3.3.57 \(\int \frac {(1+x^3) \log (x)}{2+x^4} \, dx\) [257]

Optimal. Leaf size=227 \[ \frac {1}{8} \left (2+i \sqrt [4]{-2}\right ) \log (x) \log \left (1-\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{16} \left (4+(1-i) 2^{3/4}\right ) \log (x) \log \left (1+\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2+\sqrt [4]{-2}\right ) \log (x) \log \left (1-\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{8} \left (2-\sqrt [4]{-2}\right ) \log (x) \log \left (1+\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{16} \left (4+(1-i) 2^{3/4}\right ) \text {Li}_2\left (-\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2+i \sqrt [4]{-2}\right ) \text {Li}_2\left (\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2-\sqrt [4]{-2}\right ) \text {Li}_2\left (-\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{8} \left (2+\sqrt [4]{-2}\right ) \text {Li}_2\left (\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right ) \]

[Out]

1/8*(2+I*(-2)^(1/4))*ln(x)*ln(1-(1/2+1/2*I)*x*2^(1/4))+1/16*(4+(1-I)*2^(3/4))*ln(x)*ln(1+(1/2+1/2*I)*x*2^(1/4)
)+1/8*(2+(-2)^(1/4))*ln(x)*ln(1-1/2*(-1)^(3/4)*x*2^(3/4))+1/8*(2-(-2)^(1/4))*ln(x)*ln(1+1/2*(-1)^(3/4)*x*2^(3/
4))+1/16*(4+(1-I)*2^(3/4))*polylog(2,(-1/2-1/2*I)*x*2^(1/4))+1/8*(2+I*(-2)^(1/4))*polylog(2,(1/2+1/2*I)*x*2^(1
/4))+1/8*(2-(-2)^(1/4))*polylog(2,-1/2*(-1)^(3/4)*x*2^(3/4))+1/8*(2+(-2)^(1/4))*polylog(2,1/2*(-1)^(3/4)*x*2^(
3/4))

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Rubi [A]
time = 0.14, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2404, 2354, 2438} \begin {gather*} \frac {1}{16} \left (4+(1-i) 2^{3/4}\right ) \text {PolyLog}\left (2,-\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2+i \sqrt [4]{-2}\right ) \text {PolyLog}\left (2,\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2-\sqrt [4]{-2}\right ) \text {PolyLog}\left (2,-\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{8} \left (2+\sqrt [4]{-2}\right ) \text {PolyLog}\left (2,\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{8} \left (2+i \sqrt [4]{-2}\right ) \log (x) \log \left (1-\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{16} \left (4+(1-i) 2^{3/4}\right ) \log (x) \log \left (1+\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2+\sqrt [4]{-2}\right ) \log (x) \log \left (1-\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{8} \left (2-\sqrt [4]{-2}\right ) \log (x) \log \left (\frac {(-1)^{3/4} x}{\sqrt [4]{2}}+1\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((1 + x^3)*Log[x])/(2 + x^4),x]

[Out]

((2 + I*(-2)^(1/4))*Log[x]*Log[1 - ((1 + I)*x)/2^(3/4)])/8 + ((4 + (1 - I)*2^(3/4))*Log[x]*Log[1 + ((1 + I)*x)
/2^(3/4)])/16 + ((2 + (-2)^(1/4))*Log[x]*Log[1 - ((-1)^(3/4)*x)/2^(1/4)])/8 + ((2 - (-2)^(1/4))*Log[x]*Log[1 +
((-1)^(3/4)*x)/2^(1/4)])/8 + ((4 + (1 - I)*2^(3/4))*PolyLog[2, ((-1 - I)*x)/2^(3/4)])/16 + ((2 + I*(-2)^(1/4)
)*PolyLog[2, ((1 + I)*x)/2^(3/4)])/8 + ((2 - (-2)^(1/4))*PolyLog[2, -(((-1)^(3/4)*x)/2^(1/4))])/8 + ((2 + (-2)
^(1/4))*PolyLog[2, ((-1)^(3/4)*x)/2^(1/4)])/8

Rule 2354

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[Log[1 + e*(x/d)]*((a +
b*Log[c*x^n])^p/e), x] - Dist[b*n*(p/e), Int[Log[1 + e*(x/d)]*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[
{a, b, c, d, e, n}, x] && IGtQ[p, 0]

