3.2839 \(\int \frac{1}{(\frac{c}{(a+b x)^{3/2}})^{2/3}} \, dx\)

Optimal. Leaf size=27 \[ \frac{a+b x}{2 b \left (\frac{c}{(a+b x)^{3/2}}\right )^{2/3}} \]

[Out]

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

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Rubi [A]  time = 0.0101676, antiderivative size = 27, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {247, 15, 30} \[ \frac{a+b x}{2 b \left (\frac{c}{(a+b x)^{3/2}}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x)^(3/2))^(-2/3),x]

[Out]

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

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

Mathematica [A]  time = 0.0206451, size = 34, normalized size = 1.26 \[ \frac{x (2 a+b x)}{2 (a+b x) \left (\frac{c}{(a+b x)^{3/2}}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x)^(3/2))^(-2/3),x]

[Out]

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

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Maple [A]  time = 0.002, size = 29, normalized size = 1.1 \begin{align*}{\frac{x \left ( bx+2\,a \right ) }{2\,bx+2\,a} \left ({c \left ( bx+a \right ) ^{-{\frac{3}{2}}}} \right ) ^{-{\frac{2}{3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.957195, size = 28, normalized size = 1.04 \begin{align*} \frac{b x + a}{2 \, b \left (\frac{c}{{\left (b x + a\right )}^{\frac{3}{2}}}\right )^{\frac{2}{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.6807, size = 117, normalized size = 4.33 \begin{align*} \frac{{\left (b x^{2} + 2 \, a x\right )} \sqrt{b x + a} \left (\frac{\sqrt{b x + a} c}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )^{\frac{1}{3}}}{2 \, c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )^{\frac{3}{2}}}\right )^{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((c/(a + b*x)**(3/2))**(-2/3), 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 )}^{\frac{3}{2}}}\right )^{\frac{2}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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