### 3.59 $$\int \frac{1}{x^3 (b x+c x^2)^{3/2}} \, dx$$

Optimal. Leaf size=103 $\frac{128 c^3 (b+2 c x)}{35 b^5 \sqrt{b x+c x^2}}-\frac{32 c^2}{35 b^3 x \sqrt{b x+c x^2}}+\frac{16 c}{35 b^2 x^2 \sqrt{b x+c x^2}}-\frac{2}{7 b x^3 \sqrt{b x+c x^2}}$

[Out]

-2/(7*b*x^3*Sqrt[b*x + c*x^2]) + (16*c)/(35*b^2*x^2*Sqrt[b*x + c*x^2]) - (32*c^2)/(35*b^3*x*Sqrt[b*x + c*x^2])
+ (128*c^3*(b + 2*c*x))/(35*b^5*Sqrt[b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.0394125, antiderivative size = 103, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {658, 613} $\frac{128 c^3 (b+2 c x)}{35 b^5 \sqrt{b x+c x^2}}-\frac{32 c^2}{35 b^3 x \sqrt{b x+c x^2}}+\frac{16 c}{35 b^2 x^2 \sqrt{b x+c x^2}}-\frac{2}{7 b x^3 \sqrt{b x+c x^2}}$

Antiderivative was successfully veriﬁed.

[In]

Int[1/(x^3*(b*x + c*x^2)^(3/2)),x]

[Out]

-2/(7*b*x^3*Sqrt[b*x + c*x^2]) + (16*c)/(35*b^2*x^2*Sqrt[b*x + c*x^2]) - (32*c^2)/(35*b^3*x*Sqrt[b*x + c*x^2])
+ (128*c^3*(b + 2*c*x))/(35*b^5*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^3 \left (b x+c x^2\right )^{3/2}} \, dx &=-\frac{2}{7 b x^3 \sqrt{b x+c x^2}}-\frac{(8 c) \int \frac{1}{x^2 \left (b x+c x^2\right )^{3/2}} \, dx}{7 b}\\ &=-\frac{2}{7 b x^3 \sqrt{b x+c x^2}}+\frac{16 c}{35 b^2 x^2 \sqrt{b x+c x^2}}+\frac{\left (48 c^2\right ) \int \frac{1}{x \left (b x+c x^2\right )^{3/2}} \, dx}{35 b^2}\\ &=-\frac{2}{7 b x^3 \sqrt{b x+c x^2}}+\frac{16 c}{35 b^2 x^2 \sqrt{b x+c x^2}}-\frac{32 c^2}{35 b^3 x \sqrt{b x+c x^2}}-\frac{\left (64 c^3\right ) \int \frac{1}{\left (b x+c x^2\right )^{3/2}} \, dx}{35 b^3}\\ &=-\frac{2}{7 b x^3 \sqrt{b x+c x^2}}+\frac{16 c}{35 b^2 x^2 \sqrt{b x+c x^2}}-\frac{32 c^2}{35 b^3 x \sqrt{b x+c x^2}}+\frac{128 c^3 (b+2 c x)}{35 b^5 \sqrt{b x+c x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0144757, size = 62, normalized size = 0.6 $\frac{2 \left (-16 b^2 c^2 x^2+8 b^3 c x-5 b^4+64 b c^3 x^3+128 c^4 x^4\right )}{35 b^5 x^3 \sqrt{x (b+c x)}}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(2*(-5*b^4 + 8*b^3*c*x - 16*b^2*c^2*x^2 + 64*b*c^3*x^3 + 128*c^4*x^4))/(35*b^5*x^3*Sqrt[x*(b + c*x)])

________________________________________________________________________________________

Maple [A]  time = 0.045, size = 66, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -128\,{c}^{4}{x}^{4}-64\,{x}^{3}{c}^{3}b+16\,{c}^{2}{x}^{2}{b}^{2}-8\,cx{b}^{3}+5\,{b}^{4} \right ) }{35\,{x}^{2}{b}^{5}} \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^3/(c*x^2+b*x)^(3/2),x)

[Out]

-2/35*(c*x+b)*(-128*c^4*x^4-64*b*c^3*x^3+16*b^2*c^2*x^2-8*b^3*c*x+5*b^4)/x^2/b^5/(c*x^2+b*x)^(3/2)

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.93629, size = 151, normalized size = 1.47 \begin{align*} \frac{2 \,{\left (128 \, c^{4} x^{4} + 64 \, b c^{3} x^{3} - 16 \, b^{2} c^{2} x^{2} + 8 \, b^{3} c x - 5 \, b^{4}\right )} \sqrt{c x^{2} + b x}}{35 \,{\left (b^{5} c x^{5} + b^{6} x^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*(128*c^4*x^4 + 64*b*c^3*x^3 - 16*b^2*c^2*x^2 + 8*b^3*c*x - 5*b^4)*sqrt(c*x^2 + b*x)/(b^5*c*x^5 + b^6*x^4)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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^{3}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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