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]

-2/7*c*(c/(b*x+a)^3)^(1/2)/b/(b*x+a)^2

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Rubi [A]  time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, 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 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 \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*}

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Mathematica [A]  time = 0.01, 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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fricas [B]  time = 0.85, size = 58, normalized size = 2.07 \[ -\frac {2 \, c \sqrt {\frac {c}{b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}}}}{7 \, {\left (b^{3} x^{2} + 2 \, a b^{2} x + a^{2} b\right )}} \]

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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giac [B]  time = 0.18, size = 52, normalized size = 1.86 \[ -\frac {2 \, c^{5} \mathrm {sgn}\left (b^{3} x^{3} + 3 \, a b^{2} x^{2} + 3 \, a^{2} b x + a^{3}\right ) \mathrm {sgn}\left (b x + a\right )}{7 \, {\left (b c x + a c\right )}^{\frac {7}{2}} b} \]

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)

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maple [A]  time = 0.00, size = 22, normalized size = 0.79 \[ -\frac {2 \left (b x +a \right ) \left (\frac {c}{\left (b x +a \right )^{3}}\right )^{\frac {3}{2}}}{7 b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/7*(b*x+a)*(1/(b*x+a)^3*c)^(3/2)/b

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maxima [A]  time = 0.60, size = 24, normalized size = 0.86 \[ -\frac {2 \, {\left (b c^{\frac {3}{2}} x + a c^{\frac {3}{2}}\right )}}{7 \, {\left (b x + a\right )}^{\frac {9}{2}} b} \]

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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mupad [B]  time = 1.20, size = 24, normalized size = 0.86 \[ -\frac {2\,c\,\sqrt {\frac {c}{{\left (a+b\,x\right )}^3}}}{7\,b\,{\left (a+b\,x\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/(a + b*x)^3)^(3/2),x)

[Out]

-(2*c*(c/(a + b*x)^3)^(1/2))/(7*b*(a + b*x)^2)

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sympy [A]  time = 8.92, size = 357, normalized size = 12.75 \[ \begin {cases} - \frac {67 a^{3} c^{\frac {3}{2}} \left (\frac {1}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\right )^{\frac {3}{2}}}{14 a^{2} b + 28 a b^{2} x + 14 b^{3} x^{2}} - \frac {201 a^{2} b c^{\frac {3}{2}} x \left (\frac {1}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\right )^{\frac {3}{2}}}{14 a^{2} b + 28 a b^{2} x + 14 b^{3} x^{2}} - \frac {201 a b^{2} c^{\frac {3}{2}} x^{2} \left (\frac {1}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\right )^{\frac {3}{2}}}{14 a^{2} b + 28 a b^{2} x + 14 b^{3} x^{2}} - \frac {67 b^{3} c^{\frac {3}{2}} x^{3} \left (\frac {1}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}\right )^{\frac {3}{2}}}{14 a^{2} b + 28 a b^{2} x + 14 b^{3} x^{2}} + \frac {63 c^{\frac {3}{2}} \sqrt {\frac {1}{a^{3} + 3 a^{2} b x + 3 a b^{2} x^{2} + b^{3} x^{3}}}}{14 a^{2} b + 28 a b^{2} x + 14 b^{3} x^{2}} & \text {for}\: b \neq 0 \\x \left (\frac {c}{a^{3}}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \]

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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