Rule 2404

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[(a + b*Log[c*x^
n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, n}, x] && RationalFunctionQ[RFx, x] && IGtQ[p, 0]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin {align*} \int \frac {\left (1+x^3\right ) \log (x)}{2+x^4} \, dx &=\int \left (\frac {\left (-2+\sqrt [4]{-2}\right ) \log (x)}{8 \left (\sqrt [4]{-2}-x\right )}+\frac {\left (-2 i+\sqrt [4]{-2}\right ) \log (x)}{8 \left (\sqrt [4]{-2}-i x\right )}+\frac {\left (2 i+\sqrt [4]{-2}\right ) \log (x)}{8 \left (\sqrt [4]{-2}+i x\right )}+\frac {\left (2+\sqrt [4]{-2}\right ) \log (x)}{8 \left (\sqrt [4]{-2}+x\right )}\right ) \, dx\\ &=\frac {1}{8} \left (-2+\sqrt [4]{-2}\right ) \int \frac {\log (x)}{\sqrt [4]{-2}-x} \, dx+\frac {1}{8} \left (-2 i+\sqrt [4]{-2}\right ) \int \frac {\log (x)}{\sqrt [4]{-2}-i x} \, dx+\frac {1}{8} \left (2 i+\sqrt [4]{-2}\right ) \int \frac {\log (x)}{\sqrt [4]{-2}+i x} \, dx+\frac {1}{8} \left (2+\sqrt [4]{-2}\right ) \int \frac {\log (x)}{\sqrt [4]{-2}+x} \, dx\\ &=\frac {1}{8} \left (2+i \sqrt [4]{-2}\right ) \log (x) \log \left (1-\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2-i \sqrt [4]{-2}\right ) \log (x) \log \left (1+\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2+\sqrt [4]{-2}\right ) \log (x) \log \left (1-\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{8} \left (2-\sqrt [4]{-2}\right ) \log (x) \log \left (1+\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{8} \left (-2-\sqrt [4]{-2}\right ) \int \frac {\log \left (1-\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )}{x} \, dx+\frac {1}{8} \left (i \left (2 i-\sqrt [4]{-2}\right )\right ) \int \frac {\log \left (1-\sqrt [4]{-\frac {1}{2}} x\right )}{x} \, dx+\frac {1}{8} \left (-2+i \sqrt [4]{-2}\right ) \int \frac {\log \left (1+\sqrt [4]{-\frac {1}{2}} x\right )}{x} \, dx+\frac {1}{8} \left (-2+\sqrt [4]{-2}\right ) \int \frac {\log \left (1+\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )}{x} \, dx\\ &=\frac {1}{8} \left (2+i \sqrt [4]{-2}\right ) \log (x) \log \left (1-\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2-i \sqrt [4]{-2}\right ) \log (x) \log \left (1+\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2+\sqrt [4]{-2}\right ) \log (x) \log \left (1-\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{8} \left (2-\sqrt [4]{-2}\right ) \log (x) \log \left (1+\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{16} \left (4+(1-i) 2^{3/4}\right ) \text {Li}_2\left (-\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2+i \sqrt [4]{-2}\right ) \text {Li}_2\left (\frac {(1+i) x}{2^{3/4}}\right )+\frac {1}{8} \left (2-\sqrt [4]{-2}\right ) \text {Li}_2\left (-\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )+\frac {1}{8} \left (2+\sqrt [4]{-2}\right ) \text {Li}_2\left (\frac {(-1)^{3/4} x}{\sqrt [4]{2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.19, size = 194, normalized size = 0.85 \begin {gather*} \frac {1}{8} \left (\left (2+i \sqrt [4]{-2}\right ) \log (x) \log \left (1-\sqrt [4]{-\frac {1}{2}} x\right )+\left (2+\frac {1-i}{\sqrt [4]{2}}\right ) \log (x) \log \left (1+\sqrt [4]{-\frac {1}{2}} x\right )-\left (-2+\sqrt [4]{-2}\right ) \log (x) \log \left (1-\frac {(1-i) x}{2^{3/4}}\right )+\left (2+\sqrt [4]{-2}\right ) \log (x) \log \left (1+\frac {(1-i) x}{2^{3/4}}\right )+\left (2+\frac {1-i}{\sqrt [4]{2}}\right ) \text {Li}_2\left (-\frac {(1+i) x}{2^{3/4}}\right )+\left (2+\sqrt [4]{-2}\right ) \text {Li}_2\left (-\frac {(1-i) x}{2^{3/4}}\right )-\left (-2+\sqrt [4]{-2}\right ) \text {Li}_2\left (\frac {(1-i) x}{2^{3/4}}\right )+\left (2+i \sqrt [4]{-2}\right ) \text {Li}_2\left (\frac {(1+i) x}{2^{3/4}}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((1 + x^3)*Log[x])/(2 + x^4),x]

