\(\int \frac {(c x^2)^{3/2}}{x^3 (a+b x)^2} \, dx\) [905]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 25 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^3 (a+b x)^2} \, dx=-\frac {c \sqrt {c x^2}}{b x (a+b x)} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 32} \[ \int \frac {\left (c x^2\right )^{3/2}}{x^3 (a+b x)^2} \, dx=-\frac {c \sqrt {c x^2}}{b x (a+b x)} \]

[In]

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

[Out]

-((c*Sqrt[c*x^2])/(b*x*(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*} \text {integral}& = \frac {\left (c \sqrt {c x^2}\right ) \int \frac {1}{(a+b x)^2} \, dx}{x} \\ & = -\frac {c \sqrt {c x^2}}{b x (a+b x)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^3 (a+b x)^2} \, dx=-\frac {\left (c x^2\right )^{3/2}}{b x^3 (a+b x)} \]

[In]

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

[Out]

-((c*x^2)^(3/2)/(b*x^3*(a + b*x)))

Maple [A] (verified)

Time = 0.37 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

method result size
gosper \(-\frac {\left (c \,x^{2}\right )^{\frac {3}{2}}}{\left (b x +a \right ) b \,x^{3}}\) \(23\)
default \(-\frac {\left (c \,x^{2}\right )^{\frac {3}{2}}}{\left (b x +a \right ) b \,x^{3}}\) \(23\)
risch \(-\frac {c \sqrt {c \,x^{2}}}{b x \left (b x +a \right )}\) \(24\)
trager \(\frac {c \left (-1+x \right ) \sqrt {c \,x^{2}}}{\left (b x +a \right ) \left (a +b \right ) x}\) \(28\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^3 (a+b x)^2} \, dx=-\frac {\sqrt {c x^{2}} c}{b^{2} x^{2} + a b x} \]

[In]

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

[Out]

-sqrt(c*x^2)*c/(b^2*x^2 + a*b*x)

Sympy [A] (verification not implemented)

Time = 1.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.48 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^3 (a+b x)^2} \, dx=\begin {cases} - \frac {\left (c x^{2}\right )^{\frac {3}{2}}}{a b x^{3} + b^{2} x^{4}} & \text {for}\: b \neq 0 \\\frac {\left (c x^{2}\right )^{\frac {3}{2}}}{a^{2} x^{2}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.64 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^3 (a+b x)^2} \, dx=-\frac {c^{\frac {3}{2}}}{b^{2} x + a b} \]

[In]

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

[Out]

-c^(3/2)/(b^2*x + a*b)

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.16 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^3 (a+b x)^2} \, dx=-c^{\frac {3}{2}} {\left (\frac {\mathrm {sgn}\left (x\right )}{{\left (b x + a\right )} b} - \frac {\mathrm {sgn}\left (x\right )}{a b}\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {\left (c x^2\right )^{3/2}}{x^3 (a+b x)^2} \, dx=-\frac {c^{3/2}\,\sqrt {x^2}}{b^2\,x^2+a\,b\,x} \]

[In]

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

[Out]

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