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

Optimal. Leaf size=53 \[ \frac {111}{49 (3 x+2)}+\frac {3}{14 (3 x+2)^2}-\frac {8 \log (1-2 x)}{3773}-\frac {3897}{343} \log (3 x+2)+\frac {125}{11} \log (5 x+3) \]

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Rubi [A]  time = 0.02, antiderivative size = 53, 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 {111}{49 (3 x+2)}+\frac {3}{14 (3 x+2)^2}-\frac {8 \log (1-2 x)}{3773}-\frac {3897}{343} \log (3 x+2)+\frac {125}{11} \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

3/(14*(2 + 3*x)^2) + 111/(49*(2 + 3*x)) - (8*Log[1 - 2*x])/3773 - (3897*Log[2 + 3*x])/343 + (125*Log[3 + 5*x])
/11

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 (3+5 x)} \, dx &=\int \left (-\frac {16}{3773 (-1+2 x)}-\frac {9}{7 (2+3 x)^3}-\frac {333}{49 (2+3 x)^2}-\frac {11691}{343 (2+3 x)}+\frac {625}{11 (3+5 x)}\right ) \, dx\\ &=\frac {3}{14 (2+3 x)^2}+\frac {111}{49 (2+3 x)}-\frac {8 \log (1-2 x)}{3773}-\frac {3897}{343} \log (2+3 x)+\frac {125}{11} \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 49, normalized size = 0.92 \begin {gather*} \frac {\frac {8547}{3 x+2}+\frac {1617}{2 (3 x+2)^2}-8 \log (1-2 x)-42867 \log (6 x+4)+42875 \log (10 x+6)}{3773} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(1617/(2*(2 + 3*x)^2) + 8547/(2 + 3*x) - 8*Log[1 - 2*x] - 42867*Log[4 + 6*x] + 42875*Log[6 + 10*x])/3773

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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 (3+5 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.87, size = 73, normalized size = 1.38 \begin {gather*} \frac {85750 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (5 \, x + 3\right ) - 85734 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) - 16 \, {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (2 \, x - 1\right ) + 51282 \, x + 35805}{7546 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7546*(85750*(9*x^2 + 12*x + 4)*log(5*x + 3) - 85734*(9*x^2 + 12*x + 4)*log(3*x + 2) - 16*(9*x^2 + 12*x + 4)*
log(2*x - 1) + 51282*x + 35805)/(9*x^2 + 12*x + 4)

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giac [A]  time = 0.96, size = 42, normalized size = 0.79 \begin {gather*} \frac {3 \, {\left (222 \, x + 155\right )}}{98 \, {\left (3 \, x + 2\right )}^{2}} + \frac {125}{11} \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - \frac {3897}{343} \, \log \left ({\left | 3 \, x + 2 \right |}\right ) - \frac {8}{3773} \, \log \left ({\left | 2 \, x - 1 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/98*(222*x + 155)/(3*x + 2)^2 + 125/11*log(abs(5*x + 3)) - 3897/343*log(abs(3*x + 2)) - 8/3773*log(abs(2*x -
1))

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maple [A]  time = 0.01, size = 44, normalized size = 0.83 \begin {gather*} -\frac {8 \ln \left (2 x -1\right )}{3773}-\frac {3897 \ln \left (3 x +2\right )}{343}+\frac {125 \ln \left (5 x +3\right )}{11}+\frac {3}{14 \left (3 x +2\right )^{2}}+\frac {111}{49 \left (3 x +2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

125/11*ln(5*x+3)+3/14/(3*x+2)^2+111/49/(3*x+2)-3897/343*ln(3*x+2)-8/3773*ln(2*x-1)

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maxima [A]  time = 0.48, size = 44, normalized size = 0.83 \begin {gather*} \frac {3 \, {\left (222 \, x + 155\right )}}{98 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} + \frac {125}{11} \, \log \left (5 \, x + 3\right ) - \frac {3897}{343} \, \log \left (3 \, x + 2\right ) - \frac {8}{3773} \, \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/(3+5*x),x, algorithm="maxima")

[Out]

3/98*(222*x + 155)/(9*x^2 + 12*x + 4) + 125/11*log(5*x + 3) - 3897/343*log(3*x + 2) - 8/3773*log(2*x - 1)

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mupad [B]  time = 0.04, size = 35, normalized size = 0.66 \begin {gather*} \frac {125\,\ln \left (x+\frac {3}{5}\right )}{11}-\frac {3897\,\ln \left (x+\frac {2}{3}\right )}{343}-\frac {8\,\ln \left (x-\frac {1}{2}\right )}{3773}+\frac {\frac {37\,x}{49}+\frac {155}{294}}{x^2+\frac {4\,x}{3}+\frac {4}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(125*log(x + 3/5))/11 - (3897*log(x + 2/3))/343 - (8*log(x - 1/2))/3773 + ((37*x)/49 + 155/294)/((4*x)/3 + x^2
+ 4/9)

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sympy [A]  time = 0.21, size = 46, normalized size = 0.87 \begin {gather*} - \frac {- 666 x - 465}{882 x^{2} + 1176 x + 392} - \frac {8 \log {\left (x - \frac {1}{2} \right )}}{3773} + \frac {125 \log {\left (x + \frac {3}{5} \right )}}{11} - \frac {3897 \log {\left (x + \frac {2}{3} \right )}}{343} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-666*x - 465)/(882*x**2 + 1176*x + 392) - 8*log(x - 1/2)/3773 + 125*log(x + 3/5)/11 - 3897*log(x + 2/3)/343

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