3.1651 \(\int \frac{(2+3 x) (3+5 x)^2}{(1-2 x)^3} \, dx\)

Optimal. Leaf size=38 \[ -\frac{75 x}{8}-\frac{1133}{16 (1-2 x)}+\frac{847}{32 (1-2 x)^2}-\frac{505}{16} \log (1-2 x) \]

[Out]

847/(32*(1 - 2*x)^2) - 1133/(16*(1 - 2*x)) - (75*x)/8 - (505*Log[1 - 2*x])/16

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Rubi [A]  time = 0.0153867, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {77} \[ -\frac{75 x}{8}-\frac{1133}{16 (1-2 x)}+\frac{847}{32 (1-2 x)^2}-\frac{505}{16} \log (1-2 x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

847/(32*(1 - 2*x)^2) - 1133/(16*(1 - 2*x)) - (75*x)/8 - (505*Log[1 - 2*x])/16

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)^2}{(1-2 x)^3} \, dx &=\int \left (-\frac{75}{8}-\frac{847}{8 (-1+2 x)^3}-\frac{1133}{8 (-1+2 x)^2}-\frac{505}{8 (-1+2 x)}\right ) \, dx\\ &=\frac{847}{32 (1-2 x)^2}-\frac{1133}{16 (1-2 x)}-\frac{75 x}{8}-\frac{505}{16} \log (1-2 x)\\ \end{align*}

Mathematica [A]  time = 0.0217107, size = 34, normalized size = 0.89 \[ \frac{1}{32} \left (\frac{600 x^2+3932 x-1269}{(1-2 x)^2}-300 x-1010 \log (1-2 x)\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-300*x + (-1269 + 3932*x + 600*x^2)/(1 - 2*x)^2 - 1010*Log[1 - 2*x])/32

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Maple [A]  time = 0.005, size = 31, normalized size = 0.8 \begin{align*} -{\frac{75\,x}{8}}-{\frac{505\,\ln \left ( 2\,x-1 \right ) }{16}}+{\frac{847}{32\, \left ( 2\,x-1 \right ) ^{2}}}+{\frac{1133}{32\,x-16}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-75/8*x-505/16*ln(2*x-1)+847/32/(2*x-1)^2+1133/16/(2*x-1)

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Maxima [A]  time = 1.33125, size = 42, normalized size = 1.11 \begin{align*} -\frac{75}{8} \, x + \frac{11 \,{\left (412 \, x - 129\right )}}{32 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac{505}{16} \, \log \left (2 \, x - 1\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-75/8*x + 11/32*(412*x - 129)/(4*x^2 - 4*x + 1) - 505/16*log(2*x - 1)

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Fricas [A]  time = 1.54826, size = 136, normalized size = 3.58 \begin{align*} -\frac{1200 \, x^{3} - 1200 \, x^{2} + 1010 \,{\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 4232 \, x + 1419}{32 \,{\left (4 \, x^{2} - 4 \, x + 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(1200*x^3 - 1200*x^2 + 1010*(4*x^2 - 4*x + 1)*log(2*x - 1) - 4232*x + 1419)/(4*x^2 - 4*x + 1)

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Sympy [A]  time = 0.11723, size = 29, normalized size = 0.76 \begin{align*} - \frac{75 x}{8} + \frac{4532 x - 1419}{128 x^{2} - 128 x + 32} - \frac{505 \log{\left (2 x - 1 \right )}}{16} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-75*x/8 + (4532*x - 1419)/(128*x**2 - 128*x + 32) - 505*log(2*x - 1)/16

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Giac [A]  time = 3.04248, size = 36, normalized size = 0.95 \begin{align*} -\frac{75}{8} \, x + \frac{11 \,{\left (412 \, x - 129\right )}}{32 \,{\left (2 \, x - 1\right )}^{2}} - \frac{505}{16} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-75/8*x + 11/32*(412*x - 129)/(2*x - 1)^2 - 505/16*log(abs(2*x - 1))