3.95.99 \(\int \frac {45-24 x-84 x^2-24 x^3+(36+36 x) \log (-\frac {1}{1+x})}{16 x^4+16 x^5+(48 x^2+48 x^3) \log (-\frac {1}{1+x})+(36+36 x) \log ^2(-\frac {1}{1+x})} \, dx\)

Optimal. Leaf size=26 \[ \frac {\frac {5}{4}+x}{\frac {2 x^2}{3}+\log \left (-1+\frac {x}{1+x}\right )} \]

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Rubi [F]  time = 0.59, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {45-24 x-84 x^2-24 x^3+(36+36 x) \log \left (-\frac {1}{1+x}\right )}{16 x^4+16 x^5+\left (48 x^2+48 x^3\right ) \log \left (-\frac {1}{1+x}\right )+(36+36 x) \log ^2\left (-\frac {1}{1+x}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(45 - 24*x - 84*x^2 - 24*x^3 + (36 + 36*x)*Log[-(1 + x)^(-1)])/(16*x^4 + 16*x^5 + (48*x^2 + 48*x^3)*Log[-(
1 + x)^(-1)] + (36 + 36*x)*Log[-(1 + x)^(-1)]^2),x]

[Out]

9*Defer[Int][(2*x^2 + 3*Log[-(1 + x)^(-1)])^(-2), x] - 15*Defer[Int][x/(2*x^2 + 3*Log[-(1 + x)^(-1)])^2, x] -
12*Defer[Int][x^2/(2*x^2 + 3*Log[-(1 + x)^(-1)])^2, x] + (9*Defer[Int][1/((1 + x)*(2*x^2 + 3*Log[-(1 + x)^(-1)
])^2), x])/4 + 3*Defer[Int][(2*x^2 + 3*Log[-(1 + x)^(-1)])^(-1), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-3 \left (-15+8 x+28 x^2+8 x^3\right )+36 (1+x) \log \left (-\frac {1}{1+x}\right )}{4 (1+x) \left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2} \, dx\\ &=\frac {1}{4} \int \frac {-3 \left (-15+8 x+28 x^2+8 x^3\right )+36 (1+x) \log \left (-\frac {1}{1+x}\right )}{(1+x) \left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2} \, dx\\ &=\frac {1}{4} \int \left (-\frac {3 \left (-15+8 x+36 x^2+16 x^3\right )}{(1+x) \left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2}+\frac {12}{2 x^2+3 \log \left (-\frac {1}{1+x}\right )}\right ) \, dx\\ &=-\left (\frac {3}{4} \int \frac {-15+8 x+36 x^2+16 x^3}{(1+x) \left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2} \, dx\right )+3 \int \frac {1}{2 x^2+3 \log \left (-\frac {1}{1+x}\right )} \, dx\\ &=-\left (\frac {3}{4} \int \left (-\frac {12}{\left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2}+\frac {20 x}{\left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2}+\frac {16 x^2}{\left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2}-\frac {3}{(1+x) \left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2}\right ) \, dx\right )+3 \int \frac {1}{2 x^2+3 \log \left (-\frac {1}{1+x}\right )} \, dx\\ &=\frac {9}{4} \int \frac {1}{(1+x) \left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2} \, dx+3 \int \frac {1}{2 x^2+3 \log \left (-\frac {1}{1+x}\right )} \, dx+9 \int \frac {1}{\left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2} \, dx-12 \int \frac {x^2}{\left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2} \, dx-15 \int \frac {x}{\left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.86, size = 27, normalized size = 1.04 \begin {gather*} \frac {3 (5+4 x)}{4 \left (2 x^2+3 \log \left (-\frac {1}{1+x}\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(45 - 24*x - 84*x^2 - 24*x^3 + (36 + 36*x)*Log[-(1 + x)^(-1)])/(16*x^4 + 16*x^5 + (48*x^2 + 48*x^3)*
Log[-(1 + x)^(-1)] + (36 + 36*x)*Log[-(1 + x)^(-1)]^2),x]

[Out]

(3*(5 + 4*x))/(4*(2*x^2 + 3*Log[-(1 + x)^(-1)]))

