\(\int \frac {1}{(\frac {c}{(a+b x)^{3/2}})^{2/3}} \, dx\) [2839]

Optimal result

Integrand size = 15, antiderivative size = 27 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3}} \, dx=\frac {a+b x}{2 b \left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3}} \]

[Out]

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

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} \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3}} \, dx=\frac {a+b x}{2 b \left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3}} \]

[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 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 \frac {1}{\left (\frac {c}{x^{3/2}}\right )^{2/3}} \, dx,x,a+b x\right )}{b} \\ & = \frac {\text {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] (verified)

Time = 0.02 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.26 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3}} \, dx=\frac {x (2 a+b x)}{2 \left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3} (a+b x)} \]

[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))

Maple [A] (verified)

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

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

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 51 vs. \(2 (21) = 42\).

Time = 0.33 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.89 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3}} \, dx=\frac {{\left (b x^{2} + 2 \, a x\right )} \sqrt {b x + a} \left (\frac {\sqrt {b x + a} c}{b^{2} x^{2} + 2 \, a b x + a^{2}}\right )^{\frac {1}{3}}}{2 \, c} \]

[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

Sympy [A] (verification not implemented)

Time = 0.96 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.19 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3}} \, dx=\begin {cases} \frac {a + b x}{2 b \left (\frac {c}{\left (a + b x\right )^{\frac {3}{2}}}\right )^{\frac {2}{3}}} & \text {for}\: b \neq 0 \\\frac {x}{\left (\frac {c}{a^{\frac {3}{2}}}\right )^{\frac {2}{3}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3}} \, dx=\frac {b x + a}{2 \, b \left (\frac {c}{{\left (b x + a\right )}^{\frac {3}{2}}}\right )^{\frac {2}{3}}} \]

[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))

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 1, normalized size of antiderivative = 0.04 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3}} \, dx=+\infty \]

[In]

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

[Out]

+Infinity

Mupad [B] (verification not implemented)

Time = 5.85 (sec) , antiderivative size = 31, normalized size of antiderivative = 1.15 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{3/2}}\right )^{2/3}} \, dx=\frac {x\,{\left (\frac {c}{{\left (a+b\,x\right )}^{3/2}}\right )}^{1/3}\,\left (2\,a+b\,x\right )\,\sqrt {a+b\,x}}{2\,c} \]

[In]

int(1/(c/(a + b*x)^(3/2))^(2/3),x)

[Out]

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