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

Optimal. Leaf size=25 \[ -\frac {x}{b c \sqrt {c x^2} (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 32} \[ -\frac {x}{b c \sqrt {c x^2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((c*x^2)^(3/2)*(a + b*x)^2),x]

[Out]

-(x/(b*c*Sqrt[c*x^2]*(a + b*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 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 24, normalized size = 0.96 \[ -\frac {x^3}{b \left (c x^2\right )^{3/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3/((c*x^2)^(3/2)*(a + b*x)^2),x]

[Out]

-(x^3/(b*(c*x^2)^(3/2)*(a + b*x)))

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fricas [A]  time = 0.42, size = 29, normalized size = 1.16 \[ -\frac {\sqrt {c x^{2}}}{b^{2} c^{2} x^{2} + a b c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.16, size = 38, normalized size = 1.52 \[ \frac {1}{{\left (b x + a\right )} b c^{\frac {3}{2}} \mathrm {sgn}\left (-\frac {b}{b x + a} + \frac {a b}{{\left (b x + a\right )}^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 23, normalized size = 0.92 \[ -\frac {x^{3}}{\left (b x +a \right ) \left (c \,x^{2}\right )^{\frac {3}{2}} b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.48, size = 47, normalized size = 1.88 \[ \frac {a}{\sqrt {c x^{2}} b^{3} c x + \sqrt {c x^{2}} a b^{2} c} - \frac {1}{\sqrt {c x^{2}} b^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.17, size = 25, normalized size = 1.00 \[ -\frac {\sqrt {c\,x^2}}{b\,c^2\,x\,\left (a+b\,x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*x^2)^(1/2)/(b*c^2*x*(a + b*x))

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sympy [A]  time = 1.94, size = 90, normalized size = 3.60 \[ \begin {cases} \frac {\tilde {\infty } x^{2}}{c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\\frac {\tilde {\infty } x^{4}}{c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}}} & \text {for}\: a = - b x \\\frac {x^{4}}{a^{2} c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}}} & \text {for}\: b = 0 \\- \frac {x^{3}}{a b c^{\frac {3}{2}} \left (x^{2}\right )^{\frac {3}{2}} + b^{2} c^{\frac {3}{2}} x \left (x^{2}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x**2/(c**(3/2)*(x**2)**(3/2)), Eq(a, 0) & Eq(b, 0)), (zoo*x**4/(c**(3/2)*(x**2)**(3/2)), Eq(a,
-b*x)), (x**4/(a**2*c**(3/2)*(x**2)**(3/2)), Eq(b, 0)), (-x**3/(a*b*c**(3/2)*(x**2)**(3/2) + b**2*c**(3/2)*x*(
x**2)**(3/2)), True))

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