3.2834 \(\int \frac{1}{\sqrt{\frac{c}{(a+b x)^3}}} \, dx\)

Optimal. Leaf size=25 \[ \frac{2 (a+b x)}{5 b \sqrt{\frac{c}{(a+b x)^3}}} \]

[Out]

(2*(a + b*x))/(5*b*Sqrt[c/(a + b*x)^3])

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Rubi [A]  time = 0.0082432, antiderivative size = 25, 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} \[ \frac{2 (a+b x)}{5 b \sqrt{\frac{c}{(a+b x)^3}}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[c/(a + b*x)^3],x]

[Out]

(2*(a + b*x))/(5*b*Sqrt[c/(a + b*x)^3])

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

Mathematica [A]  time = 0.0152308, size = 25, normalized size = 1. \[ \frac{2 (a+b x)}{5 b \sqrt{\frac{c}{(a+b x)^3}}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[c/(a + b*x)^3],x]

[Out]

(2*(a + b*x))/(5*b*Sqrt[c/(a + b*x)^3])

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Maple [A]  time = 0.002, size = 22, normalized size = 0.9 \begin{align*}{\frac{2\,bx+2\,a}{5\,b}{\frac{1}{\sqrt{{\frac{c}{ \left ( bx+a \right ) ^{3}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b*x+a)/b/(c/(b*x+a)^3)^(1/2)

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Maxima [A]  time = 1.5605, size = 36, normalized size = 1.44 \begin{align*} \frac{2 \,{\left (b \sqrt{c} x + a \sqrt{c}\right )}{\left (b x + a\right )}^{\frac{3}{2}}}{5 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(b*sqrt(c)*x + a*sqrt(c))*(b*x + a)^(3/2)/(b*c)

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Fricas [B]  time = 1.31795, size = 161, normalized size = 6.44 \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^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}}}{5 \, b c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/5*(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^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)
)/(b*c)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{\frac{c}{\left (a + b x\right )^{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(c/(a + b*x)**3), x)

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Giac [B]  time = 1.15508, size = 194, normalized size = 7.76 \begin{align*} \frac{2 \,{\left (15 \, \sqrt{b c x + a c} a^{2} - \frac{10 \,{\left (3 \, \sqrt{b c x + a c} a c -{\left (b c x + a c\right )}^{\frac{3}{2}}\right )} a}{c} + \frac{15 \, \sqrt{b c x + a c} a^{2} c^{2} - 10 \,{\left (b c x + a c\right )}^{\frac{3}{2}} a c + 3 \,{\left (b c x + a c\right )}^{\frac{5}{2}}}{c^{2}}\right )}}{15 \, b c \mathrm{sgn}\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right ) \mathrm{sgn}\left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(15*sqrt(b*c*x + a*c)*a^2 - 10*(3*sqrt(b*c*x + a*c)*a*c - (b*c*x + a*c)^(3/2))*a/c + (15*sqrt(b*c*x + a*c
)*a^2*c^2 - 10*(b*c*x + a*c)^(3/2)*a*c + 3*(b*c*x + a*c)^(5/2))/c^2)/(b*c*sgn(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*
x + a^3)*sgn(b*x + a))