\(\int \frac {x (a+b x)}{\sqrt {c x^2}} \, dx\) [782]

Optimal result

Integrand size = 16, antiderivative size = 32 \[ \int \frac {x (a+b x)}{\sqrt {c x^2}} \, dx=\frac {a x^2}{\sqrt {c x^2}}+\frac {b x^3}{2 \sqrt {c x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {15} \[ \int \frac {x (a+b x)}{\sqrt {c x^2}} \, dx=\frac {a x^2}{\sqrt {c x^2}}+\frac {b x^3}{2 \sqrt {c x^2}} \]

[In]

Int[(x*(a + b*x))/Sqrt[c*x^2],x]

[Out]

(a*x^2)/Sqrt[c*x^2] + (b*x^3)/(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]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.72 \[ \int \frac {x (a+b x)}{\sqrt {c x^2}} \, dx=\frac {x^2 (2 a+b x)}{2 \sqrt {c x^2}} \]

[In]

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

[Out]

(x^2*(2*a + b*x))/(2*Sqrt[c*x^2])

Maple [A] (verified)

Time = 0.03 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.62

method	result	size
gosper	\(\frac {x^{2} \left (b x +2 a \right )}{2 \sqrt {c \,x^{2}}}\)	\(20\)
default	\(\frac {x^{2} \left (b x +2 a \right )}{2 \sqrt {c \,x^{2}}}\)	\(20\)
trager	\(\frac {\left (b x +2 a +b \right ) \left (-1+x \right ) \sqrt {c \,x^{2}}}{2 c x}\)	\(27\)
risch	\(\frac {a \,x^{2}}{\sqrt {c \,x^{2}}}+\frac {b \,x^{3}}{2 \sqrt {c \,x^{2}}}\)	\(27\)

[In]

int(x*(b*x+a)/(c*x^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.59 \[ \int \frac {x (a+b x)}{\sqrt {c x^2}} \, dx=\frac {\sqrt {c x^{2}} {\left (b x + 2 \, a\right )}}{2 \, c} \]

[In]

integrate(x*(b*x+a)/(c*x^2)^(1/2),x, algorithm="fricas")

[Out]

1/2*sqrt(c*x^2)*(b*x + 2*a)/c

Sympy [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.84 \[ \int \frac {x (a+b x)}{\sqrt {c x^2}} \, dx=\frac {a x^{2}}{\sqrt {c x^{2}}} + \frac {b x^{3}}{2 \sqrt {c x^{2}}} \]

[In]

integrate(x*(b*x+a)/(c*x**2)**(1/2),x)

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 22, normalized size of antiderivative = 0.69 \[ \int \frac {x (a+b x)}{\sqrt {c x^2}} \, dx=\frac {b x^{2}}{2 \, \sqrt {c}} + \frac {\sqrt {c x^{2}} a}{c} \]

[In]

integrate(x*(b*x+a)/(c*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/2*b*x^2/sqrt(c) + sqrt(c*x^2)*a/c

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.59 \[ \int \frac {x (a+b x)}{\sqrt {c x^2}} \, dx=\frac {b x^{2} + 2 \, a x}{2 \, \sqrt {c} \mathrm {sgn}\left (x\right )} \]

[In]

integrate(x*(b*x+a)/(c*x^2)^(1/2),x, algorithm="giac")

[Out]

1/2*(b*x^2 + 2*a*x)/(sqrt(c)*sgn(x))

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.59 \[ \int \frac {x (a+b x)}{\sqrt {c x^2}} \, dx=\frac {2\,a\,\left |x\right |+b\,x\,\sqrt {x^2}}{2\,\sqrt {c}} \]

[In]

int((x*(a + b*x))/(c*x^2)^(1/2),x)

[Out]

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