3.2151 \(\int \frac {(3+5 x)^2}{(1-2 x)^{5/2}} \, dx\)

Optimal. Leaf size=40 \[ -\frac {25}{4} \sqrt {1-2 x}-\frac {55}{2 \sqrt {1-2 x}}+\frac {121}{12 (1-2 x)^{3/2}} \]

[Out]

121/12/(1-2*x)^(3/2)-55/2/(1-2*x)^(1/2)-25/4*(1-2*x)^(1/2)

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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 = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {43} \[ -\frac {25}{4} \sqrt {1-2 x}-\frac {55}{2 \sqrt {1-2 x}}+\frac {121}{12 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(3 + 5*x)^2/(1 - 2*x)^(5/2),x]

[Out]

121/(12*(1 - 2*x)^(3/2)) - 55/(2*Sqrt[1 - 2*x]) - (25*Sqrt[1 - 2*x])/4

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

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Mathematica [A]  time = 0.02, size = 23, normalized size = 0.58 \[ -\frac {75 x^2-240 x+71}{3 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(71 - 240*x + 75*x^2)/(1 - 2*x)^(3/2)

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fricas [A]  time = 0.99, size = 31, normalized size = 0.78 \[ -\frac {{\left (75 \, x^{2} - 240 \, x + 71\right )} \sqrt {-2 \, x + 1}}{3 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(75*x^2 - 240*x + 71)*sqrt(-2*x + 1)/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.22, size = 31, normalized size = 0.78 \[ -\frac {25}{4} \, \sqrt {-2 \, x + 1} - \frac {11 \, {\left (60 \, x - 19\right )}}{12 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/4*sqrt(-2*x + 1) - 11/12*(60*x - 19)/((2*x - 1)*sqrt(-2*x + 1))

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maple [A]  time = 0.00, size = 20, normalized size = 0.50 \[ -\frac {75 x^{2}-240 x +71}{3 \left (-2 x +1\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(75*x^2-240*x+71)/(-2*x+1)^(3/2)

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maxima [A]  time = 0.63, size = 24, normalized size = 0.60 \[ -\frac {25}{4} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (60 \, x - 19\right )}}{12 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-25/4*sqrt(-2*x + 1) + 11/12*(60*x - 19)/(-2*x + 1)^(3/2)

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mupad [B]  time = 0.03, size = 29, normalized size = 0.72 \[ \frac {75\,{\left (2\,x-1\right )}^2-660\,x+209}{\sqrt {1-2\,x}\,\left (24\,x-12\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x + 3)^2/(1 - 2*x)^(5/2),x)

[Out]

(75*(2*x - 1)^2 - 660*x + 209)/((1 - 2*x)^(1/2)*(24*x - 12))

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sympy [B]  time = 0.61, size = 75, normalized size = 1.88 \[ \frac {75 x^{2}}{6 x \sqrt {1 - 2 x} - 3 \sqrt {1 - 2 x}} - \frac {240 x}{6 x \sqrt {1 - 2 x} - 3 \sqrt {1 - 2 x}} + \frac {71}{6 x \sqrt {1 - 2 x} - 3 \sqrt {1 - 2 x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

75*x**2/(6*x*sqrt(1 - 2*x) - 3*sqrt(1 - 2*x)) - 240*x/(6*x*sqrt(1 - 2*x) - 3*sqrt(1 - 2*x)) + 71/(6*x*sqrt(1 -
2*x) - 3*sqrt(1 - 2*x))

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