\(\int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx\) [76]

Optimal result

Integrand size = 19, antiderivative size = 25 \[ \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx=\frac {2 \left (b x+c x^2\right )^{3/2}}{3 c x^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {662} \[ \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx=\frac {2 \left (b x+c x^2\right )^{3/2}}{3 c x^{3/2}} \]

[In]

Int[Sqrt[b*x + c*x^2]/Sqrt[x],x]

[Out]

(2*(b*x + c*x^2)^(3/2))/(3*c*x^(3/2))

Rule 662

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[e*(d + e*x)^(m - 1)*
((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c
*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \left (b x+c x^2\right )^{3/2}}{3 c x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx=\frac {2 (x (b+c x))^{3/2}}{3 c x^{3/2}} \]

[In]

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

[Out]

(2*(x*(b + c*x))^(3/2))/(3*c*x^(3/2))

Maple [A] (verified)

Time = 1.99 (sec) , antiderivative size = 23, normalized size of antiderivative = 0.92

method	result	size
default	\(\frac {2 \left (c x +b \right ) \sqrt {x \left (c x +b \right )}}{3 c \sqrt {x}}\)	\(23\)
gosper	\(\frac {2 \left (c x +b \right ) \sqrt {c \,x^{2}+b x}}{3 c \sqrt {x}}\)	\(25\)
risch	\(\frac {2 \left (c x +b \right )^{2} \sqrt {x}}{3 \sqrt {x \left (c x +b \right )}\, c}\)	\(25\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.96 \[ \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx=\frac {2 \, \sqrt {c x^{2} + b x} {\left (c x + b\right )}}{3 \, c \sqrt {x}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx=\int \frac {\sqrt {x \left (b + c x\right )}}{\sqrt {x}}\, dx \]

[In]

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

[Out]

Integral(sqrt(x*(b + c*x))/sqrt(x), x)

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.48 \[ \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx=\frac {2 \, {\left (c x + b\right )}^{\frac {3}{2}}}{3 \, c} \]

[In]

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

[Out]

2/3*(c*x + b)^(3/2)/c

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.84 \[ \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx=\frac {2 \, {\left (c x + b\right )}^{\frac {3}{2}}}{3 \, c} - \frac {2 \, b^{\frac {3}{2}}}{3 \, c} \]

[In]

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

[Out]

2/3*(c*x + b)^(3/2)/c - 2/3*b^(3/2)/c

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx=\int \frac {\sqrt {c\,x^2+b\,x}}{\sqrt {x}} \,d x \]

[In]

int((b*x + c*x^2)^(1/2)/x^(1/2),x)

[Out]

int((b*x + c*x^2)^(1/2)/x^(1/2), x)