3.17.51 \(\int \frac {(2+3 x) (3+5 x)^2}{(1-2 x)^3} \, dx\) [1651]

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

[Out]

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

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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 = {78} \begin {gather*} -\frac {75 x}{8}-\frac {1133}{16 (1-2 x)}+\frac {847}{32 (1-2 x)^2}-\frac {505}{16} \log (1-2 x) \end {gather*}

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 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)^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*}

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Mathematica [A]
time = 0.02, size = 34, normalized size = 0.89 \begin {gather*} \frac {1}{32} \left (-300 x+\frac {-1269+3932 x+600 x^2}{(1-2 x)^2}-1010 \log (1-2 x)\right ) \end {gather*}

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 = 0.11, size = 31, normalized size = 0.82

method	result	size
risch	\(-\frac {75 x}{8}+\frac {\frac {1133 x}{8}-\frac {1419}{32}}{\left (-1+2 x \right )^{2}}-\frac {505 \ln \left (-1+2 x \right )}{16}\)	\(27\)
default	\(-\frac {75 x}{8}+\frac {1133}{16 \left (-1+2 x \right )}+\frac {847}{32 \left (-1+2 x \right )^{2}}-\frac {505 \ln \left (-1+2 x \right )}{16}\)	\(31\)
norman	\(\frac {-\frac {361}{8} x +\frac {1719}{8} x^{2}-\frac {75}{2} x^{3}}{\left (-1+2 x \right )^{2}}-\frac {505 \ln \left (-1+2 x \right )}{16}\)	\(32\)
meijerg	\(\frac {9 x \left (2-2 x \right )}{\left (1-2 x \right )^{2}}+\frac {87 x^{2}}{2 \left (1-2 x \right )^{2}}-\frac {35 x \left (-18 x +6\right )}{6 \left (1-2 x \right )^{2}}-\frac {505 \ln \left (1-2 x \right )}{16}-\frac {75 x \left (16 x^{2}-36 x +12\right )}{32 \left (1-2 x \right )^{2}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 31, normalized size = 0.82 \begin {gather*} -\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 {gather*}

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 = 0.42, size = 47, normalized size = 1.24 \begin {gather*} -\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 {gather*}

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 = 0.05, size = 31, normalized size = 0.82 \begin {gather*} - \frac {75 x}{8} - \frac {1419 - 4532 x}{128 x^{2} - 128 x + 32} - \frac {505 \log {\left (2 x - 1 \right )}}{16} \end {gather*}

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

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Giac [A]
time = 0.49, size = 27, normalized size = 0.71 \begin {gather*} -\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 {gather*}

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))

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Mupad [B]
time = 0.03, size = 26, normalized size = 0.68 \begin {gather*} \frac {\frac {1133\,x}{32}-\frac {1419}{128}}{x^2-x+\frac {1}{4}}-\frac {505\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {75\,x}{8} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1133*x)/32 - 1419/128)/(x^2 - x + 1/4) - (505*log(x - 1/2))/16 - (75*x)/8

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