\(\int \frac {x^2 (a+b x)^2}{(c x^2)^{5/2}} \, dx\) [845]

Optimal result

Integrand size = 20, antiderivative size = 58 \[ \int \frac {x^2 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {2 a b}{c^2 \sqrt {c x^2}}-\frac {a^2}{2 c^2 x \sqrt {c x^2}}+\frac {b^2 x \log (x)}{c^2 \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {15, 45} \[ \int \frac {x^2 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {a^2}{2 c^2 x \sqrt {c x^2}}-\frac {2 a b}{c^2 \sqrt {c x^2}}+\frac {b^2 x \log (x)}{c^2 \sqrt {c x^2}} \]

[In]

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

[Out]

(-2*a*b)/(c^2*Sqrt[c*x^2]) - a^2/(2*c^2*x*Sqrt[c*x^2]) + (b^2*x*Log[x])/(c^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 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.59 \[ \int \frac {x^2 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=\frac {-\frac {1}{2} a x^3 (a+4 b x)+b^2 x^5 \log (x)}{\left (c x^2\right )^{5/2}} \]

[In]

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

[Out]

(-1/2*(a*x^3*(a + 4*b*x)) + b^2*x^5*Log[x])/(c*x^2)^(5/2)

Maple [A] (verified)

Time = 0.12 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.59

method	result	size
default	\(\frac {x^{3} \left (2 b^{2} \ln \left (x \right ) x^{2}-4 a b x -a^{2}\right )}{2 \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(34\)
risch	\(\frac {-\frac {1}{2} a^{2}-2 a b x}{c^{2} x \sqrt {c \,x^{2}}}+\frac {b^{2} x \ln \left (x \right )}{c^{2} \sqrt {c \,x^{2}}}\)	\(44\)

[In]

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

[Out]

1/2*x^3*(2*b^2*ln(x)*x^2-4*a*b*x-a^2)/(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.62 \[ \int \frac {x^2 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=\frac {{\left (2 \, b^{2} x^{2} \log \left (x\right ) - 4 \, a b x - a^{2}\right )} \sqrt {c x^{2}}}{2 \, c^{3} x^{3}} \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 1.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.88 \[ \int \frac {x^2 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=- \frac {a^{2} x^{3}}{2 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {2 a b x^{4}}{\left (c x^{2}\right )^{\frac {5}{2}}} + \frac {b^{2} x^{5} \log {\left (x \right )}}{\left (c x^{2}\right )^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.66 \[ \int \frac {x^2 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=-\frac {2 \, a b x^{2}}{\left (c x^{2}\right )^{\frac {3}{2}} c} + \frac {b^{2} \log \left (x\right )}{c^{\frac {5}{2}}} - \frac {a^{2}}{2 \, c^{\frac {5}{2}} x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.42 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.62 \[ \int \frac {x^2 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=\frac {b^{2} \log \left ({\left | x \right |}\right )}{c^{\frac {5}{2}} \mathrm {sgn}\left (x\right )} - \frac {4 \, a b x + a^{2}}{2 \, c^{\frac {5}{2}} x^{2} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

b^2*log(abs(x))/(c^(5/2)*sgn(x)) - 1/2*(4*a*b*x + a^2)/(c^(5/2)*x^2*sgn(x))

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2 (a+b x)^2}{\left (c x^2\right )^{5/2}} \, dx=\int \frac {x^2\,{\left (a+b\,x\right )}^2}{{\left (c\,x^2\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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