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

Optimal. Leaf size=153 \[ \frac {\sqrt {a+b x+c x^2} \left (-8 a B c-2 A b c+3 b^2 B\right )}{c^2 \left (b^2-4 a c\right )}-\frac {2 x \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {(3 b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{5/2}} \]

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Rubi [A]  time = 0.14, antiderivative size = 153, 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 = {818, 640, 621, 206} \begin {gather*} \frac {\sqrt {a+b x+c x^2} \left (-8 a B c-2 A b c+3 b^2 B\right )}{c^2 \left (b^2-4 a c\right )}-\frac {2 x \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {(3 b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x*(a*(b*B - 2*A*c) + (b^2*B - A*b*c - 2*a*B*c)*x))/(c*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + ((3*b^2*B - 2
*A*b*c - 8*a*B*c)*Sqrt[a + b*x + c*x^2])/(c^2*(b^2 - 4*a*c)) - ((3*b*B - 2*A*c)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]
*Sqrt[a + b*x + c*x^2])])/(2*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 640

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

Rule 818

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])

Rubi steps

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

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Mathematica [A]  time = 0.23, size = 148, normalized size = 0.97 \begin {gather*} \frac {\frac {2 \sqrt {c} \left (8 a^2 B c+a \left (2 b c (A+5 B x)+4 c^2 x (B x-A)-3 b^2 B\right )-b^2 x (-2 A c+3 b B+B c x)\right )}{\sqrt {a+x (b+c x)}}+\left (b^2-4 a c\right ) (3 b B-2 A c) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{2 c^{5/2} \left (4 a c-b^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.71, size = 148, normalized size = 0.97 \begin {gather*} \frac {8 a^2 B c+2 a A b c-4 a A c^2 x-3 a b^2 B+10 a b B c x+4 a B c^2 x^2+2 A b^2 c x-3 b^3 B x-b^2 B c x^2}{c^2 \left (4 a c-b^2\right ) \sqrt {a+b x+c x^2}}+\frac {(3 b B-2 A c) \log \left (-2 \sqrt {c} \sqrt {a+b x+c x^2}+b+2 c x\right )}{2 c^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*b^2*B + 2*a*A*b*c + 8*a^2*B*c - 3*b^3*B*x + 2*A*b^2*c*x + 10*a*b*B*c*x - 4*a*A*c^2*x - b^2*B*c*x^2 + 4*a
*B*c^2*x^2)/(c^2*(-b^2 + 4*a*c)*Sqrt[a + b*x + c*x^2]) + ((3*b*B - 2*A*c)*Log[b + 2*c*x - 2*Sqrt[c]*Sqrt[a + b
*x + c*x^2]])/(2*c^(5/2))

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fricas [B]  time = 0.64, size = 603, normalized size = 3.94 \begin {gather*} \left [-\frac {{\left (3 \, B a b^{3} + 8 \, A a^{2} c^{2} + {\left (3 \, B b^{3} c + 8 \, A a c^{3} - 2 \, {\left (6 \, B a b + A b^{2}\right )} c^{2}\right )} x^{2} - 2 \, {\left (6 \, B a^{2} b + A a b^{2}\right )} c + {\left (3 \, B b^{4} + 8 \, A a b c^{2} - 2 \, {\left (6 \, B a b^{2} + A b^{3}\right )} c\right )} x\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 (3 \, B a b^{2} c - 2 \, {\left (4 \, B a^{2} + A a b\right )} c^{2} + {\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} x^{2} + {\left (3 \, B b^{3} c + 4 \, A a c^{3} - 2 \, {\left (5 \, B a b + A b^{2}\right )} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{4 \, {\left (a b^{2} c^{3} - 4 \, a^{2} c^{4} + {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{2} + {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x\right )}}, \frac {{\left (3 \, B a b^{3} + 8 \, A a^{2} c^{2} + {\left (3 \, B b^{3} c + 8 \, A a c^{3} - 2 \, {\left (6 \, B a b + A b^{2}\right )} c^{2}\right )} x^{2} - 2 \, {\left (6 \, B a^{2} b + A a b^{2}\right )} c + {\left (3 \, B b^{4} + 8 \, A a b c^{2} - 2 \, {\left (6 \, B a b^{2} + A b^{3}\right )} c\right )} x\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 (3 \, B a b^{2} c - 2 \, {\left (4 \, B a^{2} + A a b\right )} c^{2} + {\left (B b^{2} c^{2} - 4 \, B a c^{3}\right )} x^{2} + {\left (3 \, B b^{3} c + 4 \, A a c^{3} - 2 \, {\left (5 \, B a b + A b^{2}\right )} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (a b^{2} c^{3} - 4 \, a^{2} c^{4} + {\left (b^{2} c^{4} - 4 \, a c^{5}\right )} x^{2} + {\left (b^{3} c^{3} - 4 \, a b c^{4}\right )} x\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

