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

Optimal. Leaf size=127 \[ -\frac{b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{7/2}}+\frac{b \sqrt{b x+c x^2} (5 b B-6 A c)}{8 c^3}-\frac{x \sqrt{b x+c x^2} (5 b B-6 A c)}{12 c^2}+\frac{B x^2 \sqrt{b x+c x^2}}{3 c} \]

[Out]

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

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Rubi [A]  time = 0.117979, antiderivative size = 127, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.227, Rules used = {794, 670, 640, 620, 206} \[ -\frac{b^2 (5 b B-6 A c) \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{7/2}}+\frac{b \sqrt{b x+c x^2} (5 b B-6 A c)}{8 c^3}-\frac{x \sqrt{b x+c x^2} (5 b B-6 A c)}{12 c^2}+\frac{B x^2 \sqrt{b x+c x^2}}{3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 794

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

Rule 670

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

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

Mathematica [A]  time = 0.0878908, size = 116, normalized size = 0.91 \[ \frac{\sqrt{c} x (b+c x) \left (-2 b c (9 A+5 B x)+4 c^2 x (3 A+2 B x)+15 b^2 B\right )-3 b^{5/2} \sqrt{x} \sqrt{\frac{c x}{b}+1} (5 b B-6 A c) \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{24 c^{7/2} \sqrt{x (b+c x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c]*x*(b + c*x)*(15*b^2*B + 4*c^2*x*(3*A + 2*B*x) - 2*b*c*(9*A + 5*B*x)) - 3*b^(5/2)*(5*b*B - 6*A*c)*Sqrt
[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(24*c^(7/2)*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.009, size = 163, normalized size = 1.3 \begin{align*}{\frac{{x}^{2}B}{3\,c}\sqrt{c{x}^{2}+bx}}-{\frac{5\,bBx}{12\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,{b}^{2}B}{8\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,{b}^{3}B}{16}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}}+{\frac{Ax}{2\,c}\sqrt{c{x}^{2}+bx}}-{\frac{3\,Ab}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,A{b}^{2}}{8}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{5}{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)^(1/2),x)

[Out]

1/3*B*x^2*(c*x^2+b*x)^(1/2)/c-5/12*B*b/c^2*x*(c*x^2+b*x)^(1/2)+5/8*B*b^2/c^3*(c*x^2+b*x)^(1/2)-5/16*B*b^3/c^(7
/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x)^(1/2))+1/2*A*x/c*(c*x^2+b*x)^(1/2)-3/4*A*b/c^2*(c*x^2+b*x)^(1/2)+3/8*A*
b^2/c^(5/2)*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)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.97101, size = 483, normalized size = 3.8 \begin{align*} \left [-\frac{3 \,{\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \,{\left (8 \, B c^{3} x^{2} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \,{\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{48 \, c^{4}}, \frac{3 \,{\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (8 \, B c^{3} x^{2} + 15 \, B b^{2} c - 18 \, A b c^{2} - 2 \,{\left (5 \, B b c^{2} - 6 \, A c^{3}\right )} x\right )} \sqrt{c x^{2} + b x}}{24 \, 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)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*(5*B*b^3 - 6*A*b^2*c)*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) - 2*(8*B*c^3*x^2 + 15*B*b
^2*c - 18*A*b*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x)*sqrt(c*x^2 + b*x))/c^4, 1/24*(3*(5*B*b^3 - 6*A*b^2*c)*sqrt(-c)*
arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (8*B*c^3*x^2 + 15*B*b^2*c - 18*A*b*c^2 - 2*(5*B*b*c^2 - 6*A*c^3)*x)
*sqrt(c*x^2 + b*x))/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{x \left (b + c x\right )}}\, 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)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.20846, size = 147, normalized size = 1.16 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (\frac{4 \, B x}{c} - \frac{5 \, B b c - 6 \, A c^{2}}{c^{3}}\right )} x + \frac{3 \,{\left (5 \, B b^{2} - 6 \, A b c\right )}}{c^{3}}\right )} + \frac{{\left (5 \, B b^{3} - 6 \, A b^{2} c\right )} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\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)^(1/2),x, algorithm="giac")

[Out]

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