3.2139 \(\int \frac {(2+3 x) (3+5 x)}{(1-2 x)^{5/2}} \, dx\)

Optimal. Leaf size=38 \[ -\frac {15}{4} \sqrt {1-2 x}-\frac {17}{\sqrt {1-2 x}}+\frac {77}{12 (1-2 x)^{3/2}} \]

[Out]

77/12/(1-2*x)^(3/2)-17/(1-2*x)^(1/2)-15/4*(1-2*x)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 38, 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 = {77} \[ -\frac {15}{4} \sqrt {1-2 x}-\frac {17}{\sqrt {1-2 x}}+\frac {77}{12 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

77/(12*(1 - 2*x)^(3/2)) - 17/Sqrt[1 - 2*x] - (15*Sqrt[1 - 2*x])/4

Rule 77

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

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Mathematica [A]  time = 0.01, size = 23, normalized size = 0.61 \[ -\frac {45 x^2-147 x+43}{3 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*(43 - 147*x + 45*x^2)/(1 - 2*x)^(3/2)

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fricas [A]  time = 0.89, size = 31, normalized size = 0.82 \[ -\frac {{\left (45 \, x^{2} - 147 \, x + 43\right )} \sqrt {-2 \, x + 1}}{3 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(45*x^2 - 147*x + 43)*sqrt(-2*x + 1)/(4*x^2 - 4*x + 1)

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giac [A]  time = 1.19, size = 31, normalized size = 0.82 \[ -\frac {15}{4} \, \sqrt {-2 \, x + 1} - \frac {408 \, x - 127}{12 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*sqrt(-2*x + 1) - 1/12*(408*x - 127)/((2*x - 1)*sqrt(-2*x + 1))

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maple [A]  time = 0.00, size = 20, normalized size = 0.53 \[ -\frac {45 x^{2}-147 x +43}{3 \left (-2 x +1\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(45*x^2-147*x+43)/(-2*x+1)^(3/2)

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maxima [A]  time = 0.54, size = 24, normalized size = 0.63 \[ -\frac {15}{4} \, \sqrt {-2 \, x + 1} + \frac {408 \, x - 127}{12 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-15/4*sqrt(-2*x + 1) + 1/12*(408*x - 127)/(-2*x + 1)^(3/2)

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mupad [B]  time = 1.22, size = 29, normalized size = 0.76 \[ \frac {45\,{\left (2\,x-1\right )}^2-408\,x+127}{\sqrt {1-2\,x}\,\left (24\,x-12\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(45*(2*x - 1)^2 - 408*x + 127)/((1 - 2*x)^(1/2)*(24*x - 12))

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sympy [B]  time = 0.66, size = 75, normalized size = 1.97 \[ \frac {45 x^{2}}{6 x \sqrt {1 - 2 x} - 3 \sqrt {1 - 2 x}} - \frac {147 x}{6 x \sqrt {1 - 2 x} - 3 \sqrt {1 - 2 x}} + \frac {43}{6 x \sqrt {1 - 2 x} - 3 \sqrt {1 - 2 x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

45*x**2/(6*x*sqrt(1 - 2*x) - 3*sqrt(1 - 2*x)) - 147*x/(6*x*sqrt(1 - 2*x) - 3*sqrt(1 - 2*x)) + 43/(6*x*sqrt(1 -
2*x) - 3*sqrt(1 - 2*x))

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