3.1202 \(\int \frac{1-2 x}{(2+3 x)^3 (3+5 x)} \, dx\)

Optimal. Leaf size=37 \[ \frac{11}{3 x+2}+\frac{7}{6 (3 x+2)^2}-55 \log (3 x+2)+55 \log (5 x+3) \]

[Out]

7/(6*(2 + 3*x)^2) + 11/(2 + 3*x) - 55*Log[2 + 3*x] + 55*Log[3 + 5*x]

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Rubi [A]  time = 0.015847, antiderivative size = 37, 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} \[ \frac{11}{3 x+2}+\frac{7}{6 (3 x+2)^2}-55 \log (3 x+2)+55 \log (5 x+3) \]

Antiderivative was successfully verified.

[In]

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

[Out]

7/(6*(2 + 3*x)^2) + 11/(2 + 3*x) - 55*Log[2 + 3*x] + 55*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)^3 (3+5 x)} \, dx &=\int \left (-\frac{7}{(2+3 x)^3}-\frac{33}{(2+3 x)^2}-\frac{165}{2+3 x}+\frac{275}{3+5 x}\right ) \, dx\\ &=\frac{7}{6 (2+3 x)^2}+\frac{11}{2+3 x}-55 \log (2+3 x)+55 \log (3+5 x)\\ \end{align*}

Mathematica [A]  time = 0.0200619, size = 35, normalized size = 0.95 \[ \frac{198 x+139}{6 (3 x+2)^2}-55 \log (3 x+2)+55 \log (-3 (5 x+3)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(139 + 198*x)/(6*(2 + 3*x)^2) - 55*Log[2 + 3*x] + 55*Log[-3*(3 + 5*x)]

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Maple [A]  time = 0.007, size = 36, normalized size = 1. \begin{align*}{\frac{7}{6\, \left ( 2+3\,x \right ) ^{2}}}+11\, \left ( 2+3\,x \right ) ^{-1}-55\,\ln \left ( 2+3\,x \right ) +55\,\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)^3/(3+5*x),x)

[Out]

7/6/(2+3*x)^2+11/(2+3*x)-55*ln(2+3*x)+55*ln(3+5*x)

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Maxima [A]  time = 1.15115, size = 49, normalized size = 1.32 \begin{align*} \frac{198 \, x + 139}{6 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} + 55 \, \log \left (5 \, x + 3\right ) - 55 \, \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)^3/(3+5*x),x, algorithm="maxima")

[Out]

1/6*(198*x + 139)/(9*x^2 + 12*x + 4) + 55*log(5*x + 3) - 55*log(3*x + 2)

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Fricas [A]  time = 1.69812, size = 154, normalized size = 4.16 \begin{align*} \frac{330 \,{\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (5 \, x + 3\right ) - 330 \,{\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (3 \, x + 2\right ) + 198 \, x + 139}{6 \,{\left (9 \, x^{2} + 12 \, x + 4\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(330*(9*x^2 + 12*x + 4)*log(5*x + 3) - 330*(9*x^2 + 12*x + 4)*log(3*x + 2) + 198*x + 139)/(9*x^2 + 12*x +
4)

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Sympy [A]  time = 0.127344, size = 31, normalized size = 0.84 \begin{align*} \frac{198 x + 139}{54 x^{2} + 72 x + 24} + 55 \log{\left (x + \frac{3}{5} \right )} - 55 \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)**3/(3+5*x),x)

[Out]

(198*x + 139)/(54*x**2 + 72*x + 24) + 55*log(x + 3/5) - 55*log(x + 2/3)

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Giac [A]  time = 2.43496, size = 45, normalized size = 1.22 \begin{align*} \frac{198 \, x + 139}{6 \,{\left (3 \, x + 2\right )}^{2}} + 55 \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - 55 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(198*x + 139)/(3*x + 2)^2 + 55*log(abs(5*x + 3)) - 55*log(abs(3*x + 2))