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

Optimal. Leaf size=72 \[ \frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 \sqrt{c}}+\frac{3}{4} b \sqrt{b x+c x^2}+\frac{\left (b x+c x^2\right )^{3/2}}{2 x} \]

[Out]

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

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Rubi [A]  time = 0.0279387, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {664, 620, 206} \[ \frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 \sqrt{c}}+\frac{3}{4} b \sqrt{b x+c x^2}+\frac{\left (b x+c x^2\right )^{3/2}}{2 x} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(3/2)/x^2,x]

[Out]

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

Rule 664

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[(p*(2*c*d - b*e))/(e^2*(m + 2*p + 1)), Int[(d + e*x)^(m + 1)*
(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a
*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*p + 1, 0] && IntegerQ[2*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 \frac{\left (b x+c x^2\right )^{3/2}}{x^2} \, dx &=\frac{\left (b x+c x^2\right )^{3/2}}{2 x}+\frac{1}{4} (3 b) \int \frac{\sqrt{b x+c x^2}}{x} \, dx\\ &=\frac{3}{4} b \sqrt{b x+c x^2}+\frac{\left (b x+c x^2\right )^{3/2}}{2 x}+\frac{1}{8} \left (3 b^2\right ) \int \frac{1}{\sqrt{b x+c x^2}} \, dx\\ &=\frac{3}{4} b \sqrt{b x+c x^2}+\frac{\left (b x+c x^2\right )^{3/2}}{2 x}+\frac{1}{4} \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 )\\ &=\frac{3}{4} b \sqrt{b x+c x^2}+\frac{\left (b x+c x^2\right )^{3/2}}{2 x}+\frac{3 b^2 \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )}{4 \sqrt{c}}\\ \end{align*}

Mathematica [A]  time = 0.106998, size = 69, normalized size = 0.96 \[ \frac{1}{4} \sqrt{x (b+c x)} \left (\frac{3 b^{3/2} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{x}}{\sqrt{b}}\right )}{\sqrt{c} \sqrt{x} \sqrt{\frac{c x}{b}+1}}+5 b+2 c x\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^2,x]

[Out]

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

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Maple [A]  time = 0.046, size = 99, normalized size = 1.4 \begin{align*} 2\,{\frac{ \left ( c{x}^{2}+bx \right ) ^{5/2}}{b{x}^{2}}}-2\,{\frac{c \left ( c{x}^{2}+bx \right ) ^{3/2}}{b}}-{\frac{3\,cx}{2}\sqrt{c{x}^{2}+bx}}-{\frac{3\,b}{4}\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 ){\frac{1}{\sqrt{c}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x)^(3/2)/x^2,x)

[Out]

2/b/x^2*(c*x^2+b*x)^(5/2)-2*c/b*(c*x^2+b*x)^(3/2)-3/2*c*(c*x^2+b*x)^(1/2)*x-3/4*b*(c*x^2+b*x)^(1/2)+3/8/c^(1/2
)*b^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((c*x^2+b*x)^(3/2)/x^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.8205, size = 292, normalized size = 4.06 \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 + 5 \, b c\right )} \sqrt{c x^{2} + b x}}{8 \, c}, -\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 + 5 \, b c\right )} \sqrt{c x^{2} + b x}}{4 \, c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^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 + 5*b*c)*sqrt(c*x^2 + b*x))/c, -
1/4*(3*b^2*sqrt(-c)*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2*c^2*x + 5*b*c)*sqrt(c*x^2 + b*x))/c]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**2,x)

[Out]

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

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Giac [A]  time = 1.38224, size = 81, normalized size = 1.12 \begin{align*} -\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 \, \sqrt{c}} + \frac{1}{4} \, \sqrt{c x^{2} + b x}{\left (2 \, c x + 5 \, b\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^2,x, algorithm="giac")

[Out]

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