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/(b*x^2+a*x)^(3/2)+16/15*b*(2*b*x+a)/a^3/(b*x^2+a*x)^(3/2)-128/15*b^2*(2*b*x+a)/a^5/(b*x^2+a*x)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 80, normalized size of antiderivative = 1.00, 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 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]

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 \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*}

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Mathematica [A]  time = 0.02, size = 62, normalized size = 0.78 \[ -\frac {2 \left (3 a^4-8 a^3 b x+48 a^2 b^2 x^2+192 a b^3 x^3+128 b^4 x^4\right )}{15 a^5 x (x (a+b x))^{3/2}} \]

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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fricas [A]  time = 0.65, size = 83, normalized size = 1.04 \[ -\frac {2 \, {\left (128 \, b^{4} x^{4} + 192 \, a b^{3} x^{3} + 48 \, a^{2} b^{2} x^{2} - 8 \, a^{3} b x + 3 \, a^{4}\right )} \sqrt {b x^{2} + a x}}{15 \, {\left (a^{5} b^{2} x^{5} + 2 \, a^{6} b x^{4} + a^{7} x^{3}\right )}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a x\right )}^{\frac {5}{2}} x}\,{d x} \]

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)

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maple [A]  time = 0.04, size = 63, normalized size = 0.79 \[ -\frac {2 \left (b x +a \right ) \left (128 b^{4} x^{4}+192 a \,b^{3} x^{3}+48 b^{2} x^{2} a^{2}-8 b x \,a^{3}+3 a^{4}\right )}{15 \left (b \,x^{2}+a x \right )^{\frac {5}{2}} a^{5}} \]

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 [A]  time = 1.41, size = 96, normalized size = 1.20 \[ \frac {32 \, b^{2} x}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{3}} - \frac {256 \, b^{3} x}{15 \, \sqrt {b x^{2} + a x} a^{5}} + \frac {16 \, b}{15 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a^{2}} - \frac {128 \, b^{2}}{15 \, \sqrt {b x^{2} + a x} a^{4}} - \frac {2}{5 \, {\left (b x^{2} + a x\right )}^{\frac {3}{2}} a x} \]

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]

32/15*b^2*x/((b*x^2 + a*x)^(3/2)*a^3) - 256/15*b^3*x/(sqrt(b*x^2 + a*x)*a^5) + 16/15*b/((b*x^2 + a*x)^(3/2)*a^
2) - 128/15*b^2/(sqrt(b*x^2 + a*x)*a^4) - 2/5/((b*x^2 + a*x)^(3/2)*a*x)

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mupad [B]  time = 0.30, size = 67, normalized size = 0.84 \[ -\frac {2\,\sqrt {b\,x^2+a\,x}\,\left (3\,a^4-8\,a^3\,b\,x+48\,a^2\,b^2\,x^2+192\,a\,b^3\,x^3+128\,b^4\,x^4\right )}{15\,a^5\,x^3\,{\left (a+b\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*(a*x + b*x^2)^(5/2)),x)

[Out]

-(2*(a*x + b*x^2)^(1/2)*(3*a^4 + 128*b^4*x^4 + 192*a*b^3*x^3 + 48*a^2*b^2*x^2 - 8*a^3*b*x))/(15*a^5*x^3*(a + b
*x)^2)

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

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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