3.2824 \(\int \frac{1}{(\frac{c}{a+b x})^{5/2}} \, dx\)

Optimal. Leaf size=30 \[ \frac{2 (a+b x)^3}{7 b c^2 \sqrt{\frac{c}{a+b x}}} \]

[Out]

(2*(a + b*x)^3)/(7*b*c^2*Sqrt[c/(a + b*x)])

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Rubi [A]  time = 0.0087011, antiderivative size = 30, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {247, 15, 30} \[ \frac{2 (a+b x)^3}{7 b c^2 \sqrt{\frac{c}{a+b x}}} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x))^(-5/2),x]

[Out]

(2*(a + b*x)^3)/(7*b*c^2*Sqrt[c/(a + b*x)])

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin{align*} \int \frac{1}{\left (\frac{c}{a+b x}\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{\left (\frac{c}{x}\right )^{5/2}} \, dx,x,a+b x\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int x^{5/2} \, dx,x,a+b x\right )}{b c^2 \sqrt{\frac{c}{a+b x}} \sqrt{a+b x}}\\ &=\frac{2 (a+b x)^3}{7 b c^2 \sqrt{\frac{c}{a+b x}}}\\ \end{align*}

Mathematica [A]  time = 0.0174388, size = 21, normalized size = 0.7 \[ \frac{2 c}{7 b \left (\frac{c}{a+b x}\right )^{7/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x))^(-5/2),x]

[Out]

(2*c)/(7*b*(c/(a + b*x))^(7/2))

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Maple [A]  time = 0.003, size = 22, normalized size = 0.7 \begin{align*}{\frac{2\,bx+2\,a}{7\,b} \left ({\frac{c}{bx+a}} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b*x+a)/b/(c/(b*x+a))^(5/2)

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Maxima [A]  time = 1.30175, size = 23, normalized size = 0.77 \begin{align*} \frac{2 \, c}{7 \, b \left (\frac{c}{b x + a}\right )^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*c/(b*(c/(b*x + a))^(7/2))

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Fricas [B]  time = 1.20559, size = 120, normalized size = 4. \begin{align*} \frac{2 \,{\left (b^{4} x^{4} + 4 \, a b^{3} x^{3} + 6 \, a^{2} b^{2} x^{2} + 4 \, a^{3} b x + a^{4}\right )} \sqrt{\frac{c}{b x + a}}}{7 \, b c^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2*x^2 + 4*a^3*b*x + a^4)*sqrt(c/(b*x + a))/(b*c^3)

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Sympy [A]  time = 1.7963, size = 49, normalized size = 1.63 \begin{align*} \begin{cases} \frac{2 a}{7 b c^{\frac{5}{2}} \left (\frac{1}{a + b x}\right )^{\frac{5}{2}}} + \frac{2 x}{7 c^{\frac{5}{2}} \left (\frac{1}{a + b x}\right )^{\frac{5}{2}}} & \text{for}\: b \neq 0 \\\frac{x}{\left (\frac{c}{a}\right )^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*a/(7*b*c**(5/2)*(1/(a + b*x))**(5/2)) + 2*x/(7*c**(5/2)*(1/(a + b*x))**(5/2)), Ne(b, 0)), (x/(c/a
)**(5/2), True))

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Giac [A]  time = 1.12226, size = 35, normalized size = 1.17 \begin{align*} \frac{2 \,{\left (b x + a\right )}^{3}}{7 \, b c^{2} \sqrt{\frac{c}{b x + a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*(b*x + a)^3/(b*c^2*sqrt(c/(b*x + a)))