3.1496 \(\int \frac{1}{(1-2 x) (2+3 x)^3 (3+5 x)} \, dx\)

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

[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

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Rubi [A]  time = 0.022631, antiderivative size = 53, normalized size of antiderivative = 1., 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} \[ \frac{111}{49 (3 x+2)}+\frac{3}{14 (3 x+2)^2}-\frac{8 \log (1-2 x)}{3773}-\frac{3897}{343} \log (3 x+2)+\frac{125}{11} \log (5 x+3) \]

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*}

Mathematica [A]  time = 0.0206484, size = 49, normalized size = 0.92 \[ \frac{\frac{8547}{3 x+2}+\frac{1617}{2 (3 x+2)^2}-8 \log (1-2 x)-42867 \log (6 x+4)+42875 \log (10 x+6)}{3773} \]

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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Maple [A]  time = 0.009, size = 44, normalized size = 0.8 \begin{align*} -{\frac{8\,\ln \left ( 2\,x-1 \right ) }{3773}}+{\frac{3}{14\, \left ( 2+3\,x \right ) ^{2}}}+{\frac{111}{98+147\,x}}-{\frac{3897\,\ln \left ( 2+3\,x \right ) }{343}}+{\frac{125\,\ln \left ( 3+5\,x \right ) }{11}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.04625, size = 59, normalized size = 1.11 \begin{align*} \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{align*}

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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Fricas [A]  time = 1.43646, size = 219, normalized size = 4.13 \begin{align*} \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{align*}

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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Sympy [A]  time = 0.181788, size = 44, normalized size = 0.83 \begin{align*} \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{align*}

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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Giac [A]  time = 3.09301, size = 57, normalized size = 1.08 \begin{align*} \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{align*}

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))