[-1/4*((3*B*a*b^3 + 8*A*a^2*c^2 + (3*B*b^3*c + 8*A*a*c^3 - 2*(6*B*a*b + A*b^2)*c^2)*x^2 - 2*(6*B*a^2*b + A*a*b
^2)*c + (3*B*b^4 + 8*A*a*b*c^2 - 2*(6*B*a*b^2 + A*b^3)*c)*x)*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*(3*B*a*b^2*c - 2*(4*B*a^2 + A*a*b)*c^2 + (B*b^2*c^2 - 4*B*a*c
^3)*x^2 + (3*B*b^3*c + 4*A*a*c^3 - 2*(5*B*a*b + A*b^2)*c^2)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^3 - 4*a^2*c^4 +
(b^2*c^4 - 4*a*c^5)*x^2 + (b^3*c^3 - 4*a*b*c^4)*x), 1/2*((3*B*a*b^3 + 8*A*a^2*c^2 + (3*B*b^3*c + 8*A*a*c^3 -
2*(6*B*a*b + A*b^2)*c^2)*x^2 - 2*(6*B*a^2*b + A*a*b^2)*c + (3*B*b^4 + 8*A*a*b*c^2 - 2*(6*B*a*b^2 + A*b^3)*c)*x
)*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*(3*B*a*b^2*c - 2
*(4*B*a^2 + A*a*b)*c^2 + (B*b^2*c^2 - 4*B*a*c^3)*x^2 + (3*B*b^3*c + 4*A*a*c^3 - 2*(5*B*a*b + A*b^2)*c^2)*x)*sq
rt(c*x^2 + b*x + a))/(a*b^2*c^3 - 4*a^2*c^4 + (b^2*c^4 - 4*a*c^5)*x^2 + (b^3*c^3 - 4*a*b*c^4)*x)]

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giac [A]  time = 0.26, size = 177, normalized size = 1.16 \begin {gather*} \frac {{\left (\frac {{\left (B b^{2} c - 4 \, B a c^{2}\right )} x}{b^{2} c^{2} - 4 \, a c^{3}} + \frac {3 \, B b^{3} - 10 \, B a b c - 2 \, A b^{2} c + 4 \, A a c^{2}}{b^{2} c^{2} - 4 \, a c^{3}}\right )} x + \frac {3 \, B a b^{2} - 8 \, B a^{2} c - 2 \, A a b c}{b^{2} c^{2} - 4 \, a c^{3}}}{\sqrt {c x^{2} + b x + a}} + \frac {{\left (3 \, B b - 2 \, A c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {5}{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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.06, size = 382, normalized size = 2.50 \begin {gather*} \frac {A \,b^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {4 B a b x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {3 B \,b^{3} x}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {A \,b^{3}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {2 B a \,b^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 B \,b^{4}}{4 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{3}}+\frac {B \,x^{2}}{\sqrt {c \,x^{2}+b x +a}\, c}-\frac {A x}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {3 B b x}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {A \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}-\frac {3 B b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{2 c^{\frac {5}{2}}}+\frac {A b}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}}+\frac {2 B a}{\sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {3 B \,b^{2}}{4 \sqrt {c \,x^{2}+b x +a}\, 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)^(3/2),x)

[Out]

B*x^2/c/(c*x^2+b*x+a)^(1/2)+3/2*B*b/c^2*x/(c*x^2+b*x+a)^(1/2)-3/4*B*b^2/c^3/(c*x^2+b*x+a)^(1/2)-3/2*B*b^3/c^2/
(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-3/4*B*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-3/2*B*b/c^(5/2)*ln((c*x+1/2*b)
/c^(1/2)+(c*x^2+b*x+a)^(1/2))+2*B*a/c^2/(c*x^2+b*x+a)^(1/2)+4*B*a/c*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+2*B*a/
c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-A*x/c/(c*x^2+b*x+a)^(1/2)+1/2*A*b/c^2/(c*x^2+b*x+a)^(1/2)+A*b^2/c/(4*a
*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+1/2*A*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+A/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)^(3/2),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 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2\,\left (A+B\,x\right )}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \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)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2} \left (A + B x\right )}{\left (a + b x + c x^{2}\right )^{\frac {3}{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)**(3/2),x)

[Out]

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

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