### 3.58 $$\int \frac{1}{x^2 (b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=77 $-\frac{16 c^2 (b+2 c x)}{5 b^4 \sqrt{b x+c x^2}}+\frac{4 c}{5 b^2 x \sqrt{b x+c x^2}}-\frac{2}{5 b x^2 \sqrt{b x+c x^2}}$

[Out]

-2/(5*b*x^2*Sqrt[b*x + c*x^2]) + (4*c)/(5*b^2*x*Sqrt[b*x + c*x^2]) - (16*c^2*(b + 2*c*x))/(5*b^4*Sqrt[b*x + c*
x^2])

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Rubi [A]  time = 0.0244401, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {658, 613} $-\frac{16 c^2 (b+2 c x)}{5 b^4 \sqrt{b x+c x^2}}+\frac{4 c}{5 b^2 x \sqrt{b x+c x^2}}-\frac{2}{5 b x^2 \sqrt{b x+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(x^2*(b*x + c*x^2)^(3/2)),x]

[Out]

-2/(5*b*x^2*Sqrt[b*x + c*x^2]) + (4*c)/(5*b^2*x*Sqrt[b*x + c*x^2]) - (16*c^2*(b + 2*c*x))/(5*b^4*Sqrt[b*x + c*
x^2])

Rule 658

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c
, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && ILtQ[Simplify[m + 2*p + 2], 0]

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rubi steps

\begin{align*} \int \frac{1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{2}{5 b x^2 \sqrt{b x+c x^2}}-\frac{(6 c) \int \frac{1}{x \left (b x+c x^2\right )^{3/2}} \, dx}{5 b}\\ &=-\frac{2}{5 b x^2 \sqrt{b x+c x^2}}+\frac{4 c}{5 b^2 x \sqrt{b x+c x^2}}+\frac{\left (8 c^2\right ) \int \frac{1}{\left (b x+c x^2\right )^{3/2}} \, dx}{5 b^2}\\ &=-\frac{2}{5 b x^2 \sqrt{b x+c x^2}}+\frac{4 c}{5 b^2 x \sqrt{b x+c x^2}}-\frac{16 c^2 (b+2 c x)}{5 b^4 \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0122747, size = 49, normalized size = 0.64 $-\frac{2 \left (-2 b^2 c x+b^3+8 b c^2 x^2+16 c^3 x^3\right )}{5 b^4 x^2 \sqrt{x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[1/(x^2*(b*x + c*x^2)^(3/2)),x]

[Out]

(-2*(b^3 - 2*b^2*c*x + 8*b*c^2*x^2 + 16*c^3*x^3))/(5*b^4*x^2*Sqrt[x*(b + c*x)])

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Maple [A]  time = 0.045, size = 53, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( 16\,{x}^{3}{c}^{3}+8\,b{x}^{2}{c}^{2}-2\,{b}^{2}xc+{b}^{3} \right ) }{5\,x{b}^{4}} \left ( c{x}^{2}+bx \right ) ^{-{\frac{3}{2}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(c*x^2+b*x)^(3/2),x)

[Out]

-2/5*(c*x+b)*(16*c^3*x^3+8*b*c^2*x^2-2*b^2*c*x+b^3)/x/b^4/(c*x^2+b*x)^(3/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.86682, size = 123, normalized size = 1.6 \begin{align*} -\frac{2 \,{\left (16 \, c^{3} x^{3} + 8 \, b c^{2} x^{2} - 2 \, b^{2} c x + b^{3}\right )} \sqrt{c x^{2} + b x}}{5 \,{\left (b^{4} c x^{4} + b^{5} x^{3}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^2+b*x)^(3/2),x, algorithm="fricas")

[Out]

-2/5*(16*c^3*x^3 + 8*b*c^2*x^2 - 2*b^2*c*x + b^3)*sqrt(c*x^2 + b*x)/(b^4*c*x^4 + b^5*x^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \left (x \left (b + c x\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(c*x**2+b*x)**(3/2),x)

[Out]

Integral(1/(x**2*(x*(b + c*x))**(3/2)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (c x^{2} + b x\right )}^{\frac{3}{2}} x^{2}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(c*x^2+b*x)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((c*x^2 + b*x)^(3/2)*x^2), x)