\(\int \frac {1}{(\frac {c}{(a+b x)^{2/3}})^{3/2}} \, dx\) [2840]

Optimal result

Integrand size = 15, antiderivative size = 34 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{2/3}}\right )^{3/2}} \, dx=\frac {(a+b x)^{5/3}}{2 b c \sqrt {\frac {c}{(a+b x)^{2/3}}}} \]

[Out]

1/2*(b*x+a)^(5/3)/b/c/(c/(b*x+a)^(2/3))^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 34, 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} \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{2/3}}\right )^{3/2}} \, dx=\frac {(a+b x)^{5/3}}{2 b c \sqrt {\frac {c}{(a+b x)^{2/3}}}} \]

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

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{2/3}}\right )^{3/2}} \, dx=\frac {x (2 a+b x)}{2 \left (\frac {c}{(a+b x)^{2/3}}\right )^{3/2} (a+b x)} \]

[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))

Maple [A] (verified)

Time = 0.07 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.65

method	result	size
derivativedivides	\(\frac {b x +a}{2 b \left (\frac {c}{\left (b x +a \right )^{\frac {2}{3}}}\right )^{\frac {3}{2}}}\)	\(22\)
gosper	\(\frac {x \left (b x +2 a \right )}{2 \left (b x +a \right ) \left (\frac {c}{\left (b x +a \right )^{\frac {2}{3}}}\right )^{\frac {3}{2}}}\)	\(29\)
default	\(\frac {x \left (b x +2 a \right )}{2 \left (b x +a \right ) \left (\frac {c}{\left (b x +a \right )^{\frac {2}{3}}}\right )^{\frac {3}{2}}}\)	\(29\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{2/3}}\right )^{3/2}} \, dx=\frac {b x^{2} + 2 \, a x}{2 \, c^{\frac {3}{2}}} \]

[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)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 80 vs. \(2 (26) = 52\).

Time = 0.44 (sec) , antiderivative size = 80, normalized size of antiderivative = 2.35 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{2/3}}\right )^{3/2}} \, dx=\frac {2 a x}{2 a \left (\frac {c}{\left (a + b x\right )^{\frac {2}{3}}}\right )^{\frac {3}{2}} + 2 b x \left (\frac {c}{\left (a + b x\right )^{\frac {2}{3}}}\right )^{\frac {3}{2}}} + \frac {b x^{2}}{2 a \left (\frac {c}{\left (a + b x\right )^{\frac {2}{3}}}\right )^{\frac {3}{2}} + 2 b x \left (\frac {c}{\left (a + b x\right )^{\frac {2}{3}}}\right )^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{2/3}}\right )^{3/2}} \, dx=\frac {b x^{2} + 2 \, a x}{2 \, c^{\frac {3}{2}}} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.44 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{2/3}}\right )^{3/2}} \, dx=\frac {b x^{2} + 2 \, a x}{2 \, c^{\frac {3}{2}}} \]

[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)

Mupad [B] (verification not implemented)

Time = 5.86 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int \frac {1}{\left (\frac {c}{(a+b x)^{2/3}}\right )^{3/2}} \, dx=\sqrt {\frac {c}{{\left (a+b\,x\right )}^{2/3}}}\,\left (\frac {a\,x\,{\left (a+b\,x\right )}^{1/3}}{c^2}+\frac {b\,x^2\,{\left (a+b\,x\right )}^{1/3}}{2\,c^2}\right ) \]

[In]

int(1/(c/(a + b*x)^(2/3))^(3/2),x)

[Out]

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