B*b^5 - 40*B*a*b^3*c - 10*A*b^4*c + 48*B*a^2*b*
c^2 + 48*A*a*b^2*c^2 - 32*A*a^2*c^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(9/2)

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maple [B]  time = 0.06, size = 497, normalized size = 2.42 \begin {gather*} -\frac {A \,a^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{8 c^{\frac {3}{2}}}+\frac {3 A a \,b^{2} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}-\frac {5 A \,b^{4} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{128 c^{\frac {7}{2}}}+\frac {3 B \,a^{2} b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {5}{2}}}-\frac {5 B a \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32 c^{\frac {7}{2}}}+\frac {7 B \,b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {9}{2}}}-\frac {\sqrt {c \,x^{2}+b x +a}\, A a x}{8 c}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,b^{2} x}{32 c^{2}}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B a b x}{16 c^{2}}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, B \,b^{3} x}{64 c^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,x^{2}}{5 c}-\frac {\sqrt {c \,x^{2}+b x +a}\, A a b}{16 c^{2}}+\frac {5 \sqrt {c \,x^{2}+b x +a}\, A \,b^{3}}{64 c^{3}}+\frac {\left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A x}{4 c}+\frac {3 \sqrt {c \,x^{2}+b x +a}\, B a \,b^{2}}{32 c^{3}}-\frac {7 \sqrt {c \,x^{2}+b x +a}\, B \,b^{4}}{128 c^{4}}-\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B b x}{40 c^{2}}-\frac {5 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} A b}{24 c^{2}}-\frac {2 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B a}{15 c^{2}}+\frac {7 \left (c \,x^{2}+b x +a \right )^{\frac {3}{2}} B \,b^{2}}{48 c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(x^2*(B*x+A)*(c*x^2+b*x+a)^(1/2),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 [B]  time = 1.72, size = 463, normalized size = 2.26 \begin {gather*} \frac {B\,x^2\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{5\,c}-\frac {A\,a\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}-\frac {5\,A\,b\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}-\frac {2\,B\,a\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{5\,c}+\frac {A\,x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}+\frac {7\,B\,b\,\left (\frac {5\,b\,\left (\frac {\ln \left (\frac {b+2\,c\,x}{\sqrt {c}}+2\,\sqrt {c\,x^2+b\,x+a}\right )\,\left (b^3-4\,a\,b\,c\right )}{16\,c^{5/2}}+\frac {\left (-3\,b^2+2\,c\,x\,b+8\,c\,\left (c\,x^2+a\right )\right )\,\sqrt {c\,x^2+b\,x+a}}{24\,c^2}\right )}{8\,c}-\frac {x\,{\left (c\,x^2+b\,x+a\right )}^{3/2}}{4\,c}+\frac {a\,\left (\left (\frac {x}{2}+\frac {b}{4\,c}\right )\,\sqrt {c\,x^2+b\,x+a}+\frac {\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )\,\left (a\,c-\frac {b^2}{4}\right )}{2\,c^{3/2}}\right )}{4\,c}\right )}{10\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(B*x^2*(a + b*x + c*x^2)^(3/2))/(5*c) - (A*a*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^(1/
2) + (a + b*x + c*x^2)^(1/2))*(a*c - b^2/4))/(2*c^(3/2))))/(4*c) - (5*A*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b
*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2
))/(24*c^2)))/(8*c) - (2*B*a*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/
2)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/(5*c) + (A*x*(a + b*x + c*x^2)^
(3/2))/(4*c) + (7*B*b*((5*b*((log((b + 2*c*x)/c^(1/2) + 2*(a + b*x + c*x^2)^(1/2))*(b^3 - 4*a*b*c))/(16*c^(5/2
)) + ((8*c*(a + c*x^2) - 3*b^2 + 2*b*c*x)*(a + b*x + c*x^2)^(1/2))/(24*c^2)))/(8*c) - (x*(a + b*x + c*x^2)^(3/
2))/(4*c) + (a*((x/2 + b/(4*c))*(a + b*x + c*x^2)^(1/2) + (log((b/2 + c*x)/c^(1/2) + (a + b*x + c*x^2)^(1/2))*
(a*c - b^2/4))/(2*c^(3/2))))/(4*c)))/(10*c)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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