[Out]

((2 + I*(-2)^(1/4))*Log[x]*Log[1 - (-1/2)^(1/4)*x] + (2 + (1 - I)/2^(1/4))*Log[x]*Log[1 + (-1/2)^(1/4)*x] - (-
2 + (-2)^(1/4))*Log[x]*Log[1 - ((1 - I)*x)/2^(3/4)] + (2 + (-2)^(1/4))*Log[x]*Log[1 + ((1 - I)*x)/2^(3/4)] + (
2 + (1 - I)/2^(1/4))*PolyLog[2, ((-1 - I)*x)/2^(3/4)] + (2 + (-2)^(1/4))*PolyLog[2, ((-1 + I)*x)/2^(3/4)] - (-
2 + (-2)^(1/4))*PolyLog[2, ((1 - I)*x)/2^(3/4)] + (2 + I*(-2)^(1/4))*PolyLog[2, ((1 + I)*x)/2^(3/4)])/8

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 393 vs. \(2 (171 ) = 342\).
time = 0.04, size = 394, normalized size = 1.74

method	result	size
default	\(\frac {\left (\left (\frac {2^{\frac {3}{4}}}{2}+\frac {i 2^{\frac {3}{4}}}{2}\right )^{3}+1\right ) \left (\ln \left (x \right ) \ln \left (\frac {\frac {2^{\frac {3}{4}}}{2}+\frac {i 2^{\frac {3}{4}}}{2}-x}{\frac {2^{\frac {3}{4}}}{2}+\frac {i 2^{\frac {3}{4}}}{2}}\right )+\dilog \left (\frac {\frac {2^{\frac {3}{4}}}{2}+\frac {i 2^{\frac {3}{4}}}{2}-x}{\frac {2^{\frac {3}{4}}}{2}+\frac {i 2^{\frac {3}{4}}}{2}}\right )\right )}{4 \left (\frac {2^{\frac {3}{4}}}{2}+\frac {i 2^{\frac {3}{4}}}{2}\right )^{3}}+\frac {\left (\left (\frac {i 2^{\frac {3}{4}}}{2}-\frac {2^{\frac {3}{4}}}{2}\right )^{3}+1\right ) \left (\ln \left (x \right ) \ln \left (\frac {\frac {i 2^{\frac {3}{4}}}{2}-\frac {2^{\frac {3}{4}}}{2}-x}{\frac {i 2^{\frac {3}{4}}}{2}-\frac {2^{\frac {3}{4}}}{2}}\right )+\dilog \left (\frac {\frac {i 2^{\frac {3}{4}}}{2}-\frac {2^{\frac {3}{4}}}{2}-x}{\frac {i 2^{\frac {3}{4}}}{2}-\frac {2^{\frac {3}{4}}}{2}}\right )\right )}{4 \left (\frac {i 2^{\frac {3}{4}}}{2}-\frac {2^{\frac {3}{4}}}{2}\right )^{3}}+\frac {\left (\left (-\frac {2^{\frac {3}{4}}}{2}-\frac {i 2^{\frac {3}{4}}}{2}\right )^{3}+1\right ) \left (\ln \left (x \right ) \ln \left (\frac {-\frac {2^{\frac {3}{4}}}{2}-\frac {i 2^{\frac {3}{4}}}{2}-x}{-\frac {2^{\frac {3}{4}}}{2}-\frac {i 2^{\frac {3}{4}}}{2}}\right )+\dilog \left (\frac {-\frac {2^{\frac {3}{4}}}{2}-\frac {i 2^{\frac {3}{4}}}{2}-x}{-\frac {2^{\frac {3}{4}}}{2}-\frac {i 2^{\frac {3}{4}}}{2}}\right )\right )}{4 \left (-\frac {2^{\frac {3}{4}}}{2}-\frac {i 2^{\frac {3}{4}}}{2}\right )^{3}}+\frac {\left (\left (-\frac {i 2^{\frac {3}{4}}}{2}+\frac {2^{\frac {3}{4}}}{2}\right )^{3}+1\right ) \left (\ln \left (x \right ) \ln \left (\frac {-\frac {i 2^{\frac {3}{4}}}{2}+\frac {2^{\frac {3}{4}}}{2}-x}{-\frac {i 2^{\frac {3}{4}}}{2}+\frac {2^{\frac {3}{4}}}{2}}\right )+\dilog \left (\frac {-\frac {i 2^{\frac {3}{4}}}{2}+\frac {2^{\frac {3}{4}}}{2}-x}{-\frac {i 2^{\frac {3}{4}}}{2}+\frac {2^{\frac {3}{4}}}{2}}\right )\right )}{4 \left (-\frac {i 2^{\frac {3}{4}}}{2}+\frac {2^{\frac {3}{4}}}{2}\right )^{3}}\)	\(394\)
risch	\(\text {Expression too large to display}\)	\(1210\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3+1)*ln(x)/(x^4+2),x,method=_RETURNVERBOSE)

