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

Optimal. Leaf size=43 \[ -\frac {7}{121 (5 x+3)}-\frac {1}{110 (5 x+3)^2}-\frac {14 \log (1-2 x)}{1331}+\frac {14 \log (5 x+3)}{1331} \]

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Rubi [A]  time = 0.02, antiderivative size = 43, 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 = {77} \begin {gather*} -\frac {7}{121 (5 x+3)}-\frac {1}{110 (5 x+3)^2}-\frac {14 \log (1-2 x)}{1331}+\frac {14 \log (5 x+3)}{1331} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/(110*(3 + 5*x)^2) - 7/(121*(3 + 5*x)) - (14*Log[1 - 2*x])/1331 + (14*Log[3 + 5*x])/1331

Rule 77

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}{(1-2 x) (3+5 x)^3} \, dx &=\int \left (-\frac {28}{1331 (-1+2 x)}+\frac {1}{11 (3+5 x)^3}+\frac {35}{121 (3+5 x)^2}+\frac {70}{1331 (3+5 x)}\right ) \, dx\\ &=-\frac {1}{110 (3+5 x)^2}-\frac {7}{121 (3+5 x)}-\frac {14 \log (1-2 x)}{1331}+\frac {14 \log (3+5 x)}{1331}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 35, normalized size = 0.81 \begin {gather*} \frac {-\frac {11 (350 x+221)}{(5 x+3)^2}-140 \log (5-10 x)+140 \log (5 x+3)}{13310} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-11*(221 + 350*x))/(3 + 5*x)^2 - 140*Log[5 - 10*x] + 140*Log[3 + 5*x])/13310

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.66, size = 55, normalized size = 1.28 \begin {gather*} \frac {140 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 140 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (2 \, x - 1\right ) - 3850 \, x - 2431}{13310 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13310*(140*(25*x^2 + 30*x + 9)*log(5*x + 3) - 140*(25*x^2 + 30*x + 9)*log(2*x - 1) - 3850*x - 2431)/(25*x^2
+ 30*x + 9)

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giac [A]  time = 1.05, size = 33, normalized size = 0.77 \begin {gather*} -\frac {350 \, x + 221}{1210 \, {\left (5 \, x + 3\right )}^{2}} + \frac {14}{1331} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {14}{1331} \, \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)/(1-2*x)/(3+5*x)^3,x, algorithm="giac")

[Out]

-1/1210*(350*x + 221)/(5*x + 3)^2 + 14/1331*log(abs(5*x + 3)) - 14/1331*log(abs(2*x - 1))

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maple [A]  time = 0.01, size = 36, normalized size = 0.84 \begin {gather*} -\frac {14 \ln \left (2 x -1\right )}{1331}+\frac {14 \ln \left (5 x +3\right )}{1331}-\frac {1}{110 \left (5 x +3\right )^{2}}-\frac {7}{121 \left (5 x +3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/110/(5*x+3)^2-7/121/(5*x+3)+14/1331*ln(5*x+3)-14/1331*ln(2*x-1)

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maxima [A]  time = 0.61, size = 36, normalized size = 0.84 \begin {gather*} -\frac {350 \, x + 221}{1210 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + \frac {14}{1331} \, \log \left (5 \, x + 3\right ) - \frac {14}{1331} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1210*(350*x + 221)/(25*x^2 + 30*x + 9) + 14/1331*log(5*x + 3) - 14/1331*log(2*x - 1)

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mupad [B]  time = 0.04, size = 26, normalized size = 0.60 \begin {gather*} \frac {28\,\mathrm {atanh}\left (\frac {20\,x}{11}+\frac {1}{11}\right )}{1331}-\frac {\frac {7\,x}{605}+\frac {221}{30250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(28*atanh((20*x)/11 + 1/11))/1331 - ((7*x)/605 + 221/30250)/((6*x)/5 + x^2 + 9/25)

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sympy [A]  time = 0.15, size = 34, normalized size = 0.79 \begin {gather*} - \frac {350 x + 221}{30250 x^{2} + 36300 x + 10890} - \frac {14 \log {\left (x - \frac {1}{2} \right )}}{1331} + \frac {14 \log {\left (x + \frac {3}{5} \right )}}{1331} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(350*x + 221)/(30250*x**2 + 36300*x + 10890) - 14*log(x - 1/2)/1331 + 14*log(x + 3/5)/1331

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