3.1807 \(\int \sqrt {1-2 x} (3+5 x)^2 \, dx\)

Optimal. Leaf size=40 \[ -\frac {25}{28} (1-2 x)^{7/2}+\frac {11}{2} (1-2 x)^{5/2}-\frac {121}{12} (1-2 x)^{3/2} \]

[Out]

-121/12*(1-2*x)^(3/2)+11/2*(1-2*x)^(5/2)-25/28*(1-2*x)^(7/2)

________________________________________________________________________________________

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 = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {43} \[ -\frac {25}{28} (1-2 x)^{7/2}+\frac {11}{2} (1-2 x)^{5/2}-\frac {121}{12} (1-2 x)^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[1 - 2*x]*(3 + 5*x)^2,x]

[Out]

(-121*(1 - 2*x)^(3/2))/12 + (11*(1 - 2*x)^(5/2))/2 - (25*(1 - 2*x)^(7/2))/28

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 \sqrt {1-2 x} (3+5 x)^2 \, dx &=\int \left (\frac {121}{4} \sqrt {1-2 x}-\frac {55}{2} (1-2 x)^{3/2}+\frac {25}{4} (1-2 x)^{5/2}\right ) \, dx\\ &=-\frac {121}{12} (1-2 x)^{3/2}+\frac {11}{2} (1-2 x)^{5/2}-\frac {25}{28} (1-2 x)^{7/2}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.01, size = 23, normalized size = 0.58 \[ -\frac {1}{21} (1-2 x)^{3/2} \left (75 x^2+156 x+115\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/21*((1 - 2*x)^(3/2)*(115 + 156*x + 75*x^2))

________________________________________________________________________________________

fricas [A]  time = 0.92, size = 24, normalized size = 0.60 \[ \frac {1}{21} \, {\left (150 \, x^{3} + 237 \, x^{2} + 74 \, x - 115\right )} \sqrt {-2 \, x + 1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/21*(150*x^3 + 237*x^2 + 74*x - 115)*sqrt(-2*x + 1)

________________________________________________________________________________________

giac [A]  time = 1.21, size = 42, normalized size = 1.05 \[ \frac {25}{28} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {11}{2} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {121}{12} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/28*(2*x - 1)^3*sqrt(-2*x + 1) + 11/2*(2*x - 1)^2*sqrt(-2*x + 1) - 121/12*(-2*x + 1)^(3/2)

________________________________________________________________________________________

maple [A]  time = 0.00, size = 20, normalized size = 0.50 \[ -\frac {\left (75 x^{2}+156 x +115\right ) \left (-2 x +1\right )^{\frac {3}{2}}}{21} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/21*(75*x^2+156*x+115)*(-2*x+1)^(3/2)

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 28, normalized size = 0.70 \[ -\frac {25}{28} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {11}{2} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {121}{12} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/28*(-2*x + 1)^(7/2) + 11/2*(-2*x + 1)^(5/2) - 121/12*(-2*x + 1)^(3/2)

________________________________________________________________________________________

mupad [B]  time = 0.03, size = 23, normalized size = 0.58 \[ -\frac {{\left (1-2\,x\right )}^{3/2}\,\left (924\,x+75\,{\left (2\,x-1\right )}^2+385\right )}{84} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1 - 2*x)^(1/2)*(5*x + 3)^2,x)

[Out]

-((1 - 2*x)^(3/2)*(924*x + 75*(2*x - 1)^2 + 385))/84

________________________________________________________________________________________

sympy [B]  time = 1.43, size = 187, normalized size = 4.68 \[ \begin {cases} \frac {10 \sqrt {5} i \left (x + \frac {3}{5}\right )^{3} \sqrt {10 x - 5}}{7} - \frac {11 \sqrt {5} i \left (x + \frac {3}{5}\right )^{2} \sqrt {10 x - 5}}{35} - \frac {242 \sqrt {5} i \left (x + \frac {3}{5}\right ) \sqrt {10 x - 5}}{525} - \frac {2662 \sqrt {5} i \sqrt {10 x - 5}}{2625} & \text {for}\: \frac {10 \left |{x + \frac {3}{5}}\right |}{11} > 1 \\\frac {10 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )^{3}}{7} - \frac {11 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )^{2}}{35} - \frac {242 \sqrt {5} \sqrt {5 - 10 x} \left (x + \frac {3}{5}\right )}{525} - \frac {2662 \sqrt {5} \sqrt {5 - 10 x}}{2625} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((10*sqrt(5)*I*(x + 3/5)**3*sqrt(10*x - 5)/7 - 11*sqrt(5)*I*(x + 3/5)**2*sqrt(10*x - 5)/35 - 242*sqrt
(5)*I*(x + 3/5)*sqrt(10*x - 5)/525 - 2662*sqrt(5)*I*sqrt(10*x - 5)/2625, 10*Abs(x + 3/5)/11 > 1), (10*sqrt(5)*
sqrt(5 - 10*x)*(x + 3/5)**3/7 - 11*sqrt(5)*sqrt(5 - 10*x)*(x + 3/5)**2/35 - 242*sqrt(5)*sqrt(5 - 10*x)*(x + 3/
5)/525 - 2662*sqrt(5)*sqrt(5 - 10*x)/2625, True))

________________________________________________________________________________________