3.16.61 \(\int \frac {(2+3 x) (3+5 x)^2}{(1-2 x)^2} \, dx\) [1561]

Optimal. Leaf size=34 \[ \frac {847}{16 (1-2 x)}+\frac {215 x}{4}+\frac {75 x^2}{8}+\frac {1133}{16} \log (1-2 x) \]

[Out]

847/16/(1-2*x)+215/4*x+75/8*x^2+1133/16*ln(1-2*x)

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Rubi [A]
time = 0.01, antiderivative size = 34, 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^2}{8}+\frac {215 x}{4}+\frac {847}{16 (1-2 x)}+\frac {1133}{16} \log (1-2 x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

847/(16*(1 - 2*x)) + (215*x)/4 + (75*x^2)/8 + (1133*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)^2} \, dx &=\int \left (\frac {215}{4}+\frac {75 x}{4}+\frac {847}{8 (-1+2 x)^2}+\frac {1133}{8 (-1+2 x)}\right ) \, dx\\ &=\frac {847}{16 (1-2 x)}+\frac {215 x}{4}+\frac {75 x^2}{8}+\frac {1133}{16} \log (1-2 x)\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 36, normalized size = 1.06 \begin {gather*} \frac {-759-3590 x+3140 x^2+600 x^3+2266 (-1+2 x) \log (1-2 x)}{-32+64 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-759 - 3590*x + 3140*x^2 + 600*x^3 + 2266*(-1 + 2*x)*Log[1 - 2*x])/(-32 + 64*x)

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Maple [A]
time = 0.10, size = 27, normalized size = 0.79

method	result	size
risch	\(\frac {75 x^{2}}{8}+\frac {215 x}{4}-\frac {847}{32 \left (-\frac {1}{2}+x \right )}+\frac {1133 \ln \left (-1+2 x \right )}{16}\)	\(25\)
default	\(\frac {75 x^{2}}{8}+\frac {215 x}{4}-\frac {847}{16 \left (-1+2 x \right )}+\frac {1133 \ln \left (-1+2 x \right )}{16}\)	\(27\)
norman	\(\frac {-\frac {1277}{8} x +\frac {785}{8} x^{2}+\frac {75}{4} x^{3}}{-1+2 x}+\frac {1133 \ln \left (-1+2 x \right )}{16}\)	\(32\)
meijerg	\(\frac {123 x}{2 \left (1-2 x \right )}+\frac {1133 \ln \left (1-2 x \right )}{16}+\frac {35 x \left (-6 x +6\right )}{3 \left (1-2 x \right )}+\frac {75 x \left (-8 x^{2}-12 x +12\right )}{32 \left (1-2 x \right )}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

75/8*x^2+215/4*x-847/16/(-1+2*x)+1133/16*ln(-1+2*x)

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Maxima [A]
time = 0.38, size = 26, normalized size = 0.76 \begin {gather*} \frac {75}{8} \, x^{2} + \frac {215}{4} \, x - \frac {847}{16 \, {\left (2 \, x - 1\right )}} + \frac {1133}{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)^2,x, algorithm="maxima")

[Out]

75/8*x^2 + 215/4*x - 847/16/(2*x - 1) + 1133/16*log(2*x - 1)

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Fricas [A]
time = 1.09, size = 37, normalized size = 1.09 \begin {gather*} \frac {300 \, x^{3} + 1570 \, x^{2} + 1133 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 860 \, x - 847}{16 \, {\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)^2,x, algorithm="fricas")

[Out]

1/16*(300*x^3 + 1570*x^2 + 1133*(2*x - 1)*log(2*x - 1) - 860*x - 847)/(2*x - 1)

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Sympy [A]
time = 0.03, size = 27, normalized size = 0.79 \begin {gather*} \frac {75 x^{2}}{8} + \frac {215 x}{4} + \frac {1133 \log {\left (2 x - 1 \right )}}{16} - \frac {847}{32 x - 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)**2,x)

[Out]

75*x**2/8 + 215*x/4 + 1133*log(2*x - 1)/16 - 847/(32*x - 16)

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Giac [A]
time = 1.41, size = 48, normalized size = 1.41 \begin {gather*} \frac {5}{32} \, {\left (2 \, x - 1\right )}^{2} {\left (\frac {202}{2 \, x - 1} + 15\right )} - \frac {847}{16 \, {\left (2 \, x - 1\right )}} - \frac {1133}{16} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\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)^2,x, algorithm="giac")

[Out]

5/32*(2*x - 1)^2*(202/(2*x - 1) + 15) - 847/16/(2*x - 1) - 1133/16*log(1/2*abs(2*x - 1)/(2*x - 1)^2)

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Mupad [B]
time = 0.03, size = 24, normalized size = 0.71 \begin {gather*} \frac {215\,x}{4}+\frac {1133\,\ln \left (x-\frac {1}{2}\right )}{16}-\frac {847}{32\,\left (x-\frac {1}{2}\right )}+\frac {75\,x^2}{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)^2,x)

[Out]

(215*x)/4 + (1133*log(x - 1/2))/16 - 847/(32*(x - 1/2)) + (75*x^2)/8

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