### 3.18 $$\int \frac{(b x+c x^2)^{3/2}}{x^4} \, dx$$

Optimal. Leaf size=68 $2 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )-\frac{2 c \sqrt{b x+c x^2}}{x}-\frac{2 \left (b x+c x^2\right )^{3/2}}{3 x^3}$

[Out]

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

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Rubi [A]  time = 0.0278444, antiderivative size = 68, 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 = {662, 620, 206} $2 c^{3/2} \tanh ^{-1}\left (\frac{\sqrt{c} x}{\sqrt{b x+c x^2}}\right )-\frac{2 c \sqrt{b x+c x^2}}{x}-\frac{2 \left (b x+c x^2\right )^{3/2}}{3 x^3}$

Antiderivative was successfully veriﬁed.

[In]

Int[(b*x + c*x^2)^(3/2)/x^4,x]

[Out]

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

Mathematica [C]  time = 0.0135384, size = 48, normalized size = 0.71 $-\frac{2 b \sqrt{x (b+c x)} \, _2F_1\left (-\frac{3}{2},-\frac{3}{2};-\frac{1}{2};-\frac{c x}{b}\right )}{3 x^2 \sqrt{\frac{c x}{b}+1}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(b*x + c*x^2)^(3/2)/x^4,x]

[Out]

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

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Maple [B]  time = 0.046, size = 149, normalized size = 2.2 \begin{align*} -{\frac{2}{3\,b{x}^{4}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{4\,c}{3\,{b}^{2}{x}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}+{\frac{16\,{c}^{2}}{3\,{b}^{3}{x}^{2}} \left ( c{x}^{2}+bx \right ) ^{{\frac{5}{2}}}}-{\frac{16\,{c}^{3}}{3\,{b}^{3}} \left ( c{x}^{2}+bx \right ) ^{{\frac{3}{2}}}}-4\,{\frac{{c}^{3}\sqrt{c{x}^{2}+bx}x}{{b}^{2}}}-2\,{\frac{{c}^{2}\sqrt{c{x}^{2}+bx}}{b}}+{c}^{{\frac{3}{2}}}\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^4,x)

[Out]

-2/3/b/x^4*(c*x^2+b*x)^(5/2)-4/3*c/b^2/x^3*(c*x^2+b*x)^(5/2)+16/3*c^2/b^3/x^2*(c*x^2+b*x)^(5/2)-16/3*c^3/b^3*(
c*x^2+b*x)^(3/2)-4*c^3/b^2*(c*x^2+b*x)^(1/2)*x-2*c^2/b*(c*x^2+b*x)^(1/2)+c^(3/2)*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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.0361, size = 284, normalized size = 4.18 \begin{align*} \left [\frac{3 \, c^{\frac{3}{2}} x^{2} \log \left (2 \, c x + b + 2 \, \sqrt{c x^{2} + b x} \sqrt{c}\right ) - 2 \, \sqrt{c x^{2} + b x}{\left (4 \, c x + b\right )}}{3 \, x^{2}}, -\frac{2 \,{\left (3 \, \sqrt{-c} c x^{2} \arctan \left (\frac{\sqrt{c x^{2} + b x} \sqrt{-c}}{c x}\right ) + \sqrt{c x^{2} + b x}{\left (4 \, c x + b\right )}\right )}}{3 \, x^{2}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^4,x, algorithm="fricas")

[Out]

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

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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^{4}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x)**(3/2)/x**4,x)

[Out]

Integral((x*(b + c*x))**(3/2)/x**4, x)

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Giac [B]  time = 1.34296, size = 155, normalized size = 2.28 \begin{align*} -c^{\frac{3}{2}} \log \left ({\left | -2 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} \sqrt{c} - b \right |}\right ) + \frac{2 \,{\left (6 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{2} b c + 3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )} b^{2} \sqrt{c} + b^{3}\right )}}{3 \,{\left (\sqrt{c} x - \sqrt{c x^{2} + b x}\right )}^{3}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x)^(3/2)/x^4,x, algorithm="giac")

[Out]

-c^(3/2)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) - b)) + 2/3*(6*(sqrt(c)*x - sqrt(c*x^2 + b*x))^2*b
*c + 3*(sqrt(c)*x - sqrt(c*x^2 + b*x))*b^2*sqrt(c) + b^3)/(sqrt(c)*x - sqrt(c*x^2 + b*x))^3