3.18.24 \(\int (1-2 x)^{3/2} (3+5 x) \, dx\)

Optimal. Leaf size=27 \[ \frac {5}{14} (1-2 x)^{7/2}-\frac {11}{10} (1-2 x)^{5/2} \]

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Rubi [A]  time = 0.01, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {43} \begin {gather*} \frac {5}{14} (1-2 x)^{7/2}-\frac {11}{10} (1-2 x)^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 - 2*x)^(3/2)*(3 + 5*x),x]

[Out]

(-11*(1 - 2*x)^(5/2))/10 + (5*(1 - 2*x)^(7/2))/14

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin {align*} \int (1-2 x)^{3/2} (3+5 x) \, dx &=\int \left (\frac {11}{2} (1-2 x)^{3/2}-\frac {5}{2} (1-2 x)^{5/2}\right ) \, dx\\ &=-\frac {11}{10} (1-2 x)^{5/2}+\frac {5}{14} (1-2 x)^{7/2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.67 \begin {gather*} -\frac {1}{35} (1-2 x)^{5/2} (25 x+26) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/35*((1 - 2*x)^(5/2)*(26 + 25*x))

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IntegrateAlgebraic [A]  time = 0.01, size = 22, normalized size = 0.81 \begin {gather*} \frac {1}{70} (25 (1-2 x)-77) (1-2 x)^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-77 + 25*(1 - 2*x))*(1 - 2*x)^(5/2))/70

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fricas [A]  time = 1.59, size = 24, normalized size = 0.89 \begin {gather*} -\frac {1}{35} \, {\left (100 \, x^{3} + 4 \, x^{2} - 79 \, x + 26\right )} \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(100*x^3 + 4*x^2 - 79*x + 26)*sqrt(-2*x + 1)

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giac [A]  time = 0.92, size = 33, normalized size = 1.22 \begin {gather*} -\frac {5}{14} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {11}{10} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/14*(2*x - 1)^3*sqrt(-2*x + 1) - 11/10*(2*x - 1)^2*sqrt(-2*x + 1)

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maple [A]  time = 0.00, size = 15, normalized size = 0.56 \begin {gather*} -\frac {\left (25 x +26\right ) \left (-2 x +1\right )^{\frac {5}{2}}}{35} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(25*x+26)*(-2*x+1)^(5/2)

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maxima [A]  time = 0.48, size = 19, normalized size = 0.70 \begin {gather*} \frac {5}{14} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {11}{10} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/14*(-2*x + 1)^(7/2) - 11/10*(-2*x + 1)^(5/2)

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mupad [B]  time = 0.02, size = 14, normalized size = 0.52 \begin {gather*} -\frac {{\left (1-2\,x\right )}^{5/2}\,\left (50\,x+52\right )}{70} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(3/2)*(5*x + 3),x)

[Out]

-((1 - 2*x)^(5/2)*(50*x + 52))/70

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sympy [B]  time = 0.38, size = 54, normalized size = 2.00 \begin {gather*} - \frac {20 x^{3} \sqrt {1 - 2 x}}{7} - \frac {4 x^{2} \sqrt {1 - 2 x}}{35} + \frac {79 x \sqrt {1 - 2 x}}{35} - \frac {26 \sqrt {1 - 2 x}}{35} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-20*x**3*sqrt(1 - 2*x)/7 - 4*x**2*sqrt(1 - 2*x)/35 + 79*x*sqrt(1 - 2*x)/35 - 26*sqrt(1 - 2*x)/35

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