3.17.35 \(\int \frac {(2+3 x)^3 (3+5 x)}{(1-2 x)^3} \, dx\) [1635]

Optimal. Leaf size=45 \[ \frac {3773}{64 (1-2 x)^2}-\frac {3283}{16 (1-2 x)}-\frac {1107 x}{16}-\frac {135 x^2}{16}-\frac {1071}{8} \log (1-2 x) \]

[Out]

3773/64/(1-2*x)^2-3283/16/(1-2*x)-1107/16*x-135/16*x^2-1071/8*ln(1-2*x)

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Rubi [A]
time = 0.01, antiderivative size = 45, 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 = {78} \begin {gather*} -\frac {135 x^2}{16}-\frac {1107 x}{16}-\frac {3283}{16 (1-2 x)}+\frac {3773}{64 (1-2 x)^2}-\frac {1071}{8} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3773/(64*(1 - 2*x)^2) - 3283/(16*(1 - 2*x)) - (1107*x)/16 - (135*x^2)/16 - (1071*Log[1 - 2*x])/8

Rule 78

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 (3+5 x)}{(1-2 x)^3} \, dx &=\int \left (-\frac {1107}{16}-\frac {135 x}{8}-\frac {3773}{16 (-1+2 x)^3}-\frac {3283}{8 (-1+2 x)^2}-\frac {1071}{4 (-1+2 x)}\right ) \, dx\\ &=\frac {3773}{64 (1-2 x)^2}-\frac {3283}{16 (1-2 x)}-\frac {1107 x}{16}-\frac {135 x^2}{16}-\frac {1071}{8} \log (1-2 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 46, normalized size = 1.02 \begin {gather*} -\frac {3505-6220 x-13284 x^2+7776 x^3+1080 x^4+4284 (1-2 x)^2 \log (1-2 x)}{32 (1-2 x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(3505 - 6220*x - 13284*x^2 + 7776*x^3 + 1080*x^4 + 4284*(1 - 2*x)^2*Log[1 - 2*x])/(1 - 2*x)^2

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Maple [A]
time = 0.08, size = 36, normalized size = 0.80

method	result	size
risch	\(-\frac {135 x^{2}}{16}-\frac {1107 x}{16}+\frac {\frac {3283 x}{8}-\frac {9359}{64}}{\left (-1+2 x \right )^{2}}-\frac {1071 \ln \left (-1+2 x \right )}{8}\)	\(32\)
default	\(-\frac {135 x^{2}}{16}-\frac {1107 x}{16}+\frac {3283}{16 \left (-1+2 x \right )}+\frac {3773}{64 \left (-1+2 x \right )^{2}}-\frac {1071 \ln \left (-1+2 x \right )}{8}\)	\(36\)
norman	\(\frac {-\frac {975}{4} x +\frac {3413}{4} x^{2}-243 x^{3}-\frac {135}{4} x^{4}}{\left (-1+2 x \right )^{2}}-\frac {1071 \ln \left (-1+2 x \right )}{8}\)	\(37\)
meijerg	\(\frac {12 x \left (2-2 x \right )}{\left (1-2 x \right )^{2}}+\frac {74 x^{2}}{\left (1-2 x \right )^{2}}-\frac {57 x \left (-18 x +6\right )}{4 \left (1-2 x \right )^{2}}-\frac {1071 \ln \left (1-2 x \right )}{8}-\frac {351 x \left (16 x^{2}-36 x +12\right )}{32 \left (1-2 x \right )^{2}}-\frac {27 x \left (40 x^{3}+80 x^{2}-180 x +60\right )}{32 \left (1-2 x \right )^{2}}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-135/16*x^2-1107/16*x+3283/16/(-1+2*x)+3773/64/(-1+2*x)^2-1071/8*ln(-1+2*x)

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Maxima [A]
time = 0.38, size = 36, normalized size = 0.80 \begin {gather*} -\frac {135}{16} \, x^{2} - \frac {1107}{16} \, x + \frac {49 \, {\left (536 \, x - 191\right )}}{64 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {1071}{8} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-135/16*x^2 - 1107/16*x + 49/64*(536*x - 191)/(4*x^2 - 4*x + 1) - 1071/8*log(2*x - 1)

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Fricas [A]
time = 0.42, size = 52, normalized size = 1.16 \begin {gather*} -\frac {2160 \, x^{4} + 15552 \, x^{3} - 17172 \, x^{2} + 8568 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 21836 \, x + 9359}{64 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(2160*x^4 + 15552*x^3 - 17172*x^2 + 8568*(4*x^2 - 4*x + 1)*log(2*x - 1) - 21836*x + 9359)/(4*x^2 - 4*x +
1)

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Sympy [A]
time = 0.06, size = 37, normalized size = 0.82 \begin {gather*} - \frac {135 x^{2}}{16} - \frac {1107 x}{16} - \frac {9359 - 26264 x}{256 x^{2} - 256 x + 64} - \frac {1071 \log {\left (2 x - 1 \right )}}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-135*x**2/16 - 1107*x/16 - (9359 - 26264*x)/(256*x**2 - 256*x + 64) - 1071*log(2*x - 1)/8

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Giac [A]
time = 0.83, size = 32, normalized size = 0.71 \begin {gather*} -\frac {135}{16} \, x^{2} - \frac {1107}{16} \, x + \frac {49 \, {\left (536 \, x - 191\right )}}{64 \, {\left (2 \, x - 1\right )}^{2}} - \frac {1071}{8} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-135/16*x^2 - 1107/16*x + 49/64*(536*x - 191)/(2*x - 1)^2 - 1071/8*log(abs(2*x - 1))

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Mupad [B]
time = 0.03, size = 31, normalized size = 0.69 \begin {gather*} \frac {\frac {3283\,x}{32}-\frac {9359}{256}}{x^2-x+\frac {1}{4}}-\frac {1071\,\ln \left (x-\frac {1}{2}\right )}{8}-\frac {1107\,x}{16}-\frac {135\,x^2}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((3283*x)/32 - 9359/256)/(x^2 - x + 1/4) - (1071*log(x - 1/2))/8 - (1107*x)/16 - (135*x^2)/16

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