3.18.95 \(\int \sqrt {1-2 x} (2+3 x) (3+5 x) \, dx\) [1795]

Optimal. Leaf size=40 \[ -\frac {77}{12} (1-2 x)^{3/2}+\frac {17}{5} (1-2 x)^{5/2}-\frac {15}{28} (1-2 x)^{7/2} \]

[Out]

-77/12*(1-2*x)^(3/2)+17/5*(1-2*x)^(5/2)-15/28*(1-2*x)^(7/2)

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Rubi [A]
time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {78} \begin {gather*} -\frac {15}{28} (1-2 x)^{7/2}+\frac {17}{5} (1-2 x)^{5/2}-\frac {77}{12} (1-2 x)^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-77*(1 - 2*x)^(3/2))/12 + (17*(1 - 2*x)^(5/2))/5 - (15*(1 - 2*x)^(7/2))/28

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 \sqrt {1-2 x} (2+3 x) (3+5 x) \, dx &=\int \left (\frac {77}{4} \sqrt {1-2 x}-17 (1-2 x)^{3/2}+\frac {15}{4} (1-2 x)^{5/2}\right ) \, dx\\ &=-\frac {77}{12} (1-2 x)^{3/2}+\frac {17}{5} (1-2 x)^{5/2}-\frac {15}{28} (1-2 x)^{7/2}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 23, normalized size = 0.58 \begin {gather*} -\frac {1}{105} (1-2 x)^{3/2} \left (373+489 x+225 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/105*((1 - 2*x)^(3/2)*(373 + 489*x + 225*x^2))

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Maple [A]
time = 0.09, size = 29, normalized size = 0.72

method	result	size
gosper	\(-\frac {\left (225 x^{2}+489 x +373\right ) \left (1-2 x \right )^{\frac {3}{2}}}{105}\)	\(20\)
trager	\(\left (\frac {30}{7} x^{3}+\frac {251}{35} x^{2}+\frac {257}{105} x -\frac {373}{105}\right ) \sqrt {1-2 x}\)	\(24\)
derivativedivides	\(-\frac {77 \left (1-2 x \right )^{\frac {3}{2}}}{12}+\frac {17 \left (1-2 x \right )^{\frac {5}{2}}}{5}-\frac {15 \left (1-2 x \right )^{\frac {7}{2}}}{28}\)	\(29\)
default	\(-\frac {77 \left (1-2 x \right )^{\frac {3}{2}}}{12}+\frac {17 \left (1-2 x \right )^{\frac {5}{2}}}{5}-\frac {15 \left (1-2 x \right )^{\frac {7}{2}}}{28}\)	\(29\)
risch	\(-\frac {\left (450 x^{3}+753 x^{2}+257 x -373\right ) \left (-1+2 x \right )}{105 \sqrt {1-2 x}}\)	\(30\)
meijerg	\(\frac {2 \sqrt {\pi }-\sqrt {\pi }\, \left (2-4 x \right ) \sqrt {1-2 x}}{\sqrt {\pi }}-\frac {19 \left (-\frac {8 \sqrt {\pi }}{15}+\frac {4 \sqrt {\pi }\, \left (1-2 x \right )^{\frac {3}{2}} \left (6 x +2\right )}{15}\right )}{8 \sqrt {\pi }}+\frac {\frac {2 \sqrt {\pi }}{7}-\frac {\sqrt {\pi }\, \left (1-2 x \right )^{\frac {3}{2}} \left (60 x^{2}+24 x +8\right )}{28}}{\sqrt {\pi }}\)	\(91\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-77/12*(1-2*x)^(3/2)+17/5*(1-2*x)^(5/2)-15/28*(1-2*x)^(7/2)

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Maxima [A]
time = 0.28, size = 28, normalized size = 0.70 \begin {gather*} -\frac {15}{28} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {17}{5} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {77}{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)*(3+5*x)*(1-2*x)^(1/2),x, algorithm="maxima")

[Out]

-15/28*(-2*x + 1)^(7/2) + 17/5*(-2*x + 1)^(5/2) - 77/12*(-2*x + 1)^(3/2)

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Fricas [A]
time = 1.30, size = 24, normalized size = 0.60 \begin {gather*} \frac {1}{105} \, {\left (450 \, x^{3} + 753 \, x^{2} + 257 \, x - 373\right )} \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(450*x^3 + 753*x^2 + 257*x - 373)*sqrt(-2*x + 1)

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Sympy [A]
time = 1.00, size = 34, normalized size = 0.85 \begin {gather*} - \frac {15 \left (1 - 2 x\right )^{\frac {7}{2}}}{28} + \frac {17 \left (1 - 2 x\right )^{\frac {5}{2}}}{5} - \frac {77 \left (1 - 2 x\right )^{\frac {3}{2}}}{12} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15*(1 - 2*x)**(7/2)/28 + 17*(1 - 2*x)**(5/2)/5 - 77*(1 - 2*x)**(3/2)/12

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Giac [A]
time = 0.53, size = 42, normalized size = 1.05 \begin {gather*} \frac {15}{28} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {17}{5} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {77}{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)*(3+5*x)*(1-2*x)^(1/2),x, algorithm="giac")

[Out]

15/28*(2*x - 1)^3*sqrt(-2*x + 1) + 17/5*(2*x - 1)^2*sqrt(-2*x + 1) - 77/12*(-2*x + 1)^(3/2)

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Mupad [B]
time = 0.04, size = 23, normalized size = 0.58 \begin {gather*} -\frac {{\left (1-2\,x\right )}^{3/2}\,\left (2856\,x+225\,{\left (2\,x-1\right )}^2+1267\right )}{420} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((1 - 2*x)^(3/2)*(2856*x + 225*(2*x - 1)^2 + 1267))/420

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