3.16.7 \(\int \frac {(2+3 x)^2}{(1-2 x) (3+5 x)^2} \, dx\) [1507]

Optimal. Leaf size=32 \[ -\frac {1}{275 (3+5 x)}-\frac {49}{242} \log (1-2 x)+\frac {68 \log (3+5 x)}{3025} \]

[Out]

-1/275/(3+5*x)-49/242*ln(1-2*x)+68/3025*ln(3+5*x)

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Rubi [A]
time = 0.01, antiderivative size = 32, 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 = {90} \begin {gather*} -\frac {1}{275 (5 x+3)}-\frac {49}{242} \log (1-2 x)+\frac {68 \log (5 x+3)}{3025} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/275*1/(3 + 5*x) - (49*Log[1 - 2*x])/242 + (68*Log[3 + 5*x])/3025

Rule 90

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)^2}{(1-2 x) (3+5 x)^2} \, dx &=\int \left (-\frac {49}{121 (-1+2 x)}+\frac {1}{55 (3+5 x)^2}+\frac {68}{605 (3+5 x)}\right ) \, dx\\ &=-\frac {1}{275 (3+5 x)}-\frac {49}{242} \log (1-2 x)+\frac {68 \log (3+5 x)}{3025}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 30, normalized size = 0.94 \begin {gather*} \frac {-\frac {22}{3+5 x}-1225 \log (1-2 x)+136 \log (6+10 x)}{6050} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-22/(3 + 5*x) - 1225*Log[1 - 2*x] + 136*Log[6 + 10*x])/6050

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Maple [A]
time = 0.10, size = 27, normalized size = 0.84

method	result	size
risch	\(-\frac {1}{1375 \left (x +\frac {3}{5}\right )}-\frac {49 \ln \left (-1+2 x \right )}{242}+\frac {68 \ln \left (3+5 x \right )}{3025}\)	\(25\)
default	\(-\frac {49 \ln \left (-1+2 x \right )}{242}-\frac {1}{275 \left (3+5 x \right )}+\frac {68 \ln \left (3+5 x \right )}{3025}\)	\(27\)
norman	\(\frac {x}{495+825 x}-\frac {49 \ln \left (-1+2 x \right )}{242}+\frac {68 \ln \left (3+5 x \right )}{3025}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49/242*ln(-1+2*x)-1/275/(3+5*x)+68/3025*ln(3+5*x)

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Maxima [A]
time = 0.33, size = 26, normalized size = 0.81 \begin {gather*} -\frac {1}{275 \, {\left (5 \, x + 3\right )}} + \frac {68}{3025} \, \log \left (5 \, x + 3\right ) - \frac {49}{242} \, \log \left (2 \, x - 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/275/(5*x + 3) + 68/3025*log(5*x + 3) - 49/242*log(2*x - 1)

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Fricas [A]
time = 0.69, size = 37, normalized size = 1.16 \begin {gather*} \frac {136 \, {\left (5 \, x + 3\right )} \log \left (5 \, x + 3\right ) - 1225 \, {\left (5 \, x + 3\right )} \log \left (2 \, x - 1\right ) - 22}{6050 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6050*(136*(5*x + 3)*log(5*x + 3) - 1225*(5*x + 3)*log(2*x - 1) - 22)/(5*x + 3)

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Sympy [A]
time = 0.06, size = 26, normalized size = 0.81 \begin {gather*} - \frac {49 \log {\left (x - \frac {1}{2} \right )}}{242} + \frac {68 \log {\left (x + \frac {3}{5} \right )}}{3025} - \frac {1}{1375 x + 825} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49*log(x - 1/2)/242 + 68*log(x + 3/5)/3025 - 1/(1375*x + 825)

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Giac [A]
time = 0.99, size = 43, normalized size = 1.34 \begin {gather*} -\frac {1}{275 \, {\left (5 \, x + 3\right )}} + \frac {9}{50} \, \log \left (\frac {{\left | 5 \, x + 3 \right |}}{5 \, {\left (5 \, x + 3\right )}^{2}}\right ) - \frac {49}{242} \, \log \left ({\left | -\frac {11}{5 \, x + 3} + 2 \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/275/(5*x + 3) + 9/50*log(1/5*abs(5*x + 3)/(5*x + 3)^2) - 49/242*log(abs(-11/(5*x + 3) + 2))

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Mupad [B]
time = 0.09, size = 22, normalized size = 0.69 \begin {gather*} \frac {68\,\ln \left (x+\frac {3}{5}\right )}{3025}-\frac {49\,\ln \left (x-\frac {1}{2}\right )}{242}-\frac {1}{1375\,\left (x+\frac {3}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(68*log(x + 3/5))/3025 - (49*log(x - 1/2))/242 - 1/(1375*(x + 3/5))

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