3.2826 \(\int (\frac {c}{(a+b x)^2})^{3/2} \, dx\)

Optimal. Leaf size=28 \[ -\frac {c \sqrt {\frac {c}{(a+b x)^2}}}{2 b (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, 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} \[ -\frac {c \sqrt {\frac {c}{(a+b x)^2}}}{2 b (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x)^2)^(3/2),x]

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 0.89 \[ -\frac {(a+b x) \left (\frac {c}{(a+b x)^2}\right )^{3/2}}{2 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x)^2)^(3/2),x]

[Out]

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

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fricas [A]  time = 0.54, size = 36, normalized size = 1.29 \[ -\frac {c \sqrt {\frac {c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{2 \, {\left (b^{2} x + a b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 21, normalized size = 0.75 \[ -\frac {c^{\frac {3}{2}} \mathrm {sgn}\left (b x + a\right )}{2 \, {\left (b x + a\right )}^{2} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 22, normalized size = 0.79 \[ -\frac {\left (b x +a \right ) \left (\frac {c}{\left (b x +a \right )^{2}}\right )^{\frac {3}{2}}}{2 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.66, size = 27, normalized size = 0.96 \[ -\frac {c^{\frac {3}{2}}}{2 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.17, size = 24, normalized size = 0.86 \[ -\frac {c\,\sqrt {\frac {c}{{\left (a+b\,x\right )}^2}}}{2\,b\,\left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/(a + b*x)^2)^(3/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (\frac {c}{\left (a + b x\right )^{2}}\right )^{\frac {3}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c/(a + b*x)**2)**(3/2), x)

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