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

Optimal. Leaf size=37 \[ \frac {27 x}{20}+\frac {343}{88 (1-2 x)}+\frac {392}{121} \log (1-2 x)+\frac {\log (5 x+3)}{3025} \]

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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 = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {88} \begin {gather*} \frac {27 x}{20}+\frac {343}{88 (1-2 x)}+\frac {392}{121} \log (1-2 x)+\frac {\log (5 x+3)}{3025} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

343/(88*(1 - 2*x)) + (27*x)/20 + (392*Log[1 - 2*x])/121 + Log[3 + 5*x]/3025

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rubi steps

\begin {align*} \int \frac {(2+3 x)^3}{(1-2 x)^2 (3+5 x)} \, dx &=\int \left (\frac {27}{20}+\frac {343}{44 (-1+2 x)^2}+\frac {784}{121 (-1+2 x)}+\frac {1}{605 (3+5 x)}\right ) \, dx\\ &=\frac {343}{88 (1-2 x)}+\frac {27 x}{20}+\frac {392}{121} \log (1-2 x)+\frac {\log (3+5 x)}{3025}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 37, normalized size = 1.00 \begin {gather*} \frac {6534 (5 x+3)+\frac {94325}{1-2 x}+78400 \log (5-10 x)+8 \log (5 x+3)}{24200} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(94325/(1 - 2*x) + 6534*(3 + 5*x) + 78400*Log[5 - 10*x] + 8*Log[3 + 5*x])/24200

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

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 1.37, size = 45, normalized size = 1.22 \begin {gather*} \frac {65340 \, x^{2} + 8 \, {\left (2 \, x - 1\right )} \log \left (5 \, x + 3\right ) + 78400 \, {\left (2 \, x - 1\right )} \log \left (2 \, x - 1\right ) - 32670 \, x - 94325}{24200 \, {\left (2 \, x - 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24200*(65340*x^2 + 8*(2*x - 1)*log(5*x + 3) + 78400*(2*x - 1)*log(2*x - 1) - 32670*x - 94325)/(2*x - 1)

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giac [A]  time = 1.19, size = 47, normalized size = 1.27 \begin {gather*} \frac {27}{20} \, x - \frac {343}{88 \, {\left (2 \, x - 1\right )}} - \frac {81}{25} \, \log \left (\frac {{\left | 2 \, x - 1 \right |}}{2 \, {\left (2 \, x - 1\right )}^{2}}\right ) + \frac {1}{3025} \, \log \left ({\left | -\frac {11}{2 \, x - 1} - 5 \right |}\right ) - \frac {27}{40} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/20*x - 343/88/(2*x - 1) - 81/25*log(1/2*abs(2*x - 1)/(2*x - 1)^2) + 1/3025*log(abs(-11/(2*x - 1) - 5)) - 27
/40

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maple [A]  time = 0.01, size = 30, normalized size = 0.81 \begin {gather*} \frac {27 x}{20}+\frac {392 \ln \left (2 x -1\right )}{121}+\frac {\ln \left (5 x +3\right )}{3025}-\frac {343}{88 \left (2 x -1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/20*x+1/3025*ln(5*x+3)-343/88/(2*x-1)+392/121*ln(2*x-1)

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maxima [A]  time = 0.58, size = 29, normalized size = 0.78 \begin {gather*} \frac {27}{20} \, x - \frac {343}{88 \, {\left (2 \, x - 1\right )}} + \frac {1}{3025} \, \log \left (5 \, x + 3\right ) + \frac {392}{121} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/20*x - 343/88/(2*x - 1) + 1/3025*log(5*x + 3) + 392/121*log(2*x - 1)

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mupad [B]  time = 0.04, size = 25, normalized size = 0.68 \begin {gather*} \frac {27\,x}{20}+\frac {392\,\ln \left (x-\frac {1}{2}\right )}{121}+\frac {\ln \left (x+\frac {3}{5}\right )}{3025}-\frac {343}{176\,\left (x-\frac {1}{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(27*x)/20 + (392*log(x - 1/2))/121 + log(x + 3/5)/3025 - 343/(176*(x - 1/2))

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sympy [A]  time = 0.14, size = 29, normalized size = 0.78 \begin {gather*} \frac {27 x}{20} + \frac {392 \log {\left (x - \frac {1}{2} \right )}}{121} + \frac {\log {\left (x + \frac {3}{5} \right )}}{3025} - \frac {343}{176 x - 88} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27*x/20 + 392*log(x - 1/2)/121 + log(x + 3/5)/3025 - 343/(176*x - 88)

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