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]

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

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Rubi [A]  time = 0.01, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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]

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*}

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Mathematica [A]  time = 0.03, size = 34, normalized size = 1.00 \[ \frac {x (2 a+b x)}{2 (a+b x) \left (\frac {c}{(a+b x)^{2/3}}\right )^{3/2}} \]

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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fricas [A]  time = 0.74, size = 15, normalized size = 0.44 \[ \frac {b x^{2} + 2 \, a x}{2 \, c^{\frac {3}{2}}} \]

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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giac [A]  time = 0.15, size = 15, normalized size = 0.44 \[ \frac {b x^{2} + 2 \, a x}{2 \, c^{\frac {3}{2}}} \]

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)

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maple [A]  time = 0.00, size = 29, normalized size = 0.85 \[ \frac {\left (b x +2 a \right ) x}{2 \left (b x +a \right ) \left (\frac {c}{\left (b x +a \right )^{\frac {2}{3}}}\right )^{\frac {3}{2}}} \]

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 = 0.48, size = 15, normalized size = 0.44 \[ \frac {b x^{2} + 2 \, a x}{2 \, c^{\frac {3}{2}}} \]

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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mupad [B]  time = 1.26, size = 42, normalized size = 1.24 \[ \sqrt {\frac {c}{{\left (a+b\,x\right )}^{2/3}}}\,\left (\frac {a\,x\,{\left (a+b\,x\right )}^{1/3}}{c^2}+\frac {b\,x^2\,{\left (a+b\,x\right )}^{1/3}}{2\,c^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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sympy [A]  time = 3.15, size = 76, normalized size = 2.24 \[ \begin {cases} \frac {2 a x}{\frac {2 a c^{\frac {3}{2}}}{a + b x} + \frac {2 b c^{\frac {3}{2}} x}{a + b x}} + \frac {b x^{2}}{\frac {2 a c^{\frac {3}{2}}}{a + b x} + \frac {2 b c^{\frac {3}{2}} x}{a + b x}} & \text {for}\: a \neq 0 \vee b \neq 0 \\\frac {x}{\left (\tilde {\infty } c\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

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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