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

Optimal. Leaf size=106 \[ \frac{256 b^3 (a+2 b x)}{21 a^6 \sqrt{a x+b x^2}}-\frac{32 b^2 (a+2 b x)}{21 a^4 \left (a x+b x^2\right )^{3/2}}+\frac{4 b}{7 a^2 x \left (a x+b x^2\right )^{3/2}}-\frac{2}{7 a x^2 \left (a x+b x^2\right )^{3/2}} \]

[Out]

-2/(7*a*x^2*(a*x + b*x^2)^(3/2)) + (4*b)/(7*a^2*x*(a*x + b*x^2)^(3/2)) - (32*b^2*(a + 2*b*x))/(21*a^4*(a*x + b
*x^2)^(3/2)) + (256*b^3*(a + 2*b*x))/(21*a^6*Sqrt[a*x + b*x^2])

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Rubi [A]  time = 0.0362233, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {658, 614, 613} \[ \frac{256 b^3 (a+2 b x)}{21 a^6 \sqrt{a x+b x^2}}-\frac{32 b^2 (a+2 b x)}{21 a^4 \left (a x+b x^2\right )^{3/2}}+\frac{4 b}{7 a^2 x \left (a x+b x^2\right )^{3/2}}-\frac{2}{7 a x^2 \left (a x+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a*x + b*x^2)^(5/2)),x]

[Out]

-2/(7*a*x^2*(a*x + b*x^2)^(3/2)) + (4*b)/(7*a^2*x*(a*x + b*x^2)^(3/2)) - (32*b^2*(a + 2*b*x))/(21*a^4*(a*x + b
*x^2)^(3/2)) + (256*b^3*(a + 2*b*x))/(21*a^6*Sqrt[a*x + b*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 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]

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 (a x+b x^2\right )^{5/2}} \, dx &=-\frac{2}{7 a x^2 \left (a x+b x^2\right )^{3/2}}-\frac{(10 b) \int \frac{1}{x \left (a x+b x^2\right )^{5/2}} \, dx}{7 a}\\ &=-\frac{2}{7 a x^2 \left (a x+b x^2\right )^{3/2}}+\frac{4 b}{7 a^2 x \left (a x+b x^2\right )^{3/2}}+\frac{\left (16 b^2\right ) \int \frac{1}{\left (a x+b x^2\right )^{5/2}} \, dx}{7 a^2}\\ &=-\frac{2}{7 a x^2 \left (a x+b x^2\right )^{3/2}}+\frac{4 b}{7 a^2 x \left (a x+b x^2\right )^{3/2}}-\frac{32 b^2 (a+2 b x)}{21 a^4 \left (a x+b x^2\right )^{3/2}}-\frac{\left (128 b^3\right ) \int \frac{1}{\left (a x+b x^2\right )^{3/2}} \, dx}{21 a^4}\\ &=-\frac{2}{7 a x^2 \left (a x+b x^2\right )^{3/2}}+\frac{4 b}{7 a^2 x \left (a x+b x^2\right )^{3/2}}-\frac{32 b^2 (a+2 b x)}{21 a^4 \left (a x+b x^2\right )^{3/2}}+\frac{256 b^3 (a+2 b x)}{21 a^6 \sqrt{a x+b x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0199907, size = 73, normalized size = 0.69 \[ \frac{2 \left (-16 a^3 b^2 x^2+96 a^2 b^3 x^3+6 a^4 b x-3 a^5+384 a b^4 x^4+256 b^5 x^5\right )}{21 a^6 x^2 (x (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*(a*x + b*x^2)^(5/2)),x]

[Out]

(2*(-3*a^5 + 6*a^4*b*x - 16*a^3*b^2*x^2 + 96*a^2*b^3*x^3 + 384*a*b^4*x^4 + 256*b^5*x^5))/(21*a^6*x^2*(x*(a + b
*x))^(3/2))

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Maple [A]  time = 0.047, size = 77, normalized size = 0.7 \begin{align*} -{\frac{ \left ( 2\,bx+2\,a \right ) \left ( -256\,{b}^{5}{x}^{5}-384\,{b}^{4}{x}^{4}a-96\,{b}^{3}{x}^{3}{a}^{2}+16\,{b}^{2}{x}^{2}{a}^{3}-6\,bx{a}^{4}+3\,{a}^{5} \right ) }{21\,x{a}^{6}} \left ( b{x}^{2}+ax \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21*(b*x+a)*(-256*b^5*x^5-384*a*b^4*x^4-96*a^2*b^3*x^3+16*a^3*b^2*x^2-6*a^4*b*x+3*a^5)/x/a^6/(b*x^2+a*x)^(5/
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/(b*x^2+a*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.03942, size = 197, normalized size = 1.86 \begin{align*} \frac{2 \,{\left (256 \, b^{5} x^{5} + 384 \, a b^{4} x^{4} + 96 \, a^{2} b^{3} x^{3} - 16 \, a^{3} b^{2} x^{2} + 6 \, a^{4} b x - 3 \, a^{5}\right )} \sqrt{b x^{2} + a x}}{21 \,{\left (a^{6} b^{2} x^{6} + 2 \, a^{7} b x^{5} + a^{8} x^{4}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/21*(256*b^5*x^5 + 384*a*b^4*x^4 + 96*a^2*b^3*x^3 - 16*a^3*b^2*x^2 + 6*a^4*b*x - 3*a^5)*sqrt(b*x^2 + a*x)/(a^
6*b^2*x^6 + 2*a^7*b*x^5 + a^8*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{2} + a x\right )}^{\frac{5}{2}} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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