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

Optimal. Leaf size=143 \[ \frac{\sqrt{a+b x+c x^2} \left (-16 a B c-2 c x (5 b B-6 A c)-18 A b c+15 b^2 B\right )}{24 c^3}-\frac{\left (8 a A c^2-12 a b B c-6 A b^2 c+5 b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{7/2}}+\frac{B x^2 \sqrt{a+b x+c x^2}}{3 c} \]

[Out]

(B*x^2*Sqrt[a + b*x + c*x^2])/(3*c) + ((15*b^2*B - 18*A*b*c - 16*a*B*c - 2*c*(5*b*B - 6*A*c)*x)*Sqrt[a + b*x +
c*x^2])/(24*c^3) - ((5*b^3*B - 6*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*
x + c*x^2])])/(16*c^(7/2))

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Rubi [A]  time = 0.120068, antiderivative size = 143, normalized size of antiderivative = 1., 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 = {832, 779, 621, 206} \[ \frac{\sqrt{a+b x+c x^2} \left (-16 a B c-2 c x (5 b B-6 A c)-18 A b c+15 b^2 B\right )}{24 c^3}-\frac{\left (8 a A c^2-12 a b B c-6 A b^2 c+5 b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{16 c^{7/2}}+\frac{B x^2 \sqrt{a+b x+c x^2}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x^2*Sqrt[a + b*x + c*x^2])/(3*c) + ((15*b^2*B - 18*A*b*c - 16*a*B*c - 2*c*(5*b*B - 6*A*c)*x)*Sqrt[a + b*x +
c*x^2])/(24*c^3) - ((5*b^3*B - 6*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*
x + c*x^2])])/(16*c^(7/2))

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])

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 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 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])

Rubi steps

\begin{align*} \int \frac{x^2 (A+B x)}{\sqrt{a+b x+c x^2}} \, dx &=\frac{B x^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{\int \frac{x \left (-2 a B-\frac{1}{2} (5 b B-6 A c) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{3 c}\\ &=\frac{B x^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{\left (15 b^2 B-18 A b c-16 a B c-2 c (5 b B-6 A c) x\right ) \sqrt{a+b x+c x^2}}{24 c^3}-\frac{\left (5 b^3 B-6 A b^2 c-12 a b B c+8 a A c^2\right ) \int \frac{1}{\sqrt{a+b x+c x^2}} \, dx}{16 c^3}\\ &=\frac{B x^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{\left (15 b^2 B-18 A b c-16 a B c-2 c (5 b B-6 A c) x\right ) \sqrt{a+b x+c x^2}}{24 c^3}-\frac{\left (5 b^3 B-6 A b^2 c-12 a b B c+8 a A 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 )}{8 c^3}\\ &=\frac{B x^2 \sqrt{a+b x+c x^2}}{3 c}+\frac{\left (15 b^2 B-18 A b c-16 a B c-2 c (5 b B-6 A c) x\right ) \sqrt{a+b x+c x^2}}{24 c^3}-\frac{\left (5 b^3 B-6 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 )}{16 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.142847, size = 126, normalized size = 0.88 \[ \frac{2 \sqrt{c} \sqrt{a+x (b+c x)} \left (4 c (c x (3 A+2 B x)-4 a B)-2 b c (9 A+5 B x)+15 b^2 B\right )-3 \left (8 a A c^2-12 a b B c-6 A b^2 c+5 b^3 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{48 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[c]*Sqrt[a + x*(b + c*x)]*(15*b^2*B - 2*b*c*(9*A + 5*B*x) + 4*c*(-4*a*B + c*x*(3*A + 2*B*x))) - 3*(5*b^
3*B - 6*A*b^2*c - 12*a*b*B*c + 8*a*A*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(48*c^(7/2))

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Maple [B]  time = 0.007, size = 254, normalized size = 1.8 \begin{align*}{\frac{{x}^{2}B}{3\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{5\,bBx}{12\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{5\,{b}^{2}B}{8\,{c}^{3}}\sqrt{c{x}^{2}+bx+a}}-{\frac{5\,{b}^{3}B}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{7}{2}}}}+{\frac{3\,abB}{4}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{2\,aB}{3\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{Ax}{2\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,Ab}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,A{b}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{5}{2}}}}-{\frac{aA}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}} \end{align*}

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]

1/3*B*x^2*(c*x^2+b*x+a)^(1/2)/c-5/12*B*b/c^2*x*(c*x^2+b*x+a)^(1/2)+5/8*B*b^2/c^3*(c*x^2+b*x+a)^(1/2)-5/16*B*b^
3/c^(7/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3/4*B*b/c^(5/2)*a*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(
1/2))-2/3*B*a/c^2*(c*x^2+b*x+a)^(1/2)+1/2*A*x/c*(c*x^2+b*x+a)^(1/2)-3/4*A*b/c^2*(c*x^2+b*x+a)^(1/2)+3/8*A*b^2/
c^(5/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-1/2*A*a/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)
)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 1.99335, size = 697, normalized size = 4.87 \begin{align*} \left [\frac{3 \,{\left (5 \, B b^{3} + 8 \, A a c^{2} - 6 \,{\left (2 \, B a b + A b^{2}\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 (8 \, B c^{3} x^{2} + 15 \, B b^{2} c - 2 \,{\left (8 \, B a + 9 \, A b\right )} c^{2} - 2 \,{\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{96 \, c^{4}}, \frac{3 \,{\left (5 \, B b^{3} + 8 \, A a c^{2} - 6 \,{\left (2 \, B a b + A b^{2}\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 (8 \, B c^{3} x^{2} + 15 \, B b^{2} c - 2 \,{\left (8 \, B a + 9 \, A b\right )} c^{2} - 2 \,{\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x + a}}{48 \, c^{4}}\right ] \end{align*}

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]

[1/96*(3*(5*B*b^3 + 8*A*a*c^2 - 6*(2*B*a*b + A*b^2)*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*(8*B*c^3*x^2 + 15*B*b^2*c - 2*(8*B*a + 9*A*b)*c^2 - 2*(5*B*b*c^2 -
6*A*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^4, 1/48*(3*(5*B*b^3 + 8*A*a*c^2 - 6*(2*B*a*b + A*b^2)*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*(8*B*c^3*x^2 + 15*B*b^2*c - 2*(8*B
*a + 9*A*b)*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^4]

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

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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Giac [A]  time = 1.35297, size = 173, normalized size = 1.21 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + b x + a}{\left (2 \,{\left (\frac{4 \, B x}{c} - \frac{5 \, B b c - 6 \, A c^{2}}{c^{3}}\right )} x + \frac{15 \, B b^{2} - 16 \, B a c - 18 \, A b c}{c^{3}}\right )} + \frac{{\left (5 \, B b^{3} - 12 \, B a b c - 6 \, A b^{2} c + 8 \, A a c^{2}\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{16 \, c^{\frac{7}{2}}} \end{align*}

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]

1/24*sqrt(c*x^2 + b*x + a)*(2*(4*B*x/c - (5*B*b*c - 6*A*c^2)/c^3)*x + (15*B*b^2 - 16*B*a*c - 18*A*b*c)/c^3) +
1/16*(5*B*b^3 - 12*B*a*b*c - 6*A*b^2*c + 8*A*a*c^2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b
))/c^(7/2)