3.2837 \(\int (c (a+b x)^{3/2})^{2/3} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0095091, 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} \[ \frac{(a+b x) \left (c (a+b x)^{3/2}\right )^{2/3}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*(a + b*x)^(3/2))^(2/3),x]

[Out]

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

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

Mathematica [A]  time = 0.0151122, size = 34, normalized size = 1.26 \[ \frac{x (2 a+b x) \left (c (a+b x)^{3/2}\right )^{2/3}}{2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.003, 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((c*(b*x+a)^(3/2))^(2/3),x)

[Out]

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

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Maxima [A]  time = 1.35201, size = 28, normalized size = 1.04 \begin{align*} \frac{\left ({\left (b x + a\right )}^{\frac{3}{2}} c\right )^{\frac{2}{3}}{\left (b x + a\right )}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.8888, size = 90, normalized size = 3.33 \begin{align*} \frac{{\left (b x^{2} + 2 \, a x\right )} \left ({\left (b c x + a c\right )} \sqrt{b x + a}\right )^{\frac{2}{3}}}{2 \,{\left (b x + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (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((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 [A]  time = 1.11462, size = 20, normalized size = 0.74 \begin{align*} \frac{{\left (b x + a\right )}^{2} c^{\frac{2}{3}}}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x + a)^2*c^(2/3)/b