3.17.64 \(\int \frac {(2+3 x) (3+5 x)^3}{(1-2 x)^3} \, dx\) [1664]

Optimal. Leaf size=45 \[ \frac {9317}{64 (1-2 x)^2}-\frac {8349}{16 (1-2 x)}-\frac {2975 x}{16}-\frac {375 x^2}{16}-\frac {2805}{8} \log (1-2 x) \]

[Out]

9317/64/(1-2*x)^2-8349/16/(1-2*x)-2975/16*x-375/16*x^2-2805/8*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 {375 x^2}{16}-\frac {2975 x}{16}-\frac {8349}{16 (1-2 x)}+\frac {9317}{64 (1-2 x)^2}-\frac {2805}{8} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

9317/(64*(1 - 2*x)^2) - 8349/(16*(1 - 2*x)) - (2975*x)/16 - (375*x^2)/16 - (2805*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+5 x)^3}{(1-2 x)^3} \, dx &=\int \left (-\frac {2975}{16}-\frac {375 x}{8}-\frac {9317}{16 (-1+2 x)^3}-\frac {8349}{8 (-1+2 x)^2}-\frac {2805}{4 (-1+2 x)}\right ) \, dx\\ &=\frac {9317}{64 (1-2 x)^2}-\frac {8349}{16 (1-2 x)}-\frac {2975 x}{16}-\frac {375 x^2}{16}-\frac {2805}{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 {8877-14796 x-35700 x^2+20800 x^3+3000 x^4+11220 (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 + 5*x)^3)/(1 - 2*x)^3,x]

[Out]

-1/32*(8877 - 14796*x - 35700*x^2 + 20800*x^3 + 3000*x^4 + 11220*(1 - 2*x)^2*Log[1 - 2*x])/(1 - 2*x)^2

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Maple [A]
time = 0.11, size = 36, normalized size = 0.80

method	result	size
risch	\(-\frac {375 x^{2}}{16}-\frac {2975 x}{16}+\frac {\frac {8349 x}{8}-\frac {24079}{64}}{\left (-1+2 x \right )^{2}}-\frac {2805 \ln \left (-1+2 x \right )}{8}\)	\(32\)
default	\(-\frac {375 x^{2}}{16}-\frac {2975 x}{16}+\frac {8349}{16 \left (-1+2 x \right )}+\frac {9317}{64 \left (-1+2 x \right )^{2}}-\frac {2805 \ln \left (-1+2 x \right )}{8}\)	\(36\)
norman	\(\frac {-\frac {2589}{4} x +\frac {8901}{4} x^{2}-650 x^{3}-\frac {375}{4} x^{4}}{\left (-1+2 x \right )^{2}}-\frac {2805 \ln \left (-1+2 x \right )}{8}\)	\(37\)
meijerg	\(\frac {27 x \left (2-2 x \right )}{\left (1-2 x \right )^{2}}+\frac {351 x^{2}}{2 \left (1-2 x \right )^{2}}-\frac {285 x \left (-18 x +6\right )}{8 \left (1-2 x \right )^{2}}-\frac {2805 \ln \left (1-2 x \right )}{8}-\frac {925 x \left (16 x^{2}-36 x +12\right )}{32 \left (1-2 x \right )^{2}}-\frac {75 x \left (40 x^{3}+80 x^{2}-180 x +60\right )}{32 \left (1-2 x \right )^{2}}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-375/16*x^2-2975/16*x+8349/16/(-1+2*x)+9317/64/(-1+2*x)^2-2805/8*ln(-1+2*x)

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Maxima [A]
time = 0.30, size = 36, normalized size = 0.80 \begin {gather*} -\frac {375}{16} \, x^{2} - \frac {2975}{16} \, x + \frac {121 \, {\left (552 \, x - 199\right )}}{64 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} - \frac {2805}{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+5*x)^3/(1-2*x)^3,x, algorithm="maxima")

[Out]

-375/16*x^2 - 2975/16*x + 121/64*(552*x - 199)/(4*x^2 - 4*x + 1) - 2805/8*log(2*x - 1)

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Fricas [A]
time = 0.45, size = 52, normalized size = 1.16 \begin {gather*} -\frac {6000 \, x^{4} + 41600 \, x^{3} - 46100 \, x^{2} + 22440 \, {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (2 \, x - 1\right ) - 54892 \, x + 24079}{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+5*x)^3/(1-2*x)^3,x, algorithm="fricas")

[Out]

-1/64*(6000*x^4 + 41600*x^3 - 46100*x^2 + 22440*(4*x^2 - 4*x + 1)*log(2*x - 1) - 54892*x + 24079)/(4*x^2 - 4*x
+ 1)

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Sympy [A]
time = 0.05, size = 37, normalized size = 0.82 \begin {gather*} - \frac {375 x^{2}}{16} - \frac {2975 x}{16} - \frac {24079 - 66792 x}{256 x^{2} - 256 x + 64} - \frac {2805 \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+5*x)**3/(1-2*x)**3,x)

[Out]

-375*x**2/16 - 2975*x/16 - (24079 - 66792*x)/(256*x**2 - 256*x + 64) - 2805*log(2*x - 1)/8

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Giac [A]
time = 1.09, size = 32, normalized size = 0.71 \begin {gather*} -\frac {375}{16} \, x^{2} - \frac {2975}{16} \, x + \frac {121 \, {\left (552 \, x - 199\right )}}{64 \, {\left (2 \, x - 1\right )}^{2}} - \frac {2805}{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+5*x)^3/(1-2*x)^3,x, algorithm="giac")

[Out]

-375/16*x^2 - 2975/16*x + 121/64*(552*x - 199)/(2*x - 1)^2 - 2805/8*log(abs(2*x - 1))

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Mupad [B]
time = 0.03, size = 31, normalized size = 0.69 \begin {gather*} \frac {\frac {8349\,x}{32}-\frac {24079}{256}}{x^2-x+\frac {1}{4}}-\frac {2805\,\ln \left (x-\frac {1}{2}\right )}{8}-\frac {2975\,x}{16}-\frac {375\,x^2}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((8349*x)/32 - 24079/256)/(x^2 - x + 1/4) - (2805*log(x - 1/2))/8 - (2975*x)/16 - (375*x^2)/16

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