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

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

[Out]

-2/3*(2*c*x+b)/b^2/(c*x^2+b*x)^(3/2)+16/3*c*(2*c*x+b)/b^4/(c*x^2+b*x)^(1/2)

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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} \[ \frac {16 c (b+2 c x)}{3 b^4 \sqrt {b x+c x^2}}-\frac {2 (b+2 c x)}{3 b^2 \left (b x+c x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(b*x + c*x^2)^(-5/2),x]

[Out]

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

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

Antiderivative was successfully verified.

[In]

Integrate[(b*x + c*x^2)^(-5/2),x]

[Out]

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

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fricas [A]  time = 0.93, size = 72, normalized size = 1.33 \[ \frac {2 \, {\left (16 \, c^{3} x^{3} + 24 \, b c^{2} x^{2} + 6 \, b^{2} c x - b^{3}\right )} \sqrt {c x^{2} + b x}}{3 \, {\left (b^{4} c^{2} x^{4} + 2 \, b^{5} c x^{3} + b^{6} x^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.22, size = 50, normalized size = 0.93 \[ \frac {2 \, {\left (2 \, {\left (4 \, x {\left (\frac {2 \, c^{3} x}{b^{4}} + \frac {3 \, c^{2}}{b^{3}}\right )} + \frac {3 \, c}{b^{2}}\right )} x - \frac {1}{b}\right )}}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 51, normalized size = 0.94 \[ -\frac {2 \left (c x +b \right ) \left (-16 x^{3} c^{3}-24 b \,x^{2} c^{2}-6 b^{2} x c +b^{3}\right ) x}{3 \left (c \,x^{2}+b x \right )^{\frac {5}{2}} b^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.38, size = 72, normalized size = 1.33 \[ -\frac {4 \, c x}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b^{2}} + \frac {32 \, c^{2} x}{3 \, \sqrt {c x^{2} + b x} b^{4}} - \frac {2}{3 \, {\left (c x^{2} + b x\right )}^{\frac {3}{2}} b} + \frac {16 \, c}{3 \, \sqrt {c x^{2} + b x} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.03, size = 43, normalized size = 0.80 \[ \frac {\left (2\,b+4\,c\,x\right )\,\left (-b^2+8\,b\,c\,x+8\,c^2\,x^2\right )}{3\,b^4\,{\left (c\,x^2+b\,x\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(b*x + c*x^2)^(5/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*x + c*x**2)**(-5/2), x)

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