3.92 \(\int \sqrt{e^{a+b x}} x^2 \, dx\)

Optimal. Leaf size=53 \[ -\frac{8 x \sqrt{e^{a+b x}}}{b^2}+\frac{16 \sqrt{e^{a+b x}}}{b^3}+\frac{2 x^2 \sqrt{e^{a+b x}}}{b} \]

[Out]

(16*Sqrt[E^(a + b*x)])/b^3 - (8*Sqrt[E^(a + b*x)]*x)/b^2 + (2*Sqrt[E^(a + b*x)]*x^2)/b

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Rubi [A]  time = 0.0635734, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {2176, 2194} \[ -\frac{8 x \sqrt{e^{a+b x}}}{b^2}+\frac{16 \sqrt{e^{a+b x}}}{b^3}+\frac{2 x^2 \sqrt{e^{a+b x}}}{b} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[E^(a + b*x)]*x^2,x]

[Out]

(16*Sqrt[E^(a + b*x)])/b^3 - (8*Sqrt[E^(a + b*x)]*x)/b^2 + (2*Sqrt[E^(a + b*x)]*x^2)/b

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rubi steps

\begin{align*} \int \sqrt{e^{a+b x}} x^2 \, dx &=\frac{2 \sqrt{e^{a+b x}} x^2}{b}-\frac{4 \int \sqrt{e^{a+b x}} x \, dx}{b}\\ &=-\frac{8 \sqrt{e^{a+b x}} x}{b^2}+\frac{2 \sqrt{e^{a+b x}} x^2}{b}+\frac{8 \int \sqrt{e^{a+b x}} \, dx}{b^2}\\ &=\frac{16 \sqrt{e^{a+b x}}}{b^3}-\frac{8 \sqrt{e^{a+b x}} x}{b^2}+\frac{2 \sqrt{e^{a+b x}} x^2}{b}\\ \end{align*}

Mathematica [A]  time = 0.0126865, size = 29, normalized size = 0.55 \[ \frac{2 \left (b^2 x^2-4 b x+8\right ) \sqrt{e^{a+b x}}}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[E^(a + b*x)]*x^2,x]

[Out]

(2*Sqrt[E^(a + b*x)]*(8 - 4*b*x + b^2*x^2))/b^3

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Maple [A]  time = 0.001, size = 27, normalized size = 0.5 \begin{align*} 2\,{\frac{ \left ({x}^{2}{b}^{2}-4\,bx+8 \right ) \sqrt{{{\rm e}^{bx+a}}}}{{b}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^2*x^2-4*b*x+8)*exp(b*x+a)^(1/2)/b^3

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Maxima [A]  time = 1.02316, size = 49, normalized size = 0.92 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} e^{\left (\frac{1}{2} \, a\right )} - 4 \, b x e^{\left (\frac{1}{2} \, a\right )} + 8 \, e^{\left (\frac{1}{2} \, a\right )}\right )} e^{\left (\frac{1}{2} \, b x\right )}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^2*x^2*e^(1/2*a) - 4*b*x*e^(1/2*a) + 8*e^(1/2*a))*e^(1/2*b*x)/b^3

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Fricas [A]  time = 1.47572, size = 66, normalized size = 1.25 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} - 4 \, b x + 8\right )} e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^2*x^2 - 4*b*x + 8)*e^(1/2*b*x + 1/2*a)/b^3

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Sympy [A]  time = 0.103519, size = 34, normalized size = 0.64 \begin{align*} \begin{cases} \frac{\left (2 b^{2} x^{2} - 8 b x + 16\right ) \sqrt{e^{a + b x}}}{b^{3}} & \text{for}\: b^{3} \neq 0 \\\frac{x^{3}}{3} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise(((2*b**2*x**2 - 8*b*x + 16)*sqrt(exp(a + b*x))/b**3, Ne(b**3, 0)), (x**3/3, True))

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Giac [A]  time = 1.22556, size = 36, normalized size = 0.68 \begin{align*} \frac{2 \,{\left (b^{2} x^{2} - 4 \, b x + 8\right )} e^{\left (\frac{1}{2} \, b x + \frac{1}{2} \, a\right )}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(b^2*x^2 - 4*b*x + 8)*e^(1/2*b*x + 1/2*a)/b^3