3.1.92 \(\int \sqrt {e^{a+b x}} x^2 \, dx\) [92]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 = {2207, 2225} \begin {gather*} \frac {16 \sqrt {e^{a+b x}}}{b^3}-\frac {8 x \sqrt {e^{a+b x}}}{b^2}+\frac {2 x^2 \sqrt {e^{a+b x}}}{b} \end {gather*}

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 2207

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] &&  !TrueQ[$UseGamma]

Rule 2225

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*}

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Mathematica [A]
time = 0.04, size = 29, normalized size = 0.55 \begin {gather*} \frac {2 \sqrt {e^{a+b x}} \left (8-4 b x+b^2 x^2\right )}{b^3} \end {gather*}

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 = 0.01, size = 27, normalized size = 0.51

method	result	size
gosper	\(\frac {2 \left (b^{2} x^{2}-4 b x +8\right ) \sqrt {{\mathrm e}^{b x +a}}}{b^{3}}\)	\(27\)
risch	\(\frac {2 \left (b^{2} x^{2}-4 b x +8\right ) \sqrt {{\mathrm e}^{b x +a}}}{b^{3}}\)	\(27\)
meijerg	\(-\frac {8 \,{\mathrm e}^{-\frac {3 a}{2}-\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}} \sqrt {{\mathrm e}^{b x +a}}\, \left (2-\frac {\left (\frac {3 b^{2} x^{2} {\mathrm e}^{a}}{4}-3 b x \,{\mathrm e}^{\frac {a}{2}}+6\right ) {\mathrm e}^{\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}}}{3}\right )}{b^{3}}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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 = 0.28, size = 36, normalized size = 0.68 \begin {gather*} \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 {gather*}

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 = 0.36, size = 27, normalized size = 0.51 \begin {gather*} \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 {gather*}

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 = 0.04, size = 34, normalized size = 0.64 \begin {gather*} \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 {gather*}

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 = 1.88, size = 27, normalized size = 0.51 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 0.06, size = 29, normalized size = 0.55 \begin {gather*} \sqrt {{\mathrm {e}}^{a+b\,x}}\,\left (\frac {16}{b^3}-\frac {8\,x}{b^2}+\frac {2\,x^2}{b}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*exp(a + b*x)^(1/2),x)

[Out]

exp(a + b*x)^(1/2)*(16/b^3 - (8*x)/b^2 + (2*x^2)/b)

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