3.2028 \(\int \frac {(3+5 x)^3}{\sqrt {1-2 x}} \, dx\)

Optimal. Leaf size=53 \[ \frac {125}{56} (1-2 x)^{7/2}-\frac {165}{8} (1-2 x)^{5/2}+\frac {605}{8} (1-2 x)^{3/2}-\frac {1331}{8} \sqrt {1-2 x} \]

[Out]

605/8*(1-2*x)^(3/2)-165/8*(1-2*x)^(5/2)+125/56*(1-2*x)^(7/2)-1331/8*(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 = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {43} \[ \frac {125}{56} (1-2 x)^{7/2}-\frac {165}{8} (1-2 x)^{5/2}+\frac {605}{8} (1-2 x)^{3/2}-\frac {1331}{8} \sqrt {1-2 x} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^3/Sqrt[1 - 2*x],x]

[Out]

(-1331*Sqrt[1 - 2*x])/8 + (605*(1 - 2*x)^(3/2))/8 - (165*(1 - 2*x)^(5/2))/8 + (125*(1 - 2*x)^(7/2))/56

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 \frac {(3+5 x)^3}{\sqrt {1-2 x}} \, dx &=\int \left (\frac {1331}{8 \sqrt {1-2 x}}-\frac {1815}{8} \sqrt {1-2 x}+\frac {825}{8} (1-2 x)^{3/2}-\frac {125}{8} (1-2 x)^{5/2}\right ) \, dx\\ &=-\frac {1331}{8} \sqrt {1-2 x}+\frac {605}{8} (1-2 x)^{3/2}-\frac {165}{8} (1-2 x)^{5/2}+\frac {125}{56} (1-2 x)^{7/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 28, normalized size = 0.53 \[ -\frac {1}{7} \sqrt {1-2 x} \left (125 x^3+390 x^2+575 x+764\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/7*(Sqrt[1 - 2*x]*(764 + 575*x + 390*x^2 + 125*x^3))

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fricas [A]  time = 0.97, size = 24, normalized size = 0.45 \[ -\frac {1}{7} \, {\left (125 \, x^{3} + 390 \, x^{2} + 575 \, x + 764\right )} \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*(125*x^3 + 390*x^2 + 575*x + 764)*sqrt(-2*x + 1)

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giac [A]  time = 1.03, size = 51, normalized size = 0.96 \[ -\frac {125}{56} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {165}{8} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {605}{8} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {1331}{8} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-125/56*(2*x - 1)^3*sqrt(-2*x + 1) - 165/8*(2*x - 1)^2*sqrt(-2*x + 1) + 605/8*(-2*x + 1)^(3/2) - 1331/8*sqrt(-
2*x + 1)

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maple [A]  time = 0.00, size = 25, normalized size = 0.47 \[ -\frac {\left (125 x^{3}+390 x^{2}+575 x +764\right ) \sqrt {-2 x +1}}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/7*(125*x^3+390*x^2+575*x+764)*(-2*x+1)^(1/2)

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maxima [A]  time = 0.49, size = 37, normalized size = 0.70 \[ \frac {125}{56} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {165}{8} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {605}{8} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {1331}{8} \, \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/56*(-2*x + 1)^(7/2) - 165/8*(-2*x + 1)^(5/2) + 605/8*(-2*x + 1)^(3/2) - 1331/8*sqrt(-2*x + 1)

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mupad [B]  time = 0.03, size = 37, normalized size = 0.70 \[ \frac {605\,{\left (1-2\,x\right )}^{3/2}}{8}-\frac {1331\,\sqrt {1-2\,x}}{8}-\frac {165\,{\left (1-2\,x\right )}^{5/2}}{8}+\frac {125\,{\left (1-2\,x\right )}^{7/2}}{56} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + 3)^3/(1 - 2*x)^(1/2),x)

[Out]

(605*(1 - 2*x)^(3/2))/8 - (1331*(1 - 2*x)^(1/2))/8 - (165*(1 - 2*x)^(5/2))/8 + (125*(1 - 2*x)^(7/2))/56

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sympy [A]  time = 2.05, size = 190, normalized size = 3.58 \[ \begin {cases} - \frac {25 \sqrt {5} i \left (x + \frac {3}{5}\right )^{3} \sqrt {10 x - 5}}{7} - \frac {33 \sqrt {5} i \left (x + \frac {3}{5}\right )^{2} \sqrt {10 x - 5}}{7} - \frac {242 \sqrt {5} i \left (x + \frac {3}{5}\right ) \sqrt {10 x - 5}}{35} - \frac {2662 \sqrt {5} i \sqrt {10 x - 5}}{175} & \text {for}\: \frac {10 \left |{x + \frac {3}{5}}\right |}{11} > 1 \\- \frac {25 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )^{3}}{7} - \frac {33 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )^{2}}{7} - \frac {242 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )}{35} - \frac {2662 \sqrt {5} \sqrt {5 - 10 x}}{175} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-25*sqrt(5)*I*(x + 3/5)**3*sqrt(10*x - 5)/7 - 33*sqrt(5)*I*(x + 3/5)**2*sqrt(10*x - 5)/7 - 242*sqrt
(5)*I*(x + 3/5)*sqrt(10*x - 5)/35 - 2662*sqrt(5)*I*sqrt(10*x - 5)/175, 10*Abs(x + 3/5)/11 > 1), (-25*sqrt(5)*s
qrt(5 - 10*x)*(x + 3/5)**3/7 - 33*sqrt(5)*sqrt(5 - 10*x)*(x + 3/5)**2/7 - 242*sqrt(5)*sqrt(5 - 10*x)*(x + 3/5)
/35 - 2662*sqrt(5)*sqrt(5 - 10*x)/175, True))

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