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

Optimal. Leaf size=54 \[ -\frac {4719}{1372 (1-2 x)}+\frac {1}{1029 (3 x+2)}+\frac {1331}{392 (1-2 x)^2}-\frac {33 \log (1-2 x)}{2401}+\frac {33 \log (3 x+2)}{2401} \]

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Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {88} \begin {gather*} -\frac {4719}{1372 (1-2 x)}+\frac {1}{1029 (3 x+2)}+\frac {1331}{392 (1-2 x)^2}-\frac {33 \log (1-2 x)}{2401}+\frac {33 \log (3 x+2)}{2401} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

1331/(392*(1 - 2*x)^2) - 4719/(1372*(1 - 2*x)) + 1/(1029*(2 + 3*x)) - (33*Log[1 - 2*x])/2401 + (33*Log[2 + 3*x
])/2401

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin {align*} \int \frac {(3+5 x)^3}{(1-2 x)^3 (2+3 x)^2} \, dx &=\int \left (-\frac {1331}{98 (-1+2 x)^3}-\frac {4719}{686 (-1+2 x)^2}-\frac {66}{2401 (-1+2 x)}-\frac {1}{343 (2+3 x)^2}+\frac {99}{2401 (2+3 x)}\right ) \, dx\\ &=\frac {1331}{392 (1-2 x)^2}-\frac {4719}{1372 (1-2 x)}+\frac {1}{1029 (2+3 x)}-\frac {33 \log (1-2 x)}{2401}+\frac {33 \log (2+3 x)}{2401}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 47, normalized size = 0.87 \begin {gather*} \frac {\frac {7 \left (169916 x^2+112135 x-718\right )}{(1-2 x)^2 (3 x+2)}-792 \log (1-2 x)+792 \log (6 x+4)}{57624} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((7*(-718 + 112135*x + 169916*x^2))/((1 - 2*x)^2*(2 + 3*x)) - 792*Log[1 - 2*x] + 792*Log[4 + 6*x])/57624

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(3+5 x)^3}{(1-2 x)^3 (2+3 x)^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.21, size = 75, normalized size = 1.39 \begin {gather*} \frac {1189412 \, x^{2} + 792 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (3 \, x + 2\right ) - 792 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )} \log \left (2 \, x - 1\right ) + 784945 \, x - 5026}{57624 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/57624*(1189412*x^2 + 792*(12*x^3 - 4*x^2 - 5*x + 2)*log(3*x + 2) - 792*(12*x^3 - 4*x^2 - 5*x + 2)*log(2*x -
1) + 784945*x - 5026)/(12*x^3 - 4*x^2 - 5*x + 2)

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giac [A]  time = 1.20, size = 51, normalized size = 0.94 \begin {gather*} \frac {1}{1029 \, {\left (3 \, x + 2\right )}} - \frac {1089 \, {\left (\frac {14}{3 \, x + 2} - 15\right )}}{4802 \, {\left (\frac {7}{3 \, x + 2} - 2\right )}^{2}} - \frac {33}{2401} \, \log \left ({\left | -\frac {7}{3 \, x + 2} + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1029/(3*x + 2) - 1089/4802*(14/(3*x + 2) - 15)/(7/(3*x + 2) - 2)^2 - 33/2401*log(abs(-7/(3*x + 2) + 2))

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maple [A]  time = 0.01, size = 45, normalized size = 0.83 \begin {gather*} -\frac {33 \ln \left (2 x -1\right )}{2401}+\frac {33 \ln \left (3 x +2\right )}{2401}+\frac {1}{3087 x +2058}+\frac {1331}{392 \left (2 x -1\right )^{2}}+\frac {4719}{1372 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1029/(3*x+2)+33/2401*ln(3*x+2)+1331/392/(2*x-1)^2+4719/1372/(2*x-1)-33/2401*ln(2*x-1)

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maxima [A]  time = 0.56, size = 46, normalized size = 0.85 \begin {gather*} \frac {169916 \, x^{2} + 112135 \, x - 718}{8232 \, {\left (12 \, x^{3} - 4 \, x^{2} - 5 \, x + 2\right )}} + \frac {33}{2401} \, \log \left (3 \, x + 2\right ) - \frac {33}{2401} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8232*(169916*x^2 + 112135*x - 718)/(12*x^3 - 4*x^2 - 5*x + 2) + 33/2401*log(3*x + 2) - 33/2401*log(2*x - 1)

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mupad [B]  time = 0.04, size = 38, normalized size = 0.70 \begin {gather*} \frac {66\,\mathrm {atanh}\left (\frac {12\,x}{7}+\frac {1}{7}\right )}{2401}-\frac {\frac {42479\,x^2}{24696}+\frac {112135\,x}{98784}-\frac {359}{49392}}{-x^3+\frac {x^2}{3}+\frac {5\,x}{12}-\frac {1}{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(66*atanh((12*x)/7 + 1/7))/2401 - ((112135*x)/98784 + (42479*x^2)/24696 - 359/49392)/((5*x)/12 + x^2/3 - x^3 -
1/6)

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sympy [A]  time = 0.18, size = 44, normalized size = 0.81 \begin {gather*} - \frac {- 169916 x^{2} - 112135 x + 718}{98784 x^{3} - 32928 x^{2} - 41160 x + 16464} - \frac {33 \log {\left (x - \frac {1}{2} \right )}}{2401} + \frac {33 \log {\left (x + \frac {2}{3} \right )}}{2401} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-169916*x**2 - 112135*x + 718)/(98784*x**3 - 32928*x**2 - 41160*x + 16464) - 33*log(x - 1/2)/2401 + 33*log(x
+ 2/3)/2401

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