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fricas [A]  time = 0.60, size = 25, normalized size = 0.96 \begin {gather*} \frac {3 \, {\left (4 \, x + 5\right )}}{4 \, {\left (2 \, x^{2} + 3 \, \log \left (-\frac {1}{x + 1}\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x+36)*log(-1/(x+1))-24*x^3-84*x^2-24*x+45)/((36*x+36)*log(-1/(x+1))^2+(48*x^3+48*x^2)*log(-1/(x
+1))+16*x^5+16*x^4),x, algorithm="fricas")

[Out]

3/4*(4*x + 5)/(2*x^2 + 3*log(-1/(x + 1)))

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giac [A]  time = 0.20, size = 48, normalized size = 1.85 \begin {gather*} -\frac {3 \, {\left (\frac {4}{x + 1} + \frac {1}{{\left (x + 1\right )}^{2}}\right )}}{4 \, {\left (\frac {4}{x + 1} - \frac {3 \, \log \left (-\frac {1}{x + 1}\right )}{{\left (x + 1\right )}^{2}} - \frac {2}{{\left (x + 1\right )}^{2}} - 2\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x+36)*log(-1/(x+1))-24*x^3-84*x^2-24*x+45)/((36*x+36)*log(-1/(x+1))^2+(48*x^3+48*x^2)*log(-1/(x
+1))+16*x^5+16*x^4),x, algorithm="giac")

[Out]

-3/4*(4/(x + 1) + 1/(x + 1)^2)/(4/(x + 1) - 3*log(-1/(x + 1))/(x + 1)^2 - 2/(x + 1)^2 - 2)

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maple [A]  time = 0.10, size = 25, normalized size = 0.96

method	result	size
norman	\(\frac {\frac {15}{4}+3 x}{2 x^{2}+3 \ln \left (-\frac {1}{x +1}\right )}\)	\(25\)
risch	\(\frac {\frac {15}{4}+3 x}{2 x^{2}+3 \ln \left (-\frac {1}{x +1}\right )}\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((36*x+36)*ln(-1/(x+1))-24*x^3-84*x^2-24*x+45)/((36*x+36)*ln(-1/(x+1))^2+(48*x^3+48*x^2)*ln(-1/(x+1))+16*x
^5+16*x^4),x,method=_RETURNVERBOSE)

[Out]

(15/4+3*x)/(2*x^2+3*ln(-1/(x+1)))

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maxima [A]  time = 0.39, size = 23, normalized size = 0.88 \begin {gather*} \frac {3 \, {\left (4 \, x + 5\right )}}{4 \, {\left (2 \, x^{2} - 3 \, \log \left (-x - 1\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x+36)*log(-1/(x+1))-24*x^3-84*x^2-24*x+45)/((36*x+36)*log(-1/(x+1))^2+(48*x^3+48*x^2)*log(-1/(x
+1))+16*x^5+16*x^4),x, algorithm="maxima")

[Out]

3/4*(4*x + 5)/(2*x^2 - 3*log(-x - 1))

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mupad [B]  time = 7.63, size = 25, normalized size = 0.96 \begin {gather*} \frac {3\,\left (4\,x+5\right )}{4\,\left (3\,\ln \left (-\frac {1}{x+1}\right )+2\,x^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(24*x - log(-1/(x + 1))*(36*x + 36) + 84*x^2 + 24*x^3 - 45)/(log(-1/(x + 1))*(48*x^2 + 48*x^3) + 16*x^4 +
16*x^5 + log(-1/(x + 1))^2*(36*x + 36)),x)

[Out]

(3*(4*x + 5))/(4*(3*log(-1/(x + 1)) + 2*x^2))

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sympy [A]  time = 0.13, size = 19, normalized size = 0.73 \begin {gather*} \frac {12 x + 15}{8 x^{2} + 12 \log {\left (- \frac {1}{x + 1} \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((36*x+36)*ln(-1/(x+1))-24*x**3-84*x**2-24*x+45)/((36*x+36)*ln(-1/(x+1))**2+(48*x**3+48*x**2)*ln(-1/
(x+1))+16*x**5+16*x**4),x)

[Out]

(12*x + 15)/(8*x**2 + 12*log(-1/(x + 1)))

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