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

Optimal. Leaf size=96 \[ \frac {B \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}}-\frac {2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 96, normalized size of antiderivative = 1.00, 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 = {777, 621, 206} \[ \frac {B \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{3/2}}-\frac {2 \left (x \left (-2 a B c-A b c+b^2 B\right )+a (b B-2 A c)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 777

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

Rubi steps

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

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Mathematica [A]  time = 0.28, size = 102, normalized size = 1.06 \[ \frac {\frac {2 \sqrt {c} (-2 a c (A+B x)+a b B+b x (b B-A c))}{\sqrt {a+x (b+c x)}}-B \left (b^2-4 a c\right ) \log \left (2 \sqrt {c} \sqrt {a+x (b+c x)}+b+2 c x\right )}{c^{3/2} \left (4 a c-b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.25, size = 405, normalized size = 4.22 \[ \left [\frac {{\left (B a b^{2} - 4 \, B a^{2} c + {\left (B b^{2} c - 4 \, B a c^{2}\right )} x^{2} + {\left (B b^{3} - 4 \, B a b c\right )} x\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} - 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) - 4 \, {\left (B a b c - 2 \, A a c^{2} + {\left (B b^{2} c - {\left (2 \, B a + A b\right )} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{2 \, {\left (a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x\right )}}, -\frac {{\left (B a b^{2} - 4 \, B a^{2} c + {\left (B b^{2} c - 4 \, B a c^{2}\right )} x^{2} + {\left (B b^{3} - 4 \, B a b c\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (B a b c - 2 \, A a c^{2} + {\left (B b^{2} c - {\left (2 \, B a + A b\right )} c^{2}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{a b^{2} c^{2} - 4 \, a^{2} c^{3} + {\left (b^{2} c^{3} - 4 \, a c^{4}\right )} x^{2} + {\left (b^{3} c^{2} - 4 \, a b c^{3}\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.25, size = 110, normalized size = 1.15 \[ -\frac {2 \, {\left (\frac {{\left (B b^{2} - 2 \, B a c - A b c\right )} x}{b^{2} c - 4 \, a c^{2}} + \frac {B a b - 2 \, A a c}{b^{2} c - 4 \, a c^{2}}\right )}}{\sqrt {c x^{2} + b x + a}} - \frac {B \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(B*x+A)/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 216, normalized size = 2.25 \[ -\frac {2 A b x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {B \,b^{2} x}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}-\frac {A \,b^{2}}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c}+\frac {B \,b^{3}}{2 \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}\, c^{2}}-\frac {B x}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {B \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}-\frac {A}{\sqrt {c \,x^{2}+b x +a}\, c}+\frac {B b}{2 \sqrt {c \,x^{2}+b x +a}\, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-B*x/c/(c*x^2+b*x+a)^(1/2)+1/2*B*b/c^2/(c*x^2+b*x+a)^(1/2)+B*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x+1/2*B*b^3
/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+B/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-A/c/(c*x^2+b*x+a)^(
1/2)-2*A*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x-A*b^2/c/(4*a*c-b^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 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(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 [B]  time = 1.64, size = 111, normalized size = 1.16 \[ \frac {B\,\ln \left (\frac {\frac {b}{2}+c\,x}{\sqrt {c}}+\sqrt {c\,x^2+b\,x+a}\right )}{c^{3/2}}-\frac {A\,\left (4\,a+2\,b\,x\right )}{\left (4\,a\,c-b^2\right )\,\sqrt {c\,x^2+b\,x+a}}+\frac {B\,\left (\frac {a\,b}{2}-x\,\left (a\,c-\frac {b^2}{2}\right )\right )}{c\,\left (a\,c-\frac {b^2}{4}\right )\,\sqrt {c\,x^2+b\,x+a}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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