\(\int \frac {x \log (1+x^2) \log (x+\sqrt {1+x^2})}{\sqrt {1+x^2}} \, dx\) [11]

Optimal result

Integrand size = 29, antiderivative size = 68 \[ \int \frac {x \log \left (1+x^2\right ) \log \left (x+\sqrt {1+x^2}\right )}{\sqrt {1+x^2}} \, dx=4 x-2 \arctan (x)-x \log \left (1+x^2\right )-2 \sqrt {1+x^2} \log \left (x+\sqrt {1+x^2}\right )+\sqrt {1+x^2} \log \left (1+x^2\right ) \log \left (x+\sqrt {1+x^2}\right ) \] Output:

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.94 \[ \int \frac {x \log \left (1+x^2\right ) \log \left (x+\sqrt {1+x^2}\right )}{\sqrt {1+x^2}} \, dx=4 x-2 \arctan (x)-2 \sqrt {1+x^2} \log \left (x+\sqrt {1+x^2}\right )+\log \left (1+x^2\right ) \left (-x+\sqrt {1+x^2} \log \left (x+\sqrt {1+x^2}\right )\right ) \] Input:

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

Output:

4*x - 2*ArcTan[x] - 2*Sqrt[1 + x^2]*Log[x + Sqrt[1 + x^2]] + Log[1 + x^2]* 
(-x + Sqrt[1 + x^2]*Log[x + Sqrt[1 + x^2]])
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {3037, 27, 2898, 262, 216, 3034, 24}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x \log \left (x^2+1\right ) \log \left (\sqrt {x^2+1}+x\right )}{\sqrt {x^2+1}} \, dx\)

\(\Big \downarrow \) 3037

\(\displaystyle -\int \log \left (x^2+1\right )dx-\int \frac {2 x \log \left (x+\sqrt {x^2+1}\right )}{\sqrt {x^2+1}}dx+\sqrt {x^2+1} \log \left (x^2+1\right ) \log \left (\sqrt {x^2+1}+x\right )\)

\(\Big \downarrow \) 27

\(\displaystyle -\int \log \left (x^2+1\right )dx-2 \int \frac {x \log \left (x+\sqrt {x^2+1}\right )}{\sqrt {x^2+1}}dx+\sqrt {x^2+1} \log \left (x^2+1\right ) \log \left (\sqrt {x^2+1}+x\right )\)

\(\Big \downarrow \) 2898

\(\displaystyle 2 \int \frac {x^2}{x^2+1}dx-2 \int \frac {x \log \left (x+\sqrt {x^2+1}\right )}{\sqrt {x^2+1}}dx-x \log \left (x^2+1\right )+\sqrt {x^2+1} \log \left (\sqrt {x^2+1}+x\right ) \log \left (x^2+1\right )\)

\(\Big \downarrow \) 262

\(\displaystyle 2 \left (x-\int \frac {1}{x^2+1}dx\right )-2 \int \frac {x \log \left (x+\sqrt {x^2+1}\right )}{\sqrt {x^2+1}}dx-x \log \left (x^2+1\right )+\sqrt {x^2+1} \log \left (\sqrt {x^2+1}+x\right ) \log \left (x^2+1\right )\)

\(\Big \downarrow \) 216

\(\displaystyle -2 \int \frac {x \log \left (x+\sqrt {x^2+1}\right )}{\sqrt {x^2+1}}dx+2 (x-\arctan (x))-x \log \left (x^2+1\right )+\sqrt {x^2+1} \log \left (x^2+1\right ) \log \left (\sqrt {x^2+1}+x\right )\)

\(\Big \downarrow \) 3034

\(\displaystyle -2 \left (\sqrt {x^2+1} \log \left (\sqrt {x^2+1}+x\right )-\int 1dx\right )+2 (x-\arctan (x))-x \log \left (x^2+1\right )+\sqrt {x^2+1} \log \left (x^2+1\right ) \log \left (\sqrt {x^2+1}+x\right )\)

\(\Big \downarrow \) 24

\(\displaystyle 2 (x-\arctan (x))-x \log \left (x^2+1\right )+\sqrt {x^2+1} \log \left (x^2+1\right ) \log \left (\sqrt {x^2+1}+x\right )-2 \left (\sqrt {x^2+1} \log \left (\sqrt {x^2+1}+x\right )-x\right )\)

Input:

Int[(x*Log[1 + x^2]*Log[x + Sqrt[1 + x^2]])/Sqrt[1 + x^2],x]
 

Output:

2*(x - ArcTan[x]) - x*Log[1 + x^2] + Sqrt[1 + x^2]*Log[1 + x^2]*Log[x + Sq 
rt[1 + x^2]] - 2*(-x + Sqrt[1 + x^2]*Log[x + Sqrt[1 + x^2]])
 