[Out]

1/4*((1/2*2^(3/4)+1/2*I*2^(3/4))^3+1)/(1/2*2^(3/4)+1/2*I*2^(3/4))^3*(ln(x)*ln((1/2*2^(3/4)+1/2*I*2^(3/4)-x)/(1
/2*2^(3/4)+1/2*I*2^(3/4)))+dilog((1/2*2^(3/4)+1/2*I*2^(3/4)-x)/(1/2*2^(3/4)+1/2*I*2^(3/4))))+1/4*((1/2*I*2^(3/
4)-1/2*2^(3/4))^3+1)/(1/2*I*2^(3/4)-1/2*2^(3/4))^3*(ln(x)*ln((1/2*I*2^(3/4)-1/2*2^(3/4)-x)/(1/2*I*2^(3/4)-1/2*
2^(3/4)))+dilog((1/2*I*2^(3/4)-1/2*2^(3/4)-x)/(1/2*I*2^(3/4)-1/2*2^(3/4))))+1/4*((-1/2*2^(3/4)-1/2*I*2^(3/4))^
3+1)/(-1/2*2^(3/4)-1/2*I*2^(3/4))^3*(ln(x)*ln((-1/2*2^(3/4)-1/2*I*2^(3/4)-x)/(-1/2*2^(3/4)-1/2*I*2^(3/4)))+dil
og((-1/2*2^(3/4)-1/2*I*2^(3/4)-x)/(-1/2*2^(3/4)-1/2*I*2^(3/4))))+1/4*((-1/2*I*2^(3/4)+1/2*2^(3/4))^3+1)/(-1/2*
I*2^(3/4)+1/2*2^(3/4))^3*(ln(x)*ln((-1/2*I*2^(3/4)+1/2*2^(3/4)-x)/(-1/2*I*2^(3/4)+1/2*2^(3/4)))+dilog((-1/2*I*
2^(3/4)+1/2*2^(3/4)-x)/(-1/2*I*2^(3/4)+1/2*2^(3/4))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)*log(x)/(x^4+2),x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)*log(x)/(x^4+2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (x + 1\right ) \left (x^{2} - x + 1\right ) \log {\left (x \right )}}{x^{4} + 2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**3+1)*ln(x)/(x**4+2),x)

[Out]

Integral((x + 1)*(x**2 - x + 1)*log(x)/(x**4 + 2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^3+1)*log(x)/(x^4+2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\ln \left (x\right )\,\left (x^3+1\right )}{x^4+2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(x^3 + 1))/(x^4 + 2),x)

[Out]

int((log(x)*(x^3 + 1))/(x^4 + 2), x)

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