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

Optimal. Leaf size=37 \[ \frac {7}{5 x+3}-\frac {11}{10 (5 x+3)^2}-21 \log (3 x+2)+21 \log (5 x+3) \]

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {77} \begin {gather*} \frac {7}{5 x+3}-\frac {11}{10 (5 x+3)^2}-21 \log (3 x+2)+21 \log (5 x+3) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-11/(10*(3 + 5*x)^2) + 7/(3 + 5*x) - 21*Log[2 + 3*x] + 21*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+5 x)^3} \, dx &=\int \left (-\frac {63}{2+3 x}+\frac {11}{(3+5 x)^3}-\frac {35}{(3+5 x)^2}+\frac {105}{3+5 x}\right ) \, dx\\ &=-\frac {11}{10 (3+5 x)^2}+\frac {7}{3+5 x}-21 \log (2+3 x)+21 \log (3+5 x)\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 48, normalized size = 1.30 \begin {gather*} \frac {350 x-210 (5 x+3)^2 \log (5 (3 x+2))+210 (5 x+3)^2 \log (5 x+3)+199}{10 (5 x+3)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(199 + 350*x - 210*(3 + 5*x)^2*Log[5*(2 + 3*x)] + 210*(3 + 5*x)^2*Log[3 + 5*x])/(10*(3 + 5*x)^2)

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 55, normalized size = 1.49 \begin {gather*} \frac {210 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (5 \, x + 3\right ) - 210 \, {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (3 \, x + 2\right ) + 350 \, x + 199}{10 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(210*(25*x^2 + 30*x + 9)*log(5*x + 3) - 210*(25*x^2 + 30*x + 9)*log(3*x + 2) + 350*x + 199)/(25*x^2 + 30*
x + 9)

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giac [A]  time = 1.23, size = 33, normalized size = 0.89 \begin {gather*} \frac {350 \, x + 199}{10 \, {\left (5 \, x + 3\right )}^{2}} + 21 \, \log \left ({\left | 5 \, x + 3 \right |}\right ) - 21 \, \log \left ({\left | 3 \, x + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(350*x + 199)/(5*x + 3)^2 + 21*log(abs(5*x + 3)) - 21*log(abs(3*x + 2))

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maple [A]  time = 0.01, size = 36, normalized size = 0.97 \begin {gather*} -21 \ln \left (3 x +2\right )+21 \ln \left (5 x +3\right )-\frac {11}{10 \left (5 x +3\right )^{2}}+\frac {7}{5 x +3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-11/10/(5*x+3)^2+7/(5*x+3)-21*ln(3*x+2)+21*ln(5*x+3)

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maxima [A]  time = 0.56, size = 36, normalized size = 0.97 \begin {gather*} \frac {350 \, x + 199}{10 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} + 21 \, \log \left (5 \, x + 3\right ) - 21 \, \log \left (3 \, x + 2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*(350*x + 199)/(25*x^2 + 30*x + 9) + 21*log(5*x + 3) - 21*log(3*x + 2)

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mupad [B]  time = 0.04, size = 25, normalized size = 0.68 \begin {gather*} \frac {\frac {7\,x}{5}+\frac {199}{250}}{x^2+\frac {6\,x}{5}+\frac {9}{25}}-42\,\mathrm {atanh}\left (30\,x+19\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((7*x)/5 + 199/250)/((6*x)/5 + x^2 + 9/25) - 42*atanh(30*x + 19)

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sympy [A]  time = 0.14, size = 32, normalized size = 0.86 \begin {gather*} - \frac {- 350 x - 199}{250 x^{2} + 300 x + 90} + 21 \log {\left (x + \frac {3}{5} \right )} - 21 \log {\left (x + \frac {2}{3} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-350*x - 199)/(250*x**2 + 300*x + 90) + 21*log(x + 3/5) - 21*log(x + 2/3)

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