3.23.90 \(\int \frac {(5-x) (2+5 x+3 x^2)}{(3+2 x)^{3/2}} \, dx\)

Optimal. Leaf size=53 \[ -\frac {3}{40} (2 x+3)^{5/2}+\frac {47}{24} (2 x+3)^{3/2}-\frac {109}{8} \sqrt {2 x+3}-\frac {65}{8 \sqrt {2 x+3}} \]

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Rubi [A]  time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {771} \begin {gather*} -\frac {3}{40} (2 x+3)^{5/2}+\frac {47}{24} (2 x+3)^{3/2}-\frac {109}{8} \sqrt {2 x+3}-\frac {65}{8 \sqrt {2 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-65/(8*Sqrt[3 + 2*x]) - (109*Sqrt[3 + 2*x])/8 + (47*(3 + 2*x)^(3/2))/24 - (3*(3 + 2*x)^(5/2))/40

Rule 771

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(d + e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && N
eQ[b^2 - 4*a*c, 0] && IntegerQ[p] && (GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

\begin {align*} \int \frac {(5-x) \left (2+5 x+3 x^2\right )}{(3+2 x)^{3/2}} \, dx &=\int \left (\frac {65}{8 (3+2 x)^{3/2}}-\frac {109}{8 \sqrt {3+2 x}}+\frac {47}{8} \sqrt {3+2 x}-\frac {3}{8} (3+2 x)^{3/2}\right ) \, dx\\ &=-\frac {65}{8 \sqrt {3+2 x}}-\frac {109}{8} \sqrt {3+2 x}+\frac {47}{24} (3+2 x)^{3/2}-\frac {3}{40} (3+2 x)^{5/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.53 \begin {gather*} -\frac {9 x^3-77 x^2+117 x+501}{15 \sqrt {2 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/15*(501 + 117*x - 77*x^2 + 9*x^3)/Sqrt[3 + 2*x]

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IntegrateAlgebraic [A]  time = 0.04, size = 40, normalized size = 0.75 \begin {gather*} \frac {-9 (2 x+3)^3+235 (2 x+3)^2-1635 (2 x+3)-975}{120 \sqrt {2 x+3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-975 - 1635*(3 + 2*x) + 235*(3 + 2*x)^2 - 9*(3 + 2*x)^3)/(120*Sqrt[3 + 2*x])

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fricas [A]  time = 0.38, size = 24, normalized size = 0.45 \begin {gather*} -\frac {9 \, x^{3} - 77 \, x^{2} + 117 \, x + 501}{15 \, \sqrt {2 \, x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(9*x^3 - 77*x^2 + 117*x + 501)/sqrt(2*x + 3)

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giac [A]  time = 0.16, size = 37, normalized size = 0.70 \begin {gather*} -\frac {3}{40} \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + \frac {47}{24} \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - \frac {109}{8} \, \sqrt {2 \, x + 3} - \frac {65}{8 \, \sqrt {2 \, x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/40*(2*x + 3)^(5/2) + 47/24*(2*x + 3)^(3/2) - 109/8*sqrt(2*x + 3) - 65/8/sqrt(2*x + 3)

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maple [A]  time = 0.00, size = 25, normalized size = 0.47 \begin {gather*} -\frac {9 x^{3}-77 x^{2}+117 x +501}{15 \sqrt {2 x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(9*x^3-77*x^2+117*x+501)/(2*x+3)^(1/2)

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maxima [A]  time = 0.51, size = 37, normalized size = 0.70 \begin {gather*} -\frac {3}{40} \, {\left (2 \, x + 3\right )}^{\frac {5}{2}} + \frac {47}{24} \, {\left (2 \, x + 3\right )}^{\frac {3}{2}} - \frac {109}{8} \, \sqrt {2 \, x + 3} - \frac {65}{8 \, \sqrt {2 \, x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/40*(2*x + 3)^(5/2) + 47/24*(2*x + 3)^(3/2) - 109/8*sqrt(2*x + 3) - 65/8/sqrt(2*x + 3)

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mupad [B]  time = 0.04, size = 37, normalized size = 0.70 \begin {gather*} \frac {47\,{\left (2\,x+3\right )}^{3/2}}{24}-\frac {109\,\sqrt {2\,x+3}}{8}-\frac {65}{8\,\sqrt {2\,x+3}}-\frac {3\,{\left (2\,x+3\right )}^{5/2}}{40} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(47*(2*x + 3)^(3/2))/24 - (109*(2*x + 3)^(1/2))/8 - 65/(8*(2*x + 3)^(1/2)) - (3*(2*x + 3)^(5/2))/40

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sympy [A]  time = 19.15, size = 46, normalized size = 0.87 \begin {gather*} - \frac {3 \left (2 x + 3\right )^{\frac {5}{2}}}{40} + \frac {47 \left (2 x + 3\right )^{\frac {3}{2}}}{24} - \frac {109 \sqrt {2 x + 3}}{8} - \frac {65}{8 \sqrt {2 x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*(2*x + 3)**(5/2)/40 + 47*(2*x + 3)**(3/2)/24 - 109*sqrt(2*x + 3)/8 - 65/(8*sqrt(2*x + 3))

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