3.2829 \(\int \frac{1}{(\frac{c}{(a+b x)^2})^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0084612, 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} \[ \frac{(a+b x)^3}{4 b c \sqrt{\frac{c}{(a+b x)^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x)^2)^(-3/2),x]

[Out]

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

Mathematica [A]  time = 0.0137646, size = 25, normalized size = 0.83 \[ \frac{a+b x}{4 b \left (\frac{c}{(a+b x)^2}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x)^2)^(-3/2),x]

[Out]

(a + b*x)/(4*b*(c/(a + b*x)^2)^(3/2))

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Maple [A]  time = 0.001, size = 51, normalized size = 1.7 \begin{align*}{\frac{x \left ({b}^{3}{x}^{3}+4\,a{b}^{2}{x}^{2}+6\,{a}^{2}bx+4\,{a}^{3} \right ) }{4\, \left ( bx+a \right ) ^{3}} \left ({\frac{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]

1/4*x*(b^3*x^3+4*a*b^2*x^2+6*a^2*b*x+4*a^3)/(b*x+a)^3/(c/(b*x+a)^2)^(3/2)

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Maxima [A]  time = 1.09139, size = 50, normalized size = 1.67 \begin{align*} \frac{b^{3} x^{4} + 4 \, a b^{2} x^{3} + 6 \, a^{2} b x^{2} + 4 \, a^{3} x}{4 \, c^{\frac{3}{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]

1/4*(b^3*x^4 + 4*a*b^2*x^3 + 6*a^2*b*x^2 + 4*a^3*x)/c^(3/2)

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Fricas [B]  time = 1.24886, size = 147, normalized size = 4.9 \begin{align*} \frac{{\left (b^{4} x^{5} + 5 \, a b^{3} x^{4} + 10 \, a^{2} b^{2} x^{3} + 10 \, a^{3} b x^{2} + 4 \, a^{4} x\right )} \sqrt{\frac{c}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{4 \, c^{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="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (\frac{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*} \int \frac{1}{\left (\frac{c}{{\left (b x + a\right )}^{2}}\right )^{\frac{3}{2}}}\,{d 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]

integrate((c/(b*x + a)^2)^(-3/2), x)