3.1216 \(\int \frac{1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx\)

Optimal. Leaf size=35 \[ -\frac{7}{3 x+2}-\frac{11}{5 x+3}+68 \log (3 x+2)-68 \log (5 x+3) \]

[Out]

-7/(2 + 3*x) - 11/(3 + 5*x) + 68*Log[2 + 3*x] - 68*Log[3 + 5*x]

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Rubi [A]  time = 0.0163061, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {77} \[ -\frac{7}{3 x+2}-\frac{11}{5 x+3}+68 \log (3 x+2)-68 \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-7/(2 + 3*x) - 11/(3 + 5*x) + 68*Log[2 + 3*x] - 68*Log[3 + 5*x]

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{1-2 x}{(2+3 x)^2 (3+5 x)^2} \, dx &=\int \left (\frac{21}{(2+3 x)^2}+\frac{204}{2+3 x}+\frac{55}{(3+5 x)^2}-\frac{340}{3+5 x}\right ) \, dx\\ &=-\frac{7}{2+3 x}-\frac{11}{3+5 x}+68 \log (2+3 x)-68 \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0174377, size = 37, normalized size = 1.06 \[ -\frac{7}{3 x+2}-\frac{11}{5 x+3}+68 \log (3 x+2)-68 \log (-3 (5 x+3)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-7/(2 + 3*x) - 11/(3 + 5*x) + 68*Log[2 + 3*x] - 68*Log[-3*(3 + 5*x)]

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Maple [A]  time = 0.007, size = 36, normalized size = 1. \begin{align*} -7\, \left ( 2+3\,x \right ) ^{-1}-11\, \left ( 3+5\,x \right ) ^{-1}+68\,\ln \left ( 2+3\,x \right ) -68\,\ln \left ( 3+5\,x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-7/(2+3*x)-11/(3+5*x)+68*ln(2+3*x)-68*ln(3+5*x)

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Maxima [A]  time = 1.06285, size = 49, normalized size = 1.4 \begin{align*} -\frac{68 \, x + 43}{15 \, x^{2} + 19 \, x + 6} - 68 \, \log \left (5 \, x + 3\right ) + 68 \, \log \left (3 \, x + 2\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(68*x + 43)/(15*x^2 + 19*x + 6) - 68*log(5*x + 3) + 68*log(3*x + 2)

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Fricas [A]  time = 1.51341, size = 149, normalized size = 4.26 \begin{align*} -\frac{68 \,{\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (5 \, x + 3\right ) - 68 \,{\left (15 \, x^{2} + 19 \, x + 6\right )} \log \left (3 \, x + 2\right ) + 68 \, x + 43}{15 \, x^{2} + 19 \, x + 6} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(68*(15*x^2 + 19*x + 6)*log(5*x + 3) - 68*(15*x^2 + 19*x + 6)*log(3*x + 2) + 68*x + 43)/(15*x^2 + 19*x + 6)

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Sympy [A]  time = 0.125139, size = 31, normalized size = 0.89 \begin{align*} - \frac{68 x + 43}{15 x^{2} + 19 x + 6} - 68 \log{\left (x + \frac{3}{5} \right )} + 68 \log{\left (x + \frac{2}{3} \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(68*x + 43)/(15*x**2 + 19*x + 6) - 68*log(x + 3/5) + 68*log(x + 2/3)

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Giac [A]  time = 2.68657, size = 51, normalized size = 1.46 \begin{align*} -\frac{11}{5 \, x + 3} + \frac{105}{\frac{1}{5 \, x + 3} + 3} + 68 \, \log \left ({\left | -\frac{1}{5 \, x + 3} - 3 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/(5*x + 3) + 105/(1/(5*x + 3) + 3) + 68*log(abs(-1/(5*x + 3) - 3))