\(\int \sqrt {x} \sqrt {b x+c x^2} \, dx\) [75]

Optimal result

Integrand size = 19, antiderivative size = 52 \[ \int \sqrt {x} \sqrt {b x+c x^2} \, dx=-\frac {4 b \left (b x+c x^2\right )^{3/2}}{15 c^2 x^{3/2}}+\frac {2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}} \]

[Out]

-4/15*b*(c*x^2+b*x)^(3/2)/c^2/x^(3/2)+2/5*(c*x^2+b*x)^(3/2)/c/x^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {670, 662} \[ \int \sqrt {x} \sqrt {b x+c x^2} \, dx=\frac {2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}-\frac {4 b \left (b x+c x^2\right )^{3/2}}{15 c^2 x^{3/2}} \]

[In]

Int[Sqrt[x]*Sqrt[b*x + c*x^2],x]

[Out]

(-4*b*(b*x + c*x^2)^(3/2))/(15*c^2*x^(3/2)) + (2*(b*x + c*x^2)^(3/2))/(5*c*Sqrt[x])

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]

Rule 670

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*(m + 2*p + 1))), x] + Dist[Simplify[m + p]*((2*c*d - b*e)/(c*(m + 2*p + 1))), In
t[(d + e*x)^(m - 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && E
qQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[Simplify[m + p], 0]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}}-\frac {(2 b) \int \frac {\sqrt {b x+c x^2}}{\sqrt {x}} \, dx}{5 c} \\ & = -\frac {4 b \left (b x+c x^2\right )^{3/2}}{15 c^2 x^{3/2}}+\frac {2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.79 \[ \int \sqrt {x} \sqrt {b x+c x^2} \, dx=\frac {2 \sqrt {x (b+c x)} \left (-2 b^2+b c x+3 c^2 x^2\right )}{15 c^2 \sqrt {x}} \]

[In]

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

[Out]

(2*Sqrt[x*(b + c*x)]*(-2*b^2 + b*c*x + 3*c^2*x^2))/(15*c^2*Sqrt[x])

Maple [A] (verified)

Time = 2.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.60

method	result	size
default	\(-\frac {2 \left (c x +b \right ) \left (-3 c x +2 b \right ) \sqrt {x \left (c x +b \right )}}{15 c^{2} \sqrt {x}}\)	\(31\)
gosper	\(-\frac {2 \left (c x +b \right ) \left (-3 c x +2 b \right ) \sqrt {c \,x^{2}+b x}}{15 c^{2} \sqrt {x}}\)	\(33\)
risch	\(-\frac {2 \left (c x +b \right ) \sqrt {x}\, \left (-3 c^{2} x^{2}-b c x +2 b^{2}\right )}{15 \sqrt {x \left (c x +b \right )}\, c^{2}}\)	\(42\)

[In]

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

[Out]

-2/15*(c*x+b)*(-3*c*x+2*b)*(x*(c*x+b))^(1/2)/c^2/x^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.71 \[ \int \sqrt {x} \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (3 \, c^{2} x^{2} + b c x - 2 \, b^{2}\right )} \sqrt {c x^{2} + b x}}{15 \, c^{2} \sqrt {x}} \]

[In]

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

[Out]

2/15*(3*c^2*x^2 + b*c*x - 2*b^2)*sqrt(c*x^2 + b*x)/(c^2*sqrt(x))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.58 \[ \int \sqrt {x} \sqrt {b x+c x^2} \, dx=\frac {2 \, {\left (3 \, c^{2} x^{2} + b c x - 2 \, b^{2}\right )} \sqrt {c x + b}}{15 \, c^{2}} \]

[In]

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

[Out]

2/15*(3*c^2*x^2 + b*c*x - 2*b^2)*sqrt(c*x + b)/c^2

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.65 \[ \int \sqrt {x} \sqrt {b x+c x^2} \, dx=\frac {4 \, b^{\frac {5}{2}}}{15 \, c^{2}} + \frac {2 \, {\left (3 \, {\left (c x + b\right )}^{\frac {5}{2}} - 5 \, {\left (c x + b\right )}^{\frac {3}{2}} b\right )}}{15 \, c^{2}} \]

[In]

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

[Out]

4/15*b^(5/2)/c^2 + 2/15*(3*(c*x + b)^(5/2) - 5*(c*x + b)^(3/2)*b)/c^2

Mupad [F(-1)]

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

[In]

int(x^(1/2)*(b*x + c*x^2)^(1/2),x)

[Out]

int(x^(1/2)*(b*x + c*x^2)^(1/2), x)