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

Optimal. Leaf size=92 \[ \frac{\left (-4 a B c-4 A b c+3 b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{5/2}}-\frac{\sqrt{a+b x+c x^2} (-4 A c+3 b B-2 B c x)}{4 c^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0396828, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {779, 621, 206} \[ \frac{\left (-4 a B c-4 A b c+3 b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+b x+c x^2}}\right )}{8 c^{5/2}}-\frac{\sqrt{a+b x+c x^2} (-4 A c+3 b B-2 B c x)}{4 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.121781, size = 90, normalized size = 0.98 \[ \frac{\sqrt{a+x (b+c x)} (4 A c-3 b B+2 B c x)}{4 c^2}-\frac{\left (4 a B c+4 A b c-3 b^2 B\right ) \tanh ^{-1}\left (\frac{b+2 c x}{2 \sqrt{c} \sqrt{a+x (b+c x)}}\right )}{8 c^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.006, size = 155, normalized size = 1.7 \begin{align*}{\frac{Bx}{2\,c}\sqrt{c{x}^{2}+bx+a}}-{\frac{3\,bB}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx+a}}+{\frac{3\,{b}^{2}B}{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{aB}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx+a} \right ){c}^{-{\frac{3}{2}}}}+{\frac{A}{c}\sqrt{c{x}^{2}+bx+a}}-{\frac{Ab}{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*(B*x+A)/(c*x^2+b*x+a)^(1/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.83909, size = 517, normalized size = 5.62 \begin{align*} \left [-\frac{{\left (3 \, B b^{2} - 4 \,{\left (B a + A b\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 (2 \, B c^{2} x - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt{c x^{2} + b x + a}}{16 \, c^{3}}, -\frac{{\left (3 \, B b^{2} - 4 \,{\left (B a + A b\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 (2 \, B c^{2} x - 3 \, B b c + 4 \, A c^{2}\right )} \sqrt{c x^{2} + b x + a}}{8 \, c^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*((3*B*b^2 - 4*(B*a + A*b)*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*(2*B*c^2*x - 3*B*b*c + 4*A*c^2)*sqrt(c*x^2 + b*x + a))/c^3, -1/8*((3*B*b^2 - 4*(B*a +
A*b)*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*(2*B*c^2*x
- 3*B*b*c + 4*A*c^2)*sqrt(c*x^2 + b*x + a))/c^3]

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \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*(B*x+A)/(c*x**2+b*x+a)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.36773, size = 122, normalized size = 1.33 \begin{align*} \frac{1}{4} \, \sqrt{c x^{2} + b x + a}{\left (\frac{2 \, B x}{c} - \frac{3 \, B b - 4 \, A c}{c^{2}}\right )} - \frac{{\left (3 \, B b^{2} - 4 \, B a c - 4 \, A b c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x + a}\right )} \sqrt{c} - b \right |}\right )}{8 \, c^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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