\(\int \frac {(1+x)^{3/2} (1-x+x^2)^{3/2}}{x} \, dx\) [483]

Optimal result

Integrand size = 23, antiderivative size = 94 \[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x} \, dx=\frac {2}{3} \sqrt {1+x} \sqrt {1-x+x^2}+\frac {2}{9} \sqrt {1+x} \sqrt {1-x+x^2} \left (1+x^3\right )-\frac {2 \sqrt {1+x} \sqrt {1-x+x^2} \text {arctanh}\left (\sqrt {1+x^3}\right )}{3 \sqrt {1+x^3}} \] Output:

2/3*(1+x)^(1/2)*(x^2-x+1)^(1/2)+2/9*(1+x)^(1/2)*(x^2-x+1)^(1/2)*(x^3+1)-2/ 
3*(1+x)^(1/2)*(x^2-x+1)^(1/2)*arctanh((x^3+1)^(1/2))/(x^3+1)^(1/2)
 

Mathematica [A] (verified)

Time = 10.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.56 \[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x} \, dx=\frac {2}{9} \left (\sqrt {1+x} \sqrt {1-x+x^2} \left (4+x^3\right )-3 \text {arctanh}\left (\sqrt {1+x} \sqrt {1-x+x^2}\right )\right ) \] Input:

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

Output:

(2*(Sqrt[1 + x]*Sqrt[1 - x + x^2]*(4 + x^3) - 3*ArcTanh[Sqrt[1 + x]*Sqrt[1 
 - x + x^2]]))/9
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.73, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {1210, 798, 60, 60, 73, 220}

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.

\(\displaystyle \int \frac {(x+1)^{3/2} \left (x^2-x+1\right )^{3/2}}{x} \, dx\)

\(\Big \downarrow \) 1210

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \int \frac {\left (x^3+1\right )^{3/2}}{x}dx}{\sqrt {x^3+1}}\)

\(\Big \downarrow \) 798

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \int \frac {\left (x^3+1\right )^{3/2}}{x^3}dx^3}{3 \sqrt {x^3+1}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\int \frac {\sqrt {x^3+1}}{x^3}dx^3+\frac {2}{3} \left (x^3+1\right )^{3/2}\right )}{3 \sqrt {x^3+1}}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (\int \frac {1}{x^3 \sqrt {x^3+1}}dx^3+\frac {2}{3} \left (x^3+1\right )^{3/2}+2 \sqrt {x^3+1}\right )}{3 \sqrt {x^3+1}}\)

\(\Big \downarrow \) 73

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (2 \int \frac {1}{x^6-1}d\sqrt {x^3+1}+\frac {2}{3} \left (x^3+1\right )^{3/2}+2 \sqrt {x^3+1}\right )}{3 \sqrt {x^3+1}}\)

\(\Big \downarrow \) 220

\(\displaystyle \frac {\sqrt {x+1} \sqrt {x^2-x+1} \left (-2 \text {arctanh}\left (\sqrt {x^3+1}\right )+\frac {2}{3} \left (x^3+1\right )^{3/2}+2 \sqrt {x^3+1}\right )}{3 \sqrt {x^3+1}}\)

Input:

Int[((1 + x)^(3/2)*(1 - x + x^2)^(3/2))/x,x]
 

Output:

(Sqrt[1 + x]*Sqrt[1 - x + x^2]*(2*Sqrt[1 + x^3] + (2*(1 + x^3)^(3/2))/3 - 
2*ArcTanh[Sqrt[1 + x^3]]))/(3*Sqrt[1 + x^3])
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, x]
 

rule 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

rule 220

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

rule 798

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[1/n   Subst 
[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p, x], x, x^n], x] /; FreeQ[{a, 
b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]
 

rule 1210

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_.)*((a_) + (b_.)*(x_) 
 + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^FracPart[p]*((a + b*x + 
c*x^2)^FracPart[p]/(a*d + c*e*x^3)^FracPart[p])   Int[(f + g*x)^n*(a*d + c* 
e*x^3)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && EqQ[b*d + a*e, 
 0] && EqQ[c*d + b*e, 0] && EqQ[m, p]
 

Maple [A] (verified)

Time = 2.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.61

method	result	size
default	\(-\frac {2 \sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \left (-x^{3} \sqrt {x^{3}+1}+3 \,\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right )-4 \sqrt {x^{3}+1}\right )}{9 \sqrt {x^{3}+1}}\)	\(57\)
elliptic	\(\frac {\sqrt {x +1}\, \sqrt {x^{2}-x +1}\, \sqrt {\left (x +1\right ) \left (x^{2}-x +1\right )}\, \left (\frac {2 x^{3} \sqrt {x^{3}+1}}{9}+\frac {8 \sqrt {x^{3}+1}}{9}-\frac {2 \,\operatorname {arctanh}\left (\sqrt {x^{3}+1}\right )}{3}\right )}{x^{3}+1}\)	\(70\)

Input:

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

Output:

-2/9*(x+1)^(1/2)*(x^2-x+1)^(1/2)*(-x^3*(x^3+1)^(1/2)+3*arctanh((x^3+1)^(1/ 
2))-4*(x^3+1)^(1/2))/(x^3+1)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.69 \[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x} \, dx=\frac {2}{9} \, {\left (x^{3} + 4\right )} \sqrt {x^{2} - x + 1} \sqrt {x + 1} - \frac {1}{3} \, \log \left (\sqrt {x^{2} - x + 1} \sqrt {x + 1} + 1\right ) + \frac {1}{3} \, \log \left (\sqrt {x^{2} - x + 1} \sqrt {x + 1} - 1\right ) \] Input:

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

Output:

2/9*(x^3 + 4)*sqrt(x^2 - x + 1)*sqrt(x + 1) - 1/3*log(sqrt(x^2 - x + 1)*sq 
rt(x + 1) + 1) + 1/3*log(sqrt(x^2 - x + 1)*sqrt(x + 1) - 1)
 

Sympy [F]

\[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x} \, dx=\int \frac {\left (x + 1\right )^{\frac {3}{2}} \left (x^{2} - x + 1\right )^{\frac {3}{2}}}{x}\, dx \] Input:

integrate((1+x)**(3/2)*(x**2-x+1)**(3/2)/x,x)
 

Output:

Integral((x + 1)**(3/2)*(x**2 - x + 1)**(3/2)/x, x)
 

Maxima [F]

\[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x} \, dx=\int { \frac {{\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}}}{x} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x} \, dx=\int { \frac {{\left (x^{2} - x + 1\right )}^{\frac {3}{2}} {\left (x + 1\right )}^{\frac {3}{2}}}{x} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x} \, dx=\int \frac {{\left (x+1\right )}^{3/2}\,{\left (x^2-x+1\right )}^{3/2}}{x} \,d x \] Input:

int(((x + 1)^(3/2)*(x^2 - x + 1)^(3/2))/x,x)
 

Output:

int(((x + 1)^(3/2)*(x^2 - x + 1)^(3/2))/x, x)
 

Reduce [F]

\[ \int \frac {(1+x)^{3/2} \left (1-x+x^2\right )^{3/2}}{x} \, dx=\int \frac {\left (x +1\right )^{\frac {3}{2}} \left (x^{2}-x +1\right )^{\frac {3}{2}}}{x}d x \] Input:

int((1+x)^(3/2)*(x^2-x+1)^(3/2)/x,x)
 

Output:

int((1+x)^(3/2)*(x^2-x+1)^(3/2)/x,x)