3.1.96 \(\int \frac {2+3 x+5 x^2}{(3-x+2 x^2)^{5/2}} \, dx\) [96]

Optimal. Leaf size=47 \[ -\frac {11 (5+3 x)}{69 \left (3-x+2 x^2\right )^{3/2}}-\frac {71 (1-4 x)}{529 \sqrt {3-x+2 x^2}} \]

[Out]

-11/69*(5+3*x)/(2*x^2-x+3)^(3/2)-71/529*(1-4*x)/(2*x^2-x+3)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {1674, 12, 627} \begin {gather*} -\frac {71 (1-4 x)}{529 \sqrt {2 x^2-x+3}}-\frac {11 (3 x+5)}{69 \left (2 x^2-x+3\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 3*x + 5*x^2)/(3 - x + 2*x^2)^(5/2),x]

[Out]

(-11*(5 + 3*x))/(69*(3 - x + 2*x^2)^(3/2)) - (71*(1 - 4*x))/(529*Sqrt[3 - x + 2*x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 627

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[-2*((b + 2*c*x)/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2])), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1674

Int[(Pq_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, a + b*x + c*
x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x + c*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b
*x + c*x^2, x], x, 1]}, Simp[(b*f - 2*a*g + (2*c*f - b*g)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)
)), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)), Int[(a + b*x + c*x^2)^(p + 1)*ExpandToSum[(p + 1)*(b^2 - 4*a*c)*Q - (
2*p + 3)*(2*c*f - b*g), x], x], x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1
]

Rubi steps

\begin {align*} \int \frac {2+3 x+5 x^2}{\left (3-x+2 x^2\right )^{5/2}} \, dx &=-\frac {11 (5+3 x)}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac {2}{69} \int \frac {213}{4 \left (3-x+2 x^2\right )^{3/2}} \, dx\\ &=-\frac {11 (5+3 x)}{69 \left (3-x+2 x^2\right )^{3/2}}+\frac {71}{46} \int \frac {1}{\left (3-x+2 x^2\right )^{3/2}} \, dx\\ &=-\frac {11 (5+3 x)}{69 \left (3-x+2 x^2\right )^{3/2}}-\frac {71 (1-4 x)}{529 \sqrt {3-x+2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.35, size = 33, normalized size = 0.70 \begin {gather*} \frac {2 \left (-952+1005 x-639 x^2+852 x^3\right )}{1587 \left (3-x+2 x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 3*x + 5*x^2)/(3 - x + 2*x^2)^(5/2),x]

[Out]

(2*(-952 + 1005*x - 639*x^2 + 852*x^3))/(1587*(3 - x + 2*x^2)^(3/2))

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Maple [A]
time = 0.12, size = 69, normalized size = 1.47

method	result	size
gosper	\(\frac {\frac {568}{529} x^{3}-\frac {426}{529} x^{2}+\frac {670}{529} x -\frac {1904}{1587}}{\left (2 x^{2}-x +3\right )^{\frac {3}{2}}}\)	\(30\)
trager	\(\frac {\frac {568}{529} x^{3}-\frac {426}{529} x^{2}+\frac {670}{529} x -\frac {1904}{1587}}{\left (2 x^{2}-x +3\right )^{\frac {3}{2}}}\)	\(30\)
risch	\(\frac {\frac {568}{529} x^{3}-\frac {426}{529} x^{2}+\frac {670}{529} x -\frac {1904}{1587}}{\left (2 x^{2}-x +3\right )^{\frac {3}{2}}}\)	\(30\)
default	\(-\frac {5 x}{4 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}-\frac {29}{48 \left (2 x^{2}-x +3\right )^{\frac {3}{2}}}+\frac {\frac {71 x}{92}-\frac {71}{368}}{\left (2 x^{2}-x +3\right )^{\frac {3}{2}}}+\frac {\frac {284 x}{529}-\frac {71}{529}}{\sqrt {2 x^{2}-x +3}}\)	\(69\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x^2+3*x+2)/(2*x^2-x+3)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-5/4*x/(2*x^2-x+3)^(3/2)-29/48/(2*x^2-x+3)^(3/2)+71/368*(4*x-1)/(2*x^2-x+3)^(3/2)+71/529*(4*x-1)/(2*x^2-x+3)^(
1/2)

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Maxima [A]
time = 0.29, size = 59, normalized size = 1.26 \begin {gather*} \frac {284 \, x}{529 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {71}{529 \, \sqrt {2 \, x^{2} - x + 3}} - \frac {11 \, x}{23 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} - \frac {55}{69 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(5/2),x, algorithm="maxima")

[Out]

284/529*x/sqrt(2*x^2 - x + 3) - 71/529/sqrt(2*x^2 - x + 3) - 11/23*x/(2*x^2 - x + 3)^(3/2) - 55/69/(2*x^2 - x
+ 3)^(3/2)

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Fricas [A]
time = 2.07, size = 51, normalized size = 1.09 \begin {gather*} \frac {2 \, {\left (852 \, x^{3} - 639 \, x^{2} + 1005 \, x - 952\right )} \sqrt {2 \, x^{2} - x + 3}}{1587 \, {\left (4 \, x^{4} - 4 \, x^{3} + 13 \, x^{2} - 6 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(5/2),x, algorithm="fricas")

[Out]

2/1587*(852*x^3 - 639*x^2 + 1005*x - 952)*sqrt(2*x^2 - x + 3)/(4*x^4 - 4*x^3 + 13*x^2 - 6*x + 9)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {5 x^{2} + 3 x + 2}{\left (2 x^{2} - x + 3\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x**2+3*x+2)/(2*x**2-x+3)**(5/2),x)

[Out]

Integral((5*x**2 + 3*x + 2)/(2*x**2 - x + 3)**(5/2), x)

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Giac [A]
time = 4.19, size = 29, normalized size = 0.62 \begin {gather*} \frac {2 \, {\left (3 \, {\left (71 \, {\left (4 \, x - 3\right )} x + 335\right )} x - 952\right )}}{1587 \, {\left (2 \, x^{2} - x + 3\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x^2+3*x+2)/(2*x^2-x+3)^(5/2),x, algorithm="giac")

[Out]

2/1587*(3*(71*(4*x - 3)*x + 335)*x - 952)/(2*x^2 - x + 3)^(3/2)

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Mupad [B]
time = 0.09, size = 29, normalized size = 0.62 \begin {gather*} \frac {2\,\left (852\,x^3-639\,x^2+1005\,x-952\right )}{1587\,{\left (2\,x^2-x+3\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x + 5*x^2 + 2)/(2*x^2 - x + 3)^(5/2),x)

[Out]

(2*(1005*x - 639*x^2 + 852*x^3 - 952))/(1587*(2*x^2 - x + 3)^(3/2))

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