3.22.38 \(\int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^{5/2}} \, dx\) [2138]

Optimal. Leaf size=53 \[ \frac {539}{24 (1-2 x)^{3/2}}-\frac {707}{8 \sqrt {1-2 x}}-\frac {309}{8} \sqrt {1-2 x}+\frac {15}{8} (1-2 x)^{3/2} \]

[Out]

539/24/(1-2*x)^(3/2)+15/8*(1-2*x)^(3/2)-707/8/(1-2*x)^(1/2)-309/8*(1-2*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {78} \begin {gather*} \frac {15}{8} (1-2 x)^{3/2}-\frac {309}{8} \sqrt {1-2 x}-\frac {707}{8 \sqrt {1-2 x}}+\frac {539}{24 (1-2 x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((2 + 3*x)^2*(3 + 5*x))/(1 - 2*x)^(5/2),x]

[Out]

539/(24*(1 - 2*x)^(3/2)) - 707/(8*Sqrt[1 - 2*x]) - (309*Sqrt[1 - 2*x])/8 + (15*(1 - 2*x)^(3/2))/8

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

\begin {align*} \int \frac {(2+3 x)^2 (3+5 x)}{(1-2 x)^{5/2}} \, dx &=\int \left (\frac {539}{8 (1-2 x)^{5/2}}-\frac {707}{8 (1-2 x)^{3/2}}+\frac {309}{8 \sqrt {1-2 x}}-\frac {45}{8} \sqrt {1-2 x}\right ) \, dx\\ &=\frac {539}{24 (1-2 x)^{3/2}}-\frac {707}{8 \sqrt {1-2 x}}-\frac {309}{8} \sqrt {1-2 x}+\frac {15}{8} (1-2 x)^{3/2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 28, normalized size = 0.53 \begin {gather*} -\frac {308-960 x+396 x^2+45 x^3}{3 (1-2 x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2 + 3*x)^2*(3 + 5*x))/(1 - 2*x)^(5/2),x]

[Out]

-1/3*(308 - 960*x + 396*x^2 + 45*x^3)/(1 - 2*x)^(3/2)

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Maple [A]
time = 0.10, size = 38, normalized size = 0.72

method	result	size
gosper	\(-\frac {45 x^{3}+396 x^{2}-960 x +308}{3 \left (1-2 x \right )^{\frac {3}{2}}}\)	\(25\)
trager	\(-\frac {\left (45 x^{3}+396 x^{2}-960 x +308\right ) \sqrt {1-2 x}}{3 \left (-1+2 x \right )^{2}}\)	\(32\)
risch	\(\frac {45 x^{3}+396 x^{2}-960 x +308}{3 \left (-1+2 x \right ) \sqrt {1-2 x}}\)	\(32\)
derivativedivides	\(\frac {539}{24 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {15 \left (1-2 x \right )^{\frac {3}{2}}}{8}-\frac {707}{8 \sqrt {1-2 x}}-\frac {309 \sqrt {1-2 x}}{8}\)	\(38\)
default	\(\frac {539}{24 \left (1-2 x \right )^{\frac {3}{2}}}+\frac {15 \left (1-2 x \right )^{\frac {3}{2}}}{8}-\frac {707}{8 \sqrt {1-2 x}}-\frac {309 \sqrt {1-2 x}}{8}\)	\(38\)
meijerg	\(-\frac {8 \left (\frac {\sqrt {\pi }}{2}-\frac {\sqrt {\pi }}{2 \left (1-2 x \right )^{\frac {3}{2}}}\right )}{\sqrt {\pi }}+\frac {\frac {56 \sqrt {\pi }}{3}-\frac {7 \sqrt {\pi }\, \left (-24 x +8\right )}{3 \left (1-2 x \right )^{\frac {3}{2}}}}{\sqrt {\pi }}-\frac {29 \left (-4 \sqrt {\pi }+\frac {\sqrt {\pi }\, \left (24 x^{2}-48 x +16\right )}{4 \left (1-2 x \right )^{\frac {3}{2}}}\right )}{2 \sqrt {\pi }}+\frac {30 \sqrt {\pi }-\frac {15 \sqrt {\pi }\, \left (64 x^{3}+192 x^{2}-384 x +128\right )}{64 \left (1-2 x \right )^{\frac {3}{2}}}}{\sqrt {\pi }}\)	\(122\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2+3*x)^2*(3+5*x)/(1-2*x)^(5/2),x,method=_RETURNVERBOSE)

[Out]

539/24/(1-2*x)^(3/2)+15/8*(1-2*x)^(3/2)-707/8/(1-2*x)^(1/2)-309/8*(1-2*x)^(1/2)

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Maxima [A]
time = 0.31, size = 33, normalized size = 0.62 \begin {gather*} \frac {15}{8} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {309}{8} \, \sqrt {-2 \, x + 1} + \frac {7 \, {\left (303 \, x - 113\right )}}{12 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^2*(3+5*x)/(1-2*x)^(5/2),x, algorithm="maxima")

[Out]

15/8*(-2*x + 1)^(3/2) - 309/8*sqrt(-2*x + 1) + 7/12*(303*x - 113)/(-2*x + 1)^(3/2)

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Fricas [A]
time = 0.99, size = 36, normalized size = 0.68 \begin {gather*} -\frac {{\left (45 \, x^{3} + 396 \, x^{2} - 960 \, x + 308\right )} \sqrt {-2 \, x + 1}}{3 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^2*(3+5*x)/(1-2*x)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(45*x^3 + 396*x^2 - 960*x + 308)*sqrt(-2*x + 1)/(4*x^2 - 4*x + 1)

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Sympy [A]
time = 0.24, size = 88, normalized size = 1.66 \begin {gather*} - \frac {45 x^{3} \sqrt {1 - 2 x}}{12 x^{2} - 12 x + 3} - \frac {396 x^{2} \sqrt {1 - 2 x}}{12 x^{2} - 12 x + 3} + \frac {960 x \sqrt {1 - 2 x}}{12 x^{2} - 12 x + 3} - \frac {308 \sqrt {1 - 2 x}}{12 x^{2} - 12 x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)**2*(3+5*x)/(1-2*x)**(5/2),x)

[Out]

-45*x**3*sqrt(1 - 2*x)/(12*x**2 - 12*x + 3) - 396*x**2*sqrt(1 - 2*x)/(12*x**2 - 12*x + 3) + 960*x*sqrt(1 - 2*x
)/(12*x**2 - 12*x + 3) - 308*sqrt(1 - 2*x)/(12*x**2 - 12*x + 3)

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Giac [A]
time = 2.07, size = 40, normalized size = 0.75 \begin {gather*} \frac {15}{8} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {309}{8} \, \sqrt {-2 \, x + 1} - \frac {7 \, {\left (303 \, x - 113\right )}}{12 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^2*(3+5*x)/(1-2*x)^(5/2),x, algorithm="giac")

[Out]

15/8*(-2*x + 1)^(3/2) - 309/8*sqrt(-2*x + 1) - 7/12*(303*x - 113)/((2*x - 1)*sqrt(-2*x + 1))

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Mupad [B]
time = 1.19, size = 38, normalized size = 0.72 \begin {gather*} \frac {927\,{\left (2\,x-1\right )}^2-4242\,x+45\,{\left (2\,x-1\right )}^3+1582}{\sqrt {1-2\,x}\,\left (48\,x-24\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x + 2)^2*(5*x + 3))/(1 - 2*x)^(5/2),x)

[Out]

(927*(2*x - 1)^2 - 4242*x + 45*(2*x - 1)^3 + 1582)/((1 - 2*x)^(1/2)*(48*x - 24))

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