3.2840 \(\int \frac{1}{(\frac{c}{(a+b x)^{2/3}})^{3/2}} \, dx\)

Optimal. Leaf size=34 \[ \frac{(a+b x)^{5/3}}{2 b c \sqrt{\frac{c}{(a+b x)^{2/3}}}} \]

[Out]

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

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Rubi [A]  time = 0.0114691, antiderivative size = 34, 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} \[ \frac{(a+b x)^{5/3}}{2 b c \sqrt{\frac{c}{(a+b x)^{2/3}}}} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x)^(2/3))^(-3/2),x]

[Out]

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

Mathematica [A]  time = 0.0210627, size = 34, normalized size = 1. \[ \frac{x (2 a+b x)}{2 (a+b x) \left (\frac{c}{(a+b x)^{2/3}}\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.002, size = 29, normalized size = 0.9 \begin{align*}{\frac{x \left ( bx+2\,a \right ) }{2\,bx+2\,a} \left ({c \left ( bx+a \right ) ^{-{\frac{2}{3}}}} \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))^(3/2),x)

[Out]

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

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Maxima [A]  time = 0.954062, size = 20, normalized size = 0.59 \begin{align*} \frac{b x^{2} + 2 \, a x}{2 \, 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))^(3/2),x, algorithm="maxima")

[Out]

1/2*(b*x^2 + 2*a*x)/c^(3/2)

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Fricas [A]  time = 1.50003, size = 39, normalized size = 1.15 \begin{align*} \frac{b x^{2} + 2 \, a x}{2 \, 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))^(3/2),x, algorithm="fricas")

[Out]

1/2*(b*x^2 + 2*a*x)/c^(3/2)

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Sympy [A]  time = 2.60487, size = 76, normalized size = 2.24 \begin{align*} \begin{cases} \frac{2 a x}{\frac{2 a c^{\frac{3}{2}}}{a + b x} + \frac{2 b c^{\frac{3}{2}} x}{a + b x}} + \frac{b x^{2}}{\frac{2 a c^{\frac{3}{2}}}{a + b x} + \frac{2 b c^{\frac{3}{2}} x}{a + b x}} & \text{for}\: a \neq 0 \vee b \neq 0 \\\frac{x}{\left (\tilde{\infty } c\right )^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((2*a*x/(2*a*c**(3/2)/(a + b*x) + 2*b*c**(3/2)*x/(a + b*x)) + b*x**2/(2*a*c**(3/2)/(a + b*x) + 2*b*c*
*(3/2)*x/(a + b*x)), Ne(a, 0) | Ne(b, 0)), (x/(zoo*c)**(3/2), True))

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Giac [A]  time = 1.13091, size = 20, normalized size = 0.59 \begin{align*} \frac{b x^{2} + 2 \, a x}{2 \, 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))^(3/2),x, algorithm="giac")

[Out]

1/2*(b*x^2 + 2*a*x)/c^(3/2)