3.75 \(\int \sqrt{x} \sqrt{b x+c x^2} \, dx\)

Optimal. Leaf size=52 \[ \frac{2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt{x}}-\frac{4 b \left (b x+c x^2\right )^{3/2}}{15 c^2 x^{3/2}} \]

[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])

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Rubi [A]  time = 0.0152971, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {656, 648} \[ \frac{2 \left (b x+c x^2\right )^{3/2}}{5 c \sqrt{x}}-\frac{4 b \left (b x+c x^2\right )^{3/2}}{15 c^2 x^{3/2}} \]

Antiderivative was successfully verified.

[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 656

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]

Rule 648

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*} \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{(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]  time = 0.017424, size = 31, normalized size = 0.6 \[ \frac{2 (x (b+c x))^{3/2} (3 c x-2 b)}{15 c^2 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.046, size = 33, normalized size = 0.6 \begin{align*} -{\frac{ \left ( 2\,cx+2\,b \right ) \left ( -3\,cx+2\,b \right ) }{15\,{c}^{2}}\sqrt{c{x}^{2}+bx}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02837, size = 41, normalized size = 0.79 \begin{align*} \frac{2 \,{\left (3 \, c^{2} x^{2} + b c x - 2 \, b^{2}\right )} \sqrt{c x + b}}{15 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [A]  time = 2.18565, size = 89, normalized size = 1.71 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{x} \sqrt{x \left (b + c x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.21536, size = 46, normalized size = 0.88 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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