3.2806 \(\int \frac{1}{(c (a+b x)^2)^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0112982, 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} \[ -\frac{1}{2 b c (a+b x) \sqrt{c (a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*(a + b*x)^2)^(-3/2),x]

[Out]

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

Mathematica [A]  time = 0.0085386, size = 25, normalized size = 0.83 \[ -\frac{a+b x}{2 b \left (c (a+b x)^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*(a + b*x)^2)^(-3/2),x]

[Out]

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

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Maple [A]  time = 0.002, size = 22, normalized size = 0.7 \begin{align*} -{\frac{bx+a}{2\,b} \left ( c \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*(b*x+a)^2)^(3/2),x)

[Out]

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

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Maxima [A]  time = 0.955827, size = 24, normalized size = 0.8 \begin{align*} -\frac{1}{2 \, \left (b^{2} c\right )^{\frac{3}{2}}{\left (x + \frac{a}{b}\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*(b*x+a)^2)^(3/2),x, algorithm="maxima")

[Out]

-1/2/((b^2*c)^(3/2)*(x + a/b)^2)

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Fricas [B]  time = 1.34497, size = 140, normalized size = 4.67 \begin{align*} -\frac{\sqrt{b^{2} c x^{2} + 2 \, a b c x + a^{2} c}}{2 \,{\left (b^{4} c^{2} x^{3} + 3 \, a b^{3} c^{2} x^{2} + 3 \, a^{2} b^{2} c^{2} x + a^{3} b c^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*(b*x+a)^2)^(3/2),x, algorithm="fricas")

[Out]

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

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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{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*(b*x+a)**2)**(3/2),x)

[Out]

Integral((c*(a + b*x)**2)**(-3/2), 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)^(3/2),x, algorithm="giac")

[Out]

sage0*x