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

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

[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)

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Rubi [A]  time = 0.0887486, antiderivative size = 99, normalized size of antiderivative = 1., 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 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]

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 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 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)}{\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 = 0.0883548, 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)])

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Maple [A]  time = 0.007, size = 118, normalized size = 1.2 \begin{align*}{\frac{{x}^{2}B}{c}{\frac{1}{\sqrt{c{x}^{2}+bx}}}}+3\,{\frac{bBx}{{c}^{2}\sqrt{c{x}^{2}+bx}}}-{\frac{3\,bB}{2}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{2}}}}-2\,{\frac{Ax}{c\sqrt{c{x}^{2}+bx}}}+{A\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \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)^(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((1/2*b+c*x)/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((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99579, size = 459, normalized size = 4.64 \begin{align*} \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 ] \end{align*}

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)]

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

[Out]

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

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

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]

Exception raised: TypeError