Defintions of rubi rules used

rule 24

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 2898

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

rule 3034

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Simp[Log[u]   w, x 
] - Int[SimplifyIntegrand[w*(D[u, x]/u), x], x] /; InverseFunctionFreeQ[w, 
x]] /; InverseFunctionFreeQ[u, x]
 

rule 3037

Int[Log[v_]*Log[w_]*(u_), x_Symbol] :> With[{z = IntHide[u, x]}, Simp[Log[v 
]*Log[w]   z, x] + (-Int[SimplifyIntegrand[z*Log[w]*(D[v, x]/v), x], x] - I 
nt[SimplifyIntegrand[z*Log[v]*(D[w, x]/w), x], x]) /; InverseFunctionFreeQ[ 
z, x]] /; InverseFunctionFreeQ[v, x] && InverseFunctionFreeQ[w, x]
 

Maple [F]

\[\int \frac {x \ln \left (x^{2}+1\right ) \ln \left (x +\sqrt {x^{2}+1}\right )}{\sqrt {x^{2}+1}}d x\]

Input:

int(x*ln(x^2+1)*ln(x+(x^2+1)^(1/2))/(x^2+1)^(1/2),x)
 

Output:

int(x*ln(x^2+1)*ln(x+(x^2+1)^(1/2))/(x^2+1)^(1/2),x)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.63 \[ \int \frac {x \log \left (1+x^2\right ) \log \left (x+\sqrt {1+x^2}\right )}{\sqrt {1+x^2}} \, dx=\sqrt {x^{2} + 1} {\left (\log \left (x^{2} + 1\right ) - 2\right )} \log \left (x + \sqrt {x^{2} + 1}\right ) - x \log \left (x^{2} + 1\right ) + 4 \, x - 2 \, \arctan \left (x\right ) \] Input:

integrate(x*log(x^2+1)*log(x+(x^2+1)^(1/2))/(x^2+1)^(1/2),x, algorithm="fr 
icas")
 

Output:

sqrt(x^2 + 1)*(log(x^2 + 1) - 2)*log(x + sqrt(x^2 + 1)) - x*log(x^2 + 1) + 
 4*x - 2*arctan(x)
 

Sympy [F]

\[ \int \frac {x \log \left (1+x^2\right ) \log \left (x+\sqrt {1+x^2}\right )}{\sqrt {1+x^2}} \, dx=\int \frac {x \log {\left (x + \sqrt {x^{2} + 1} \right )} \log {\left (x^{2} + 1 \right )}}{\sqrt {x^{2} + 1}}\, dx \] Input:

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

Output:

Integral(x*log(x + sqrt(x**2 + 1))*log(x**2 + 1)/sqrt(x**2 + 1), x)
 

Maxima [F]

\[ \int \frac {x \log \left (1+x^2\right ) \log \left (x+\sqrt {1+x^2}\right )}{\sqrt {1+x^2}} \, dx=\int { \frac {x \log \left (x^{2} + 1\right ) \log \left (x + \sqrt {x^{2} + 1}\right )}{\sqrt {x^{2} + 1}} \,d x } \] Input:

integrate(x*log(x^2+1)*log(x+(x^2+1)^(1/2))/(x^2+1)^(1/2),x, algorithm="ma 
xima")
 

Output:

-(2*x^2 - (x^2 + 1)*log(x^2 + 1) + 2)*log(x + sqrt(x^2 + 1))/sqrt(x^2 + 1) 
 + integrate((log(x^2 + 1) - 2)/(x^2 + sqrt(x^2 + 1)*x), x) - integrate(-( 
2*x^2 - (x^2 + 1)*log(x^2 + 1) + 2)/(sqrt(x^2 + 1)*x), x)
 

Giac [F]

\[ \int \frac {x \log \left (1+x^2\right ) \log \left (x+\sqrt {1+x^2}\right )}{\sqrt {1+x^2}} \, dx=\int { \frac {x \log \left (x^{2} + 1\right ) \log \left (x + \sqrt {x^{2} + 1}\right )}{\sqrt {x^{2} + 1}} \,d x } \] Input:

integrate(x*log(x^2+1)*log(x+(x^2+1)^(1/2))/(x^2+1)^(1/2),x, algorithm="gi 
ac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {x \log \left (1+x^2\right ) \log \left (x+\sqrt {1+x^2}\right )}{\sqrt {1+x^2}} \, dx=\int \frac {x\,\ln \left (x^2+1\right )\,\ln \left (x+\sqrt {x^2+1}\right )}{\sqrt {x^2+1}} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \frac {x \log \left (1+x^2\right ) \log \left (x+\sqrt {1+x^2}\right )}{\sqrt {1+x^2}} \, dx=\int \frac {\mathrm {log}\left (x^{2}+1\right ) \mathrm {log}\left (\sqrt {x^{2}+1}+x \right ) x}{\sqrt {x^{2}+1}}d x \] Input:

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

Output:

int((log(x**2 + 1)*log(sqrt(x**2 + 1) + x)*x)/sqrt(x**2 + 1),x)