3.2807 \(\int \frac{1}{(c (a+b x)^2)^{5/2}} \, dx\)

Optimal. Leaf size=30 \[ -\frac{1}{4 b c^2 (a+b x)^3 \sqrt{c (a+b x)^2}} \]

[Out]

-1/(4*b*c^2*(a + b*x)^3*Sqrt[c*(a + b*x)^2])

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Rubi [A]  time = 0.0125872, 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} \[ -\frac{1}{4 b c^2 (a+b x)^3 \sqrt{c (a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*(a + b*x)^2)^(-5/2),x]

[Out]

-1/(4*b*c^2*(a + b*x)^3*Sqrt[c*(a + b*x)^2])

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

Mathematica [A]  time = 0.011869, size = 25, normalized size = 0.83 \[ -\frac{a+b x}{4 b \left (c (a+b x)^2\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*(a + b*x)^2)^(-5/2),x]

[Out]

-(a + b*x)/(4*b*(c*(a + b*x)^2)^(5/2))

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Maple [A]  time = 0.003, size = 22, normalized size = 0.7 \begin{align*} -{\frac{bx+a}{4\,b} \left ( c \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(b*x+a)/b/(c*(b*x+a)^2)^(5/2)

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Maxima [A]  time = 0.958019, size = 24, normalized size = 0.8 \begin{align*} -\frac{1}{4 \, \left (b^{2} c\right )^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/((b^2*c)^(5/2)*(x + a/b)^4)

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Fricas [B]  time = 1.35234, size = 197, normalized size = 6.57 \begin{align*} -\frac{\sqrt{b^{2} c x^{2} + 2 \, a b c x + a^{2} c}}{4 \,{\left (b^{6} c^{3} x^{5} + 5 \, a b^{5} c^{3} x^{4} + 10 \, a^{2} b^{4} c^{3} x^{3} + 10 \, a^{3} b^{3} c^{3} x^{2} + 5 \, a^{4} b^{2} c^{3} x + a^{5} b c^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*sqrt(b^2*c*x^2 + 2*a*b*c*x + a^2*c)/(b^6*c^3*x^5 + 5*a*b^5*c^3*x^4 + 10*a^2*b^4*c^3*x^3 + 10*a^3*b^3*c^3*
x^2 + 5*a^4*b^2*c^3*x + a^5*b*c^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c*(a + b*x)**2)**(-5/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x