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

Optimal. Leaf size=99 \[ -\frac {(3 b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}}+\frac {\sqrt {b x+c x^2} (3 b B-2 A c)}{b c^2}-\frac {2 x^2 (b B-A c)}{b c \sqrt {b x+c x^2}} \]

[Out]

-(-2*A*c+3*B*b)*arctanh(x*c^(1/2)/(c*x^2+b*x)^(1/2))/c^(5/2)-2*(-A*c+B*b)*x^2/b/c/(c*x^2+b*x)^(1/2)+(-2*A*c+3*
B*b)*(c*x^2+b*x)^(1/2)/b/c^2

________________________________________________________________________________________

Rubi [A]  time = 0.09, antiderivative size = 99, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {788, 640, 620, 206} \[ \frac {\sqrt {b x+c x^2} (3 b B-2 A c)}{b c^2}-\frac {(3 b B-2 A c) \tanh ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{c^{5/2}}-\frac {2 x^2 (b B-A c)}{b c \sqrt {b x+c x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*B - A*c)*x^2)/(b*c*Sqrt[b*x + c*x^2]) + ((3*b*B - 2*A*c)*Sqrt[b*x + c*x^2])/(b*c^2) - ((3*b*B - 2*A*c)*
ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^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 620

Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2
]], x] /; FreeQ[{b, c}, x]

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 788

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.08, size = 88, normalized size = 0.89 \[ \frac {\sqrt {c} x (-2 A c+3 b B+B c x)-\sqrt {b} \sqrt {x} \sqrt {\frac {c x}{b}+1} (3 b B-2 A c) \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}}\right )}{c^{5/2} \sqrt {x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*x*(3*b*B - 2*A*c + B*c*x) - Sqrt[b]*(3*b*B - 2*A*c)*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x
])/Sqrt[b]])/(c^(5/2)*Sqrt[x*(b + c*x)])

________________________________________________________________________________________

fricas [A]  time = 0.74, size = 202, normalized size = 2.04 \[ \left [-\frac {{\left (3 \, B b^{2} - 2 \, A b c + {\left (3 \, B b c - 2 \, A c^{2}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) - 2 \, {\left (B c^{2} x + 3 \, B b c - 2 \, A c^{2}\right )} \sqrt {c x^{2} + b x}}{2 \, {\left (c^{4} x + b c^{3}\right )}}, \frac {{\left (3 \, B b^{2} - 2 \, A b c + {\left (3 \, B b c - 2 \, A c^{2}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) + {\left (B c^{2} x + 3 \, B b c - 2 \, A c^{2}\right )} \sqrt {c x^{2} + b x}}{c^{4} x + b c^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.28, size = 106, normalized size = 1.07 \[ \frac {\sqrt {c x^{2} + b x} B}{c^{2}} + \frac {{\left (3 \, B b - 2 \, A c\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} - b \right |}\right )}{2 \, c^{\frac {5}{2}}} + \frac {2 \, {\left (B b^{2} - A b c\right )}}{{\left ({\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} c + b \sqrt {c}\right )} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.05, size = 118, normalized size = 1.19 \[ \frac {B \,x^{2}}{\sqrt {c \,x^{2}+b x}\, c}-\frac {2 A x}{\sqrt {c \,x^{2}+b x}\, c}+\frac {3 B b x}{\sqrt {c \,x^{2}+b x}\, c^{2}}+\frac {A \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}-\frac {3 B b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{2 c^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.88, size = 115, normalized size = 1.16 \[ \frac {B x^{2}}{\sqrt {c x^{2} + b x} c} + \frac {3 \, B b x}{\sqrt {c x^{2} + b x} c^{2}} - \frac {2 \, A x}{\sqrt {c x^{2} + b x} c} - \frac {3 \, B b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} + \frac {A \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

B*x^2/(sqrt(c*x^2 + b*x)*c) + 3*B*b*x/(sqrt(c*x^2 + b*x)*c^2) - 2*A*x/(sqrt(c*x^2 + b*x)*c) - 3/2*B*b*log(2*c*
x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) + A*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________