3.1.91 \(\int \sqrt {e^{a+b x}} x^3 \, dx\) [91]

Optimal. Leaf size=72 \[ -\frac {96 \sqrt {e^{a+b x}}}{b^4}+\frac {48 \sqrt {e^{a+b x}} x}{b^3}-\frac {12 \sqrt {e^{a+b x}} x^2}{b^2}+\frac {2 \sqrt {e^{a+b x}} x^3}{b} \]

[Out]

-96*exp(b*x+a)^(1/2)/b^4+48*x*exp(b*x+a)^(1/2)/b^3-12*x^2*exp(b*x+a)^(1/2)/b^2+2*x^3*exp(b*x+a)^(1/2)/b

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Rubi [A]
time = 0.07, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, 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 {96 \sqrt {e^{a+b x}}}{b^4}+\frac {48 x \sqrt {e^{a+b x}}}{b^3}-\frac {12 x^2 \sqrt {e^{a+b x}}}{b^2}+\frac {2 x^3 \sqrt {e^{a+b x}}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[E^(a + b*x)]*x^3,x]

[Out]

(-96*Sqrt[E^(a + b*x)])/b^4 + (48*Sqrt[E^(a + b*x)]*x)/b^3 - (12*Sqrt[E^(a + b*x)]*x^2)/b^2 + (2*Sqrt[E^(a + b
*x)]*x^3)/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^3 \, dx &=\frac {2 \sqrt {e^{a+b x}} x^3}{b}-\frac {6 \int \sqrt {e^{a+b x}} x^2 \, dx}{b}\\ &=-\frac {12 \sqrt {e^{a+b x}} x^2}{b^2}+\frac {2 \sqrt {e^{a+b x}} x^3}{b}+\frac {24 \int \sqrt {e^{a+b x}} x \, dx}{b^2}\\ &=\frac {48 \sqrt {e^{a+b x}} x}{b^3}-\frac {12 \sqrt {e^{a+b x}} x^2}{b^2}+\frac {2 \sqrt {e^{a+b x}} x^3}{b}-\frac {48 \int \sqrt {e^{a+b x}} \, dx}{b^3}\\ &=-\frac {96 \sqrt {e^{a+b x}}}{b^4}+\frac {48 \sqrt {e^{a+b x}} x}{b^3}-\frac {12 \sqrt {e^{a+b x}} x^2}{b^2}+\frac {2 \sqrt {e^{a+b x}} x^3}{b}\\ \end {align*}

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Mathematica [A]
time = 0.04, size = 37, normalized size = 0.51 \begin {gather*} \frac {2 \sqrt {e^{a+b x}} \left (-48+24 b x-6 b^2 x^2+b^3 x^3\right )}{b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[E^(a + b*x)]*x^3,x]

[Out]

(2*Sqrt[E^(a + b*x)]*(-48 + 24*b*x - 6*b^2*x^2 + b^3*x^3))/b^4

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Maple [A]
time = 0.01, size = 35, normalized size = 0.49

method	result	size
gosper	\(\frac {2 \left (b^{3} x^{3}-6 b^{2} x^{2}+24 b x -48\right ) \sqrt {{\mathrm e}^{b x +a}}}{b^{4}}\)	\(35\)
risch	\(\frac {2 \left (b^{3} x^{3}-6 b^{2} x^{2}+24 b x -48\right ) \sqrt {{\mathrm e}^{b x +a}}}{b^{4}}\)	\(35\)
meijerg	\(\frac {16 \,{\mathrm e}^{-2 a -\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}} \sqrt {{\mathrm e}^{b x +a}}\, \left (6-\frac {\left (-\frac {b^{3} x^{3} {\mathrm e}^{\frac {3 a}{2}}}{2}+3 b^{2} x^{2} {\mathrm e}^{a}-12 b x \,{\mathrm e}^{\frac {a}{2}}+24\right ) {\mathrm e}^{\frac {b x \,{\mathrm e}^{\frac {a}{2}}}{2}}}{4}\right )}{b^{4}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*exp(b*x+a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*(b^3*x^3-6*b^2*x^2+24*b*x-48)*exp(b*x+a)^(1/2)/b^4

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Maxima [A]
time = 0.28, size = 48, normalized size = 0.67 \begin {gather*} \frac {2 \, {\left (b^{3} x^{3} e^{\left (\frac {1}{2} \, a\right )} - 6 \, b^{2} x^{2} e^{\left (\frac {1}{2} \, a\right )} + 24 \, b x e^{\left (\frac {1}{2} \, a\right )} - 48 \, e^{\left (\frac {1}{2} \, a\right )}\right )} e^{\left (\frac {1}{2} \, b x\right )}}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*exp(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

2*(b^3*x^3*e^(1/2*a) - 6*b^2*x^2*e^(1/2*a) + 24*b*x*e^(1/2*a) - 48*e^(1/2*a))*e^(1/2*b*x)/b^4

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Fricas [A]
time = 0.49, size = 35, normalized size = 0.49 \begin {gather*} \frac {2 \, {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 24 \, b x - 48\right )} e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*exp(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

2*(b^3*x^3 - 6*b^2*x^2 + 24*b*x - 48)*e^(1/2*b*x + 1/2*a)/b^4

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Sympy [A]
time = 0.04, size = 42, normalized size = 0.58 \begin {gather*} \begin {cases} \frac {\left (2 b^{3} x^{3} - 12 b^{2} x^{2} + 48 b x - 96\right ) \sqrt {e^{a + b x}}}{b^{4}} & \text {for}\: b^{4} \neq 0 \\\frac {x^{4}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*exp(b*x+a)**(1/2),x)

[Out]

Piecewise(((2*b**3*x**3 - 12*b**2*x**2 + 48*b*x - 96)*sqrt(exp(a + b*x))/b**4, Ne(b**4, 0)), (x**4/4, True))

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Giac [A]
time = 1.86, size = 35, normalized size = 0.49 \begin {gather*} \frac {2 \, {\left (b^{3} x^{3} - 6 \, b^{2} x^{2} + 24 \, b x - 48\right )} e^{\left (\frac {1}{2} \, b x + \frac {1}{2} \, a\right )}}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*exp(b*x+a)^(1/2),x, algorithm="giac")

[Out]

2*(b^3*x^3 - 6*b^2*x^2 + 24*b*x - 48)*e^(1/2*b*x + 1/2*a)/b^4

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Mupad [B]
time = 0.05, size = 37, normalized size = 0.51 \begin {gather*} \sqrt {{\mathrm {e}}^{a+b\,x}}\,\left (\frac {48\,x}{b^3}-\frac {96}{b^4}+\frac {2\,x^3}{b}-\frac {12\,x^2}{b^2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*exp(a + b*x)^(1/2),x)

[Out]

exp(a + b*x)^(1/2)*((48*x)/b^3 - 96/b^4 + (2*x^3)/b - (12*x^2)/b^2)

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