3.40.5 \(\int \frac {-25 x-15 x^2-2 x^3+(5+27 x+15 x^2+2 x^3) \log (1+5 x+x^2)}{(1+5 x+x^2) \log ^2(1+5 x+x^2)} \, dx\)

Optimal. Leaf size=16 \[ \frac {x (5+x)}{\log \left (x \left (5+\frac {1}{x}+x\right )\right )} \]

________________________________________________________________________________________

Rubi [F]  time = 0.44, 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 {-25 x-15 x^2-2 x^3+\left (5+27 x+15 x^2+2 x^3\right ) \log \left (1+5 x+x^2\right )}{\left (1+5 x+x^2\right ) \log ^2\left (1+5 x+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-25*x - 15*x^2 - 2*x^3 + (5 + 27*x + 15*x^2 + 2*x^3)*Log[1 + 5*x + x^2])/((1 + 5*x + x^2)*Log[1 + 5*x + x
^2]^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {x \left (25+15 x+2 x^2\right )}{\left (1+5 x+x^2\right ) \log ^2\left (1+5 x+x^2\right )}+\frac {5+2 x}{\log \left (1+5 x+x^2\right )}\right ) \, dx\\ &=-\int \frac {x \left (25+15 x+2 x^2\right )}{\left (1+5 x+x^2\right ) \log ^2\left (1+5 x+x^2\right )} \, dx+\int \frac {5+2 x}{\log \left (1+5 x+x^2\right )} \, dx\\ &=-\int \left (\frac {5}{\log ^2\left (1+5 x+x^2\right )}+\frac {2 x}{\log ^2\left (1+5 x+x^2\right )}+\frac {-5-2 x}{\left (1+5 x+x^2\right ) \log ^2\left (1+5 x+x^2\right )}\right ) \, dx+\int \left (\frac {5}{\log \left (1+5 x+x^2\right )}+\frac {2 x}{\log \left (1+5 x+x^2\right )}\right ) \, dx\\ &=-\left (2 \int \frac {x}{\log ^2\left (1+5 x+x^2\right )} \, dx\right )+2 \int \frac {x}{\log \left (1+5 x+x^2\right )} \, dx-5 \int \frac {1}{\log ^2\left (1+5 x+x^2\right )} \, dx+5 \int \frac {1}{\log \left (1+5 x+x^2\right )} \, dx-\int \frac {-5-2 x}{\left (1+5 x+x^2\right ) \log ^2\left (1+5 x+x^2\right )} \, dx\\ &=-\frac {1}{\log \left (1+5 x+x^2\right )}-2 \int \frac {x}{\log ^2\left (1+5 x+x^2\right )} \, dx+2 \int \frac {x}{\log \left (1+5 x+x^2\right )} \, dx-5 \int \frac {1}{\log ^2\left (1+5 x+x^2\right )} \, dx+5 \int \frac {1}{\log \left (1+5 x+x^2\right )} \, dx\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 16, normalized size = 1.00 \begin {gather*} \frac {x (5+x)}{\log \left (1+5 x+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-25*x - 15*x^2 - 2*x^3 + (5 + 27*x + 15*x^2 + 2*x^3)*Log[1 + 5*x + x^2])/((1 + 5*x + x^2)*Log[1 + 5
*x + x^2]^2),x]

[Out]

(x*(5 + x))/Log[1 + 5*x + x^2]

________________________________________________________________________________________

fricas [A]  time = 0.84, size = 19, normalized size = 1.19 \begin {gather*} \frac {x^{2} + 5 \, x}{\log \left (x^{2} + 5 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+15*x^2+27*x+5)*log(x^2+5*x+1)-2*x^3-15*x^2-25*x)/(x^2+5*x+1)/log(x^2+5*x+1)^2,x, algorithm="
fricas")

[Out]

(x^2 + 5*x)/log(x^2 + 5*x + 1)

________________________________________________________________________________________

giac [A]  time = 0.19, size = 19, normalized size = 1.19 \begin {gather*} \frac {x^{2} + 5 \, x}{\log \left (x^{2} + 5 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+15*x^2+27*x+5)*log(x^2+5*x+1)-2*x^3-15*x^2-25*x)/(x^2+5*x+1)/log(x^2+5*x+1)^2,x, algorithm="
giac")

[Out]

(x^2 + 5*x)/log(x^2 + 5*x + 1)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 17, normalized size = 1.06

method	result	size
risch	\(\frac {\left (5+x \right ) x}{\ln \left (x^{2}+5 x +1\right )}\)	\(17\)
norman	\(\frac {x^{2}+5 x}{\ln \left (x^{2}+5 x +1\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^3+15*x^2+27*x+5)*ln(x^2+5*x+1)-2*x^3-15*x^2-25*x)/(x^2+5*x+1)/ln(x^2+5*x+1)^2,x,method=_RETURNVERBOS
E)

[Out]

(5+x)*x/ln(x^2+5*x+1)

________________________________________________________________________________________

maxima [A]  time = 0.38, size = 19, normalized size = 1.19 \begin {gather*} \frac {x^{2} + 5 \, x}{\log \left (x^{2} + 5 \, x + 1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^3+15*x^2+27*x+5)*log(x^2+5*x+1)-2*x^3-15*x^2-25*x)/(x^2+5*x+1)/log(x^2+5*x+1)^2,x, algorithm="
maxima")

[Out]

(x^2 + 5*x)/log(x^2 + 5*x + 1)

________________________________________________________________________________________

mupad [B]  time = 0.22, size = 16, normalized size = 1.00 \begin {gather*} \frac {x\,\left (x+5\right )}{\ln \left (x^2+5\,x+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(25*x - log(5*x + x^2 + 1)*(27*x + 15*x^2 + 2*x^3 + 5) + 15*x^2 + 2*x^3)/(log(5*x + x^2 + 1)^2*(5*x + x^2
+ 1)),x)

[Out]

(x*(x + 5))/log(5*x + x^2 + 1)

________________________________________________________________________________________

sympy [A]  time = 0.12, size = 15, normalized size = 0.94 \begin {gather*} \frac {x^{2} + 5 x}{\log {\left (x^{2} + 5 x + 1 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**3+15*x**2+27*x+5)*ln(x**2+5*x+1)-2*x**3-15*x**2-25*x)/(x**2+5*x+1)/ln(x**2+5*x+1)**2,x)

[Out]

(x**2 + 5*x)/log(x**2 + 5*x + 1)

________________________________________________________________________________________