3.2832 \(\int (\frac{c}{(a+b x)^3})^{3/2} \, dx\)

Optimal. Leaf size=28 \[ -\frac{2 c \sqrt{\frac{c}{(a+b x)^3}}}{7 b (a+b x)^2} \]

[Out]

(-2*c*Sqrt[c/(a + b*x)^3])/(7*b*(a + b*x)^2)

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Rubi [A]  time = 0.0087339, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {247, 15, 30} \[ -\frac{2 c \sqrt{\frac{c}{(a+b x)^3}}}{7 b (a+b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[(c/(a + b*x)^3)^(3/2),x]

[Out]

(-2*c*Sqrt[c/(a + b*x)^3])/(7*b*(a + b*x)^2)

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 (\frac{c}{(a+b x)^3}\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \left (\frac{c}{x^3}\right )^{3/2} \, dx,x,a+b x\right )}{b}\\ &=\frac{\left (c \sqrt{\frac{c}{(a+b x)^3}} (a+b x)^{3/2}\right ) \operatorname{Subst}\left (\int \frac{1}{x^{9/2}} \, dx,x,a+b x\right )}{b}\\ &=-\frac{2 c \sqrt{\frac{c}{(a+b x)^3}}}{7 b (a+b x)^2}\\ \end{align*}

Mathematica [A]  time = 0.009371, size = 25, normalized size = 0.89 \[ -\frac{2 (a+b x) \left (\frac{c}{(a+b x)^3}\right )^{3/2}}{7 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c/(a + b*x)^3)^(3/2),x]

[Out]

(-2*(c/(a + b*x)^3)^(3/2)*(a + b*x))/(7*b)

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Maple [A]  time = 0.001, size = 22, normalized size = 0.8 \begin{align*} -{\frac{2\,bx+2\,a}{7\,b} \left ({\frac{c}{ \left ( bx+a \right ) ^{3}}} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*(b*x+a)*(c/(b*x+a)^3)^(3/2)/b

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Maxima [A]  time = 1.37796, size = 32, normalized size = 1.14 \begin{align*} -\frac{2 \,{\left (b c^{\frac{3}{2}} x + a c^{\frac{3}{2}}\right )}}{7 \,{\left (b x + a\right )}^{\frac{9}{2}} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*(b*c^(3/2)*x + a*c^(3/2))/((b*x + a)^(9/2)*b)

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Fricas [B]  time = 1.20741, size = 119, normalized size = 4.25 \begin{align*} -\frac{2 \, c \sqrt{\frac{c}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}}}{7 \,{\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*c*sqrt(c/(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3))/(b^3*x^2 + 2*a*b^2*x + a^2*b)

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Sympy [A]  time = 10.7178, size = 357, normalized size = 12.75 \begin{align*} \begin{cases} - \frac{67 a^{3} c^{\frac{3}{2}} \left (\frac{1}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\right )^{\frac{3}{2}}}{14 a^{2} b + 28 a b^{2} x + 14 b^{3} x^{2}} - \frac{201 a^{2} b c^{\frac{3}{2}} x \left (\frac{1}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\right )^{\frac{3}{2}}}{14 a^{2} b + 28 a b^{2} x + 14 b^{3} x^{2}} - \frac{201 a b^{2} c^{\frac{3}{2}} x^{2} \left (\frac{1}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\right )^{\frac{3}{2}}}{14 a^{2} b + 28 a b^{2} x + 14 b^{3} x^{2}} - \frac{67 b^{3} c^{\frac{3}{2}} x^{3} \left (\frac{1}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\right )^{\frac{3}{2}}}{14 a^{2} b + 28 a b^{2} x + 14 b^{3} x^{2}} + \frac{63 c^{\frac{3}{2}} \sqrt{\frac{1}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}}}{14 a^{2} b + 28 a b^{2} x + 14 b^{3} x^{2}} & \text{for}\: b \neq 0 \\x \left (\frac{c}{a^{3}}\right )^{\frac{3}{2}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-67*a**3*c**(3/2)*(1/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))**(3/2)/(14*a**2*b + 28*a*b**2
*x + 14*b**3*x**2) - 201*a**2*b*c**(3/2)*x*(1/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3))**(3/2)/(14*a**2
*b + 28*a*b**2*x + 14*b**3*x**2) - 201*a*b**2*c**(3/2)*x**2*(1/(a**3 + 3*a**2*b*x + 3*a*b**2*x**2 + b**3*x**3)
)**(3/2)/(14*a**2*b + 28*a*b**2*x + 14*b**3*x**2) - 67*b**3*c**(3/2)*x**3*(1/(a**3 + 3*a**2*b*x + 3*a*b**2*x**
2 + b**3*x**3))**(3/2)/(14*a**2*b + 28*a*b**2*x + 14*b**3*x**2) + 63*c**(3/2)*sqrt(1/(a**3 + 3*a**2*b*x + 3*a*
b**2*x**2 + b**3*x**3))/(14*a**2*b + 28*a*b**2*x + 14*b**3*x**2), Ne(b, 0)), (x*(c/a**3)**(3/2), True))

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Giac [B]  time = 1.11565, size = 70, normalized size = 2.5 \begin{align*} -\frac{2 \, c^{5} \mathrm{sgn}\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right ) \mathrm{sgn}\left (b x + a\right )}{7 \,{\left (b c x + a c\right )}^{\frac{7}{2}} b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*c^5*sgn(b^3*x^3 + 3*a*b^2*x^2 + 3*a^2*b*x + a^3)*sgn(b*x + a)/((b*c*x + a*c)^(7/2)*b)