\(\int \frac {x (a+b x)^2}{(c x^2)^{5/2}} \, dx\) [846]

Optimal result

Integrand size = 18, antiderivative size = 29 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {(a+b x)^3}{3 a c^2 x^2 \sqrt {c x^2}} \]

[Out]

-1/3*(b*x+a)^3/a/c^2/x^2/(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} \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {(a+b x)^3}{3 a c^2 x^2 \sqrt {c x^2}} \]

[In]

Int[(x*(a + b*x)^2)/(c*x^2)^(5/2),x]

[Out]

-1/3*(a + b*x)^3/(a*c^2*x^2*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)^2}{x^4} \, dx}{c^2 \sqrt {c x^2}} \\ & = -\frac {(a+b x)^3}{3 a c^2 x^2 \sqrt {c x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {x^2 \left (a^2+3 a b x+3 b^2 x^2\right )}{3 \left (c x^2\right )^{5/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.03

method	result	size
gosper	\(-\frac {x^{2} \left (3 b^{2} x^{2}+3 a b x +a^{2}\right )}{3 \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(30\)
default	\(-\frac {x^{2} \left (3 b^{2} x^{2}+3 a b x +a^{2}\right )}{3 \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(30\)
risch	\(\frac {-b^{2} x^{2}-a b x -\frac {1}{3} a^{2}}{c^{2} x^{2} \sqrt {c \,x^{2}}}\)	\(34\)
trager	\(\frac {\left (-1+x \right ) \left (a^{2} x^{2}+3 a b \,x^{2}+3 b^{2} x^{2}+a^{2} x +3 a b x +a^{2}\right ) \sqrt {c \,x^{2}}}{3 c^{3} x^{4}}\)	\(55\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.10 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {{\left (3 \, b^{2} x^{2} + 3 \, a b x + a^{2}\right )} \sqrt {c x^{2}}}{3 \, c^{3} x^{4}} \]

[In]

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

[Out]

-1/3*(3*b^2*x^2 + 3*a*b*x + a^2)*sqrt(c*x^2)/(c^3*x^4)

Sympy [A] (verification not implemented)

Time = 0.66 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.66 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=- \frac {a^{2} x^{2}}{3 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {a b x^{3}}{\left (c x^{2}\right )^{\frac {5}{2}}} - \frac {b^{2} x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.52 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {b^{2} x^{2}}{\left (c x^{2}\right )^{\frac {3}{2}} c} - \frac {a^{2}}{3 \, \left (c x^{2}\right )^{\frac {3}{2}} c} - \frac {a b}{c^{\frac {5}{2}} x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 29, normalized size of antiderivative = 1.00 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {3 \, b^{2} x^{2} + 3 \, a b x + a^{2}}{3 \, c^{\frac {5}{2}} x^{3} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 33, normalized size of antiderivative = 1.14 \[ \int \frac {x (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {a^2\,x^2+3\,a\,b\,x^3+3\,b^2\,x^4}{3\,c^{5/2}\,{\left (x^2\right )}^{5/2}} \]

[In]

int((x*(a + b*x)^2)/(c*x^2)^(5/2),x)

[Out]

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