3.1.65 \(\int \frac {x}{(a x+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=48 \[ \frac {2 x}{3 a \left (a x+b x^2\right )^{3/2}}-\frac {8 (a+2 b x)}{3 a^3 \sqrt {a x+b x^2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {638, 613} \begin {gather*} \frac {2 x}{3 a \left (a x+b x^2\right )^{3/2}}-\frac {8 (a+2 b x)}{3 a^3 \sqrt {a x+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(a*x + b*x^2)^(5/2),x]

[Out]

(2*x)/(3*a*(a*x + b*x^2)^(3/2)) - (8*(a + 2*b*x))/(3*a^3*Sqrt[a*x + b*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 638

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b*d - 2*a*e + (2*c*d -
b*e)*x)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(b^2 - 4*a*c)), x] - Dist[((2*p + 3)*(2*c*d - b*e))/((p + 1)*(b^2
- 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^
2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 38, normalized size = 0.79 \begin {gather*} -\frac {2 x \left (3 a^2+12 a b x+8 b^2 x^2\right )}{3 a^3 (x (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x/(a*x + b*x^2)^(5/2),x]

[Out]

(-2*x*(3*a^2 + 12*a*b*x + 8*b^2*x^2))/(3*a^3*(x*(a + b*x))^(3/2))

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IntegrateAlgebraic [A]  time = 0.32, size = 49, normalized size = 1.02 \begin {gather*} -\frac {2 \sqrt {a x+b x^2} \left (3 a^2+12 a b x+8 b^2 x^2\right )}{3 a^3 x (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x/(a*x + b*x^2)^(5/2),x]

[Out]

(-2*Sqrt[a*x + b*x^2]*(3*a^2 + 12*a*b*x + 8*b^2*x^2))/(3*a^3*x*(a + b*x)^2)

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fricas [A]  time = 0.41, size = 59, normalized size = 1.23 \begin {gather*} -\frac {2 \, {\left (8 \, b^{2} x^{2} + 12 \, a b x + 3 \, a^{2}\right )} \sqrt {b x^{2} + a x}}{3 \, {\left (a^{3} b^{2} x^{3} + 2 \, a^{4} b x^{2} + a^{5} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^2+a*x)^(5/2),x, algorithm="fricas")

[Out]

-2/3*(8*b^2*x^2 + 12*a*b*x + 3*a^2)*sqrt(b*x^2 + a*x)/(a^3*b^2*x^3 + 2*a^4*b*x^2 + a^5*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{{\left (b x^{2} + a x\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^2+a*x)^(5/2),x, algorithm="giac")

[Out]

integrate(x/(b*x^2 + a*x)^(5/2), x)

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maple [A]  time = 0.05, size = 44, normalized size = 0.92 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (8 b^{2} x^{2}+12 a b x +3 a^{2}\right ) x^{2}}{3 \left (b \,x^{2}+a x \right )^{\frac {5}{2}} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(b*x^2+a*x)^(5/2),x)

[Out]

-2/3*(b*x+a)*x^2*(8*b^2*x^2+12*a*b*x+3*a^2)/a^3/(b*x^2+a*x)^(5/2)

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maxima [A]  time = 1.37, size = 52, normalized size = 1.08 \begin {gather*} \frac {2 \, x}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a} - \frac {16 \, b x}{3 \, \sqrt {b x^{2} + a x} a^{3}} - \frac {8}{3 \, \sqrt {b x^{2} + a x} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x^2+a*x)^(5/2),x, algorithm="maxima")

[Out]

2/3*x/((b*x^2 + a*x)^(3/2)*a) - 16/3*b*x/(sqrt(b*x^2 + a*x)*a^3) - 8/3/(sqrt(b*x^2 + a*x)*a^2)

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mupad [B]  time = 0.24, size = 45, normalized size = 0.94 \begin {gather*} -\frac {2\,\sqrt {b\,x^2+a\,x}\,\left (3\,a^2+12\,a\,b\,x+8\,b^2\,x^2\right )}{3\,a^3\,x\,{\left (a+b\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(a*x + b*x^2)^(5/2),x)

[Out]

-(2*(a*x + b*x^2)^(1/2)*(3*a^2 + 8*b^2*x^2 + 12*a*b*x))/(3*a^3*x*(a + b*x)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(b*x**2+a*x)**(5/2),x)

[Out]

Integral(x/(x*(a + b*x))**(5/2), x)

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