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

Optimal. Leaf size=42 \[ -\frac {5}{11 (5 x+3)}-\frac {4}{847} \log (1-2 x)+\frac {9}{7} \log (3 x+2)-\frac {155}{121} \log (5 x+3) \]

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Rubi [A]  time = 0.02, antiderivative size = 42, 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 = {72} \begin {gather*} -\frac {5}{11 (5 x+3)}-\frac {4}{847} \log (1-2 x)+\frac {9}{7} \log (3 x+2)-\frac {155}{121} \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-5/(11*(3 + 5*x)) - (4*Log[1 - 2*x])/847 + (9*Log[2 + 3*x])/7 - (155*Log[3 + 5*x])/121

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {1}{(1-2 x) (2+3 x) (3+5 x)^2} \, dx &=\int \left (-\frac {8}{847 (-1+2 x)}+\frac {27}{7 (2+3 x)}+\frac {25}{11 (3+5 x)^2}-\frac {775}{121 (3+5 x)}\right ) \, dx\\ &=-\frac {5}{11 (3+5 x)}-\frac {4}{847} \log (1-2 x)+\frac {9}{7} \log (2+3 x)-\frac {155}{121} \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 38, normalized size = 0.90 \begin {gather*} \frac {1}{847} \left (-\frac {385}{5 x+3}-4 \log (1-2 x)+1089 \log (6 x+4)-1085 \log (10 x+6)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-385/(3 + 5*x) - 4*Log[1 - 2*x] + 1089*Log[4 + 6*x] - 1085*Log[6 + 10*x])/847

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.32, size = 50, normalized size = 1.19 \begin {gather*} -\frac {1085 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1089 \, {\left (5 \, x + 3\right )} \log \left (3 \, x + 2\right ) + 4 \, {\left (5 \, x + 3\right )} \log \left (2 \, x - 1\right ) + 385}{847 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/847*(1085*(5*x + 3)*log(5*x + 3) - 1089*(5*x + 3)*log(3*x + 2) + 4*(5*x + 3)*log(2*x - 1) + 385)/(5*x + 3)

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giac [A]  time = 0.94, size = 40, normalized size = 0.95 \begin {gather*} -\frac {5}{11 \, {\left (5 \, x + 3\right )}} + \frac {9}{7} \, \log \left ({\left | -\frac {1}{5 \, x + 3} - 3 \right |}\right ) - \frac {4}{847} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/11/(5*x + 3) + 9/7*log(abs(-1/(5*x + 3) - 3)) - 4/847*log(abs(-11/(5*x + 3) + 2))

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maple [A]  time = 0.01, size = 35, normalized size = 0.83 \begin {gather*} -\frac {4 \ln \left (2 x -1\right )}{847}+\frac {9 \ln \left (3 x +2\right )}{7}-\frac {155 \ln \left (5 x +3\right )}{121}-\frac {5}{11 \left (5 x +3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/11/(5*x+3)-155/121*ln(5*x+3)+9/7*ln(3*x+2)-4/847*ln(2*x-1)

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maxima [A]  time = 0.58, size = 34, normalized size = 0.81 \begin {gather*} -\frac {5}{11 \, {\left (5 \, x + 3\right )}} - \frac {155}{121} \, \log \left (5 \, x + 3\right ) + \frac {9}{7} \, \log \left (3 \, x + 2\right ) - \frac {4}{847} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/11/(5*x + 3) - 155/121*log(5*x + 3) + 9/7*log(3*x + 2) - 4/847*log(2*x - 1)

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mupad [B]  time = 1.11, size = 28, normalized size = 0.67 \begin {gather*} \frac {9\,\ln \left (x+\frac {2}{3}\right )}{7}-\frac {4\,\ln \left (x-\frac {1}{2}\right )}{847}-\frac {155\,\ln \left (x+\frac {3}{5}\right )}{121}-\frac {1}{11\,\left (x+\frac {3}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*log(x + 2/3))/7 - (4*log(x - 1/2))/847 - (155*log(x + 3/5))/121 - 1/(11*(x + 3/5))

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sympy [A]  time = 0.18, size = 36, normalized size = 0.86 \begin {gather*} - \frac {4 \log {\left (x - \frac {1}{2} \right )}}{847} - \frac {155 \log {\left (x + \frac {3}{5} \right )}}{121} + \frac {9 \log {\left (x + \frac {2}{3} \right )}}{7} - \frac {5}{55 x + 33} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*log(x - 1/2)/847 - 155*log(x + 3/5)/121 + 9*log(x + 2/3)/7 - 5/(55*x + 33)

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