3.1.66 \(\int \frac {1}{(a x+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=54 \[ \frac {16 b (a+2 b x)}{3 a^4 \sqrt {a x+b x^2}}-\frac {2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.01, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {614, 613} \begin {gather*} \frac {16 b (a+2 b x)}{3 a^4 \sqrt {a x+b x^2}}-\frac {2 (a+2 b x)}{3 a^2 \left (a x+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^2)^(-5/2),x]

[Out]

(-2*(a + 2*b*x))/(3*a^2*(a*x + b*x^2)^(3/2)) + (16*b*(a + 2*b*x))/(3*a^4*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 614

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

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 48, normalized size = 0.89 \begin {gather*} \frac {-2 a^3+12 a^2 b x+48 a b^2 x^2+32 b^3 x^3}{3 a^4 (x (a+b x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^2)^(-5/2),x]

[Out]

(-2*a^3 + 12*a^2*b*x + 48*a*b^2*x^2 + 32*b^3*x^3)/(3*a^4*(x*(a + b*x))^(3/2))

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IntegrateAlgebraic [A]  time = 0.32, size = 60, normalized size = 1.11 \begin {gather*} \frac {2 \sqrt {a x+b x^2} \left (-a^3+6 a^2 b x+24 a b^2 x^2+16 b^3 x^3\right )}{3 a^4 x^2 (a+b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a*x + b*x^2]*(-a^3 + 6*a^2*b*x + 24*a*b^2*x^2 + 16*b^3*x^3))/(3*a^4*x^2*(a + b*x)^2)

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fricas [A]  time = 0.41, size = 72, normalized size = 1.33 \begin {gather*} \frac {2 \, {\left (16 \, b^{3} x^{3} + 24 \, a b^{2} x^{2} + 6 \, a^{2} b x - a^{3}\right )} \sqrt {b x^{2} + a x}}{3 \, {\left (a^{4} b^{2} x^{4} + 2 \, a^{5} b x^{3} + a^{6} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a*x)^(5/2),x, algorithm="fricas")

[Out]

2/3*(16*b^3*x^3 + 24*a*b^2*x^2 + 6*a^2*b*x - a^3)*sqrt(b*x^2 + a*x)/(a^4*b^2*x^4 + 2*a^5*b*x^3 + a^6*x^2)

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giac [A]  time = 0.24, size = 50, normalized size = 0.93 \begin {gather*} \frac {2 \, {\left (2 \, {\left (4 \, x {\left (\frac {2 \, b^{3} x}{a^{4}} + \frac {3 \, b^{2}}{a^{3}}\right )} + \frac {3 \, b}{a^{2}}\right )} x - \frac {1}{a}\right )}}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a*x)^(5/2),x, algorithm="giac")

[Out]

2/3*(2*(4*x*(2*b^3*x/a^4 + 3*b^2/a^3) + 3*b/a^2)*x - 1/a)/(b*x^2 + a*x)^(3/2)

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maple [A]  time = 0.05, size = 51, normalized size = 0.94 \begin {gather*} -\frac {2 \left (b x +a \right ) \left (-16 b^{3} x^{3}-24 a \,b^{2} x^{2}-6 a^{2} b x +a^{3}\right ) x}{3 \left (b \,x^{2}+a x \right )^{\frac {5}{2}} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x^2+a*x)^(5/2),x)

[Out]

-2/3*(b*x+a)*x*(-16*b^3*x^3-24*a*b^2*x^2-6*a^2*b*x+a^3)/a^4/(b*x^2+a*x)^(5/2)

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maxima [A]  time = 1.32, size = 72, normalized size = 1.33 \begin {gather*} -\frac {4 \, b x}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} + \frac {32 \, b^{2} x}{3 \, \sqrt {b x^{2} + a x} a^{4}} - \frac {2}{3 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a} + \frac {16 \, b}{3 \, \sqrt {b x^{2} + a x} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x^2+a*x)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.04, size = 43, normalized size = 0.80 \begin {gather*} \frac {\left (2\,a+4\,b\,x\right )\,\left (-a^2+8\,a\,b\,x+8\,b^2\,x^2\right )}{3\,a^4\,{\left (b\,x^2+a\,x\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a*x + b*x^2)^(5/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a x + b x^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(b*x**2+a*x)**(5/2),x)

[Out]

Integral((a*x + b*x**2)**(-5/2), x)

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