Integrand size = 32, antiderivative size = 150 \[ \int \frac {x^3 \sqrt {x+x^2}}{\sqrt {x^2+x \sqrt {x+x^2}}} \, dx=\frac {\sqrt {x+x^2} \left (495495-264264 x+201344 x^2-168960 x^3-1146880 x^4\right ) \sqrt {x \left (x+\sqrt {x+x^2}\right )}}{5160960 x}+\sqrt {x \left (x+\sqrt {x+x^2}\right )} \left (\frac {-165165+37752 x-18304 x^2+1387520 x^3+1146880 x^4}{5160960}-\frac {1573 \sqrt {-x+\sqrt {x+x^2}} \text {arctanh}\left (\sqrt {2} \sqrt {-x+\sqrt {x+x^2}}\right )}{16384 \sqrt {2} x}\right ) \]
1/5160960*(x^2+x)^(1/2)*(-1146880*x^4-168960*x^3+201344*x^2-264264*x+49549 5)*(x*(x+(x^2+x)^(1/2)))^(1/2)/x+(x*(x+(x^2+x)^(1/2)))^(1/2)*(2/9*x^4+271/ 1008*x^3-143/40320*x^2+1573/215040*x-1573/49152-1573/32768*2^(1/2)*(-x+(x^ 2+x)^(1/2))^(1/2)*arctanh(2^(1/2)*(-x+(x^2+x)^(1/2))^(1/2))/x)
Time = 3.34 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.19 \[ \int \frac {x^3 \sqrt {x+x^2}}{\sqrt {x^2+x \sqrt {x+x^2}}} \, dx=\frac {(1+x) \sqrt {x \left (x+\sqrt {x (1+x)}\right )} \left (2 x \left (-1146880 x^5-165165 \left (-3+\sqrt {x (1+x)}\right )+4719 x \left (49+8 \sqrt {x (1+x)}\right )-1144 x^2 \left (55+16 \sqrt {x (1+x)}\right )+5120 x^4 \left (-257+224 \sqrt {x (1+x)}\right )+128 x^3 \left (253+10840 \sqrt {x (1+x)}\right )\right )-495495 \sqrt {2} \sqrt {x (1+x)} \sqrt {-x+\sqrt {x (1+x)}} \text {arctanh}\left (\sqrt {-2 x+2 \sqrt {x (1+x)}}\right )\right )}{10321920 (x (1+x))^{3/2}} \]
((1 + x)*Sqrt[x*(x + Sqrt[x*(1 + x)])]*(2*x*(-1146880*x^5 - 165165*(-3 + S qrt[x*(1 + x)]) + 4719*x*(49 + 8*Sqrt[x*(1 + x)]) - 1144*x^2*(55 + 16*Sqrt [x*(1 + x)]) + 5120*x^4*(-257 + 224*Sqrt[x*(1 + x)]) + 128*x^3*(253 + 1084 0*Sqrt[x*(1 + x)])) - 495495*Sqrt[2]*Sqrt[x*(1 + x)]*Sqrt[-x + Sqrt[x*(1 + x)]]*ArcTanh[Sqrt[-2*x + 2*Sqrt[x*(1 + x)]]]))/(10321920*(x*(1 + x))^(3/2 ))
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.
\(\displaystyle \int \frac {x^3 \sqrt {x^2+x}}{\sqrt {x^2+\sqrt {x^2+x} x}} \, dx\) |
\(\Big \downarrow \) 2467 |
\(\displaystyle \frac {\sqrt {x^2+x} \int \frac {x^{7/2} \sqrt {x+1}}{\sqrt {x^2+\sqrt {x^2+x} x}}dx}{\sqrt {x} \sqrt {x+1}}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle \frac {2 \sqrt {x^2+x} \int \frac {x^4 \sqrt {x+1}}{\sqrt {x^2+\sqrt {x^2+x} x}}d\sqrt {x}}{\sqrt {x} \sqrt {x+1}}\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \frac {2 \sqrt {x^2+x} \int \frac {x^4 \sqrt {x+1}}{\sqrt {x^2+\sqrt {x^2+x} x}}d\sqrt {x}}{\sqrt {x} \sqrt {x+1}}\) |
3.21.84.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Fx_.)*(Px_)^(p_), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[Px^F racPart[p]/(x^(r*FracPart[p])*ExpandToSum[Px/x^r, x]^FracPart[p]) Int[x^( p*r)*ExpandToSum[Px/x^r, x]^p*Fx, x], x] /; IGtQ[r, 0]] /; FreeQ[p, x] && P olyQ[Px, x] && !IntegerQ[p] && !MonomialQ[Px, x] && !PolyQ[Fx, x]
\[\int \frac {x^{3} \sqrt {x^{2}+x}}{\sqrt {x^{2}+x \sqrt {x^{2}+x}}}d x\]
Time = 0.25 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.92 \[ \int \frac {x^3 \sqrt {x+x^2}}{\sqrt {x^2+x \sqrt {x+x^2}}} \, dx=\frac {495495 \, \sqrt {2} x \log \left (\frac {4 \, x^{2} - 2 \, \sqrt {x^{2} + \sqrt {x^{2} + x} x} {\left (\sqrt {2} x + \sqrt {2} \sqrt {x^{2} + x}\right )} + 4 \, \sqrt {x^{2} + x} x + x}{x}\right ) + 4 \, {\left (1146880 \, x^{5} + 1387520 \, x^{4} - 18304 \, x^{3} + 37752 \, x^{2} - {\left (1146880 \, x^{4} + 168960 \, x^{3} - 201344 \, x^{2} + 264264 \, x - 495495\right )} \sqrt {x^{2} + x} - 165165 \, x\right )} \sqrt {x^{2} + \sqrt {x^{2} + x} x}}{20643840 \, x} \]
1/20643840*(495495*sqrt(2)*x*log((4*x^2 - 2*sqrt(x^2 + sqrt(x^2 + x)*x)*(s qrt(2)*x + sqrt(2)*sqrt(x^2 + x)) + 4*sqrt(x^2 + x)*x + x)/x) + 4*(1146880 *x^5 + 1387520*x^4 - 18304*x^3 + 37752*x^2 - (1146880*x^4 + 168960*x^3 - 2 01344*x^2 + 264264*x - 495495)*sqrt(x^2 + x) - 165165*x)*sqrt(x^2 + sqrt(x ^2 + x)*x))/x
\[ \int \frac {x^3 \sqrt {x+x^2}}{\sqrt {x^2+x \sqrt {x+x^2}}} \, dx=\int \frac {x^{3} \sqrt {x \left (x + 1\right )}}{\sqrt {x \left (x + \sqrt {x^{2} + x}\right )}}\, dx \]
\[ \int \frac {x^3 \sqrt {x+x^2}}{\sqrt {x^2+x \sqrt {x+x^2}}} \, dx=\int { \frac {\sqrt {x^{2} + x} x^{3}}{\sqrt {x^{2} + \sqrt {x^{2} + x} x}} \,d x } \]
\[ \int \frac {x^3 \sqrt {x+x^2}}{\sqrt {x^2+x \sqrt {x+x^2}}} \, dx=\int { \frac {\sqrt {x^{2} + x} x^{3}}{\sqrt {x^{2} + \sqrt {x^{2} + x} x}} \,d x } \]
Timed out. \[ \int \frac {x^3 \sqrt {x+x^2}}{\sqrt {x^2+x \sqrt {x+x^2}}} \, dx=\int \frac {x^3\,\sqrt {x^2+x}}{\sqrt {x^2+x\,\sqrt {x^2+x}}} \,d x \]