\(\int \frac {x (a+b x)}{(c x^2)^{5/2}} \, dx\) [798]

Optimal result

Integrand size = 16, antiderivative size = 41 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {a}{3 c^2 x^2 \sqrt {c x^2}}-\frac {b}{2 c^2 x \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {15, 45} \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {a}{3 c^2 x^2 \sqrt {c x^2}}-\frac {b}{2 c^2 x \sqrt {c x^2}} \]

[In]

Int[(x*(a + b*x))/(c*x^2)^(5/2),x]

[Out]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.59 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {x^2 (2 a+3 b x)}{6 \left (c x^2\right )^{5/2}} \]

[In]

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

[Out]

-1/6*(x^2*(2*a + 3*b*x))/(c*x^2)^(5/2)

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.51

method	result	size
gosper	\(-\frac {x^{2} \left (3 b x +2 a \right )}{6 \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(21\)
default	\(-\frac {x^{2} \left (3 b x +2 a \right )}{6 \left (c \,x^{2}\right )^{\frac {5}{2}}}\)	\(21\)
risch	\(\frac {-\frac {b x}{2}-\frac {a}{3}}{c^{2} x^{2} \sqrt {c \,x^{2}}}\)	\(23\)
trager	\(\frac {\left (-1+x \right ) \left (2 a \,x^{2}+3 b \,x^{2}+2 a x +3 b x +2 a \right ) \sqrt {c \,x^{2}}}{6 c^{3} x^{4}}\)	\(43\)

[In]

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

[Out]

-1/6*x^2*(3*b*x+2*a)/(c*x^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.56 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c x^{2}} {\left (3 \, b x + 2 \, a\right )}}{6 \, c^{3} x^{4}} \]

[In]

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

[Out]

-1/6*sqrt(c*x^2)*(3*b*x + 2*a)/(c^3*x^4)

Sympy [A] (verification not implemented)

Time = 0.60 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.76 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=- \frac {a x^{2}}{3 \left (c x^{2}\right )^{\frac {5}{2}}} - \frac {b x^{3}}{2 \left (c x^{2}\right )^{\frac {5}{2}}} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.56 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {a}{3 \, \left (c x^{2}\right )^{\frac {3}{2}} c} - \frac {b}{2 \, c^{\frac {5}{2}} x^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.49 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {3 \, b x + 2 \, a}{6 \, c^{\frac {5}{2}} x^{3} \mathrm {sgn}\left (x\right )} \]

[In]

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

[Out]

-1/6*(3*b*x + 2*a)/(c^(5/2)*x^3*sgn(x))

Mupad [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.63 \[ \int \frac {x (a+b x)}{\left (c x^2\right )^{5/2}} \, dx=-\frac {2\,a\,\sqrt {x^2}+3\,b\,x\,\sqrt {x^2}}{6\,c^{5/2}\,x^4} \]

[In]

int((x*(a + b*x))/(c*x^2)^(5/2),x)

[Out]

-(2*a*(x^2)^(1/2) + 3*b*x*(x^2)^(1/2))/(6*c^(5/2)*x^4)