3.2 \(\int x^2 \sqrt{b x+c x^2} \, dx\)

Optimal. Leaf size=105 \[ \frac{5 b^2 (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}-\frac{5 b^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{7/2}}-\frac{5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{x \left (b x+c x^2\right )^{3/2}}{4 c} \]

[Out]

(5*b^2*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) - (5*b*(b*x + c*x^2)^(3/2))/(24*c^2) + (x*(b*x + c*x^2)^(3/2))/
(4*c) - (5*b^4*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(7/2))

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Rubi [A]  time = 0.0374605, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {670, 640, 612, 620, 206} \[ \frac{5 b^2 (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}-\frac{5 b^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{7/2}}-\frac{5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{x \left (b x+c x^2\right )^{3/2}}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[b*x + c*x^2],x]

[Out]

(5*b^2*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(64*c^3) - (5*b*(b*x + c*x^2)^(3/2))/(24*c^2) + (x*(b*x + c*x^2)^(3/2))/
(4*c) - (5*b^4*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c*x^2]])/(64*c^(7/2))

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 612

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

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 x^2 \sqrt{b x+c x^2} \, dx &=\frac{x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac{(5 b) \int x \sqrt{b x+c x^2} \, dx}{8 c}\\ &=-\frac{5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{x \left (b x+c x^2\right )^{3/2}}{4 c}+\frac{\left (5 b^2\right ) \int \sqrt{b x+c x^2} \, dx}{16 c^2}\\ &=\frac{5 b^2 (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}-\frac{5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac{\left (5 b^4\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx}{128 c^3}\\ &=\frac{5 b^2 (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}-\frac{5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac{\left (5 b^4\right ) \operatorname{Subst}\left (\int \frac{1}{1-c x^2} \, dx,x,\frac{x}{\sqrt{b x+c x^2}}\right )}{64 c^3}\\ &=\frac{5 b^2 (b+2 c x) \sqrt{b x+c x^2}}{64 c^3}-\frac{5 b \left (b x+c x^2\right )^{3/2}}{24 c^2}+\frac{x \left (b x+c x^2\right )^{3/2}}{4 c}-\frac{5 b^4 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{64 c^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.13163, size = 98, normalized size = 0.93 \[ \frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-10 b^2 c x+15 b^3+8 b c^2 x^2+48 c^3 x^3\right )-\frac{15 b^{7/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{192 c^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(15*b^3 - 10*b^2*c*x + 8*b*c^2*x^2 + 48*c^3*x^3) - (15*b^(7/2)*ArcSinh[(Sqrt[c]*Sq
rt[x])/Sqrt[b]])/(Sqrt[x]*Sqrt[1 + (c*x)/b])))/(192*c^(7/2))

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Maple [A]  time = 0.044, size = 107, normalized size = 1. \begin{align*}{\frac{x}{4\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{5\,b}{24\,{c}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}+{\frac{5\,{b}^{2}x}{32\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{5\,{b}^{3}}{64\,{c}^{3}}\sqrt{c{x}^{2}+bx}}-{\frac{5\,{b}^{4}}{128}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ){c}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(c*x^2+b*x)^(1/2),x)

[Out]

1/4*x*(c*x^2+b*x)^(3/2)/c-5/24*b*(c*x^2+b*x)^(3/2)/c^2+5/32*b^2/c^2*(c*x^2+b*x)^(1/2)*x+5/64*b^3/c^3*(c*x^2+b*
x)^(1/2)-5/128*b^4/c^(7/2)*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*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.0657, size = 398, normalized size = 3.79 \begin{align*} \left [\frac{15 \, b^{4} \sqrt{c} \log \left (2 \, c x + b - 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} - 10 \, b^{2} c^{2} x + 15 \, b^{3} c\right )} \sqrt{c x^{2} + b x}}{384 \, c^{4}}, \frac{15 \, b^{4} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) +{\left (48 \, c^{4} x^{3} + 8 \, b c^{3} x^{2} - 10 \, b^{2} c^{2} x + 15 \, b^{3} c\right )} \sqrt{c x^{2} + b x}}{192 \, c^{4}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(15*b^4*sqrt(c)*log(2*c*x + b - 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(48*c^4*x^3 + 8*b*c^3*x^2 - 10*b^2*c^2
*x + 15*b^3*c)*sqrt(c*x^2 + b*x))/c^4, 1/192*(15*b^4*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) + (48*c
^4*x^3 + 8*b*c^3*x^2 - 10*b^2*c^2*x + 15*b^3*c)*sqrt(c*x^2 + b*x))/c^4]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \sqrt{x \left (b + c x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(c*x**2+b*x)**(1/2),x)

[Out]

Integral(x**2*sqrt(x*(b + c*x)), x)

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Giac [A]  time = 1.35702, size = 115, normalized size = 1.1 \begin{align*} \frac{1}{192} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \,{\left (6 \, x + \frac{b}{c}\right )} x - \frac{5 \, b^{2}}{c^{2}}\right )} x + \frac{15 \, b^{3}}{c^{3}}\right )} + \frac{5 \, b^{4} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{128 \, c^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*sqrt(c*x^2 + b*x)*(2*(4*(6*x + b/c)*x - 5*b^2/c^2)*x + 15*b^3/c^3) + 5/128*b^4*log(abs(-2*(sqrt(c)*x - s
qrt(c*x^2 + b*x))*sqrt(c) - b))/c^(7/2)