### 3.44 $$\int \frac{x^2}{\sqrt{b x+c x^2}} \, dx$$

Optimal. Leaf size=76 $\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{5/2}}-\frac{3 b \sqrt{b x+c x^2}}{4 c^2}+\frac{x \sqrt{b x+c x^2}}{2 c}$

[Out]

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

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Rubi [A]  time = 0.0264775, antiderivative size = 76, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.235, Rules used = {670, 640, 620, 206} $\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 c^{5/2}}-\frac{3 b \sqrt{b x+c x^2}}{4 c^2}+\frac{x \sqrt{b x+c x^2}}{2 c}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^2/Sqrt[b*x + c*x^2],x]

[Out]

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

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

Mathematica [A]  time = 0.0536501, size = 88, normalized size = 1.16 $\frac{\sqrt{c} x \left (-3 b^2-b c x+2 c^2 x^2\right )+3 b^{5/2} \sqrt{x} \sqrt{\frac{c x}{b}+1} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{4 c^{5/2} \sqrt{x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^2/Sqrt[b*x + c*x^2],x]

[Out]

(Sqrt[c]*x*(-3*b^2 - b*c*x + 2*c^2*x^2) + 3*b^(5/2)*Sqrt[x]*Sqrt[1 + (c*x)/b]*ArcSinh[(Sqrt[c]*Sqrt[x])/Sqrt[b
]])/(4*c^(5/2)*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.045, size = 68, normalized size = 0.9 \begin{align*}{\frac{x}{2\,c}\sqrt{c{x}^{2}+bx}}-{\frac{3\,b}{4\,{c}^{2}}\sqrt{c{x}^{2}+bx}}+{\frac{3\,{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/(c*x^2+b*x)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.01287, size = 297, normalized size = 3.91 \begin{align*} \left [\frac{3 \, b^{2} \sqrt{c} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \,{\left (2 \, c^{2} x - 3 \, b c\right )} \sqrt{c x^{2} + b x}}{8 \, c^{3}}, -\frac{3 \, b^{2} \sqrt{-c} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) -{\left (2 \, c^{2} x - 3 \, b c\right )} \sqrt{c x^{2} + b x}}{4 \, c^{3}}\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]

[1/8*(3*b^2*sqrt(c)*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*(2*c^2*x - 3*b*c)*sqrt(c*x^2 + b*x))/c^3,
-1/4*(3*b^2*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2*c^2*x - 3*b*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 \frac{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 = 1.26556, size = 88, normalized size = 1.16 \begin{align*} \frac{1}{4} \, \sqrt{c x^{2} + b x}{\left (\frac{2 \, x}{c} - \frac{3 \, b}{c^{2}}\right )} - \frac{3 \, b^{2} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right )}{8 \, c^{\frac{5}{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]

1/4*sqrt(c*x^2 + b*x)*(2*x/c - 3*b/c^2) - 3/8*b^2*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b))/c^(
5/2)