\(\int (c (a+b x)^{3/2})^{2/3} \, dx\) [2837]

Optimal result

Integrand size = 15, antiderivative size = 27 \[ \int \left (c (a+b x)^{3/2}\right )^{2/3} \, dx=\frac {(a+b x) \left (c (a+b x)^{3/2}\right )^{2/3}}{2 b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {253, 15, 30} \[ \int \left (c (a+b x)^{3/2}\right )^{2/3} \, dx=\frac {(a+b x) \left (c (a+b x)^{3/2}\right )^{2/3}}{2 b} \]

[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 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]

Rule 253

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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} \text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \left (c (a+b x)^{3/2}\right )^{2/3} \, dx=\frac {x \left (c (a+b x)^{3/2}\right )^{2/3} (2 a+b x)}{2 (a+b x)} \]

[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))

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.07

method	result	size
gosper	\(\frac {x \left (b x +2 a \right ) \left (c \left (b x +a \right )^{\frac {3}{2}}\right )^{\frac {2}{3}}}{2 b x +2 a}\)	\(29\)

[In]

int((c*(b*x+a)^(3/2))^(2/3),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.37 \[ \int \left (c (a+b x)^{3/2}\right )^{2/3} \, dx=\frac {{\left (b x^{2} + 2 \, a x\right )} \left ({\left (b c x + a c\right )} \sqrt {b x + a}\right )^{\frac {2}{3}}}{2 \, {\left (b x + a\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 2.54 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \left (c (a+b x)^{3/2}\right )^{2/3} \, dx=\begin {cases} \frac {\left (c \left (a + b x\right )^{\frac {3}{2}}\right )^{\frac {2}{3}} \left (a + b x\right )}{2 b} & \text {for}\: b \neq 0 \\x \left (a^{\frac {3}{2}} c\right )^{\frac {2}{3}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \left (c (a+b x)^{3/2}\right )^{2/3} \, dx=\frac {\left ({\left (b x + a\right )}^{\frac {3}{2}} c\right )^{\frac {2}{3}} {\left (b x + a\right )}}{2 \, b} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.22 \[ \int \left (c (a+b x)^{3/2}\right )^{2/3} \, dx=\frac {{\left (\sqrt {b x + a} b c x + \sqrt {b x + a} a c\right )}^{\frac {2}{3}} {\left (b x + a\right )}}{b} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \left (c (a+b x)^{3/2}\right )^{2/3} \, dx=\int {\left (c\,{\left (a+b\,x\right )}^{3/2}\right )}^{2/3} \,d x \]

[In]

int((c*(a + b*x)^(3/2))^(2/3),x)

[Out]

int((c*(a + b*x)^(3/2))^(2/3), x)