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

Optimal. Leaf size=80 \[ -\frac{128 b^2 (a+2 b x)}{15 a^5 \sqrt{a x+b x^2}}+\frac{16 b (a+2 b x)}{15 a^3 \left (a x+b x^2\right )^{3/2}}-\frac{2}{5 a x \left (a x+b x^2\right )^{3/2}} \]

[Out]

-2/(5*a*x*(a*x + b*x^2)^(3/2)) + (16*b*(a + 2*b*x))/(15*a^3*(a*x + b*x^2)^(3/2)) - (128*b^2*(a + 2*b*x))/(15*a
^5*Sqrt[a*x + b*x^2])

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Rubi [A]  time = 0.0220632, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.176, Rules used = {658, 614, 613} \[ -\frac{128 b^2 (a+2 b x)}{15 a^5 \sqrt{a x+b x^2}}+\frac{16 b (a+2 b x)}{15 a^3 \left (a x+b x^2\right )^{3/2}}-\frac{2}{5 a x \left (a x+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x*(a*x + b*x^2)^(5/2)),x]

[Out]

-2/(5*a*x*(a*x + b*x^2)^(3/2)) + (16*b*(a + 2*b*x))/(15*a^3*(a*x + b*x^2)^(3/2)) - (128*b^2*(a + 2*b*x))/(15*a
^5*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 \left (a x+b x^2\right )^{5/2}} \, dx &=-\frac{2}{5 a x \left (a x+b x^2\right )^{3/2}}-\frac{(8 b) \int \frac{1}{\left (a x+b x^2\right )^{5/2}} \, dx}{5 a}\\ &=-\frac{2}{5 a x \left (a x+b x^2\right )^{3/2}}+\frac{16 b (a+2 b x)}{15 a^3 \left (a x+b x^2\right )^{3/2}}+\frac{\left (64 b^2\right ) \int \frac{1}{\left (a x+b x^2\right )^{3/2}} \, dx}{15 a^3}\\ &=-\frac{2}{5 a x \left (a x+b x^2\right )^{3/2}}+\frac{16 b (a+2 b x)}{15 a^3 \left (a x+b x^2\right )^{3/2}}-\frac{128 b^2 (a+2 b x)}{15 a^5 \sqrt{a x+b x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0183507, size = 62, normalized size = 0.78 \[ -\frac{2 \left (48 a^2 b^2 x^2-8 a^3 b x+3 a^4+192 a b^3 x^3+128 b^4 x^4\right )}{15 a^5 x (x (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(3*a^4 - 8*a^3*b*x + 48*a^2*b^2*x^2 + 192*a*b^3*x^3 + 128*b^4*x^4))/(15*a^5*x*(x*(a + b*x))^(3/2))

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Maple [A]  time = 0.044, size = 63, normalized size = 0.8 \begin{align*} -{\frac{ \left ( 2\,bx+2\,a \right ) \left ( 128\,{b}^{4}{x}^{4}+192\,a{b}^{3}{x}^{3}+48\,{b}^{2}{x}^{2}{a}^{2}-8\,x{a}^{3}b+3\,{a}^{4} \right ) }{15\,{a}^{5}} \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/(b*x^2+a*x)^(5/2),x)

[Out]

-2/15*(b*x+a)*(128*b^4*x^4+192*a*b^3*x^3+48*a^2*b^2*x^2-8*a^3*b*x+3*a^4)/a^5/(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/(b*x^2+a*x)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.95703, size = 176, normalized size = 2.2 \begin{align*} -\frac{2 \,{\left (128 \, b^{4} x^{4} + 192 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} - 8 \, a^{3} b x + 3 \, a^{4}\right )} \sqrt{b x^{2} + a x}}{15 \,{\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(128*b^4*x^4 + 192*a*b^3*x^3 + 48*a^2*b^2*x^2 - 8*a^3*b*x + 3*a^4)*sqrt(b*x^2 + a*x)/(a^5*b^2*x^5 + 2*a^
6*b*x^4 + a^7*x^3)

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

[Out]

Integral(1/(x*(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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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