3.84.80 \(\int \frac {(2+2 x^2) \log (1+x^2)+16 x^2 \log ^7(\log (1+x^2))+(1+x^2) \log (1+x^2) \log ^8(\log (1+x^2))}{(1+x^2) \log (1+x^2)} \, dx\)

Optimal. Leaf size=15 \[ 2 x+x \log ^8\left (\log \left (1+x^2\right )\right ) \]

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Rubi [F]  time = 0.62, 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 {\left (2+2 x^2\right ) \log \left (1+x^2\right )+16 x^2 \log ^7\left (\log \left (1+x^2\right )\right )+\left (1+x^2\right ) \log \left (1+x^2\right ) \log ^8\left (\log \left (1+x^2\right )\right )}{\left (1+x^2\right ) \log \left (1+x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[((2 + 2*x^2)*Log[1 + x^2] + 16*x^2*Log[Log[1 + x^2]]^7 + (1 + x^2)*Log[1 + x^2]*Log[Log[1 + x^2]]^8)/((1 +
x^2)*Log[1 + x^2]),x]

[Out]

2*x + 16*Defer[Int][Log[Log[1 + x^2]]^7/Log[1 + x^2], x] - (8*I)*Defer[Int][Log[Log[1 + x^2]]^7/((I - x)*Log[1
+ x^2]), x] - (8*I)*Defer[Int][Log[Log[1 + x^2]]^7/((I + x)*Log[1 + x^2]), x] + Defer[Int][Log[Log[1 + x^2]]^
8, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (2+\frac {16 x^2 \log ^7\left (\log \left (1+x^2\right )\right )}{\left (1+x^2\right ) \log \left (1+x^2\right )}+\log ^8\left (\log \left (1+x^2\right )\right )\right ) \, dx\\ &=2 x+16 \int \frac {x^2 \log ^7\left (\log \left (1+x^2\right )\right )}{\left (1+x^2\right ) \log \left (1+x^2\right )} \, dx+\int \log ^8\left (\log \left (1+x^2\right )\right ) \, dx\\ &=2 x+16 \int \left (\frac {\log ^7\left (\log \left (1+x^2\right )\right )}{\log \left (1+x^2\right )}-\frac {\log ^7\left (\log \left (1+x^2\right )\right )}{\left (1+x^2\right ) \log \left (1+x^2\right )}\right ) \, dx+\int \log ^8\left (\log \left (1+x^2\right )\right ) \, dx\\ &=2 x+16 \int \frac {\log ^7\left (\log \left (1+x^2\right )\right )}{\log \left (1+x^2\right )} \, dx-16 \int \frac {\log ^7\left (\log \left (1+x^2\right )\right )}{\left (1+x^2\right ) \log \left (1+x^2\right )} \, dx+\int \log ^8\left (\log \left (1+x^2\right )\right ) \, dx\\ &=2 x+16 \int \frac {\log ^7\left (\log \left (1+x^2\right )\right )}{\log \left (1+x^2\right )} \, dx-16 \int \left (\frac {i \log ^7\left (\log \left (1+x^2\right )\right )}{2 (i-x) \log \left (1+x^2\right )}+\frac {i \log ^7\left (\log \left (1+x^2\right )\right )}{2 (i+x) \log \left (1+x^2\right )}\right ) \, dx+\int \log ^8\left (\log \left (1+x^2\right )\right ) \, dx\\ &=2 x-8 i \int \frac {\log ^7\left (\log \left (1+x^2\right )\right )}{(i-x) \log \left (1+x^2\right )} \, dx-8 i \int \frac {\log ^7\left (\log \left (1+x^2\right )\right )}{(i+x) \log \left (1+x^2\right )} \, dx+16 \int \frac {\log ^7\left (\log \left (1+x^2\right )\right )}{\log \left (1+x^2\right )} \, dx+\int \log ^8\left (\log \left (1+x^2\right )\right ) \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 15, normalized size = 1.00 \begin {gather*} 2 x+x \log ^8\left (\log \left (1+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((2 + 2*x^2)*Log[1 + x^2] + 16*x^2*Log[Log[1 + x^2]]^7 + (1 + x^2)*Log[1 + x^2]*Log[Log[1 + x^2]]^8)
/((1 + x^2)*Log[1 + x^2]),x]

[Out]

2*x + x*Log[Log[1 + x^2]]^8

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fricas [A]  time = 0.57, size = 15, normalized size = 1.00 \begin {gather*} x \log \left (\log \left (x^{2} + 1\right )\right )^{8} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+1)*log(x^2+1)*log(log(x^2+1))^8+16*x^2*log(log(x^2+1))^7+(2*x^2+2)*log(x^2+1))/(x^2+1)/log(x^2
+1),x, algorithm="fricas")

[Out]

x*log(log(x^2 + 1))^8 + 2*x

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giac [A]  time = 0.76, size = 15, normalized size = 1.00 \begin {gather*} x \log \left (\log \left (x^{2} + 1\right )\right )^{8} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+1)*log(x^2+1)*log(log(x^2+1))^8+16*x^2*log(log(x^2+1))^7+(2*x^2+2)*log(x^2+1))/(x^2+1)/log(x^2
+1),x, algorithm="giac")

[Out]

x*log(log(x^2 + 1))^8 + 2*x

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maple [A]  time = 0.34, size = 16, normalized size = 1.07

method	result	size
risch	\(2 x +x \ln \left (\ln \left (x^{2}+1\right )\right )^{8}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+1)*ln(x^2+1)*ln(ln(x^2+1))^8+16*x^2*ln(ln(x^2+1))^7+(2*x^2+2)*ln(x^2+1))/(x^2+1)/ln(x^2+1),x,method=
_RETURNVERBOSE)

[Out]

2*x+x*ln(ln(x^2+1))^8

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maxima [A]  time = 0.53, size = 15, normalized size = 1.00 \begin {gather*} x \log \left (\log \left (x^{2} + 1\right )\right )^{8} + 2 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+1)*log(x^2+1)*log(log(x^2+1))^8+16*x^2*log(log(x^2+1))^7+(2*x^2+2)*log(x^2+1))/(x^2+1)/log(x^2
+1),x, algorithm="maxima")

[Out]

x*log(log(x^2 + 1))^8 + 2*x

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mupad [B]  time = 5.37, size = 13, normalized size = 0.87 \begin {gather*} x\,\left ({\ln \left (\ln \left (x^2+1\right )\right )}^8+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x^2 + 1)*(2*x^2 + 2) + 16*x^2*log(log(x^2 + 1))^7 + log(log(x^2 + 1))^8*log(x^2 + 1)*(x^2 + 1))/(log(
x^2 + 1)*(x^2 + 1)),x)

[Out]

x*(log(log(x^2 + 1))^8 + 2)

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sympy [A]  time = 0.48, size = 14, normalized size = 0.93 \begin {gather*} x \log {\left (\log {\left (x^{2} + 1 \right )} \right )}^{8} + 2 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+1)*ln(x**2+1)*ln(ln(x**2+1))**8+16*x**2*ln(ln(x**2+1))**7+(2*x**2+2)*ln(x**2+1))/(x**2+1)/ln(
x**2+1),x)

[Out]

x*log(log(x**2 + 1))**8 + 2*x

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