### 3.17 $$\int \frac{(b x+c x^2)^{3/2}}{x^3} \, dx$$

Optimal. Leaf size=64 $-\frac{2 \left (b x+c x^2\right )^{3/2}}{x^2}+3 c \sqrt{b x+c x^2}+3 b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )$

[Out]

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

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Rubi [A]  time = 0.027478, antiderivative size = 64, 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 = {662, 664, 620, 206} $-\frac{2 \left (b x+c x^2\right )^{3/2}}{x^2}+3 c \sqrt{b x+c x^2}+3 b \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*x + c*x^2)^(3/2)/x^3,x]

[Out]

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

Rule 662

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 + p + 1)), x] - Dist[(c*p)/(e^2*(m + p + 1)), Int[(d + e*x)^(m + 2)*(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] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

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

Mathematica [C]  time = 0.0120287, size = 46, normalized size = 0.72 $-\frac{2 b \sqrt{x (b+c x)} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};-\frac{c x}{b}\right )}{x \sqrt{\frac{c x}{b}+1}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^3,x]

[Out]

(-2*b*Sqrt[x*(b + c*x)]*Hypergeometric2F1[-3/2, -1/2, 1/2, -((c*x)/b)])/(x*Sqrt[1 + (c*x)/b])

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Maple [B]  time = 0.045, size = 124, normalized size = 1.9 \begin{align*} -2\,{\frac{ \left ( c{x}^{2}+bx \right ) ^{5/2}}{b{x}^{3}}}+8\,{\frac{c \left ( c{x}^{2}+bx \right ) ^{5/2}}{{b}^{2}{x}^{2}}}-8\,{\frac{{c}^{2} \left ( c{x}^{2}+bx \right ) ^{3/2}}{{b}^{2}}}-6\,{\frac{{c}^{2}\sqrt{c{x}^{2}+bx}x}{b}}-3\,c\sqrt{c{x}^{2}+bx}+{\frac{3\,b}{2}\sqrt{c}\ln \left ({ \left ({\frac{b}{2}}+cx \right ){\frac{1}{\sqrt{c}}}}+\sqrt{c{x}^{2}+bx} \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/b/x^3*(c*x^2+b*x)^(5/2)+8*c/b^2/x^2*(c*x^2+b*x)^(5/2)-8*c^2/b^2*(c*x^2+b*x)^(3/2)-6*c^2/b*(c*x^2+b*x)^(1/2)
*x-3*c*(c*x^2+b*x)^(1/2)+3/2*c^(1/2)*b*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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.97435, size = 270, normalized size = 4.22 \begin{align*} \left [\frac{3 \, b \sqrt{c} x \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) + 2 \, \sqrt{c x^{2} + b x}{\left (c x - 2 \, b\right )}}{2 \, x}, -\frac{3 \, b \sqrt{-c} x \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) - \sqrt{c x^{2} + b x}{\left (c x - 2 \, b\right )}}{x}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*b*sqrt(c)*x*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c)) + 2*sqrt(c*x^2 + b*x)*(c*x - 2*b))/x, -(3*b*s
qrt(-c)*x*arctan(sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - sqrt(c*x^2 + b*x)*(c*x - 2*b))/x]

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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^{3}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.45613, size = 103, normalized size = 1.61 \begin{align*} -\frac{3}{2} \, b \sqrt{c} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right ) + \sqrt{c x^{2} + b x} c + \frac{2 \, b^{2}}{\sqrt{c} x - \sqrt{c x^{2} + b x}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/2*b*sqrt(c)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b)) + sqrt(c*x^2 + b*x)*c + 2*b^2/(sqrt(c)
*x - sqrt(c*x^2 + b*x))