### 3.3 $$\int x \sqrt{b x+c x^2} \, dx$$

Optimal. Leaf size=81 $\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{5/2}}-\frac{b (b+2 c x) \sqrt{b x+c x^2}}{8 c^2}+\frac{\left (b x+c x^2\right )^{3/2}}{3 c}$

[Out]

-(b*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(8*c^2) + (b*x + c*x^2)^(3/2)/(3*c) + (b^3*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c
*x^2]])/(8*c^(5/2))

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Rubi [A]  time = 0.0238282, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.267, Rules used = {640, 612, 620, 206} $\frac{b^3 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{8 c^{5/2}}-\frac{b (b+2 c x) \sqrt{b x+c x^2}}{8 c^2}+\frac{\left (b x+c x^2\right )^{3/2}}{3 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[x*Sqrt[b*x + c*x^2],x]

[Out]

-(b*(b + 2*c*x)*Sqrt[b*x + c*x^2])/(8*c^2) + (b*x + c*x^2)^(3/2)/(3*c) + (b^3*ArcTanh[(Sqrt[c]*x)/Sqrt[b*x + c
*x^2]])/(8*c^(5/2))

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

Mathematica [A]  time = 0.112991, size = 87, normalized size = 1.07 $\frac{\sqrt{x (b+c x)} \left (\sqrt{c} \left (-3 b^2+2 b c x+8 c^2 x^2\right )+\frac{3 b^{5/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{x} \sqrt{\frac{c x}{b}+1}}\right )}{24 c^{5/2}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x*Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[x*(b + c*x)]*(Sqrt[c]*(-3*b^2 + 2*b*c*x + 8*c^2*x^2) + (3*b^(5/2)*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b]])/(S
qrt[x]*Sqrt[1 + (c*x)/b])))/(24*c^(5/2))

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Maple [A]  time = 0.053, size = 87, normalized size = 1.1 \begin{align*}{\frac{1}{3\,c} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-{\frac{bx}{4\,c}\sqrt{c{x}^{2}+bx}}-{\frac{{b}^{2}}{8\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{{b}^{3}}{16}\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*(c*x^2+b*x)^(1/2),x)

[Out]

1/3*(c*x^2+b*x)^(3/2)/c-1/4*b/c*x*(c*x^2+b*x)^(1/2)-1/8*b^2/c^2*(c*x^2+b*x)^(1/2)+1/16*b^3/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*(c*x^2+b*x)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07372, size = 343, normalized size = 4.23 \begin{align*} \left [\frac{3 \, b^{3} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c\right )} \sqrt{c x^{2} + b x}}{48 \, c^{3}}, -\frac{3 \, b^{3} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (8 \, c^{3} x^{2} + 2 \, b c^{2} x - 3 \, b^{2} c\right )} \sqrt{c x^{2} + b x}}{24 \, c^{3}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*b^3*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(8*c^3*x^2 + 2*b*c^2*x - 3*b^2*c)*sqrt(c
*x^2 + b*x))/c^3, -1/24*(3*b^3*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (8*c^3*x^2 + 2*b*c^2*x - 3*
b^2*c)*sqrt(c*x^2 + b*x))/c^3]

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x*sqrt(x*(b + c*x)), x)

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Giac [A]  time = 1.31978, size = 99, normalized size = 1.22 \begin{align*} \frac{1}{24} \, \sqrt{c x^{2} + b x}{\left (2 \,{\left (4 \, x + \frac{b}{c}\right )} x - \frac{3 \, b^{2}}{c^{2}}\right )} - \frac{b^{3} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{16 \, c^{\frac{5}{2}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*sqrt(c*x^2 + b*x)*(2*(4*x + b/c)*x - 3*b^2/c^2) - 1/16*b^3*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqr
t(c) - b))/c^(5/2)