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

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

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Rubi [A]  time = 0.02, antiderivative size = 77, normalized size of antiderivative = 1.00, 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} \begin {gather*} -\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}} \end {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.01, size = 49, normalized size = 0.64 \begin {gather*} -\frac {2 \left (b^3-2 b^2 c x+8 b c^2 x^2+16 c^3 x^3\right )}{5 b^4 x^2 \sqrt {x (b+c x)}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.29, size = 58, normalized size = 0.75 \begin {gather*} -\frac {2 \sqrt {b x+c x^2} \left (b^3-2 b^2 c x+8 b c^2 x^2+16 c^3 x^3\right )}{5 b^4 x^3 (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 59, normalized size = 0.77 \begin {gather*} -\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 {gather*}

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (c x^{2} + b x\right )}^{\frac {3}{2}} x^{2}}\,{d x} \end {gather*}

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)

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maple [A]  time = 0.05, size = 53, normalized size = 0.69 \begin {gather*} -\frac {2 \left (c x +b \right ) \left (16 x^{3} c^{3}+8 b \,x^{2} c^{2}-2 b^{2} x c +b^{3}\right )}{5 \left (c \,x^{2}+b x \right )^{\frac {3}{2}} b^{4} x} \end {gather*}

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 [A]  time = 1.31, size = 79, normalized size = 1.03 \begin {gather*} -\frac {32 \, c^{3} x}{5 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {16 \, c^{2}}{5 \, \sqrt {c x^{2} + b x} b^{3}} + \frac {4 \, c}{5 \, \sqrt {c x^{2} + b x} b^{2} x} - \frac {2}{5 \, \sqrt {c x^{2} + b x} b x^{2}} \end {gather*}

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]

-32/5*c^3*x/(sqrt(c*x^2 + b*x)*b^4) - 16/5*c^2/(sqrt(c*x^2 + b*x)*b^3) + 4/5*c/(sqrt(c*x^2 + b*x)*b^2*x) - 2/5
/(sqrt(c*x^2 + b*x)*b*x^2)

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mupad [B]  time = 0.28, size = 54, normalized size = 0.70 \begin {gather*} -\frac {2\,\sqrt {c\,x^2+b\,x}\,\left (b^3-2\,b^2\,c\,x+8\,b\,c^2\,x^2+16\,c^3\,x^3\right )}{5\,b^4\,x^3\,\left (b+c\,x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(b*x + c*x^2)^(3/2)),x)

[Out]

-(2*(b*x + c*x^2)^(1/2)*(b^3 + 16*c^3*x^3 + 8*b*c^2*x^2 - 2*b^2*c*x))/(5*b^4*x^3*(b + c*x))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{x^{2} \left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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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