\(\int \frac {x^2 (a+b x)}{(c x^2)^{5/2}} \, dx\) [797]

   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 = 18, antiderivative size = 29 \[ \int \frac {x^2 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {(a+b x)^2}{2 a c^2 x \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {15, 37} \[ \int \frac {x^2 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {(a+b x)^2}{2 a c^2 x \sqrt {c x^2}} \]

[In]

Int[(x^2*(a + b*x))/(c*x^2)^(5/2),x]

[Out]

-1/2*(a + b*x)^2/(a*c^2*x*Sqrt[c*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 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {x \int \frac {a+b x}{x^3} \, dx}{c^2 \sqrt {c x^2}} \\ & = -\frac {(a+b x)^2}{2 a c^2 x \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.76 \[ \int \frac {x^2 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {x^3 (a+2 b x)}{2 \left (c x^2\right )^{5/2}} \]

[In]

Integrate[(x^2*(a + b*x))/(c*x^2)^(5/2),x]

[Out]

-1/2*(x^3*(a + 2*b*x))/(c*x^2)^(5/2)

Maple [A] (verified)

Time = 0.05 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.66

method result size
gosper \(-\frac {x^{3} \left (2 b x +a \right )}{2 \left (c \,x^{2}\right )^{\frac {5}{2}}}\) \(19\)
default \(-\frac {x^{3} \left (2 b x +a \right )}{2 \left (c \,x^{2}\right )^{\frac {5}{2}}}\) \(19\)
risch \(\frac {-b x -\frac {a}{2}}{c^{2} x \sqrt {c \,x^{2}}}\) \(23\)
trager \(\frac {\left (-1+x \right ) \left (a x +2 b x +a \right ) \sqrt {c \,x^{2}}}{2 c^{3} x^{3}}\) \(28\)

[In]

int(x^2*(b*x+a)/(c*x^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.72 \[ \int \frac {x^2 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c x^{2}} {\left (2 \, b x + a\right )}}{2 \, c^{3} x^{3}} \]

[In]

integrate(x^2*(b*x+a)/(c*x^2)^(5/2),x, algorithm="fricas")

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.76 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {x^2 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=- \frac {a x^{3}}{2 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {b x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}} \]

[In]

integrate(x**2*(b*x+a)/(c*x**2)**(5/2),x)

[Out]

-a*x**3/(2*(c*x**2)**(5/2)) - b*x**4/(c*x**2)**(5/2)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.90 \[ \int \frac {x^2 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {b x^{2}}{\left (c x^{2}\right )^{\frac {3}{2}} c} - \frac {a}{2 \, c^{\frac {5}{2}} x^{2}} \]

[In]

integrate(x^2*(b*x+a)/(c*x^2)^(5/2),x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.62 \[ \int \frac {x^2 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {2 \, b x + a}{2 \, c^{\frac {5}{2}} x^{2} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x^2*(b*x+a)/(c*x^2)^(5/2),x, algorithm="giac")

[Out]

-1/2*(2*b*x + a)/(c^(5/2)*x^2*sgn(x))

Mupad [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.86 \[ \int \frac {x^2 (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {2\,b\,x^3+a\,x^2}{2\,c^{5/2}\,x\,{\left (x^2\right )}^{3/2}} \]

[In]

int((x^2*(a + b*x))/(c*x^2)^(5/2),x)

[Out]

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